QUESTION:
How tall can a steel cylinder be before it collapses under its own weight?

ANSWER:
There is a classic estimate by Victor Weisskopf of the
maximum height a mountain can be which you can see
here .
The height h must be smaller than the height where the material down
low would start to melt. This inequality is given by h<E /(Am _{p} g )
where E is the energy per molecule necessary to melt the material, A
is the molecular weight, m _{p} is the mass of a proton, and
g is the acceleration due to gravity. For your steel cylinder I used
the latent heat of fusion of iron (A =56) to calculate E =2.5x10^{-20}
J/molecule. Then you find h <30,000 m. Keep in mind that this is just
a rough estimate; a more accurate calculation would have to be done by an
engineer. Also, this assumes you can build it on something which can support
it and that would not be possible since rock requires less energy to melt
(4.8x10^{-20} J/molecule, according to the
article referenced above) than iron.

QUESTION:
a little girl sits in a car at a traffic light holding a hellium filled balloon.the windows of the car are air tight . when the light turns green the car accelerates forward,her head pitches backward, but the balloons pitched forward. why?

ANSWER:
See an earlier answer .

QUESTION:
If you have two objects with exactly the same volume and surface area, and if you
apply an equal force on both of them, the one with less mass will have a greater acceleration. This is because F=ma. But, if you do this experiment in any fluid, which one will go farther? The surface areas are the same so neither will have a greater form drag. But because acceleration due to force, the object with less mass will slow down much more than the other It is also known that the drag acting on an object is directly proportional to the square of its speed. So, even if the object with less mass has a greater acceleration in the beginning, it will slow down much more than the one with greater mass. So which one will travel farther?

NOTE
ADDED:
After several communications, the question is this: If the same force
F acts on both objects for the same time and then stops, which object
goes farther? There is no gravity, so this is a one-dimensional problem.
Both objects start at rest. This is one of those quite technical answers
although, without understanding all the math, you can see the qualitative
behavior. This is going to be a lengthy answer, so maybe you just
might want to skip it!

ANSWER
(CORRECTED, 11/11/12):
I will assume that the drag force is given by F _{drag} =-cv ^{2}
where c is the same for both objects (as stipulated by the
questioner).

For the first part of the problem, Newton's second law is
m dv /dt =F -cv ^{2} .
At large times the velocity will become essentially equal to a constant
because F ≈cv ^{2} ;
this speed is called the terminal velocity v _{t} =√(F /c ).
Without going into any detail the solution is
v =v _{t}
tanh[(cv _{t} t /m )]. This is plotted on the first
figure above; the larger the mass, the longer it takes to reach the terminal
velocity so, as noted by the questioner, the lighter mass will go farther in
a given time. To get the position as a function of time, we need to solve v =dx /dt =v _{t} tanh[(cv _{t} t /m )];
the solution is x =(m /c ) ln[cosh(cv _{t} t /m )].
This is plotted in the second figure above; the smaller mass goes farther
but after a long enough time, the two keep a constant distance apart.

For the second part
of the problem, Newton's second law is m dv /dt =-cv ^{2} .
Now, the velocity will begin with a speed v _{0} which is
the speed at the end of the time interval in the first part.
Without going into any detail the solution is
v =v _{0} /[1+(cv _{0} t /m )]. This is plotted on the
third
figure above; the smaller mass clearly loses velocity more quickly as
expected. To get the position as a function of time, we need to solve v =dx /dt =v _{0} /[1+(cv _{0} t /m )];
the solution is x =(m /c ) ln[1+(cv _{0} t /m )].
This is plotted in the fourth figure above; if the two start with equal
speeds at the same place (which they do not), the larger mass would
always be ahead of the smaller mass. However, there is a curious result
if the drag force is of the form -cv ^{2} �because
the position is logarithmic, the objects never stop moving so the
question
"which one will travel farther?" has no answer, each will go infinitely
far. The "fix" for this situation is that the nature of the drag force
changes for very low velocities and is better described as
F _{drag} =-bv which results in a
solution like x=D (1-e^{-bt/m} ) which does
have the maximum range of D for large t . I am not going to
make this a three-step problem because at large enough times the
velocities become so small that we might as well consider the objects to
have stopped, knowing that they will certainly stop due to the changed
nature of the drag.

To solve this problem analytically from here is probably not the most
illuminating thing to do. I am going to just calculate a few numerical
examples where I will choose

F =1 N, c =1 N� s^{2} /m^{2} ,
v _{t} =1 m/s.

For the first part of the problem I will choose
two different times, t _{1} =0.5 s and t _{1} =10
s; the first will be before terminal velocity is reached, the second has
terminal velocity nearly reached by all.

For the second part of the problem I will
choose a time such that v= 0.001 m/s; these times turn out to be
approximately equal numerically to the 1000m for my choice of parameters,
e.g. it takes t ≈1000 s for a 1
kg mass to reach v =0.001 m/s regardless of initial velocity.

I will calculate for three masses, 1 kg, 2 kg,
and 5 kg.

The
results are shown in the following table:

m (kg)
t _{1
} (s)
v (m/s)
x _{1}
(m)
t _{2
} (s)
x _{2
} (m)
x _{1} +x _{2}
(m)
1
0.5
0.462
0.120
1000
4.796
4.916
1
10
1
9.307
1000
6.909
16.216
2
0.5
0.245
0.062
2000
11.011
11.073
2
10
1
8.614
2000
13.818
22.432
5
0.5
0.0997
0.025
5000
23.061
23.086
5
10
0.964
6.625
5000
34.361
40.986

My earlier answer had an
error, namely the time it takes to reach v =1 mm/s is not
t ≈m
but rather t ≈1000m .
Now the answer to the question seems to clearly be that the more massive
object goes farthest. Plotting the three cases for different masses
below (choosing a t _{1} large enough that all masses have
the terminal velocity), it is clear that at long enough times the
largest mass will always be the farthest ahead at any given time. This
conclusion is independent of the values of c or v _{t} .

ALTERNATIVE
SOLUTION:
An alternative solution, with F=-bv , may be seen
here . For a linear drag force it is shown
that the total distance traveled is independent of mass.

QUESTION:
There is an electron.it is at rest with respect to me.i have a detector.which detects a magnetic and electric field.first when i am at rest i will only detect electric field.but when "I" move i should detect both electric and magnetic field.because in my view electron is moving and moving electron creates magnetic field.....Am i true till now? if yes,it seems to me that my movement makes an electron to generate magnetic field.How can this happen?How can an observer state influence the electron?

ANSWER:
See a recent answer . That question
is essentially identical to yours. The frame of the electron is not special,
so there is no difference whether you move by the electron or the electron
moves by you. The electron has an electromagnetic field and how it looks (in
terms of electric and magnetic fields) to you depends on the relative motion
between you and the electron.

QUESTION:
If i'm riding at 100mph how much pressure in lbs is exerted onto the bridge of my nose by my glasses? Is there a sliding scale conversion i can keep in mind as i'm riding?

ANSWER:
There is a very handy approximate formula for the force F
on an object of area A in a wind of speed v , F ≈�Av ^{2} ;
this formula must use SI units, v in m/s, A in m^{2} ,
and F in N. I will convert your numbers to SI, do the calculation,
then convert back. For the speed, v =100 mph=45 m/s; I will estimate
your glasses to have an area of two lenses, each about 2.5"x2.5"=6.25 in^{2} =0.004
m^{2} , so A =0.008 m^{2} . So, I find the force is F≈4
N=0.9 lb. So, something like a pound of force will be applied over the area
the glasses frame touches you. Notice that the force is proportional to the
square of the speed, so if you go half the speed (50 mph) you will only
experience � the force. The scale you should keep in mind is that the force
increases like the square of the speed.

QUESTION:
Using Einstein's thought experiment of a photon bouncing between two mirrors, what would the behavior of the photon be in the instances where:

the mirrors were pushed ever closer together?

the 'light clock' were oriented parallel to direction of the motion of the vehicle, and the vehicle were accelerated ever-closer to the speed of light thus forcing the mirrors closer together due to Lorentz contraction?

Would it be possible to ever move the mirrors close enough that the photon is essentially non-moving?

ANSWER:

If you are pushing the mirrors closer together,
it is no longer a clock.

It would still be a perfectly good clock if
oriented so that the light paths were parallel to the velocity because,
as you note, the distance the light has to move would be shortened by a
factor of
√[1-(v /c )^{2} ]. The times for the two trips the
photon takes would not be equal, though, even though their sum would be
the correct elapsed time of the whole trip.

I am afraid that
this question doesn't make much sense to me.

QUESTION:
This is a question about lunar gravity and terminal velocity: Can a man on the moon fall from a platform hundreds of feet high and land on his feet without harm (assuming all other realistic conditions such as he's wearing a space suit are satisfied)? I don't know the math, but is there a certain height that this platform could be built so that the jump is possible?

ANSWER:
What determines whether you will get hurt on impact is how
fast you are going. Since gravitational acceleration on the moon is about
1/6 of what it is on earth, you can obviously jump from a higher platform on
the moon. The equation from which you can calculate the height h you
can jump from to have given velocity v is h=v ^{2} /(2g )
where g is the acceleration due to gravity. How high can you jump
from on earth, maybe 7 m (about 20 ft)? The corresponding speed using g _{earth} =9.8
m/s^{2} would be about v ≈12
m/s. The corresponding height using g _{moon} =1.6 m/s^{2} would be h =12^{2} /(2x1.6)≈45
m. Certainly not "hundreds of feet", more like 135 ft.

QUESTION:
i am inside a train,,the train is moving..and there is one electron inside a train ..and i am looking at that electron..in my perspective the electron is not moving and (as my high school teacher told) only moving charged particle creates magnetic field...so i will not see any magnetic field..but suppose my friend is looking that electron sitting outside the train..then he will see a magnetic field is created.because he will see that electron..then here is a problem...what is hapenning in reality..is there a magnetic field or not

ANSWER:
What you have discovered is that electric and magnetic fields
are not two different things. They sure look that way on first inspection
and the reason they are taught to be two different things is because of
history: it took a long time to find the connections between them. Two
centuries of experiments finally led to
Maxwell's equations , four
equations which show all the ways the electric and magnetic fields are
related. Maxwell's equations comprise a relativistically correct theory of
electromagnetism; it was actually questions like you ask which led Einstein
to discover the theory of special relativity. In its most sophisticated
form, we think not about electric and magnetic fields but the
electromagnetic field. The electromagnetic field is not a vector field,
rather a tensor field, and any field can be decomposed into its electric and
magnetic parts. Your friend on the ground will also see an electric field
but it will be slightly different from the field you see, which you may want
to interpret as some of the electric field transformed into magnetic field.

QUESTION:
why is it not feasible to send nuclear waste into the sun? is it just a cost issue or is there something else making it undoable?

ANSWER:
The first reason is the one you anticipate, that it is too
costly. I found an estimate of the current amount of nuclear waste, about
70,000 tons. The shuttle had a weight of about 2000 tons and our biggest
rockets could put it in a near-earth orbit, not totally escaping earth's
gravity. So, it would take something like 50 launches just to get rid of
what we have. The second, and perhaps more important, reason is that it
could be an environmental disaster if one of the launches failed.

QUESTION:
This is the problem: we define force as F=m*a. The body on Earth that is
not moving has acceleration zero. So, consequently, if a body is at rest
F=m*0 that is F=0. So, how does 'F=ma' formula fits in concept of weight
when that body is not moving relatively to Earth (there's no
acceleration)? It seems to me we can't mathematically consistent apply
equation 'F=m*a' to calculate weight of bodies at rest because there's
no acceleration. It seems like obvious contradiction.

ANSWER:
Suppose we ask what all the forces are on the body at rest,
say a box on a table. One force is the force that the earth exerts down on
the box; that is called the weight which I will denote as W .
Another force on the box is the force the table exerts up on it; that is
usually called the normal force and I will denote that force as N .
Now, Newton's first law says that the sum of all forces on the box must be
zero because the acceleration is zero, N +W =0 or
N =-W which tell us that the table exerts a force
with equal magnitude but opposite direction as the weight. The only problem
is that we do not know what the weight is. In order to do that, we need to
devise an experiment to deduce what the magnitude of the weight is. We could
do that by arranging to have the weight as the only force present and
measure the acceleration of the object�in
other words, we could simply drop the box and measure its acceleration. When
we do that we find that the acceleration has a magnitude g =9.8 m/s^{2}
and has a direction straight down which tells us that the weight has a
magnitude of W=mg . Unless you think the box has a different weight
when it is moving than when it is not moving, there is no "obvious
contradiction". Incidentally, once you have done this you never need to do
it again because all objects have the same acceleration due to their
weights, that is W=mg regardless of what m is.

FOLLOWUP QUESTION:
You say: acceleration is zero because sum of all forces are zero. But when box is at state of free fall we have the same forces: weight (W) and normal force (N) although there's no physical contact between earth and the box. Wright? Because, if we have weight (that is, force in one direction) we have to have opposite force with equal magnitude. So, it is obvious contradiction because Newton is explaining two different state's (free fall and body at rest) with same forces. In other words, from this perspective, free fall would be impossible because acceleration would be zero.

ANSWER:
There is no force N if the box is not in
contact with the table. When falling, the weight is the only force on the
box . It is clear that you misunderstand Newton's third law which states
that "if one object exerts a force on another, the other exerts an equal and
opposite force on the first." The Newton's third law pair of the weight (the
force the earth exerts on the box) is the force the box exerts on the earth.
So, if the box feels a force of, say, 5 N toward the earth, the box exerts a
5 N force on the earth, toward the box. But, in calculating the acceleration
of the box, you only consider forces on the box and the only one is
its weight. Once again, there is no "obvious contradiction".

QUESTION:
If light in earths atmosphere goes a little slower than light in a vacum, then how come when you turn on a light or laser pointer you don't hear a sonic boom?

ANSWER:
A sonic boom occurs when a source of sound (airplane, for
example, or bull whip) moves through a medium faster than the speed of sound
in that medium. A light source (photons) is not a source of sound. There is a similar situation when a charged particle
moves through a medium faster than light moves through that medium�I
guess you might call it an optic boom. This phenomenon is called Čerenkov
radiation and happens often in water-moderated nuclear reactors which
exhibit a blue glow because electrons (β radiation) go faster than the speed
of light in water; that blue light is your optic boom, actually billions of
booms from billions of electrons.

QUESTION:
Why do rifles not slightly fire high? The split second a bullet explodes
the recoil force should start to spin the gun slightly so that the
muzzle lifts up. This is because the recoil action of the gun is not
directed at the center of gravity of the gun but above it. (I just
checked my rifle and the center of gravity is almost a inch below the
center of the barrel).

ANSWER:
You are certainly right, an unrestrained rifle would both
recoil backwards and acquire an angular velocity about the center of mass.
However, and I will not do any calculations here, because the rifle has such
a large mass compared to the bullet, I am confident that the angle through
which the rifle would rotate during the extremely short time the bullet is
in the barrel would be trivially small. Even so, the sight mechanism is
always adjusted for a certain distance and the effect you site would be
included in the calibrations of the sight if it were not totally negligible.

NOTE
ADDED:
The following occurs to me: the angular momentum of the bullet plus gun
before firing is zero and must be zero afterwards. The bullet is confined to
move along the barrel and the center of mass (of the bullet plus gun) does
not move significantly during the time the bullet is in the barrel.
Therefore, I do not expect the gun to acquire an angular velocity until the
bullet leaves. Again, this is contingent on the bullet having much less mass
than the gun.

QUESTION:
Doesn't the fact that a black hole can bend light prove that something can travel faster than the speed of light? As light is pulled toward the black hole it would accelerate, since it is already traveling at the speed of light the moment it started moving toward the black hole it would be going faster than the speed of light would it not? Just curious.

ANSWER:
No, the light does not speed up as it falls into the black
hole. What happens is that, as you would expect, it gains kinetic energy as
it falls but light's energy is all kinetic. But, your idea of kinetic energy
is probably
�mv ^{2} , but this obviously cannot be true for light since
it has no mass. The energy of a photon is hf where h is
Planck's constant and f is the corresponding frequency of the
electromagnetic wave. So, what happens when a photon gains energy is that
the frequency increases; this is known as a gravitational blue shift (the
color of the light moves to shorter wavelengths) and happens when a photon
approaches any massive body, not just a black hole.

QUESTION:
I'm an audiobook narrator about to record a book of essays by Einstein. I normally do all of my own research regarding how things are pronounced and such, but in this case, I'm a bit stumped. Here's the deal: within the book, Einstein goes into a bit of math, and uses the lower case sigma (which I believe means the Stefan-Boltzmann constant) and the upper case sigma (which I think is self-energy in this case) in equations in the same section. Now, these two symbols are quite distinct when written down, but is there anyway to convey them purely verbally, or do I just need to specify "upper-case sigma" or "lower-case" sigma each time?

ANSWER:
You have come to the right person since I spend several hours
a week recording physics and astronomy textbooks for the blind as a
volunteer. The most important part of the training is learning how to
unambiguously read mathematical equations and symbols. It is conventional to
read lower-case letters as just the letter and upper-case letters as
"capital" or some prefer "cap". In your case, I would read
σ as "sigma" and Σ as "capital sigma".

NOTE
ADDED:
Σ is often used to signify a summation as in X = Σ x _{i}
which should be read "capital x equals the sum of x sub i".

QUESTION:
Why the energy corresponding to the most probable speed is not the most probable energy?

ANSWER:
What is the most probable speed? If the distribution of
speeds is N (v )dv , then the most probable speed is the
location of the greatest maximum of this function determined by setting dN /dv =0
and solving for the maximum which I will call v _{1} . Now,
what is the distribution of E =�mv ^{2} ?
N (v )dv =N (√(2E /m ))(dv /dE )dE ≡N '(E )dE ,
so N '(E )=N (√(2E /m ))(dv /dE )
and the solutions to
dN' /dE =0 do not necessarly lead to the same values of v
as the solutions of dN /dv =0 do; I will call the speed of the
most probable energy v _{2} . I will do one simple example.
Suppose N (v )=cv ^{2} exp(-cv ^{2} )
where c is a constant with units s^{2} /m^{2} (see the
graph to the right). Then dN /dv =0=2c � exp(-cv ^{2} )[v -cv ^{3} ]
which has three solutions (for positive v ), v = ∞
(minimum), v =0 (minimum), and v=v _{1} =1/√c
(maximum). Now, I will leave it to you to show that N '(E )=cE ^{1/2} � exp(-2cE /m ))/(2^{1/2} m^{3/2} ).
Then
dN' /dE =0=[ �E ^{-1/2} �(2c /m )E ^{1/2} ]exp(-2cE /m ),
with two solutions, E =∞ (minimum), and E=m /4c
(maximum); so, �mv _{2} ^{2} =m /4c and
v _{2} =1/√(2c )≠v _{1} .

QUESTION:
So, if the world were to stop spinning (or even slow down in angular velocity), would the people and everything on it be considered 'lighter' or 'heavier'? If any change were to take place? And does centrifugal force have an effect on gravity?

ANSWER:
First, let us get some terminology straight. Weight is the
force which the earth exerts on you; it has nothing to do with your motion,
it is a constant force which points toward the center of the earth. Hence,
if you are standing on the equator, your weight is, say, 200 lb (which is
about 900 N corresponding to a mass of about 90 kg) whether the earth is
spinning or not. To keep things simple, let's suppose that you are standing
on the equator. You are moving in a circle whose radius is the distance to
the center of the earth and with a speed which is the circumference of the
earth divided by one day (which turns out to be about 465 m/s≈ 1040
mph). Because you are moving in a circle, you are accelerating and
elementary physics tells us that that acceleration is equal to the speed
squared divided by the radius of the circle, about 0.034 m/s^{2} , and
the direction is toward the center of the circle. Now, assuming that you
know Newton's first and second laws, the sum of all the forces on you is
equal to your mass times your acceleration. The sum of all forces is your weight down and the force which the floor exerts up on
you F , so -900+F=Ma =-90x0.034=-3.1 N. So, the
force the floor exerts up on you is F =900-3.1=896.9 N; if you are
standing on a scale, it measures the force you push down with which is the
same as the force it exerts up on you, and so the scale will read less than
your actual weight by 100x3.1/900=0.34%. If you were sensitive enough, you
would feel lighter. If the earth were to spin 10 times faster (so 1 day=2.4
hours), your acceleration would be 100 times bigger, about 3.4 m/s2 ,
and you would feel 34% lighter; this you would be able to feel. If the earth
were to stop spinning, you would feel 0.34% heavier, not really noticeable,
but now the scale would read exactly your weight. If you were to stand at
the north or south pole, there would no effect due to the spinning and the
scale would read your weight regardless of the spin rate. Technically, there
is no such thing as centrifugal force; the force which causes you to move in
a circle (gravity in this case) is called the centripetal force. If you want
to learn more about what a centrifugal force is, see an
earlier answer . Even if you do introduce a
centrifugal force, it has no effect on gravity which is constant, but it has
an effect on how you perceive gravity.

QUESTION:
How do all objects exert a force on all other objects? Wouldn't that involve a force being applied, which would mean the object would lose energy to create this force?

ANSWER:
It takes no energy to exert a force. For example, a mass
hanging from a string has two forces on it, the weight down (the force the
earth exerts on the mass) and the force upwards which the string exerts on
the mass. If the mass hangs at rest, it can hang there forever without
supplying any energy to the string. Energy is required only if the force
acts over some distance; if that is the case, we say that the force does
work and work creates or destroys energy.

QUESTION:
I am a physics teacher and had a question from a student that I can't answer. If bathroom scales already have the force of gravity accounted for in their reading, why would we use w=mg to solve for mass. The w already has g in it so we would be accounting for it twice?

ANSWER:
(This whole answer assumes that the scale is at rest, not
accelerating.) It depends on where you live. If your scale measures pounds,
then it is measuring the force (weight) directly. If you want to find the
mass of a 200 lb man, m=w /g =200 lb/32 ft/s^{2} =6.3 lb�s^{2} /ft=6.3
slugs. If your scale measures kilograms, then it is measuring the force and
converting it into mass assuming that the acceleration due to gravity is
g =9.8 m/s^{2} . If you want to find the weight of a 100 kg man,
w=mg= 100 kg�9.8 m/s^{2} =980 kg�m/s^{2} =980 N.

QUESTION:
This lady lost her legal battle to have the Large Hadron Collider shut down. She believed that the atom smasher could create a black hole and suck up the earth. The courts sided with the scientists. The scientists said it's not possible, and even if it did create a black hole, it would be a micro black hole and collapse in on itself. That seems to go against the laws of physics and quantum physics. We now know that matter sucked up by a black hole is permenantly imprinted on the black holes surface and the size of the black hole increases. Did I miss something?

ANSWER:
I do not normally answer astrophysics questions, but I can
deal with this one, I think. First, some cosmic rays (radiation which
strikes earth from space) have energies much greater than the proton
energies in the LHC and if a black hole could be created and have sucked in
the whole earth, that would have happened long before we evolved. Second,
you did miss something�Hawking
radiation whereby a black hole can radiate energy. For something to be
"sucked into" the black hole, it must come within the
Schwartzchild
radius which is 2Gm /c ^{2} =6.67x10^{-11} m /(3x10^{8} )^{2} ≈1.5x10^{-27} m
where m is the mass of the black hole. The maximum kinetic energy of
each proton in the collider is 7 TeV which is about equal to the mass energy
of 7000 protons, so the heaviest black hole they could make would have a
mass of about 14,000 protons, about 2.3x10^{-23} kg; the
corresponding Schwartzchild radius is 3.5x10^{-50 } m! It seems to me
that even if this black hole never evaporated, it could go a really long
time before it got close enough to anything to suck it in.

QUESTION:
Assume there is a buoy in the ocean, anchored to the ocean floor. If the tide is low, there is little to no pressure on the chain holding the buoy to the floor. When the tide is high (the depth of the water = the length of the chain) the chain is completely taut, so let's say the pressure on the chain is X. But what if there is a storm and the water level rises over the buoy. Is the pressure on the chain still X or is it greater than X? Does the pressure continue to increase as the water level rises or does it stop at X?
In other words: does buoyancy increase as you dive deeper or does it have a set value?

ANSWER:
(We do not talk about "pressure" in the chain, rather the
force by the chain on what it is attached to, usually called tension;
pressure is force per unit area.) The buoyant force is equal to the weight
of the displaced fluid. To make things clear, let's choose a specific
example. Suppose the buoy has a density which is 1/10 the density of water.
Then, if untethered, it will float with 1/10 its volume submerged so the
buoyant force just counteracts the weight. Now, at low tide, attach a chain
to the ocean floor which is straight but slack. As the tide rises, more and
more of the buoy is under the surface. For example, when half the volume of
the buoy is submerged the buoyant force is 5 times the weight of the buoy.
So all the forces on the buoy are the weight W down, the tension T in the chain
down, and the buoyant force B =5W up: B-W-T=0= 5W-W-T= 4W-T ,
so T =4W . Suppose the water keeps rising until the buoy is
totally submerged so the buoyant force is now 10W; so now T =9W . Now,
if the water keeps rising, the same amount of the buoy is always submerged
and so the tension stops increasing no matter how deep the water becomes.
(Although we think of water as incompressible, at extreme depths the density
does get a tiny bit bigger so the buoyant force at the bottom of the ocean
will be a tiny bit bigger, but a trivially small difference.)

QUESTION:
Lets say you have a tennis ball machine. The tennis balls must travel though a 5ft tube before leaving the machine. Lets say that the machine is spitting out 60 balls a second at 50 mph. The machine is on a swivel and a person changes the position of the machine by moving the 5ft tube left and right by pushing or pulling with his or her hand. If the speed of the balls were increased to 500 mph would it be harder for the person to move the tube? And if so why?

ANSWER:
This question is very similar to one I answered a long time
ago. If you are interested in the mathematical details, you can
link there . Here is the old question: Suppose that a mass m is
tied to a string of length R _{0} and is going in a circle
with angular velocity
ω _{0} . Now, the string breaks. A force which has no radial
component (the direction of the force is always perpendicular to the line
drawn from the original axis of rotation to the mass) is applied to keep the
mass's angular velocity constant; what is the force? How is this like your
question? As your ball travels down the tube, assuming the barrel has no
friction, the tube exerts a force like the one described above if the tube
rotates with constant angular speed. How is it different from your question?
The original question has the ball at rest in the radial direction at the
beginning but your ball starts out with some radial velocity (your 50 or 500
mph). The earlier solution has two constants a and b which
will be different for your problem. Suppose we call the initial ball
velocity v _{0} ; then where the earlier solution says "�where
a and b are constants to be determined� "
we will get different values because dr /dt =v _{0}
instead of zero at t =0. If you really care and are good at math, you
can find the new values of the constants yourself, but I suspect you just
want an answer. The answer for the force the tube exerts on the ball (which
is the force you must exert on the tube) as a function of time turns out to
be F (t )=2mω _{0} ^{2} [R _{0} sinh(ω _{0} t )+ω _{0} v _{0} cosh(ω _{0} t )].
(Don't lose heart, this will get much less technical soon; it is just that I
want to include details for my many readers who like the math!) To finally
answer your question, I need to put in some numbers. Since physicists do not
like English units, I will use comparable metric system units:

I will choose
ω _{0} =1 s^{-1} . This corresponds to about 50^{0}
sweep of your gun in one second.

The two
velocities I will use are v _{0} =20 m/s (about 45 mph) and
v _{0} =200 m/s (about 450 mph).

The tube length
will be 2 m (about 6 ft).

I will use R _{0} =0,
that is, the bullet is launched from the point where the gun is pivoted;
hence F (t )=2mω _{0} ^{3} v _{0} cosh(ω _{0} t ).

I will choose
m =0.5 kg, a little heavy for a tennis ball, but it really doesn't
matter because I am just interested in the relative forces for the two
v _{0} values.

As you
stipulated, I will choose 60 s^{-1} as the rate that the balls
are launched, but first of all I will just look at what happens for a
single ball.

Note that the
slower ball (20 m/s) will take 0.1 s to reach the end of the tube and
the faster ball 0.01 s.

So, putting this in,
we get F (t )=v _{0} cosh(t ). Now, the
hyperbolic cosine function (cosh) at t =0 is 1, at t =0.1 is
1.005, and at t =0.01 is 1.0001. This is what I meant when I said
things would get simpler. For all intents and purposes the force is a
constant because the cosh does not vary significantly for the time the ball
is in the tube. So, finally, F≈v _{0} . The force you have to
exert while a single fast ball is in the tube is 200 N; the force you have
to exert while a single slow ball is in the tube is 20 N. But, that is not
the whole story because there may be more than one ball in the tube. The
distance between fast balls is (200 m/s)/(60 s^{-1} )=3.3 m, so there
is only one ball at a time in the tube. But, the distance between slow balls
is (20 m/s)/(60 s^{-1} )=1/3 m so there are 6 balls in the tube at a
time. The final answer: you need a force of 200 N for the fast ball gun, 120
N for the slow ball gun, the fast ball gun is harder to rotate.

NOTE
ADDED:
While the fast ball is in the tube, the required force is 200 N, but
there is only a ball in the tube (4/3)/2=2/3 of the time; so the average
force over many balls is more like 133 N, still bigger than for the small
balls but not so much bigger. It also occurs to me that for the real-world
application we should be thinking of the torque we have to apply which
changes as a ball moves down the tube; oh well, I'm getting tired of this
problem!

QUESTION:
What caused Felix' Baumgartner to go into a flat spin? I thought at that altitude he was in space and there was no atmosphere. Someone told me it was an effect of gravity but I just don't see that way as I think about it.

ANSWER:
If there were "no atmosphere", his helium balloon would not
have gotten there in the first place. Although the density of the air up
there is way less than at sea level, there is still some.

QUESTION:
We are travelling forward in a bus at 55 mph. Standing at the front of
the bus we stand in the aisle and face the back of the bus. If we throw
a ball 55 mph to a person in the back of the bus what is the overall
speed of the ball since it is moving in two different directions at the
same speed?

ANSWER:
I wish I had a $5 donation for every time I have answered
questions just like this one! I always emphasize that velocity has no
meaning unless you specify velocity relative to what . In your case,
somebody on the bus says the ball is going 55 mph whereas somebody by the
side of the road says the ball is at rest.

QUESTION:
I am a writer putting together a science fiction screenplay. Those who know me say I have an attention to detail--to a fault. There is one particular element I would like to be as accurate as possible. I'm hoping you might be able to help me.
Here is the scenario: A spacecraft leaves earth on course to the moon. In order to create an Earth-like gravity inside the ship; the ship accelerates at a constant rate exerting a force on the occupants equal to one G. Half way through the trip the craft will flip, then decelerate for the remainder of the journey. This would give the same sensation of false gravity to the occupants of the craft.
So here is the question: If this were possible; how long would it take to actually reach the moon?

ANSWER:
Since you are such a stickler for detail, I will give you
detail, probably far more than you want! Your scheme of having an
acceleration with "constant rate" would work in empty space but not between
the earth and the moon because the force causing the acceleration is not the
only force on you, the earth's and moon's gravity are also acting. As you go
away from the earth the earths gravity gets smaller like 1/r ^{2}
where r is the distance from the earth's center, and the moon's gets
bigger as you get closer. So, it becomes a complicated problem as to how
much force must be applied to keep the acceleration just right for where you
are. The picture to the right shows you in your rocket ship. Let's call your
mass M . Then there are two forces on you, your weight W
down and the force the scale you are standing on exerts on you, F .
W gets smaller as you get farther and farther away and you
want F to always be what your weight would be on the earth's surface,
Mg . So, Newton's second law says that F-W=Ma=Mg-W where a
is the acceleration you must have. Note that, for the time being, I am
ignoring the moon; that would just complicate things and its force is much
smaller than the earth's, at least for the first half of the trip. I want
you to understand the complication caused by the fact that W changes
as you go farther away. Now, how does W change? W=MM _{E} G /r ^{2} =Mg (R /r )^{2
} Where R is the radius of the earth and M _{E} is
the mass of the earth. We can now solve for the acceleration the spacecraft
would have to have: a=g (1-(R /r )^{2} ). I have
plotted this in red on the graph above. (The distance to the moon is about
60 earth radii.) Note that for most of the trip the acceleration is just
about g . I also calculated the effect the moon would have, blue
dashed line, and, except for the very end of the trip, it is pretty
negligible. Now that we have taken care of the always-important details, we
can try to answer your question. To calculate the time exactly would be very
complicated, but, since the required acceleration is g for almost the
whole trip, it looks like we can get a real good approximation by just
assuming a=g for the first half and a=-g for the second half;
your perceived weight (F ) will just decrease from twice its usual
value when you take off to about normal when you get to about 5 earth radii
in altitude. The symmetry of the situation is such that I need only
calculate the time for the first half of the trip and double it. The
appropriate equation to use is r=r _{0} +v _{0} t +�at ^{2}
where
r _{0} =R is where you start and v _{0} =0
is the speed you start with. Halfway to the moon is about r =30R=R + �at ^{2}
and so, putting in the numbers, I find t ≈1.69 hours and so the time
to the moon would be about 3.4 hours. You can also calculate the maximum
speed you would have to be about 140,000 mph halfway.

QUESTION:
I am not entirely sure how to word my question. From what I have been taught, gravity is determined by the density of an object d=m/v. I picture the force of gravity like that of a magnetic field. So I am wondering if gravity is an additive force? Like if two sphere's that were the same size and density came together, would the range of the gravitational pull increase or just in strength?

ANSWER:
You have been taught wrong, I guess, because density has
nothing to do with it. The gravitational force on a small mass m a
distance r from the center of a sphere (and outside the sphere
itself) depends on m andthe mass of the sphere, M ; the
dependence on the distance is 1/r ^{2} , so, for example if you
go a distance twice as far away, the force gets 4 times smaller. So,
F=GMm /r ^{2 } where G is just a constant; if you increase
the mass M , the force increases proportionally. But the "range" does
not change; usually, by range we mean how far the force extends and this
force extends all the way to infinity. Gravity is not at all like magnetism.
You ask if it is "additive" and, yes, this force is additive but beyond your
example of doubling the mass to double the force. You need not bring your
two spheres together to add their forces. The way you add the forces from
two different spheres is not an arithmetic sum but a vector sum, taking into
account that the gravitational force from a sphere is always toward the
center. So, if you had two identical spheres separated by some distance and
put another mass halfway between, it would experience zero force.

QUESTION:
suppose their exist a planet that went around the sun twice as fast as that of earth.what would be its orbital size as compared to earth.

ANSWER:
You can do this using Kepler's third law, the square of the
period is proportional to cube of the radius. T _{earth} =1
year, T _{new} =� year, so
( T _{new} /T _{earth} )^{2} =( R _{new} /R _{earth} )^{3}
=(�/1)^{2} =�, so
R _{new} =(^{3} √�) R _{earth} =0.63R _{earth} .

QUESTION:
My co-workers and I have been in a very heated discussion. They tell me I am wrong and my theory is stupid. Will a bullet fired (perfectly horizontal) and a bullet dropped hit the ground at the same time? My answer is yes, because gravity is constant.

ANSWER:
One picture is worth a thousand words. The figure to the
right is an actual strobed photograph of a ball (red) launched horizontally
and another (yellow) dropped. If the red ball had been going faster, both
would still have moved together in their vertical positions. The reason is
that each starts with the same vertical velocity (zero) and experience the
same acceleration in that direction (32 ft/s/s). Because a bullet has such a
high velocity, air drag will have a significant effect on it but this will
affect its motion in the horizontal direction, almost not at all in the
vertical direction. This is not a "theory", it is an experimental fact as
the photograph shows.

QUESTION:
Imagine a race of beings on a distant planet observed our star and detected a small rocky planet with an oxygen rich atmosphere and decided to send an expedition to our solar system which flies in a straight line from their home world to our system at a velocity of 0.5c. Now when the aliens are about 50LY from their destination they begin to intercept our first radio broadcasts (assuming their computers can decode our signals). Then the aliens on the moving vessel record two video messages, one to transmit back to their home world saying we found life, and one to transmit forward to earth saying we found you.
Which parties would need to set their videoplayback devices to either fast forward or slow motion to observe the videos that they are receiving (once the transmission got to its destination years later).

ANSWER:
I think you can answer this for yourself if you refer to an
earlier answer . Note that there is a difference between how things
are and how things appear to be in relativity; see my FAQ page.

QUESTION:
If I take any random event that occurs on the surface of the Earth (a match being struck for example), the light emitted from that event will leave the source and travel out into the universe. If my friend could observe that event from a very distant location, say 1 LY away, he would observe it 1 year after it took place (I think that is all reasonable so far, right?). Now, consider if I left the Earth at something approaching the speed of light at the same time as the match was struck. Would the observable event appear to "slow" down as I attempted to keep pace with it as it "travelled" away from the source? Given that the speed of light is supposed to be constant, I suppose the event would still appear to play out in real/regular time. But if this is the case, what state would the event be in when I arrived at my friend's location 1 LY away? I mean, if it took me just over one year to get to him, he would have only just seen the event which means it must have "travelled" along with me (or near me). To this end, would the event have been effectively almost frozen in time from my perspective? But then I know that can't be the case because the speed of light is constant! HELP!

ANSWER:
Let's have your speed be v =0.8c , 80% the speed
of light. Moving at this speed, you measure the distance to be shorter by a
factor of
γ =√[1-(v /c )^{2} ]=0.6 so you only have to go
0.6 LY. That means that your clock only has ticked off 0.75 years when you
arrived. On your trip, you will deduce that the light gets to your friend at
the end of 0.6 years; in other words, you arrive at the planet 0.15 years
after your friend has seen the event. That is all as measured by a clock on
your spaceship. Your friend, on the other hand, observes the event one year
after it happened (not 0.6 years) and he observes you to have taken 0.8
years to get there (not 0.75 years). You cannot observe the event happening
(a match flaring, then dying) because all that light is ahead of you. If you
are interested in how things would look if you were observing the light (for
example if you were going toward the earth observing or going away from the
earth having left before the light), see the answer just before yours.

QUESTION:
why is that Weight and reaction force never act along the same line?
Pls refer to the attached figure for the question that I asked.

ANSWER:
There is not a single "action-reaction pair" shown in this
figure. What is shown are all the forces on the block: its weight, a pulling
force, a frictional force of the table horizontally, and an upward normal
force from the table. You never see a reaction force if you look at forces
on a body because it is not a force on the body, it is a force by
the body. And, your statement that the action-reaction forces never act
along the same line is totally incorrect, they always act in exactly
opposite directions. The weight is the force which the earth exerts on the
block, so the reaction force is the force the block exerts on the earth
which has magnitude W but points straight up. The normal force is the
force which the table exerts up on the block, so the reaction force is the
force the block exerts on the table which has magnitude N but points
straight down. The friction is the force which the table exerts to the left
on the block, so the reaction force is the force the block exerts on the
table which has the same magnitude but points to the right. The force F is
the force which a string exerts on the block, so the reaction force is the
force the block exerts on the earth which has magnitude F but points
to the left. None of these reaction forces are shown in your free-body
diagram which is as it should be because they are not forces on the block.
Welcome to Newton's third law. (By the way, in case you thought the equal
and opposite forces W and N are an action-reaction pair, they are
not; the reason they are equal and opposite is because of Newton's first
law.)

QUESTION:
Earth has a nucleus which cannot be reach or dig to it simply because oh the high temperatures with lava. The question is, if I dig a hole down across the earth but avoiding the nucleus, than means instead of digging a hole straight down my feet, the hole I dig will be about 35 degrees away from the straight line across the earth. In other words, the hole will be diagonally so it will avoid the nucleus and get to the other side of the earth. If I drop a ball down the hole, will it come out from the other side? Will it get stuck in the middle since gravity pulls from the center of the earth? Does it violates the gravity law?

ANSWER:
To discuss this you have to make two assumptions. First, the
earth's mass is uniformly distributed, that is, a cubic meter taken from
anywhere has the same mass. This, of course, is not true, but it is a pretty
good approximation and is the only way you can do a simple calculation of
the motion in your tunnel. Second, you must assume that the ball moves with
no friction, again not possible but you can make the friction pretty small
with magnetic levitation and/or compensate by having a small driving force.
The motion turns out to be amazingly simple. No matter what straight line
tunnel you bore through the earth, it takes the same time to travel from one
end to the other, about 42 minutes.

QUESTION:
My question is about tension: what exactly is it? Now, I know how
tension is defined, but my problem comes when trying to reason it out
with force diagrams or drawings. Take this classic example: There is a
tug-of-war match going on between 2 contestants. One contestant pulls to
the right with a force of 300 N. The other pulls to the left with a
force of 300 N. What is the tension in the rope? Now, I know the tension
in the string is 300 N just by definition/recognition...but why? The
example proposes a situation in which the net force is zero, so there is
no acceleration of the rope. The "force of tension" resists the pull
from the contestant on the right with 300 N, and the same goes for the
contestant on the left. So, if we represent these forces of tension with
vectors, they would both be pointing toward the "midpoint" of the rope.
That being said, why don't these tension vectors add to make 600 N net
force? Or conversely, why don't these vector forces subtract to make a 0
N net force. Also a side note: In a lot of classical mechanics problems,
we always see problems involving "massless" ropes. What exactly does
that mean and why do we need to define a rope as "massless" in order to
carry out problems involving strings/ropes?

ANSWER:
The tension is nothing more than the force which a rope
exerts on something it is tied to. Tension has some important properties. It
is always tangent to the rope and a rope can never push, it only pulls. Of
course, ropes are never massless in the real world, but we can often
approximate them as if they were if they were if their mass is much less
than everything else. I can see that your main problem is that you have not
been properly instructed on how to attack these simple statics problems. The
first thing you have to do is to choose a body on which to focus and
look only at that; you are attempting to look at the rope and both men all
at once.

Let me start with a simpler example, a 1 kg
mass hanging from the ceiling on a massless rope. (I will use 10 m/s^{2}
to approximate acceleration due to gravity). So, I will choose a body,
the 1 kg mass. What are all the forces on it? Its own weight, 10 N
straight down and the force which the rope exerts up. That force by the
rope is the tension. Pretty simple, the tension at the bottom of the
rope must be 10 N.

Now I will choose a different body, the 1 kg
mass plus half the rope. What are all the forces on the body? The weight
of the body, still 10 N, and the force which the upper half of the rope
exerts on the body (at the point where the upper and lower halves of the
rope touch). The unknown force again must be 10 N and so the tension in
the middle of the rope is 10 N.

Now I will choose the 1 kg mass and the whole
rope as the body. What are all the forces on it? The weight of the body,
still 10 N, and the force which the ceiling exerts up on the top end of
the rope. So now we find that the ceiling must exert an upward force of
10 N on the top of the rope. But, Newton's third law says that if the
ceiling exerts a force on the rope, the rope exerts an equal and
opposite force on the ceiling. So, once again we find the tension in the
rope, now at the top, is 10 N. (This is very much like your tug-o-war
problem if you think about it: the weight pulls down on the rope with a
force of 10 N and the ceiling pulls up on the rope with a force of 10 N,
but the tension in the rope, I have just shown, is 10 N everywhere, not
20 N.)

Next, do the same problem but now let the mass of
the rope be 1 kg also, clearly not even approximately massless.

So, I will choose a body, the 1 kg mass. What
are all the forces on it? Its own weight, 10 N straight down and the
force which the rope exerts up. That force by the rope is the tension.
Pretty simple, the tension at the bottom of the rope must be 10 N.

Now I will choose a different body, the 1 kg
mass plus half the rope. What are all the forces on the body? The weight
of the body, now 15 N, and the force which the upper half of the rope
exerts on the body (at the point where the upper and lower halves of the
rope touch). The unknown force must now be 15 N and so the tension in
the middle of the rope is 15 N.

Now I will choose the 1 kg mass and the whole
rope as the body. What are all the forces on it? The weight of the body,
now 20 N, and the force which the ceiling exerts up on the top end of
the rope. So now we find that the ceiling must exert an upward force of
20 N on the top of the rope. But, Newton's third law says that if the
ceiling exerts a force on the rope, the rope exerts an equal and
opposite force on the ceiling. So, now we find the tension in the rope
at the top is 20 N.

Finally, let's do your tug-o-war problem. Part of
your confusion is saying that each man exerts a force of 300 N. It is much
better to simply say that it is a tie right now so the men must be exerting
equal forces in magnitude.

I will choose man #1 as the body. What are all
the (horizontal) forces on the man? The rope pulls him with an unknown
force and some other force must pull him in the opposite direction so
that he remains in equilibrium; that other force is simply the friction
between his feet and the ground which we will take to be 300 N.
Therefore the tension at end #1 will be 300 N. Of course, since the rope
exerts a force of 300 N on the man, the man must be exerting a 300 N
force on the rope as was originally stated in the problem, but that is
not a force on the chosen body .

I will next choose man #2 as the body. What are
all the (horizontal) forces on the man? The rope pulls him with an
unknown force and some other force must pull him in the opposite
direction so that he remains in equilibrium; that other force is simply
the friction between his feet and the ground which we will take to be
300 N. Therefore the tension at end #2 will be 300 N.

I will choose man #1 plus half the rope as the
body. What are all the (horizontal) forces on the body? The friction
pulls one direction with 300 N and the #2 half or the rope pulls on the
#1 half of the rope with the same force, so the tension in the middle of
the rope is 300 N.

Suppose we choose both men plus the rope as the
body. Then friction pulls to the left with 300 N and to the right with
300 N and the rope can be totally ignored because it does not exert any
net force on the body being an internal force.

FOLLOWUP QUESTION:
I see where I got confused. I didn't understand how to apply Newton's 3rd Law to the problem. So the force of the man on the rope (his pull on the
rope) is an action-reaction pair with the force of the rope on the man (tension)? Also, do we always consider action-reaction pairs as forces that "cancel" each other out or sum to zero?
Lastly, if the forces of friction between their feet and the ground are the ones that add up to a zero net force, what would happen to the tension in the case that one man was able to exert a greater force of friction in one direction? How could we calculate the tension in the rope then?

ANSWER:
Yes, Newton's third law states that if A exerts a force on
B, B exerts an equal and opposite force on A . There is never an occasion
to add the action and reaction forces because they are forces on different
things. They do sum to zero, but that means nothing because one is on the man and one is on
the rope, so why would you add them? If the men pull with different forces,
the system is no longer in equilibrium and accelerates in the direction of
the net force. But still, the tension in the rope is the same throughout its
length. Here is how that goes, using Newton's second law (N2) (it is a bit
more complicated). In the picture, the man on the left in the blue shirt has
a frictional force of 400 N to the left and the man on the right in the red
shirt has a frictional force 300 N to the right. Each man has a mass of 100
kg and the rope is massless.

First, choose both men plus the rope as the
body. All the forces are 400 N to the left and 300 N to the right, so
the net force is -400+300=-100 N to the left. N2 says this is the mass
of the body (200 kg) times the acceleration a , so -100=200a and
a =-0.5 m/s^{2} . The acceleration of both men will be 0.5
m/s^{2} to the left.

Next, choose the left man as the body. All the
forces are 400 to the left and the force the rope exerts on the blue
shirt man, call it T _{blue} . Now N2 says that T _{blue} -400=100a =-50
so T _{blue} =350 N.

Next, choose the right man as the body. All the
forces are 300 to the right and the force the rope exerts on the red
shirt man, call it T _{red} . Now N2 says that -T _{red} +300=100a =-50
so T _{red} =350 N.

So, the tension in the rope is still the same
everywhere (you could do the above exercise of choosing one man plus half
the rope to show that it is also 350 N in the middle) and, for this problem,
it turns out to be just the average of the pulls at each end.

QUESTION:
Consider two satellites in orbit around earth following each other in the very same orbital path, but say 2000 km apart. And consider that they be connected together by a very thin string. Here is the question:
Can this string be gently pulled until it makes an absolutely straight line? Or will it follow the so-called curvature of space that is supposed to exist around earth, between the two satellites� positions?

ANSWER:
This is a tricky question with lots of possible secenarios
for what might happen for various conditions.

First, suppose the string is not a string but a
rigid stick but of negligible mass which we miraculously make appear
between the two satellites happily orbiting in a circular orbit, one
following the other. Now, we no longer have two satellites but one
instead. One of the laws of classical physics says that if you want to
know how a rigid body behaves under the influence of external forces,
find the net force on the body and the center of mass of the body will
move as if it were a point mass experiencing that force. Now, the center
of mass of the pair is below the orbit they are in and therefore this
single satellite is moving too slowly to move in a circular orbit and
the center of mass will change to moving in an elliptical orbit.

What if you put a string of negligible mass
between them in a straight line? Since it has no mass, it experiences no
force and everything will behave like it was not there.

Now, if you connect them with a string which
has mass, it also experiences gravity and each piece of string is really
wanting to be at a larger orbit so, I believe, the tendency will be for
the string to curve to fit the circular orbit of the two heavier
satellites. This is an approximation assuming the string has a much
smaller mass than the satellites. In reality, the string will exert
forces on the satellites and alter their orbits.

Under your "gently pulled" secenario, the
string would pull gently also on you and the other satellite thereby
altering both orbits.

QUESTION:
i'd like to preface my question by first stating that my knowledge of physics is "rudimentary" at it's worst, "comic book" at it's best. with that said my question involves the limits of the human sensorium to effectively interpret it's environment. it's been said, although we "exist" in a physical universe, we "live" in a holographic universe, due to the limits of our sense perception apparatus [ sight, hearing, taste, touch, smell ] . so my question is if we could perceive 100% of our surroundings, what ,if anything, would it look like. or is their no relative reference to determine the answer.

ANSWER:
This is really not physics, it is metaphysics. Holographic universe? What does that even mean? Certainly we cannot perceive all of reality with our senses but our brain is remarkably good at devising devices which can. Think of the microscope, the telescope, radio receivers, infrared detectors, photographic film to detect x-rays,
etc ., etc ., etc . Science is, in some sense, a quest
to perceive more of what our senses cannot. Nobody has yet come up with
a device to "see" the proposed "dark matter" in the universe, but
astrophysicists are working on it.

QUESTION:
I study the quantum theories as a lay person.
My question relates to the nature of the electromagnetic spectrum and the nature of photons. If my understanding that the spectrum encompasses radio frequencies and visible light then I am suddenly confused. I have never thought of a radio signal as a photon phenomena but only an electromagnetic one. How and where do photons arise in the spectrum? Or perhaps; what is the nature of photons that they only come into existence in part of that spectrum? Or even; is this a question of a discrete versus a continuous nature of the so called spectrum?

ANSWER:
If, as you say, you "study the quantum theories", you must have come
across the concept of duality . The nature of particles is that
they are also, at the same time, waves; the nature of waves is that they
are also, at the same time, particles. Any electromagnetic wave is
better described as a wave/particle where the particle is a photon.
There is a correspondence between the wavelength of the wave (very long
for radio, pretty short for light, very short for x-rays, etc .)
and the energies of the corresponding photons, E=hc /λ
where E is photon energy, h is Planck's constant, λ
is the wavelength, and c is the speed of light. So photons for
visible light, for example, have more energy than photons for radio
waves. The energies of radio frequency photons is so small that they do
not affect atoms when passing through material, and interaction with
atoms is usually the way we observe individual photons (see the
photoelectric effect ); be assured that they are there, though. To
get a more complete explanation of electromagnetic waves, see an
earlier answer .

QUESTION:
How can our 3 dimensional world exist in a flat universe. Does flat ( in this case) mean 2 dimensional or a 3 dimensional world without curved edges?

ANSWER:
Sometimes words used in science have different meanings in specific
contexts than they do in everyday life. The definition of a flat
universe at dictionary.com is: "If there is just enough matter in the universe for its gravitational force to bring the expansion associated with the big bang to a stop in an infinitely long time, the universe is said to be flat. The flat universe is the dividing line between an open universe and a closed universe."

QUESTION:
If an electron can be in two places at once, why not 4, or eight or...
Has anyone ever tried to split the spit so to speak to see if after it is spit into two could each of the spit be split again?

ANSWER:
Every electron is in an infinite number of locations until you make
a measurement to determine where it is. For example, an electron in an
atom is best thought of as a "cloud" around the atom and the atom can be
found anywhere in that cloud if you go to look for it. Some places are
more likely to have the electron found there, but in any finite volume
there are an infinite number of points (just like there are an infinite
number of numbers between 1 and 1.5).

QUESTION:
Is it the orbital velocity of an object that enters the earth's atmosphere that causes that object to typically burn up and disentigrate? Could you, in theory, freefall
from the International Space Station safely if you had zero orbital velocity?

ANSWER:
It is certainly high speed through the air which causes an object to
get hot. The effect of high speeds on the body and the effectiveness of
space suits in protecting it at high speeds will be tested in
a jump from about 23 miles any day now. Felix Baumgartner is
expected to exceed sound speed before his parachute opens. However, the
altitude of the space station is about 250 miles and I suspect that
falling from that height through almost no atmosphere would give the
jumper too much speed when he encountered the atmosphere to be able to
survive.

RELATED QUESTION:
I have a question about Felix Baumgartner and his skydive attempt. How is it possible for him to break the speed of sound in his jump? My thinking is that objects reach a terminal velocity - jumping from a higher distance from earth would therefore not make any more difference in the maximum velocity. Also, I heard that he won't be able to breathe for a certain period of time. Is that because of how fast he's going, or because of the atmospheric conditions at such a distance from earth?

ANSWER:
The reason there is a terminal velocity is that the air drag on a
moving object increases approximately like the speed squared (if you go
twice as fast, you experience four times the drag). But what the
terminal speed is depends on how much air there is and at high
altitudes, there is almost no air. There is some terminal speed, but it
is much bigger than it is at lower altitudes. The terminal velocity is
inversely proportional to the density, 1/ √ρ ;
the density of air at 23 miles is less than 1/100 that at sea level, so
the terminal velocity is ten times bigger. The speed of sound, though,
also changes with altitude; it is about 740 mph at sea level and about
670 mph at 20 miles up. I do not really know what your question about " won't
be able to breathe" means. If he did not have his "space suit" he would
certainly not be able to breathe until he got down to an altitude around
5 miles up (just not enough air).

QUESTION:
What exactly is a charge. Can a particle have more than one unit of charge. Is all movement that a particle has is due to charge. What Im trying to ask is can a electron produce a photon if you make the electron rotate due to its spin property. Example: if you had a electron moving in a particle accelrator and you had magnets set up to change fields back and forth so when that electron passed though every field the spin of that electron would line up with that field, thus rotating had having movement. Then you use that process to produce photons. Would those electrons produce photons or does this movement mean nothing. That only a charged particle can produce a photon.

ANSWER:
This is not exactly a single, concise, well-focused question as
stipulated by site groundrules. Asking what charge "exactly is" is like
asking what mass is: it is simply a property that matter may have, see
an earlier answer .
It depends on what you call a particle, but one example would be an
alpha particle which has a charge 2e . "Is all movement...due to
charge"? Certainly not. Certainly if you cause a classical dipole
(little bar magnet) to spin it will radiate, so I suspect that an
electron placed in an AC magnetic field would radiate (although it is
certainly not classical, it is a dipole). Here it is the magnetic
property of the electron responsible for the radiation, not the electron
charge. The fact that a magnetic dipole can radiate and has no charge
shows that charge is not necessary to create photons. Please, in future,
try to abide by site groundrules.

QUESTION:
It is widely accepted that the various elements we find on Earth originated from stars, and that the rarer elements originate from Supernovas. Well my question is this: When a star finally explodes it would send all the various elements out into space, scattering them across the whole solar system in question. When a planet such as the Earth is formed, how is it that these elements accumalate in small pockets on the planet. Surely there would be evenly scattered atoms across the whole planet? I am thinking of gold in my example, this theory explains the rarity of gold, but how do the gold atoms find one another to be able to accumulate into gold deposits on Earth?

ANSWER:
I am not a geolgist and so I cannot give details, but I believe the
general idea is that the earth is not static but ever changing. Over
billions of years all the material of which the earth is composed has
been constantly pushed around, melted, dissolved, deposited, etc .
and different materials react in different ways when exposed to these
geoligical activities.

QUESTION:
We calculate the speed in which we drive a car, as if the Earth was stationary.
We calculate the speed of the Earth as if the Sun is stationary.
I am sure we calculate the speed the Sun is moving as if the galactic center were stationary, and I have no idea how we would calculate how fast the galaxy is moving.
So, I guess my question is, how fast are we really moving? Can we calculate the speed of a car, taking into account the speed of the Earth, Sun, and Galaxy? Or when we take those things into account, is the speed of the car, or even the Earth for that matter negligable.

ANSWER:
As I have said repeatedly when answering questions like this one,
there is no such thing as "absolute velocity", velocity has meaning only
as relative to something else. An equivalent statement is that there is
no such thing as "absolute rest". So, every one of your statements ("we
calculate the speed [of something]� as
if� [something else] � is
stationary") is correct and meaninful.

QUESTION:
Why when a cyclist is turning round a bend, he tends to lean inwards with the bike?

ANSWER:
The picture at the right shows the forces (real and fictitious) on
the cyclist. (I copied this from the Wikepedia article on bicycle
dynamics.) The circle represents the center of mass of the system. Since
he is moving in a circle of radius r and with speed v , he
experiences a centripetal acceleration a _{c} =v ^{2} /r
to the left. The forces on him are his own weight mg , the normal
force N up from the road, and the frictional force F _{f}
which is the force providing the acceleration. If you want to apply
Newton's second law in the frame of reference of the cyclist, which is
not an inertial frame, you must add the fictitious centrifugal force -ma _{c}
as shown in the figure. Note that if he were not leaning, there would be
an unbalanced torque about the point where the tire touches the ground,
τ=mLv ^{2} /r where L is the distance to the
center of mass, which would cause him to rotate clockwise, that is to
fall over. When he leans, though, the weight also exerts a torque, so
the two torques can balance if the angle is just right: mgL sinθ =mLv ^{2} cosθ /r ,
or θ= tan^{-1} (v ^{2} /rg ).

FOLLOWUP QUESTION:
Can it be explained without invoking the fictitious force because it may be too technical? Can I say the bike has a tendency to move in a tangential direction to the bend, this causes the bike to topple outward, so leaning inward will prevent the bike from toppling?

ANSWER:
You can say "the bike has a tendency
to move in a tangential direction to the bend, this causes the bike to
topple outward" but that does not explain why it "topples outward", does
it? You certainly can do this problem using Newtonian mechanics without
introducing the centrifugal force, but you must calculate any torques
about the center of mass, nowhere else will work. The new figure looks
just the same but with the centrifugal force removed. Now, the sum of
all the vertically directed forces must add to zero, -mg+N =0
which tells you that N=mg . And, the sum of all the horizontally
directed forces must equal mass times acceleration, F _{f} =mv ^{2} /r .
So, given m , v , and r , you now know all the forces.
But you still need to know the angle of lean for the cyclist to not
topple over. This is achieved by summing all torques (about the center
of mass) and setting it equal to zero (so that it does not start to
rotate in the plane of the page),
Σ τ= 0=NL sinθ-F _{f} L cosθ=mg L sinθ- (mv^{2} L /r )cosθ.
And so, solving for θ , we get the same answer as above, θ= tan^{-1} (v ^{2} /rg ).

QUESTION:
There seems to me that there is something missing concerning the nature of light and electromagnetic radiation. Is it
the nature and properties of radiation that govern its speed in the universe or is it
the nature of the cosmos?

ANSWER:
What is the difference between
"the nature of the cosmos" and "the nature and properties of radiation"?
The laws of physics are the nature of the
cosmos! To understand why the speed of light
is a universal constant and why it has
the value it does, see my FAQ
page.

QUESTION:
I am not entirely sure how to word my question. From what I have been
taught, gravity is determined by the density of an object d=m/v. I
picture the force of gravity like that of a magnetic field. So I am
wondering if gravity is an additive force? Like if two sphere's that
were the same size and density came together, would the range of the
gravitational pull increase or just in strength?

ANSWER:
I am afraid you have it all wrong. Gravity does not depend on
density. Let me summarize what the gravitational force between two
masses is; I assume that the two masses are either both spherical or
much farther apart than their sizes. The force each feels is simply
F=Gm _{1} m _{2} /r ^{2}
where m _{1} is the mass of one mass, m _{2}
is the mass of the other, r is the distance between them, and
G is a constant. If you had a sphere and doubled its mass, you would
double the force you would feel anywhere.

QUESTION:
why do electric current flow opposite to flow of electrons

ANSWER:
Simply because of the definition of what we mean by current (rate of
change of charge), the accidental choice of negative for electrons, and
the fact that it is usually electrons which carry electric current.
Nothing more profound than that.

QUESTION:
Suppose you are in the vacuum of space and nothing else except you and your supplies exist there. (no planets (okay maybe Earth) etc.)
Suppose you have just created 2 magnets in space from metal or ceramic or brought them from Earth.
If you had two two-sided magnets or two balanced magnet setups/arrays that repelled each other and you could keep them from flipping around, would they repel forever?
More importantly, If they would repel forever, then, by having a variable continuous (repelling) force over a distance, would they then create infinite energy and break the Law of Conservation of Energy?
Can the energy/repelling force win/beat out the against the opposing forces of gravity and possibly the mechanism to harvest that energy?

ANSWER:
First item of business is conservation of energy. If these two
magnets are close together, some agent had to put them there and would
have had to do work, that is put energy into the pair of magnets; you
can never get any more energy out of them than has been put into them.
Second, yes, in principle, they would repel forever. But this does not
mean that if you let them go forever that they would acquire infinite
energy. As they get farther and farther apart the force between them
falls off like 1/r ^{3} where r is the distance
between them; they end up within a few meters going nearly as fast as
they ever will because the force gets so small.

QUESTION:
Up is considered to be which direction?

ANSWER:
The opposite direction from down. I am being facetious. I would say
that the words up and down only have a well-defined meaning in a
gravitational force field where down is the direction the force points
and up is opposite that. Otherwise you can call up whatever you want.

QUESTION:
can something be accelerating but still be stationary for any period of
time

ANSWER:
If "any period of time" excludes an instant (zero time duration),
then no.

QUESTION:
I have just put a mirror up in my house. Next to the mirror there is a light fitting. When the light is switched on there is an effect of having two lights, the real light and the reflection of the light. Does this mean that the light intensity in the room has increased?

ANSWER:
Without the mirror, light from the lamp will be absorbed depending
on the color of the wall. A white wall will reflect or absorb and then
reemit a good fraction of the light and a black wall will absorb most of
the light. The mirror will reflect most of the light which hits it. So,
yes, the mirror will result in there being more light to illuminate the
room.

QUESTION:
What about the structure of alpha particles makes them such a common form of radiation? Why aren't, for instance, lithium nuclei the common form of ionizing radiation?

ANSWER:
Alpha radiation is something which happens mostly for very heavy
nuclei. Think of all the protons and neutrons buzzing around inside the
nucleus, all interacting with each other. It turns out that the alpha
particle (^{4} He, 2 protons and 2 neutrons) is the most tightly
bound of the light nuclei. The figure to the right shows the binding
energy per nucleon (the average energy it takes to remove a nucleon from
the nucleus, E /A , where E is the energy to totally
disassemble the nucleus and A is the atomic number) of stable
nuclei. So the alpha is tightly bound and has few particles and the
result is that it has a relatively high probability of spontaneously
forming inside the nucleus. The energetics are such that the total
energy of the original nucleus ^{A} X^{Z} _{N } s
greater than the energy of an alpha particle plus the daughter nucleus
^{A-4} W^{Z-2} _{N-2 } and so, alpha particle decay
will happen if a mechanism for decay can be found. It gets a little
complicated, here because the alpha particle, if very close to the
daughter nucleus (which being inside certainly is) is strongly bound;
but if it could "figure out" a way to get to a distance a little outside
the surface of the daughter nucleus, the electrical repulsion would be
bigger than the nuclear attraction and it would shoot out. The mechanism
for getting away is called quantum tunneling and there is also a
probability that an alpha particle will tunnel out and escape. Heavier
nuclei (like Li) have a lower probability of formation and a lower
probability of tunneling out if they do form. Roughly speaking, the
probability of alpha decay is the product of the probabilities of
formation and tunneling. You could also think of alpha decay as a very
asymmetric fission of a heavy nucleus

QUESTION:
what is the velocity of a person if he is on a train that is traveling
at 50m/s east and he is running at 2m/s west?

ANSWER:
Every time I get this question (which is often) I have to emphasize �who
is measuring the velocity? If it is somebody in the train, the velocity
is 2 m/s west. If it is somebody on the ground, it is 48 m/s east.

QUESTION:
Is the sense of touch because of newtons third law? is that why we have the sense of touch?

ANSWER:
I am not really sure what you are asking. Certainly, if Newton's
third law (N3) were not valid our sense of touch would not work. N3 says
that when you touch something you exert a force on it and that something
exerts an equal force back on you. Neurons have evolved to be sensitive
to forces and to convey that information to the brain. Regarding "why
we have the sense of touch", I am sure that the need for it was what
drove evolution.
(As in the following q&a, I emphasize that I am a
physicist and not a biologist.)

QUESTION:
Why and how do we feel the hotness of an object? I know that when an
object is heated it absorbs heat energy which is converted to kinetic
energy and particles will be moving faster. How does this give us the
hot sensation when we touch the object? Is hotness related to kinetic
energy of particles?

ANSWER:
I am not a neuro-biologist, so I am probably the wrong person to
ask, but I suspect you do not need to use the microscopic definition of
temperature to answer your question. I would suspect there are simply
neurons which are sensitive to temperature changes and relay the
pertinent information to the brain and central nervous system.

QUESTION:
My friend and I had a drunken argument. I would like independent council to weigh in (there's $300 on the line)
I was given a unique bottle opener by a friend who is a brewer for craft brewery in the northeast. It is a flat piece of wood with a smooth screw embedded in one end ( think ______T_ )
Its measurements in 1/16ths of inches are as follows:
88 long. screw centered 11 from one end. side of screw to end is 9 short side. 75 long side screw lip is 4 from bar bottle cap is 4 high 17 across.
The argument is as follows.
Person A: There is less force required to open the bottle pressing down with the cap positioned _________XXXT___ (between the screw and the user)
Person B: There is less force required to open the bottle pulling up with the cap postioned __________TXXX .
With my horrible description of the problem. Can you prove either arguement successfully?

ANSWER:
I have redrawn your attempt to picture this to make it clearer. The
picture to the right shows the two situations with person A below and
person B above. To answer the question I will compute the force which
the nail exerts on the bottle top for equal forces by the user.
Whichever of these is the biggest is the winner. Doing this is a simple
first-semester physics statics problems, most easily done by summing the
torques in each case about the point on the bottle cap just opposite the
nail; that point is a distance R from the end where F
is applied for person B and a distance R- 2d for person
A. I find that the nail exerts a force of A =F [(R /d )-2]
for A and a force of B =F (R /d ) for B;
B is the winner of the bet. For your numbers, d =17/16" and R =(77+17)/16"=94/16",
so R /d =5.53 and the ratio of the forces is B /A =5.53/3.53=1.57,
making option B 57% bigger, quite definitive. (If a $300 bet is really
on the line, don't forget to reward The Physicist !)

QUESTION:
Hello, I am a U.S. Navy Diver and my buddies and I like to sit around and talk about the physics relating to our job. One scenario that has come up is regarding the straw in the glass of water trick where you place your finger over the top of the straw creating a suction or vacuum and then are able to lift the water yada yada. My question is that if you can replicate the
scenario to be large enough for a man to swim up the straw could he? and if so what would happen when he reached the vacuum at the top of the straw. We were thinking that if you dipped a large tube in to a swimming pool, capped off the top and then lifted the tube 20 or so feet in the air and left just enough of the tube (2 ft) in the pool could a man swim up the tube? If he could what depth would his depth gage read at the mouth of the tube and apon ascent up the tube so forth. Also, I believe there is a vacuum of space at the very top of the tube. What would happen if the man got all the way up the tube and tried to take a breath?

ANSWER:
The pressure at the surface of the uncontained water is atmospheric
(P _{A} ) and the deeper (d ) down you go, the
greater the pressure (P ) becomes because of the weight of the
water above you, P=P _{A} + ρgd
where ρ is the mass density of water and g is the
acceleration due to gravity (so ρg is the weight density). Now,
the pressure at the top of the contained water is zero, and the pressure
you feel in the tube will therefore decrease as you go up to some height
h , P= P _{A} -ρgh.
When h is about 10 m (where the top surface is) the pressure
in the water (and therefore on you) is zero and this would be just like
being ejected into space from the space station, very uncomfortable to
say the least. You would not have to go very high up the tube before
feeling very uncomfortable. You would, however, experience the same
buoyancy in the tube because this does not depend on the pressure, just
the pressure gradient which is the same everywhere in the water. If your
head popped out the top surface in the tube, there would be no breathing
since there is no air, but you would be unconscious and likely dead by
then anyway. (To get a vacuum at the top and a top surface, your tube
must be longer than 10.3 m, about 34 ft.)

QUESTION:
In plenty of Sci-fi movies, we see scenes in space where if there is a crack in a hull of a ship every person or object seems to be sucked through (except the protagonist.) What is actually happening in this scenario and what is the force exerted on these objects that are pulling them?

ANSWER:
Same thing happens when a door is opened or a window broken in an
airliner. The air pressure is much greater inside the ship (about
atmospheric) than outside (close to zero), so the air rushes to get out
the opening, blowing everything not tied down with it. So keep your
seatbelt on!

QUESTION:
This is a two part question. Starting with a cylinder that is the length of 1 light second. Emitting light from each end of a 'still' cylinder toward the center should result in the light emitted from each source arriving at the center of the cylinder at the same time (in 1/2 second).
Repeating this experiment with a cylinder traveling through space at a high rate of speed, would you expect the light to meet in the center of the cylinder as it did when the cylinder was still?
I would expect not, since the cylinder is moving independent of the light.
Just after the light is emitted, the end of the cylinder in the direction of the cylinder is traveling will be closer to the light source. The opposite end will be farther from its light source.
If you agree that the light would not meet in the center of the cylinder, would you agree that if the observer inside the cylinder were to calculate the speed of light using the time traveled in the cylinder and the distance traveled in the cylinder, the calculation would result in two different speeds for the light?
I understand you don't have much time for these questions, but I think this has significant implications against the theory of relativiity.

ANSWER:
No, this does not have "significant
implications against the theory of relativiity"; indeed this example
is, in essence, one of the classic examples illustrating special
relativity. Normally a long train is used as the moving object, not a
cylinder, but the idea is the same. I will just give you a brief
overview. This is such a classic example, you can find it in just about
any textbook which covers special relativity. Refer to the picture. An
observer stands on the side of the track and an observer stands on the
center of the train traveling to the right. At the instant that they
pass, lightning strikes the ends of the train as seen by the observer
on the ground . Each observer will see the light wave fronts moving
with the same speed (an experimental fact) so, as shown, the observer on
the ground sees the two flashes simultaneously and the observer on the
train does not. Therefore, the two observers do not agree on the
simultaneity of two events and so the whole notion of time must be
rethought.

QUESTION:
My question deals with CERN and the LHC . Typically protons are
accelerated to near light speed with repeated pushes by bursts of
magnetic fields (or this is how I understand what they do there) and
collide in opposing directions. Einstein says at light speed you attain
infinite mass and energy. So at near light speed matter would attain
"near" infinite mass and energy as well. The question: how can they
accelerate "anything" to a high fraction of light speed and it not
literally expand to a point where the LHC or Earth for that matter could
not possibly contain it ???? Do protons act as an "energy wave" like
photons ? That's the only way I can picture this as being able to be
achieved without circumstances as I described above.

ANSWER:
Something does not have to "expand" to increase its mass. I take
that your question is when a proton gets very close to the speed of
light, why does its mass not get bigger than, say, the earth? This is a
question I have had before and the answer can be read
here . The idea is that because a proton is so small to start with,
even if it increases its mass millions of times, it is still has very
little mass.

QUESTION:
Let's imagine a huge circuit: a battery on the Earth, two wires going to Moon and a switch with a light bulb on the Moon. When one closes the switch, the power from the battery comes to the Moon and makes the lightbulb glow. If it takes about 1.278 seconds for light (and information, and power) to travel from the Earth to the Moon, what will be delay between events "switch closed" and "lightbulb glowing"? Will it be 1.278 seconds or 2*1.278=2.556 seconds (first the information about closed switch goes to the power source and then the power comes back)? And what's more important - why?

ANSWER:
What drives the current in the wire is the electric field. When you
close the switch, the electric field propogates from the battery
location along the wire with approximately the speed of light (in vacuum
it is exactly the speed of light). So, it takes about 1.278 seconds for
the current to begin passing through the bulb at which time it lights
up. But, if you are watching it, the light takes another 1.278 seconds
to reach you so you see it glow.

QUESTION:
If nuclear fusion in stars continually converts matter into light energy
is the net mass of the universe slowly decreasing in time?

ANSWER:
It is not just light energy but also thermal energy. This process does reduce the mass of the universe. However energy sometimes can cause mass to be produced so you have to ask the total creation and destruction sums. It seems to me that the net mass of the universe is likely decreasing, but there is so much we do not yet understand about how the universe works
(dark matter, dark energy), that the best answer is, I think, nobody
knows for sure.

QUESTION:
I was taught that pressure plays a role in what temperature liquids turn to gas. Specifically that the lower the pressure, the lower temperature required to turn a liquid to gas. To me that would indicate it should work in reverse aswell. So my question is, would it be possible to cause water to become a solid(ice) using pressure? Also, if it was possible: Would it remain at room temperature once it became solid? Since ice generally takes more volume would it explode or would the structure just become more compact?

ANSWER:
The answer is not simple. Let's talk only about water. There is not
a single kind of ice as you can see on the P-T diagram to the left. All
the blue is solid water and eleven kinds of ice shown. It is best if I
let you study this diagram yourself. If you start with water at normal
conditions (say 50^{0} C and about 1 bar=1 Atm), you would have
to increase the pressure to about 10,000 bar to solidify water at 50^{0} C.

QUESTION:
what is the order of magnitude range in nature? what is the smallst thing we know of to the biggest thing? and why isn't this range 100 tims more than it is? or a 100 times less. is there a reason?

ANSWER:
I like the little video on
powers of ten . It goes from 10^{23} m to 10^{-16} m,
but these are not really the limits. At the large end, the video only
goes to about 10 million light years, but the visible universe is
on the order of 13 billion light years (about 10^{26} m) and the
universe is most likely larger but we cannot see the farthest out
objects because light from them has not had time to reach us (the
universe is only about 13 billion years old). On the small end, you
cannot really specify a minimum size because, due to the uncertainty
principle, any particle could have different sizes depending on how it
is measured. It has been speculated, however, that the smallest amount
of space you can have is called the
Planck length , about 10^{-35}
m.

QUESTION:
If it's impossible to travel back into the past, then wouldn't it be impossible to travel into the future and back to the time you traveled from? I know many physicists believe time travel into the future is very possible (and has been done, even if it is an astronaut in space by mere seconds)

ANSWER:
I do not understand your question. According to laws of physics as
we know them, time travel to the future is possible, time travel to the
past is not (as you seem to understand). So, yes, if you travel to the
future, you cannot go back to your past. For example, if you travel to
one hour in the future, you might encounter some friend who has
experienced that hour but the time you experienced was less. If you
travel to 1000 years in the future, you will never see your friend
again.

QUESTION:
If an object is such as a rocket is traveling straight up at a constant speed it would need a force to maintain the speed in order to overcome gravity.
Since the force is not zero and f=ma does that mean the rocket is accelerating?

ANSWER:
If the rocket is traveling with a constant velocity it necessarily
means that the net force on it is zero. The forces on a
vertically moving rocket are its own weight down, frictional drag down,
and the rocket thrust up. The sum of these three forces must be zero.

QUESTION:
I was watching the TV show, The Universe, the episode about alien planets. It talked about how planets and their stars revolve around their common center of mass and/or gravity, I forget which one. How does that work when you have more than one planet? Because, as we know, all of our planets do not revolved with the sun in a straight line. So where is the octuple fulcrum that creates the revolutions of our sun and the planets? And how did it occur? Please give me
the most detailed answer you can, as well as fact checks with my question.

ANSWER:
For simplicity, suppose the solar system is isolated, the forces it
experiences from the rest of the galaxy or the rest of the universe are
negligible; this is an excellent approximation over times long compared
to times associated with the motion of the planets. You can compute the
center of mass of the whole solar system just about as easily as for two
bodies. Since you say you want "the most
detailed answer" I can give, I will do the details. For anybody reading
the answer, you can just skip the math and read the narrative and
understand things qualitatively. Suppose we choose some point in space
as the origin of our coordinate system; each body in the solar system
has a vector position R _{i } in this system and a
mass M _{i} . Then the position of the center of mass is
defined to be R _{com} = Σ M _{i} R _{i} /Σ M _{i}
where the symbol
Σ means to sum over all the objects. For example, the
term Σ M _{i}
is the mass of the whole solar system which I will call M .
Notice, for example, that if the mass of the sun is much greater than
all the other masses, the position of the center of mass is
approximately the position of the sun. Now I am going to find the
velocity and acceleration of the center of mass by differentiating with
respect to time once and twice respectively: M v _{com} =M (dR _{com} /dt)=Σ M _{i} (dR _{i} /dt)= Σ M _{i} v _{i
} and
M a _{com} =M (dv _{com} /dt)=Σ M _{i} (dv _{i} /dt)= Σ M _{i} a _{i
} where v _{i} is the velocity of the i^{th}
object and a _{i} is its acceleration. Now, we can
apply Newton's second law to the acceleration equation:
M a _{com} = Σ M _{i} a _{i} =ΣF _{i
} where F _{i } is the net force felt by the i^{th}
object which is simply Σ_{j≠i} F _{ij}
where F _{ij} is the force on the i^{th}
particle due to the presence of the j^{th} particle and this sum
runs over j but not the term j=i because an object does not exert a
force on itself. So,
M a _{com} = ΣΣ _{j≠i} F _{ij} .
But now look at the double sum: for every term F _{ij}
there is also a term in the sum F _{ji} and F _{ij} =-F _{ji}
because of Newton's third law (the force of the i^{th} on the j^{th}
is equal and opposite of the force of the j^{th} on the i^{th} ).
So, finally we have a _{com} =0, the center of mass
moves with constant velocity and if we happen to have chosen a
coordinate system at rest with the center of mass, the center of mass
never moves. In other words, the whole solar system, no matter how
complicated its motion, orbits around the center of mass. If the total
solar system experiences a force F from the rest of the
universe, the only difference is that the center of mass now has an
acceleration a _{com} =F /M ; but
you can still say that the solar system orbits around its center of
mass, it is just that the center of mass is now "orbiting" around some
other point. Maybe that is more detail than you really wanted, but it is
a standard derivation in any introductory physics course. The bottom
line: any system of interacting objects orbits around its collective
center of mass.

QUESTION:
I have a project I am trying to put together for my middle school earth science class. It involves a physics problem. It has been over 30 years since I have taken physics, so the answer to my question escapes me. I want to have a crime scene where one of the clues to who committed the crime is a person who can throw a ball at _____ velocity. The students in the class would then each get to throw a football at a block of wood. The students measure how far back the block of wood has been pushed back.
I got the idea from the internet after reading about a teacher who did a lab where they found the velocity of a thrown baseball by measuring the weight of the ball, the weight of the block of wood, and measuring how far the wood got pushed back.
I need to know the formula(s) I need to complete the project. In other words, if I know the weight of the ball and the block of wood, and how far the wood was pushed back, how do I calculate the velocity of the football?
I know if the students used a stopwatch they could then use the simple formula for figuring velocity, but I doubt they could measure very accurately with one, plus I think this way adds much more to the mystery theme.

ANSWER:
What you propose is pretty tricky and you have not given me enough information. Does the ball stick to the wood or bounce off? Is the block sliding on the floor? I can propose an alternative
experiment with the same general idea but the results of which would be
much more reliable and less complicated for your students. Hang a box
with a hole in it as a pendulum; situate the hole on the side of the box
so the ball could be thrown into it and get trapped inside (maybe a flap
of cloth would let the ball in but not out). Do not make the rope too
short because you want the ball+box to swing up without the rope going
slack or the box going over the top. Also, the rope should be long
compared to the size of the box for reasons I will not go into here. I
am guessing you just want a formula, not all the details. Here is the
formula which gives v in mph: v =5.45((M+m )/M ) √[L (1-cosθ )]
where M is the weight of the football, m is the weight of
the box, L is the length of the rope in feet, and θ is the
highest angle the pendulum swings to. To measure θ, one of the
kids could take a movie with his iPhone or whatever and then you could
use a protractor on a computer screen to measure the angle. For example,
for a 1 lb football, a 2 lb box, a 12' rope, and an angle of 45^{0} ,
v =5.45((2+1)/1) √[12(1-cos45^{0} )]=30.7
mph. I know middle school kids don't know what a cosine is, but you
could show them how to punch it into a calculator. You can explain to
them that the quantity L (1-cosθ ) is just the height the
ball rises above the bottom of the pendulum; so in my example, the
height the football reached was 12(1-cos45^{0} )=3.5'. If you
think measuring the angle and using the cosine is too conceptually hard,
you could measure the height directly if there was a yardstick in the
movie and just use that height h (in feet) and use
v =5.45((M+m )/M ) √h .
This experiment requires that the ball enter the box pretty close to
horizontally, but so would any similar experiment. Finally, the ultimate
simplicity would be if the box weighed much less than the football. In
that case, v ≈5.45√h.

QUESTION:
We know the Earth's rotational speed is 24 hours approximately. Using T= sqr (4 x pi^2 x R^3/GM), where R=radius of earth= 6400km, M=Mass of earth= 6x10^24 kg, I always get 1.41 hours and not 24 hours. Whats wrong with my concept?

ANSWER:
This is a perfect example of using a perfectly good equation for a
situation where it is not valid. The equation you state is Kepler's
third law for the period of a circular orbit of a mass mm going
around a mass M if m<<M . For example, if you set R
equal to the radius of the earth's orbit and M equal to the mass
of the sun, T =1 yr. Or, if you set R equal to the radius
of the moon's orbit and M equal to the mass of the earth, T ≈ 1
month. What you are doing is calculating the orbit of a satellite which
orbits near the earth's surface (like the International Space Station)
which does, indeed, orbit the earth about every hour and a half. If R =6.6
R _{E} , R _{E} being the radius of the
earth, T =24 hr; this is called a geosynchronous orbit where a satellite
over the equator appears to remain fixed in space and is where
communication satellites are placed.

QUESTION:
I would like to know what would happen to the relationship of gravity
and matter if we were to suddenly stop moving through space?

ANSWER:
One of the most fundamental laws of physics is the principle of
relativity �the laws of physics are
the same in all frames of reference. So a "stationary" earth would have
exactly the same laws of physics as a "moving" earth. I put those two
words in quotes because they really have no meaning, simply because of
the principle of relativity.

QUESTION:
All electric devices draw current as much as they need.The amount of current passing through a circuit depends upon its resistance.But I read in a book that electricity supply at our homes is 240 volts and 5 ampere.i also see that power supply circuits have maximum amperage rating.why can't a device draw more amperes than the rating?

ANSWER:
A power supply supplies power, the rate at which energy is supplied.
A power supply opperating at 240 V and supplying 5 A is delivering
240x5=1200 W. If this is the maximum rating, this means that this
particular power supply is not capable of delivering more than 1.2 kW of
power. Another consideration for your home circuits is that fuses or
circuit breakers keep the current below a certain value because of the
danger that wires carrying too much current might overheat and cause a
fire.

QUESTION:
When a charge that is free to move is released in the presence of an electric field, does it move from a region of high electric potential to a region of low electric potential or vice-versa? Does your answer depend on whether the charge is positive or negative?

ANSWER:
Electric potential decreases in the direction the electric field
points. Therefore a positive electric charge, if released from rest,
will move in the direction of the electric field, toward lower electric
potential. A negative electric charge, if released from rest, will move
opposite the direction of the electric field, toward higher electric
potential.

QUESTION:
I struggle to understand where the "c" fits into e=mc2.
As a purely mathematical formula, it must give standard solutions, therefore there must be agreed units to assign each of the values (kj, kg and kmh for example).
But I can't understand how one can say, "1kg x 1.07925285 � 109 kph" ?
It seems like saying 1 banana x 36kph.
And as the numerical value of the speed of light squared can be stated in any number of ways (kph, kps etc etc) is it not just that E=M but that there is a lot of E in any given lump of M?
Why use c2 - how does that work????

ANSWER:
A quantity can certainly have mixed units. The best known example is
velocity which has SI units of meters/second=m/s. You would not say that
this is like saying bananas per second, would you? The units of energy
(in SI units which are based on kilogram (kg), meter (m), second (s))
are kg � m^{2} /s^{2} ;
this is because energy is force times distance and force is measured in
kg � m/s^{2} ;
this is because force is mass times acceleration where acceleration is
m/s^{2} ; this is because acceleration is the rate of change of
velocity, m/s^{2} =(m/s)/s. So, we know the E part of
E=mc ^{2 } has units of kg � m^{2} /s^{2} .
The mc ^{2} part has units of kg �(m/s)^{2} = kg �m^{2} /s^{2} ,
just the same. The equation E=M would not be dimensionally correct
because you would have
kg � m^{2} /s^{2} =kg,
which would be like bananas=kg.

QUESTION:
Why is the middle part of a bar magnet has weaker magnetic field than its poles?

ANSWER:
Although not literally true, the easiest way to think about a bar
magnet is to "invent" magnetic charges + (N) and - (S) and look at this
as a dipole. As in electrostatics, the magnetic field due to a magnetic
charge is proportional to 1/r ^{2} where r is the
distance from the charge. Therefore, the regions close to the poles
(charges) have stronger fields than regions far from the poles. As I
said, there are not really magnetic charges but the concept works
perfectly fine for calculating magnetic fields, gives the same results
as a more rigorous and complicated calculation from current
distributions.

QUESTION:
i wanna ask the why reflection doesnot take place from any rough medium just like ant thing covered with dust and what happens to that light when it fall on any rough surface.

ANSWER:
The light is reflected still, but it does not form an image because
it is scattered in random directions. If no light were reflected, you
would not see the surface at all. Any light not reflected is either
transmitted through the object or is absorbed.

QUESTION:
If a metal ball was placed between a 'magnetic ceiling' and an 'elastic
floor', could you position the ball so it drops, bounces back up, and is
then attracted magnetically to the ceiling just enough so that it rises
back to the original position before falling again?

ANSWER:
What do you mean by "elastic floor"? Elastic in physics usually
means that no energy is lost on collision, so the ball would rebound to
its original height without any help. If you just mean that the ball
will bounce, then it would not return to its original position without
help. Your "magnetic ceiling" would not work unless you can turn it off
while the ball is falling and turn it back on when it is rising.

QUESTION:
Why does air friction increase as an object is falling? Say a ball is falling from a very high position. We know that its acceleration at first is 10 m/s. But after some time it will decelerate until it is moving with a constant velocity. Why is this so? I know that air friction is the reason but why does the friction increase? The surface area of the falling object remains constant but why the opposing frictional force increases?

ANSWER:
Think about the pressure in a box containing a gas with temperature
T . It is well known that the pressure on any side of inside of
the box is proportional to T . But the temperature is just
proportional to the average kinetic energy per molecule, so the force on
the wall is proportional to the square of the average speed of the
molecules. When an object is falling, it is being hit by air molecules,
and the faster it falls, the faster it sees those molecules colliding
with it. So, it is sort of like seeing the effective air temperature
getting bigger in the frame of the object, hence a bigger pressure and
force.

QUESTION:
Is it possible for a solid object, such as a cue ball, to go from a high
rate of speed, such as 100 mph, to a complete stop without slowing down
in between?

ANSWER:
So, you are asking whether an object can change its speed
instantaneously. Since the acceleration (a ) is the change in
speed (100 mph) divided by the elapsed time (0), the acceleration would
be infinite. But that would require an infinite force (F ) to stop
an object of mass m because of Newton's second law, F=ma . I think
the answer to your question is no!

FOLLOWUP
QUESTION:
yesterday, I asked the question about the cueball changing speed instantaneously. The question was whether or not a cueball could go from 100 mph to a complete stop without slowing down in betwen. You stated that you do not think this is possible. My next question: If an object can not change speed instantaneously, how does an object go from moving to a complete stop at all? At some point, the object would go from a fixed speed to a stop without slowing down in between. If it did not, and just continued to slow down, it would never come to a complete stop. At some point, the object has to change speed instantaneously in order to come to a complete stop.

ANSWER:
Your question is a variation of Zeno's paradox where it is observed
that an arrow must first travel half the distance to the target, then
half the remaining disatnce, etc . before getting there and
therefore never gets there. For your question, it is only infinite
acceleration which is forbidden. To give a simple example, consider an
object which at one time has a speed of 100 mph and is experiencing a
constant acceleration of -10 mph/second. This means that the speed of
the object decreases by 10 mph as each second goes by. So, at the end of
4 seconds, for example, it has a speed of 60 mph. At the end of 10
seconds it has a speed of 0 mph �it
is at rest.

QUESTION:
My question is this, can there be such a thing as a minus? When looking at my daughters homework she had been given equation something like 6 + -7 = -1 But surely this can not make any sense for how can we multiply, add or subtract to something that does not exist? It would not make any sense to say what is 6 apples added to -7 apples, you can not physically posses minus something. I would understand debt to be something different, but minus 10 doesn't exist.
A car can not travel minus speed, it would be simply traveling in a direction. We do not have minus temperature since temperature begins at absolute zero.
So could you please tell me, how can we discuss a minus when it doesn't exist?

ANSWER:
I am not a mathematician and this is not really a physics question.
However the concept of negative numbers is well established and to say
that they do not exist because you cannot conceptualize them is sort of
silly! You do not say how old your daughter is, but I hope you are not
doing her any favors by telling her there is no such thing as a negative
number! Having had four children go through elementary school, I know
that one of the first things they introduce children to is called the
"number line" which is a line with equally spaced marks labeled by
integers, 0, 1, 2, 3 �going to the
right and �-3, -2,-1, 0 to the left. This represents two of the most
significant intellectual advancements in arithmetic history�the concept
of zero and the concept of negative numbers. Zero apples means no
apples, so does this mean that zero does not exist? The number -2 is
that number to which you must add 2 in order to get zero. It is also the
number which you must add to 2 in order to get zero; this is what is
usually thought of, instead, as subtracting�if you take two apples away
from two apples you end up with none. Just because you cannot put your
hands on -2 apples, it does not mean that it "does not exist", it is a
concept, not a bunch of apples. Your example of speed is, technically,
correct because speed means the magnitude of the velocity, the number
which tells you the distance you will go in a certain time; so negative
speed has no meaning. On the other hand, if I tell you that a car is
going on a N-S highway 60 mph, do you know whether it is going north or
south? But if I tell you that north is the positive direction and the
velocity is -60 mph, you know both how fast it is going (60) and in
which direction (S). It is true that nothing can be colder than absolute
zero, but if you used a different scale, you certainly would not say
that -10^{0} F did not exist, would you?

QUESTION:
I have made an argument with a friend over the following,
A steel ball bearing was held just below a surface of a tank of water and released. He argued that at the instant of release, the acceleration was 9.81m/s2. His reason was there was no upthrust as the ball bearing had not moved yet. I disagreed. My argument was there was upthrust as it was in the water, be it moving or not. Hence the starting acceleration was less than 9.81m/s2.

ANSWER:
You need to ask what the forces on the ball bearing are. There is
its own weight, straight down. There is a drag force which is usually
taken to be proportional to some power of the velocity and is, at least
to a good approximation, zero if the ball is not moving; even if the
fluid were very viscous, like honey for example, the initial downward
acceleration would be g to a good approximation, but almost
immediately terminal velocity would be achieved and it would move with
constant velocity. This is what your friend has in mind, I believe.
However, there is an additional force, the buoyant force which is
upward. Therefore, the initial acceleration when v =0 would be
less than g even if the drag force were exactly zero. You should
avoid qualitative terms like "upthrust" and think about forces on the
ball bearing.

QUESTION:
In the oxygen-free environment of space, what effect does intense heat have on items that commonly burn here in our earth environment, such as paper or plastics? These have low ignition temperatures but could not combust. Would they undergo a change in their basic structure / appearance? Have there been experiments conducted on this?

ANSWER:
If you heat something in a vacuum, two things can happen �a
change of state or a chemical change. If you add a little bit of heat to
plastic, for example, it will melt and then it will vaporize if you keep
adding heat; these are changes of state. However, depending on the
material, chemical changes might occur before the object either melts or
vaporizes. I do not think paper can melt, rather the molecules from
which it is composed will break apart when enough heat is added.

QUESTION:
I have a question about acceleration due to gravity.
Does gravity cause bodies to move at constant acceleration, or is there a speed limit - perhaps imposed by the object's inertia. I guess, my question could also be : can you have terminal velocity in a vacuum?

ANSWER:
Since nothing can exceed the speed of light, there is a natural
terminal velocity for anything. Of course, to see this you have to go
really fast. I have discussed this problem in gory detail in an
earlier answer . One way to look at this is that the terminal
velocity is indeed "imposed by the
object's inertia" as you suggest, since the mass (inertia) increases as
it goes faster.

QUESTION:
given that the speed of light is constant i cannot make this reconsile this with the statement that
"Light cannot escape a black hole because the gravity is so strong that it cannot escape before getting sucked back in"
in laymans terms i lay it out as follows, if i piont a gun straight upwards from the surface of the moon ( no air resistance) and pull the trigger, from the momet the bullet leaves the barrel gravity is working to slow the bullets velosity, if the bullet has enough velosity it will escape the moons gravity and keep going(though slowed down) however if it has insufficient velosity the moons gravity will eventually slow it down , stop it and then pull it back to the surface.
the problem arises in that if the speed of light is constant ...if it cannot be slowed down then gravity would never be able to stop it escaping the black hole. the light would leave the black hole at the speed of light , going further away from the 'surface' at the speed of light , getting further and further from the black holes gravity until it escapes.

ANSWER:
Think of it this way: As the photon gets farther and farther away,
its energy decreases, just like your bullet; but the energy of a photon
is E=hf where h is Planck's constant and f is the
photon's frequency. So, as the photon gets farther away, its frequency
decreases which means its wavelength increases until eventually it has
no energy left (zero frequency, infinite wavelength), all of it having
been absorbed by the black hole.

QUESTION:
How does pressure work? Why do gas molecules repel each other, yet they are attracted to larger objects, like the earth?

ANSWER:
This is really two questions. Pressure is the result of the gas
molecules bouncing off the walls; each collision results in a momentum
change which means that a force is exerted on the wall. Gas molecules
repel each other because of the electrostatic force �when
they get close to each other, their electron clouds push each other
away. The molecules interact with the earth via the gravitational force,
that is their own weights are a force which attracts them to the earth.

QUESTION:
If a bullet is fired from a train going 500km/h in the oppposite direction to that the train is travelling will the bullet remain stationary?
A work debate where it would appear nobody has the mathmatical capacity to prove the theory. Your response would be graciously received.

ANSWER:
It depends on who is observing that bullet and how fast the train is
moving. To an observer on the train, the bullet has a speed of 500 km/h
backward. But to somebody on the ground watching the train go by, the
speed u of the bullet is u=v -500 where v is the speed of the train. So, if
the speed of the train is 500 km/h, the bullet will be stationary (that
is falling straight down). If the speed of the train is 200 km/hr, u =-300
km/h. The negative sign means that the bullet moves the opposite
direction from the train (backwards). If the speed of the train is 700
km/h, u =200 km/h, moving in the same direction as the train.

QUESTION:
If my understanding is correct, Iodine-131 is a product of the fission reaction of uranium-235. If this is correct, then by the conservation of proton number, the other product has to be an isotope Yttrium. However this means the isotope has to have a crazy mass number or a very large number of neutrons are released.
So my dilemma comes down to understanding how to complete the equation:
U-235 + 1n ---> I-131 + ? + ?neutrons.

ANSWER:
Here is what happens in nuclear fission. First the ^{235} U
splits into two nuclei which are extremely neutron rich. The two are so
neutron rich that they almost immediately eject some neutrons, typically
5-8 neutrons between them. But, after this, the remaining two nuclei are
still very neutron rich; therefore, they both undergo a series of
β decays where each decay results in a neutron turning into a proton
plus an ejected electron and neutrino. So, you can see, you cannot tell
what the initial decay products were just by looking at the final decay
products because the β decays result in a lot of negative electric
charge leaving the system.

QUESTION:
If we leave earth we would need a time scale not based off of the rotation of the earth. What would this time scale likely be and how would it progress?

ANSWER:
The unit of time is not at all based on the motion of the earth or
the solar system. Rather it is based on something called an
atomic clock .
See the definition of the
second . Just take an atomic clock with you when you leave earth.

QUESTION:
Is it possible to cause a shift in axis of rotating sphere by adding or removing mass from the sphere at either or both poles? Example; The Earth spinning (wobble) on its axis and over a short period of time lost one or both of its polar ice caps. Is that mass significant enough to cause a shift in the axis (theoretically).

ANSWER:
Certainly the mass of the earth is very large compared to the mass
of the polar ice caps, so any change at all will be very small. However,
if the earth is considered an isolated body, nothing that happens on
earth can cause the axis of rotation to change because of angular
momentum conservation. The effect would be to slow the rotational speed
because mass moves farther from the axis of rotation thereby increasing
the moment of inertia.

QUESTION:
I have a question here which I hope you can help .
If we have 2 balls on 2 different slopes inclined at different angles but same vertical height.Both of them also have a rough texture of the same material. Will the velocity of the ball be the same at the end of the slope? Why? Can you kindly explain to me the concept behind this.

ANSWER:
If an object rolls without slipping down an incline and does not
lose any energy to friction, its speed v at the bottom is v= √[2mgh /(m +(I /r ^{2} ))]
where g is the acceleration due to gravity, m is the mass, r is
the radius, h is the height of the top of the incline, and I
is the moment of inertia about the center of mass. If all those
quantities are the same for the two balls, they will have the same
speeds at the bottom; that is, if the two balls are identical their
speeds at the bottom are independent of the angle. The "concept behind
this" is conservation of energy. The equation for v is determined
by equating the potential energy at the top (mgh ) to the kinetic
energy at the bottom (�mv ^{2} +�Iω ^{2} )
where ω=v /r is the angular velocity about the center of
mass.

QUESTION:
This may be a stupid question but i was just curious about this. I know with modern technologies and advanced medicine people live longer. My question is even in the slightest amount do you think that the way people get around today; seeing as our modes of transportation are a lot faster than they were say 60 years ago, are people living longer because of it? Or does the twin paradox just relate to actual travel in space and not on earth and if so what factors keep that from being true on earth?

ANSWER:
No, it is not a stupid question. You just need to get perspective on
the magnitude of time dilation. Any time something moves there is time
dilation. The amount by which moving clocks slow down is given by
γ =√[1-(v /c )^{2} ] where v is the
speed of the clock and c is the speed of light. Consider the
speed of a near-earth satellite, about 18,000 mph≈8000 m/s. Then γ =√[1-7x10^{-10} ]≈1-3.5x10^{-10}
which means that a clock on the space station runs about 0.000000035%
slower than your clock.

QUESTION:
What I'm wondering is what would happen if a planet (assuming a solar planet) was to see a dramatic increase in mass. For the sake of simplicity, it would be an even distribution of mass, leaving the distribution of mass on the planet mostly even. I'm assuming that since it would be travelling in an orbit that the change in mass would affect that somehow as well.

ANSWER:
Assuming the mass was acquired nonviolently (not resulting from a
collision with something else massive) and the planet's mass was small
compared to the sun's mass, its orbit would be unaffected. However,
anything orbiting the planet would change its orbit because of the
greater gravitational attraction from the more massive planet.

QUESTION:
i wanted to ask that what type of lenses are used in 3D glasses and how do the 3D glasses produce such 3 dimensional images that seem so alive????

ANSWER:
There are several types of glasses; for details, see
HowStuffWorks . The essential thing to understand about depth
perception is that it depends on our
binocular vision � we
have two eyes. Each eye sees a slightly different version of what you
are looking at and your brain interprets that information with a 3D
perception. Interestingly, this is learned and if a baby has one eye not
working properly he will never have depth perception, even if the eye is
fixed later in life.

QUESTION:
I have a question about time dilation. I understand the basic principle, and I want to be clear that I'm not really doubting Einstein or the scientific community at large so much as I think I'm failing to understand a technical point. I want to believe time dilation exists, because I'm weird like that, but this particular point is hanging me up, and your help would be greatly appreciated.
It is my understanding that atomic clocks are always used to measure time in the experiments in which such dilation has been evident, and that the same atom is always used in these atomic clocks. It is also my understanding, and please pardon the crude description here, that in an atomic clock, the atom bounces up and down and that's how we measure time. So if the clock moves faster, doesn't it make sense that the atom then needs to travel at a diagonal instead of a straight vertical path, meaning it would take longer for it to complete one "up-down" trip, aka the clock would run slower? I don't see how this proves TIME is moving slower, so much as that atoms are traveling a longer distance/our atomic clocks are moving slower, and surely we can agree that while atomic clocks are well and good and precise, an atom moving up and down is our current system of measuring time, not time itself?

ANSWER:
Suppose you have a mass hanging on a spring inside a car and it
oscillates one time per second and with an amplitude of
�
m. Then, in your frame of reference, it travels a distance of 1 m every
second. Now, suppose you are driving by me with a speed of 100 mph ≈45
m/s. So the distance the mass travels in one second is just a little
larger than 45 m. So, your argument would be that an identical
mass/spring
I had would oscillate 45 times while yours oscillated once? I think you
can see that I would see your mass oscillate once a
second, just like mine. (Of course, because of time dilation, it would
run more slowly but by an almost unmeasureably small amount,
√[1-(45/3x10^{8} )^{2} ].) To get a more intuitive
understanding of time dilation, visit my
FAQ page; the light clock
discussed there does indeed depend on the path length issue you raise.
Incidentally, there are lots of examples of time dilation not dependent
on atomic clocks.

QUESTION:
How did the value of G=6*10^-11 derived???
or
How did Newton got the value of the constant G?

ANSWER:
G is a fundamental constant of nature, it cannot be derived. Newton
did not get the value of G , the best he could do was get the
product GM where M is the mass of the sun. He shows that
GM =4 π ^{2} a ^{3} /T ^{2
} where a is the semimajor axis of the orbit of a planet and
T is the period. This is a derivation of Kepler's third law and
is the real triumph of Newton. It was 70 years after Newton's death that
the first measurement of G was made by
Cavendish .
You can also get M _{earth} G =R ^{2} g
where M _{earth} is the mass of the earth, R is
the radius of the earth, and g =9.8 m/s^{2} ; so Newton
could have found the ratio M _{earth} /M without
knowing G .

QUESTION:
I read the other day that Tycho Brahe didn't think the earth could be rotating partly because if it did, then a cannonball fired in the direction of the earth's rotation should go farther than if it were fired in the opposite direction.
I'm pretty sure I understand the idea of relative motion to understand why it doesn't do so, but what I don't understand is why he thought it would travel further if it were shot in the direction of the earth's rotation. It seems if I were going to be confused about that topic, I would think the cannonball would travel further if the ball were fired in the opposite direction of the earth's rotation. So, I imagine I'm not understanding his confusion. Can you help me understand why he was confused

ANSWER:
In the simplest terms, the cannonball should go the same distance
either way because it is fired relative to the earth which is
moving. If you are in a moving train and throw a ball forward or
backward, it will go the same distance in the train. However, if someone
watches from outside the train, the ball goes faster when you thow it
forward than backward. That is why most satellites are launched in an
easterly direction �to get the extra
boost of the earth's rotational speed. Brahe was right in thinking about
this issue, though, because things behave strangely in rotating
coordinate systems and do not behave the same as in stationary systems.
Most notable is the motion of projectiles due to the
Coriolis force where cannonballs curve left or right. But the effect
was too small to be observed for cannons of his time and, since he was
born 20 years before Galileo, these effects were unknown.

QUESTION:
In quantum mechanics how does the wave function of an observable relate
to the expected value of the observable?

ANSWER:
You do not want to say "the wave
function of an observable" you just want to say "the wave function
Ψ ". The wave function of the
system may or may not be an eigenfunction of the operator associated
with the observable you are interested in. For some observable z
whose associated operator is Z , then Zφ _{n} =z _{n} φ _{n}
where z _{n} is the n^{th} eigenvalue and φ _{n
} is the eigenfunction. Only unless
Ψ= φ _{n } would
you say it is a wave function of the observable. The expectation value
of the observable z is defined as <z >≡<Ψ|Z|Ψ>=∫Ψ*ZΨ dτ
where Ψ* is the complex conjugate of Ψ and the
integral is over all space.

QUESTION:
I'm having difficulty conceptualizing the idea of electromagnetic waves (such as visible light) having a frequency and wavelength. I was sitting at my desk recently, sitting under a florescent light bulb which was flickering really fast, but slow enough for me to recognize it. I know that the light was blinking at a frequency of several times per second. Is this the "frequency" that physicist refer to when talking about the frequency of electromagnetic waves?

ANSWER:
First you should read an
earlier answer where I try to give some of
the basics of electromagnetic radiation. One important equation there is
f=c / λ where f
is the frequency, c the speed of light, and λ the
wavelength. Now, green light has a wavelength of about 555 nm=5.55x10^{-7}
m and the speed of light is about 3x10^{8} m/s. So, f =5.4x10^{14}
s^{-1} ; that is a whole lot bigger than several per second!

QUESTION:
I want to know the weight of the contents of a railcar.
I know the weight of the railcar empty.
I do not have a scale to weigh the railcar after it is loaded.
Can I calculate the weight by pulling the railcar after it is loaded? Will friction negate any accuracy.

ANSWER:
First, you have to be sure you understand the difference between
mass and weight. In SI units weight W is given by W=Mg where M is mass in kg and g =9.8 m/s^{2}
is acceleration due to gravity. So the weight of the car is measured in
Newton's (N) where 1 N=1 kg� m/s^{2} .
If you work in English units, the only differences are that g =32
ft/s^{2} and W is measured in lb. So the mass of the car
is measured in slugs where 1 slug=1 lb � s^{2} /ft.
Now, mass is inertia, resistance to acceleration, so if you can measure
the acceleration, then you can measure the mass, and then deduce the
weight. Acceleration is not easy to measure accurately, so I will try to
give you an easy thing to measure from which you can relate easily to
acceleration. The answer to your last question is that friction is going
to be quite important for a railroad car; if you ignore it you will get
a poor answer. Here is how you can find the frictional force: find the
force which you must pull on the car for it to move with a constant
speed . That will be the frictional force. Measure that speed v
and stop pulling. Now the only force on the car is the friction force
F (which you now know). Let the car roll until it stops and measure
the distance D . The mass of the car can then be calculated as
M=2FD/v ^{2} . A word of caution: v must be in either
ft/s or m/s, depending on which system you prefer. An example: suppose
you must pull with a force of 300 lb to make the car move with a
constant velocity. Suppose you get it going a speed of 20 mph=29 ft/s
and it goes a distance of 200 ft. Then the mass is M =2x300x200/29^{2} =143
slugs and so W=143x32=4576 lb for the weight of the car plus its
contents. This assumes that the frictional force does not depend on the
speed which should be a pretty good approximation.

QUESTION:
Why does metal sound differently than wood?

ANSWER:
Mainly because they have different elastic properties. When metal is
struck, it tends to have sustained vibration so that it "rings". When
wood is struck, vibrations die very quickly and you get more of a quick
"thud".

QUESTION:
How much would a 2 ton block of granite weigh under water? And how many men (approximately) would it take to place one in a specific location underwater?

ANSWER:
As I have said a thousand times or more here, weight is the force
the earth exerts on something so the answer to your question is 2 tons.
Now that I have done my nerdy-snarky thing, what you want is the
apparent weight since there is a buoyant force in the water which makes
it seem lighter. If the block if fully submerged, the apparent weight W'
is given by W'=W ( ρ _{water} /ρ _{granite} )
where W is actual weight and ρ is density. Since the
density of granite is about 2.7 times the density of water, W' =0.74
tons. That is about 1500 lb but I have no idea how many people would be
needed to carry it.

QUESTION:
I would like to know why wet sand is used for the zeer pot? Is it possible to use other material instead, for example, just simply water ?

ANSWER:
A zeer pot is essentially an evaporative refrigerator. The cooling
results from evaporation of water. The faster the water evaporates, the
cooler food in the inner pot is kept. If you just filled the space
between with water, it would evaporate slowly because there would be
only a small surface from which it could evaporate. Putting in sand, the
water wets the sand and there is a large amount of surface area from
which water can evaporate much more quickly.

QUESTION:
Is it true that at any given moment, there are up to 5 atoms in my body
that were, at one time, in Genghis Khan, or Marie Curie, ect ?

ANSWER:
You can be certain that it is considerably more than that. See my
earlier estimates for a
similar question .

QUESTION:
I read that the Andromeda Galaxy will collide with the Milky Way Galaxy in about 3 billion years. How could this occur unless the Big Bang allowed for a change of direction, OR there was more than one Big Bang?

ANSWER:
I do not normally answer astrophysics questions, but I can handle
this one. Just because the universe is expanding does not necessarily
mean that everything is moving away from everything else. Imagine two
galaxies which are originally moving away from some point in space but
fairly close to each other. Then, if they keep moving like this, each
will see the other moving away. But, they are gravitationally attracted
to each other. Therefore, as they move away from the point, they will
begin moving toward each other; so looking at one from the other, we
will see it approaching. We know that the universe is made up of local
clusters of galaxies which are gravitationally bound to each other and
therefore orbit each other, collide with each other, etc . But
if you look at that cluster, you will find that the average of all those
motions is that the cluster moves away from us.

QUESTION:
I've recently been having a heated debate over a scenario composed of very basic principles of physics, and I was wondering if you could help us figure this out. Both of us have presented different arguments, and neither of us can agree with the other. We have found very few relevant webpages, however, after scanning through some of your Q&As, we found one that I will mention. So here it is: If a man is on top of a moving object moving at a constant velocity x (in a vacuum, where air resistance is ignored) and he jumps straight up, will he land in the same position he jumped from, assuming that he doesn't flail his legs? I personally believe that he will, because first off, when released from the friction of the train, the man will still have the momentum of velocity x, and without air resistance will continue to go that velocity until gravity brings him back down. However, my brother believes that gravity will deter the motion of the man in some way that he will land off the mark.
The other example that we found is the bullet on the train example. If a bullet is fired at a speed equivalent to that of the train's from the back of the train then relative to the ground , the bullet is at a standstill and will fall straight down. I think that this example supports my answer, but we are not sure. Could you please provide an answer?

ANSWER:
You have really not given me enough information. Is this happening
on the earth? Or is it some idealized "flat" earth where the train is
traveling in a straight line and gravity is always pointing in the same
direction (perpendicular to the path of the train)? This (flat earth) is
the scenario usually portrayed in textbooks because it is usually an
excellent approximation.

In that idealized situation, you are right, you
land where you went up; gravity acts only in the direction straight down
and therefore there is no horizontal acceleration and you land where you
jumped.

Now, suppose the earth is round but not
rotating on its axis. When the man jumps, he has a velocity with two
components, one away from the center of the earth and one equal to the
train's velocity. The component parallel to the train will not change
during the flight because there is no force in that direction. But, as
he gets farther away, he covers less angular distance than the train so
that when he finally gets back down the train has traveled farther and
he lands behind it.

Finally, the real situation where the earth is
rotating. Now where the man lands depends also on the direction in which
the train is moving. If the train is in the northern hemisphere and is moving north, the man will not only
land behind but also east of the train; in the southern hemisphere he
will land west of the train. This is called the Coriolis force and
results from the fact that the coordinate system is rotating.

Overall, you win the debate with your brother
because, although only under ideal conditions does the man land where he
launched, gravity has nothing to do with it. Your bullet on the train
idea is an entirely different thing since the bullet which just drops
was not fired vertically.

QUESTION:
A chromatic musical instrument can be used to play any key, but does it play them all equally well? In other words, will it sound better it certain keys? (If you need a specific instrument, lets say a simple Helmholtz resonator, such as an ocarina.) I have read a lot about formants, the harmonic series and fundamental notes. I know that individual notes will have slightly varying timbres (harmonic series). But will the fundamental note have the richest harmonics?

ANSWER:
I guess, first of all, we should say that "sound better" is a
subjective thing. The physics of musical instruments is a field unto
itself and I certainly cannot cover it in a concise answer which is the
modus operandi of this site. A few things are worth pointing out,
though. First, terminology: the fundamental frequency is the lowest
frequency present in the note, other frequencies present are called
overtones, sometimes overtones are harmonics which are integer multiples
of the fundamental frequency. Why does the same note played on a piano
and a violin sound so different? It is determined by the timbre of the
note played on each instrument and this is, essentially, the relative
amplitudes of the fundamental and all the overtones produced by the
instrument. The amazing thing is that many musical sounds which your
brain perceives as some note, contains much more of the overtones than
of the fundamental, the frequency of the note which your brain is
hearing. In fact, if you remove the fundamental entirely from the
spectrum, your brain still will "hear" the note as the fundamental. That
is why tiny speakers which cannot put out, say f =20 Hz, bass
frquencies still sound like they have bass response. Some instruments
produce overtones which are just slightly sharper or flatter than the
corresponding harmonics, again affecting the timbre. As an example I
show the Fourier analysis for three notes played on a bassoon. Note that
at the lowest frequency, f =98 Hz, the fundamental is less loud
than most of the overtones. It gets more complicated still if you
consider time; if you look at the relative loudnesses of all the
overtones, they change from when the note begins (say plucking a string)
and when it ends (string stops vibrating).

QUESTION:
Are all the stars we see at night existing suns, or are some of them
suns that no longer exist or planet explosions? If second, how can we
still see them for so long time?

ANSWER:
The preferred terminology is that the sun is a star, not that a star
is a sun. Anyway, when you see what we generally call stars, many are
actually stars (as the sun is) but many other things can be seen.
Planets look pretty much the same as stars but are not. Some moons
around planets can be seen with a small telescope or good binoculars but
are not stars. Many objects you perceive as the same as stars are
actually very distant galaxies which are collections of billions of
stars and are very far away. And yes, many stars you are seeing today
are actually gone now; stars, like anything, eventually run out of fuel
and die. Why do we see them even though they are gone? Most stars you
see are so far away that the light you see took hundreds or thousands or
millions or even billions of years to get from there to here; so, if one
of the stars you are looking at is 1000 light years away and blows up
today, we will not know about it for 1000 years.

QUESTION:
does the universe operate mathematically, and therefore chance, free will, and possibility don't exist? If so, how does quantum physics allow for probability?

ANSWER:
I think the word you want is deterministically, not mathematically.
Most physicists believe that nature is not deterministic. Einstein
believed otherwise.

QUESTION:
What will happen if the two electrons moving parallel to each other in the same direction

ANSWER:
It depends who is observing. For example, from the frame in which
the electrons are at rest, each electron would feel a repulsive force
due solely to the electric field generated by the other. In the
laboratory frame of reference, each would have both a magnetic field and
an electric field, so they would feel a repulsion due to the electric
field and an attraction due to the magnetic field; still, the net force
would have to be repulsive regardless of the velocity.

QUESTION:
The uncertainty principle states that we can not know both position and momentum of a particle simultaneously.
What if we had a particle moving in a circle, and we hit it with a photon every so often to maintain constant momentum. The light would tell us where it was before we hit it. And if we plotted it out we could tell where it would end up. Imagine it going around the face of a clock and hitting it with a photon (aka seeing where it was) at every hour.

ANSWER:
The classic example used to support the uncertainty principle is
that if you try to use a photon to precisely determine position but, in
doing so, the momentum of the particle is necessarily changed. I do not
understand why you have it going in a circle and if it is its momentum
cannot remain constant because momentum is a vector. Your scheme will
not work.

QUESTION:
I was wondering why the collisions in the large hadron collider result in different particles produced each time. Is it due to the differences in the proton's speed , angle of impact etc? And if so, could replicating the conditions perfectly result in the same particles produced? Also, why protons? Why not electrons or neutrons or other particles?

ANSWER:
What it boils down to is that if anything can happen it will happen
with some probability. You never know for sure what will happen because
for any set of initial conditions (your "speed
, angle of impact etc "), there are innumerable things which are
allowed to happen. There have been many other kinds of colliders �proton-antiproton,
electron-electron, electron-positron. What drives the decision of what
to make collide depends on lots of considerations. Protons are good
because it is pretty easy to get a very intense beam going both
directions greatly enhancing your chances of finding what you are
looking for.

QUESTION:
If there is an absolute low temperature (absolute zero) is there such a thing as an absolute high temperature?

ANSWER:
Thinking of temperature as the average kinetic energy per particle,
the higher the temperature, the larger the speeds of the particles. But,
as velocities cannot exceed the speed of light, there would seem to be an upper
limit. However, kinetic energy is not simply ½mv ^{2} and the energy can continue to increase even
though there eventually is negligible increase in velocity. Therefore,
there is no upper limit to the temperature a collection of particles may
have. However, there is not an inifinte amount of energy in the
universe, so the upper limit would be determined by the total energy in
the rest of the universe! That said, the question is not really quite so
simple. For a much more advanced discussion of this topic, see this
article .

QUESTION:
The formula I learned in school for kinetic energy is E = mv^2 / 2. But the famous atomic energy formula is E = mc^2... why doesn't the 1/2 apply?

ANSWER:
Because they are totally different things, one is kinetic energy at
low velocities and the other is the total energy of a particle at any
velocity. In Einstein's equation m=m _{0} / √(1-v ^{2} /c ^{2} )
for a particle with velocity v ; here m _{0} is the
rest mass, the mass something has when it is at rest. I have shown it
before and will not do it again, but it can be shown that if the speed
is much smaller than c, the energy is approximately given by E ≈m _{0} c ^{2} +�mv ^{2} ,
the energy something has because of its mass plus the energy it has
because of its motion (kinetic). So relativity reduces to Newtonian
physics at everyday speeds.

QUESTION:
why doesn't your head hurt when raindrops fall on your head despite
having an accelation of 9.8 m/s?

ANSWER:
The terminal velocity for a typical raindrop is only about 20 mph (9 m/s),
so it is not accelerating all the way down from the cloud. Also, when
the drop hits your head, it deforms, splatters, and spreads out, so much of
its energy goes into this, making it not hurt like a similar sized hail
stone would.

QUESTION:
I got an undergrad in physics 10 years ago and have been wondering this question lately. What is the importance of the harmonic oscillator potential- it models large parts of classical physics as well as appearing in quantum mechanics and field theory as creation annihilation operators.
Why is this linear form important- the best answer I could think of is that it is the functionally least complex co variant potential that nature or some minimum principle might use.

ANSWER:
For any bound system, the potential energy function must have a
minumum somewhere and the system has bound states which are quantized.
An example is shown to the right. Here there is a minimum in the
potential energy function at about 0.75
�. The red lines are the energy states which you would get from solving
the Schr�dinger equation for this potential. But suppose that you did
not know what the potential energy function was for this system. Then
you could get a good first approximation by noting that the shape, near
the bottom, resembles a parabola which is the potential energy function
for a simple harmonic oscillator, V (x )=�kx ^{2} .
It is a good first approximation for low-lying states. It is remarkable
how much you can learn about the dynamics of a diatomic molecule, for
example, by modeling it as a pair of masses attached together by a tiny
spring.

QUESTION:
What's stopping entrepreneurs from using fission to turn mercury or
other elements into gold or silver? Do you think there will come a day
when we simply can create whatever elements we want? Any guesses as to
how far in the future that would be?

ANSWER:
The issue is the rate at which you can create something. The day has
already come when we can create whatever elements we want, but can we
create them in usable quantities? Fission is not what you should be
thinking about, because only a few elements (uranium and plutonium) can
sustain a chain reaction. But, using particle accelerators or neutrons
from reactors, elements may be made to transmute. So, you need a very
intense beam of particles to bombard with and a reaction with a
reasonably high probability of happening. Suppose you were able to find
a reaction where you could cause a million reactions leading to gold per
second. But how many gold atoms do you need to get a macroscopic amount?
Avagadro's number is on the order of 10^{24} . To get 10^{24}
atoms would require a time of 10^{24} /10^{6} =10^{18}
s ≈30 billion years.

QUESTION:
What is the benefit, or advantage, of a subatomic particle's position and velocity being "uncertain"? I don't mean a benefit for us (i.e. quantum computing) but for nature in general. What is the fundamental purpose for this "uncertainty"?

ANSWER:
"Benefit or advantage" are subjective adjectives, not really
meaningful in a scientific discussion. Rather you should ask about the
"consequences" of the uncertainty principle. To me, the most important
consequence is the fact that, given that we cannot know to arbitrary
precision both the position and speed of a "particle", there really is
no such thing as a particle �particles
have wave-like properties and, conversely, waves have particle-like
properties.

QUESTION:
mass increases with velocity as m=rest mass/(1-v2/c2) how is it
physically possible? As we study is amount of matter. How could it
increases with velocity? Or is it something else i dont understand. What
is that quantity that increases with velocity?

ANSWER:
You should not think of mass as the "amount of matter". In physics,
mass is inertia. What inertia is is the resistance to speeding up when
pushed on by a force. A given force can increase the speed of a baseball
in 1/10 of a second by maybe 100 mph whereas if you exert the same force
on a locomotive for 1/10 of a second, you would give it a really tiny
change in speed �a locomotive has
more inertia (mass) than a baseball does. What happens in relativity is
that the faster something goes, the harder it becomes to make it go
faster, its mass increases. [Incidentally, your equation has an error,
it should be the square root of the denominator, √(1-v ^{2} /c ^{2} ).]

QUESTION:
I studied physics for a few years at Purdue University but one question
has puzzled me. It is escape velocity for Earth's gravity. My question
is this, gravity is strongest the closer you get to the center of a mass
which, for us, would be the surface of the Earth. If a rocket has enough
constant force pushing it from the surface to break gravity at it's
strongest and it is constant all the way to say the moon why would it
need to reach a certain velocity to escape Earth's gravity? If gravity
is growing weaker the further from the surface you get and the force
pushing you doesn't change and was stronger than gravity at its
strongest point, I'm confused on why you would have to travel so fast to
escape?

ANSWER:
Escape velocity normally means the speed you must give something to
escape from the earth's surface. If you start from some higher altitude,
the escape velocity there would be lower. Of course, if you
push on something continually, it can
escape at a very low speed.

QUESTION:
In a fission reactor, where do they get the single neutron to fire at the nuclei of the chosen radioactive substance?

ANSWER:
See an earlier answer .
Also, neutron
sources can be used, for example an alpha emitter encased in
beryllium.

QUESTION:
I was wondering. Just as one can record a video in 5000 fps, how many equivilant "fps" does the Atlas-detector at LHC record at? If that comparison is even possible?

ANSWER:
The protons travel in bunches. The bunches collide at a rate of
about 40 MHz=4x10^{7} per second. Each bunch collision
results in about 25 potentially interesting proton-proton collisions per
second. So, the average event rate is about 10^{9} per second.
The detector can store about 200 events/second, so very complex and
elaborate triggoring systems must be used to distinguish the interesting
from less interesting events.

QUESTION:
What would happen if a high powered rifle were to be fired in space?
For example an M4 carbine rifle chambered in .223 / 5.56�45mm NATO
With the following characteristics
Bullet weight/type
4 g (62 gr) SS109 FMJBT
Velocity
940 m/s (3,100 ft/s)
Energy
1,767 J (1,303 ft�lbf)
How far/fast would it travel, would it's trajectory remain true, and what forces if any would cause it to slow?

ANSWER:
First let's define "space": there is no air, no gravity, so no air
friction. Therefore, the answer to your last question is that the bullet
would travel with whatever speed it left the barrel forever in a
straight line. I will assume that the velocity you state is the muzzle
velocity and that the gun, initially at rest, is fired by remote control
somehow so that I can calculate its recoil velocity knowing its mass
(which I looked up to be 2.77 kg). If we call v the speed of the
bullet and V the velocity of the rifle, then momentum
conservation requires that 0.004v =2.77V . The muzzle
velocity is the speed of the bullet with respect to the gun, so v+V =940.
Solving these two equations, v =938.6 m/s and V =1.4 m/s.
If you somehow tie the gun down, the bullet would emerge with speed of
940 m/s.

QUESTION:
consider a water wheel with mass of 600kg, rotating @ 18 RPM, wheel diameter is 3 meters and the diameter of its axle is 0.1 meter.
the wheel is fed with 10 liters of water per second from the top, causing it to rotate and eventually releasing the water at the bottom (3 meters lower).
Neglecting efficiency losses due to friction, air drag etc..., the available energy from the water should be MGH: 10 x 9.8 x 3 = 294 joules. However, considering the wheel's diameter, its mass and its RPM, the wheel's kinetic energy (1/2 x moment of inertia x angular velocity squared) is 1200 joules.
Where did the extra joules of energy come from?
Secondly, the diameter of the axle is only 0.1 meter so this hypothetical wheel has a mechanical advantage of 30 (MA= radius wheel / radius axle). Does this mean the initial energy of water is magnified 30 times at the axle? If not so, then what has been magnified 30 times?

ANSWER:
Your analysis is taking the wrong approach on several counts. First,
what you calculate as "available energy from the water" is available
power from the water, J/s=watts. How much of that energy ends up being
delivered to the wheel depends on the design of the wheel. To the right
we see water entering the wheel at some speed; if it leaves the wheel at
the same speed (which would happen if the circumference of the wheel
were moving at the same speed as the incoming water for the wheel I use
to illustrate), then your wheel is getting all that energy. If that were
the case, then this wheel would be losing energy at the same rate to
frictional forces (plus any work you were extracting from the wheel,
maybe to grind grain or something). You do not say where you got the
moment of inertia to calculate the kinetic energy of the wheel, but I
figure that the biggest it could be, if all the mass were at the
circumference (I=MR ^{2} ), would be about 5400 kg m^{2}
which, using 18 rpm=1.9 s^{-1} , would give an energy of about
9700 J, quite a bit bigger than your 1200 J. If you were able to deliver
294 J/s to the wheel, it would take about 4 seconds to get to 1200 J if
there were no friction.
The bottom line is that a discrepancy between power being delivered to
something and the energy it has is not a discrepancy at all since they
are different things. Regarding your second question, it is force which
is magnified 30 times; this is simply a lever.

FOLLOWUP QUESTION:
The formula I used for calculating the moment of inertia was 1/2 x mass x radius^2 = 1/2 x 600 x 1.5^2 = 675 kg-m^2.
Is it possible to harness the available kinetic energy to do work? i.e can this energy be used to do more work (output) than what the original input by the water allowed (10 liters/second)?
secondly, if force was magnified 30 times at the axle, is this magnification useful in terms of driving machinery such as water pumps, generators etc...? i.e can the original power of the water be used to drive a more powerful machine?

ANSWER:
First, note that I misread your original question and used 3 m as
the radius, not the diameter. So, note that the water leaves the wheel
halfway to the water below, so the available power is only 147 watts.
The moment of inertia you used was for a uniform cylinder or disk and I
would guess that the real moment of inertia is somewhere between our two
answers (mine corrected to 1350 kg m^{2} ). You can never get more
energy from a system than has been put into it. If you find a way to
start using this kinetic energy, the wheel will slow down. You can think
of the wheel as an energy storage device, having acquired that energy
from the water. So, you could use that energy but it would quickly be
used up. This is the principle of a
flywheel on a
mechanical device where you get it spinning when you have energy to
store and then later use the energy.

QUESTION:
how much weight will be on a wench when a tower weighs 300
pounds is 50 feet tall.The wench is attached 5feet above the ground.This
question refers to weight placed on the wench as the tower is lowered to a
horizontal position

QUESTION
TO QUESTIONER:
Sorry, this question does not make sense to me. Also, do you mean winch or
wrench? I think if you look up wench you will get a good laugh.

FOLLOWUP
QUESTION:
As the tower becomes more horizontal from the vertical position say
70 to 90 degrees what is the force in pounds on the winch cable if the
tower is 300 lbs?The winch cable is vertical but its attachment point on
the tower becomes horizontal as the tower is lowered to a horizontal
position.I am sorry that I have not stated the problem more clearly.I
understand that if the winch were attached higher that the weight
requirements for the winch would be much less.

ANSWER:
I am not sure still that I have the picture. Where is the winch
located if the cable is vertical when the tower is vertical and nearly
horizontal when the tower is horizontal? These cable orientations are
about as unfavorable as you could have to lower it with minimal force by
the cable. I will give you an example of lowering the tower with a cable
attached at 5 feet and always perpendicular to the tower. The tension
T in the cable would be T =1500cos θ
lb where θ is the angle the tower makes with the horizontal
and cos is the cosine function. Between 90^{0} (tower vertical)
and 0^{0} (tower horizontal) the cosine varies from 0 to 1. So,
the tension in the cable would be zero when you start to lower it and
would be 1500 lb when it is horizontal (not quite lying on the ground
yet, though).

QUESTION:
What is the reason behind right hand rule? I mean the definite current and magnetic field direction relation?
Like if current is going the thumb direction magnetic field is directed by curved fingers..why not the opposite direction?

ANSWER:
The simple answer is that it is the laws of physics, Maxwell's
equations for your question, which determines the direction of field
vectors for any situation. The right-hand rule is just something which
works correctly. Electromagnetism came first, then the right-hand rule!

QUESTION:
As I read that according to the general theory of relativity , if a guy
for example reaches the speed of light his heartbeats would gets slower
and his clock would stop...than why the heart beats don't also stop!!!
and if at this speed the mass will be completely transfer to energy !!
then the guy won't be a human or would he die...really i can't
understand...am I wrong in something??

ANSWER:
You are talking about the theory of special, not general,
relativity. I should not answer this question because site
groundrules forbid questions
where any object with mass moves at the speed of light. However, I will
talk about something moving at almost the speed of light. Maybe we would
have 1 second for me is 1 year for him. Then, I would observe his heart
beat about once a year. But, if you want to discuss whether he is dead
or not, you should ask what happens in his frame. He sees his heart
beating once every second, just as usual. You can learn more about time
dilation on my FAQ page.

QUESTION:
I have a metal tripod that I want to mount to my roof, one of the kinds like you use for mounting TV antennas. I know that if I were to actually mount any device like an antenna up on my roof that it would be wise to add a grounding wire so that if lightning should strike, at least the grounding wire would provide a path to the earth.
However, what I want to mount on this tripod (on my roof) is an anemometer that connects to a small plastic box (also mounted on the tripod) that contains a circuit board, a lithium battery and a small transmitter for transmitting the wind data to a wireless receiver in my house. My question is, will adding a ground wire to the tripod increase the likelihood of a direct lightning strike on the tripod? Or since no wires connect any part of the anemometer or the tripod to the house, is it better to leave the tripod ungrounded for purposes of reducing lightning strikes?

ANSWER:
You should definitely ground the tripod. This will protect your
house in the event of a strike. It will be no more likely to be struck
if grounded. If you put a sharply pointed rod at the top, it can even
decrease the likelihood of a strike since if a strike is immanent the
strong electric field will cause corona discharge (St. Elmo's fire) at
the point thereby decreasing the field.

QUESTION:
In physics we study that beryllium releases free electrons when exposed to alpha particles. In fact neutrons were discovered this way. But why only beryllium? What if other elements are targeted with alpha particles? Why wouldn't they release free neutrons?

ANSWER:
I think you must have meant to type "releases free neutrons". It is
not only beryllium. Bombardment of anything by alpha particles can
release neutrons. Here is the catch: because alpha particles have
charge, they are repelled by the positively charged nucleus. Therefore,
the alpha particle has to have a certain energy to get close enough to a
nucleus with a given atomic number Z; the greater Z, the greater the
energy of the alpha particle must be. When this experiment was first
done, the only source of alpha particles was radioactive alpha sources,
typically around 5 MeV of energy; this restricted nuclei which could be
studied to the very lightest.

QUESTION:
in relativity of length dose a moving measuring rod really shrink

ANSWER:
As I always emphasize, you have to be clear about who is measuring
the length and what the definition of length is. Here is the definition
of length: the distance between two points whose positions are measured
at the same time . The theory of special relativity clearly shows
that if an object is moving relative to an observer, lengths measured by
that observer (according to the definition above) along the direction of
the motion get shorter. Somebody moving along with the stick will
measure its length to not be shortened. It is also important to
distinguish how long something is as opposed to how long it
looks ; for a disucssion, see my
FAQ page .

QUESTION:
Why when a ball is hit and it moves and spins around it's own exis (ping-pong, soccer, etc.), it does (can do) an arc-like orbit? Is it because of the friction with the air? If there was almost no friction would it move in an almost straight line?

ANSWER:
In the simplest terms, a spinning ball has one side moving faster
than the other through the air. This results in the air drag being
different on one side than the other causing it to experience a froce
toward the faster moving side. This is called the Bernoulli effect. With
no air drag, the ball would follow a parabolic path moving in a vertical
plane.

QUESTION:
I understand that gel electrophoresis works by running a current through a gel. This charge difference makes charged molecules move along according to their charges.
If you spin magnets around a wire, you can produce a current in the wire: a generator.
My question: suppose an electrophoresis gel were sitting on an insulator, ungrounded. You spin the magnets around the gel. Does the electrical field now passing thru the gel cause the charged molecules to move? Or for the molecules to move, must the gel be grounded such that current can flow?

ANSWER:
What matters here is that there be an electric field in the gel. The
simplest way to achieve this is to connect a battery across two ends of
the gel. If you want to call one terminal of the battery "ground", that
is fine but not important because the magnitude of potential at a point
is unimportant, only potential differences between different points
matters. Spinning magnets do indeed create electric fields, however they
are not constant like they are for a battery. Therefore, you would cause
varying currents (AC) to flow in your gel which would not achieve moving
ions like you want, presumably.

QUESTION:
It has always bothered me that objects with greater mass will fall at
the same rate of acceleration as objects with less mass, even though
they have a greater force of attraction to the Earth. I thought to
myself that if a person were to take three cinder blocks up a tower, tie
two of them together (thus having two objects, one with twice the mass
of the other) and then drop both objects simultaneously, discounting air
resistance they will fall at the same rate of acceleration. I thought
about the two cider blocks tied together and came to the conclusion that
the reason they fall at the same acceleration as the one cinder block
objest, is because even though they are tied together they're falling
independently of each other, and all three cider blocks have the same
mass. I took this concept even further and applied it to all objects in
the universe. If we posit that all objects in the universe at their
smallest possible level are made up of the exact same particles, than
those particles during free fall to Earth would fall connected, but
independently of each other. It is therefore the attraction between that
"universal elementary particle" and the whole of the same universal
elementary particles that make up Earth that cause objects to fall at
the same acceleration and furthermore dictate at what rate of
acceleration objects fall to Earth. I know that this isn't exactly a
question, but I am presenting this idea to you in order to receive
feedback as to why it is valid or invalid.

ANSWER:
Wow, that's a little long-winded! You have, I think, discovered
Newton's third law (when applying Newton's second law). A mass M
on which there is a net force F acting experiences an
acceleration equal to F /M (that's the Newton's second law
part). So, if you focus your attention on one atom in a macroscopic
object, the forces on it are many: its own weight straight down plus all
the forces on it by all other atoms in the object, a very complicated
situation. If you ask what the net force is on the whole object, it is
simply the sum of the weights of all the individual atoms (which we call
the weight of the object) plus all the forces the atoms exert on each
other. But, Newton's third law states that if one atom exerts a force on
a second atom, then the second must exert an equal and opposite force on
the first; therefore, the force an that pair of atoms is simply the
weight of those two atoms since their forces on each other cancel out.
Similar arguments for all the other pairs of atoms would lead to the
conclusion that the net force on the whole object is simply the whole
weight of the object. Everything in your question beyond "If we posit �"
is not really necessary and the supposition that everything is made up
of some single universal particle is, as far as we know, simply not
true. I like your tying two blocks together, though. For a simple
explanation of why objects of different masses fall with equal
accelerations in the earth's gravitational field, see my
FAQ page .

QUESTION:
A yo-yo does not continue to go up and down forever. If energy is not destroyed, what causes the yo-yo to stop?

ANSWER:
Where does the yo-yo's energy go? Friction. As it moves through the
air, air drag (friction) pushes opposite the direction it is moving. To
bend or unbend the string takes a force and this force does work to
remove energy from the yo-yo. If the string rubs or slides on the yo-yo,
this force takes energy away. Where does all this energy end up? In
heat. After all is said and done, a careful measurment would find
increased temperature in the air, the string, and the yo-yo.

QUESTION:
Let's say that a man is running west on the equator. He has a device that claims he is going at seven miles per hour. However, he is running against the rotation of the Earth, which is 1037.5646 MPH at the equator. Does this mean that he is actually going negative 1030.5646 miles per hour?
I also have another question. Due to the Earth's rotation, if that man turns around 180 degrees and runs the opposite direction (east) at the same speed, would he be going 1044.5646 MPH?

ANSWER:
As I have tried to emphasize again and again, velocity has no
meaning unless you specify velocity with respect to what . With
respect to the surface of the earth, he moves with a speed 7 mph; with
respect to the earth's axis, he moves with the rotational speed of the
equator
�7 mph; relative to the sun, you would have to take into account the
velocity of the earth in its orbit also; with respect to the center of
the galaxy, you would have to take into account our velocity relative to
that point; etc ., etc ., etc .

QUESTION:
If the Big Bang is the result of everything in the universe is rushing away from each other then why is the Andromeda galaxy hurtling toward the Milky Way galaxy? Should they move in opposite directions? Did something more massive than the Andromeda galaxy fling it our way?

ANSWER:
The fact that the universe is expanding does not mean that
everything is moving away from earth. For example, sometimes Venus moves
in a direction toward the earth, sometimes away, depending on the
positions we are in our orbits. Almost everything very far away moves
away from us, but things closer, like other stars in our galaxy or
nearby galaxies, may move toward us. This can be understood in terms of
initial conditions and forces. For example, if two galaxies form
reasonably close together initially at rest with respect to each other,
their mutual gravitational attraction will pull them together.

QUESTION:
Just recently started thinking about this for some weird reason, but since light is affected by gravity, and light travels at 3*10^8 m/s, if a light source was directed straight towards the earth, would the light not accelerate for a fraction of a second at 9.8 m/s/s making it travel faster than the speed of light?

ANSWER:
The light does not accelerate. However, it is affected by gravity.
If it were a ball, it would speed up and thereby increase its kinetic
energy. For a massless particle like the photon of which light is
composed, changing energy means changing wavelength. When light
increases in energy, its wavelength gets shorter; this effect is called
gravitational blue shift. Light traveling away from a massive object
experiences gravitational red shift �decreasing
energy, increasing wavelength.

QUESTION:
why is the filament of an incandescant bulb thin ?

ANSWER:
The power P dissipated in a tungsten wire, of length L
and diameter d , and for a voltage of 120 volts is approximately
P = 2x10^{11 } d ^{2} /L .
So, suppose you wanted a 100 watt lightbulb with a 1 cm=10^{-2}
m length filament, then
d = 2x10^{-6}
m=2 microns; this is about ten times smaller than a human hair, so not
very practical. To be able to make it thicker you have to make it
longer. A typical filament is about 40 microns in diameter and so the
length would have to be about 3 meters long. So, how are you going to
fit such a long filament in a little light bulb? The photo shown to the
right shows you! Coil it. The thicker you make the filament, the longer
it has to be.

QUESTION:
What gives the the Higgs its mass, if Higgs is what gives particles mass?

ANSWER:
It is the Higgs field which is responsible for mass, not the Higgs
boson which is the quantum of the field and the means by which we can
infer the presence of the field.

QUESTION:
At some point in school I learned that sound tone is a function of frequency and loudness is a function of amplitude. If that's so, why will the same sound, while still audible, have a different loudness depending on how far you are from the source? Presumably the amplitude hasn't changed as the sound wave travels (or has it?).

ANSWER:
Yes, the amplitude has changed. Amplitude determines the energy
which the wave carries, and as you get farther and farther from the
source, the energy is spread out over a bigger and bigger area so the
amplitude necessarily decreases. If the source emits sound equally in
all directions, the intensity (which is proportional to the square of
the amplitude) falls off like 1/r ^{2} where r is the distance from the
source; so it will be 4 times quieter if you are twice as far away. By
focusing the sound to travel like a beam (like a megaphone does), you
can make the sound fall off more slowly; that is also the idea behind
the reflectors in flashlights and auto headlights.

QUESTION:
if i made a two track system 100 meters high and expanded it around the whole earth in a perfect circle and placed a giant size ball inside this 2 track system would the ball comply with the universal law and orbit earth on this track and if so, if i fired this ball at 1000 MPH would it stay at 1000mph or would it lose speed or gain ?

ANSWER:
Let's ignore any friction, either air drag or rolling friction. What
does "ball comply with the universal law"
mean? For an satellite in an orbit 100 m above the earth's surface, the
speed of the satellite would be about 18,000 mph. If you give your ball
any smaller velocity, it will roll around the world (on the track) with
that constant velocity. If you give it any larger velocity, it will
leave the track and proceed on an eliptical orbit and eventually come
back to the point where it left the track. You might understand this
better if you play with
Newton's mountain . (Of course, friction is not negligible in the
real world and, particularly at orbital speed in the atmosphere, the
ball would burn up at high speeds.)

QUESTION:
my question is about breaking the speed of light. I know that it is impossible to do so, but I thought of something that seems a bit of a paradox to me. Light moves at 299792 km per second. An object that is rotating moves faster the farther away from the center it is. So, if you were to make a tower (for lack of a better word) deep in space with no objects to slow its progress, and rotated it once per second in a circle with a diameter of 299792 km or greater would it not be possible to build it long enough to break the speed of light at the tip? If not how would the rest of the tower react?

ANSWER:
It is not possible to build it strong enough. See an
earlier answer for details.

QUESTION:
are the EMF waves mere Oscillations in field values with no movement at all , or , real undulations in the field proper as a ripple moving in space ?

ANSWER:
First, read an earlier answer
about the properties of electromagnetic waves. Now,
look at the little animation to the left. This shows a picture of what
is happening with the light wave, the red being electric fields, the
blue being magnetic fields.

QUESTION:
I want to get a tatoo (don't laugh) that expresses Quantum Theory/Quantum Mechanics. Is there a formula for this?
Originally I wanted to get the collapsible wave theory, but i've read a whole lot more about this recently and it sounds like a lot of scientists are questioning it and there may be other explanations for this phenomenon. I don't want to get a tatoo of something that may be proven wrong in 10 years if you follow me. So I was thinking I would get a tatoo of simply a formula that expresses Quantum Mechanics. Is there such a thing?

ANSWER:
Funny, I recently had another question from somebody wanting a
physics tattoo ! In your case, I can make two
suggestions. I would go with the Heisenberg Uncertainty principle which
states that you can not know exactly both the position and the momentum
(essentially velocity) of a particle; this is a real conceptual keystone
of quantum mechanics. In this equation (actually an inequality),
Δx is the uncertainty in position, Δp is the uncertainty
in momentum (mass times velocity) and ħ is the rationalized
Planck's constant (a really tiny number). A second choice, not quite so
compact or easy to understand, is Schr�dinger's equation. Without going
into too much detail, this is the equation which may be used to
determine the wave function Ψ.

QUESTION::
How long,in time and distance,would it take an object accelerating at 1G to reach the speed of light?

ANSWER:
You can never reach the speed of light. You can peruse my FAQ page
to find earlier discussions of why not. In addition, it is not possible
to have constant acceleration when you are approaching the speed of
light. For example, suppose you are 5 m/s slower than light speed with
an acceleration of g ≈10 m/s^{2} ;
that would mean that in one second you will be going 5 m/s faster than
the speed of light. However, you can ask a question like "how
long does it take to achieve 99% the speed of light with a constant
applied force and how far would you go?"
To see a full discussion of the problem where you have a constant
force acting,
see an earlier answer . The best I can do is choose F /m=g
(corresponding to an acceleration of g at velocities much smaller
than c ) and v /c =0.99 (99%) which leads to a time of
2.1x10^{8} s=6.7 years and a distance of 5.23x10^{18}
m=553 light years.

QUESTION:
My physics tutor and lecturer wouldn't answer this question, so I'll try you!
I'm hoping you'll agree a scientific theory can never be totally water-tight. Here goes:
According to length contraction as strictly understood in special relativity, the ratio of the cirumference of a circle to its radius does not always hold true, to deny c=2�r. Isn't this theory trumping common sense? I understand there is no such thing as a straight line in an non-Euclidian universe, but a circle is always a circle! Surely if a wonderfully logical and fool-proof theory conflicted with the law of addition, for example, we would doubt the theory, and not the law of addition. So why do we believe in length contraction when it conflicts with the equally sensible law of the circle? And if the law of the circle is inadmissable as a test of truth from which to judge theories, then what kind of evidence do we need to find instead?

ANSWER:
This is really more geometry than physics, but since you frame it as
a means to refute length contraction, I will comment. What you have
learned about the properties of a circle is true only in flat Euclidean
space. The ratio of the circumference of a circle to its diameter is not
3.14159 � if the space is curved.
Rather, the geometric object called a circle is defined to be that locus
of points equidistant from one point in space. In other words, you must
conform to the geometry of the space when making length measurements. It
is not true that " there
is no such thing as a straight line" in curved space, it is simply the
shortest distance between two points in that space. Your demand is
equivalent to saying that Euclidian geometry is the only "true"
geometry. Length contraction is not a scientific theory, it is a proven
fact. And, what is the "law of addition" anyway? The way things add
depends on what you are adding. You might think that 3+4=7, but if you
walk 3 miles north and then 4 miles east you end up a net distance of 5
miles from where you started. Here is the bottom line: do not trust your
"common sense" if you are trying to apply it outside your realm of
experience because experience is where common sense comes from.

QUESTION:
So light is made up of photons and photons have a spin of 1 which makes them bosons, and bosons don't obey the pauli exclusion principle. So why doesn't light just pass through walls?

ANSWER:
Whether a particle is absorbed, reflected, or scattered when it
enters a medium depends on how it interacts with the medium, not on the
exclusion principle. Photons, being the quanta of the electromagnetic
field, interact very strongly with the electrons in a medium. The
details of that interaction determine what will happen. Some media like
glass have the speed slowed as the main effect, absorption and
reflection being relatively small. Others like a wall, mainly absorb the
photons.

QUESTION:
Can physical wind affect where a sound wave is going at all? I have a fan running in my living room and with the fan running obviously it creates sound and the speakers from the TV don't sound as loud as when the fan isn't running. But does the wind produced from the fan interfere at all with sound waves to diminish the sound, or is the perception that the sound is quieter with the fan running just because the fan also creates sound?

ANSWER:
Sound moves with respect to the medium through which it is
transmitted, so if the medium moves it changes the direction the sound
travels (unless the sound and medium move in the same or opposite
directions). In your case, with the fan blowing across the direction of
the sound, the effect will be practically zero because the speed of the
air is so small compared to the speed of sound (≈330
m/s). The result would be the same as your moving maybe 1 cm. There is
certainly no "interference" going on here because there is no similarity
between the fan noise and the music (see next answer). The fan noise is
competing with the music for attention of your brain and that apparently
makes the music sound diminished.

QUESTION:
Do sound waves get affected in any way by other sound waves? As in if you have one speaker playing one sound and another speaker playing a different sound do the sound waves interfere with each other and diminish the sound?

ANSWER:
Sound waves always affect each other in the sense that they obey the
principle of superposition which says that the disturbance at any point
in space at any time is always the sum of the individual disturbances.
Unless, however, there is some simple relation between the two
interfering waves, you will never perceive any diminishment of the
sound. If the two sounds are identical, you will find places where the
sound is zero. If the two sounds are very close together in frequency,
you will hear "beats ".

QUESTION:
What is the physical significance of `radius of gyration'. Does it become zero in case of (say) ring when axis passes through center of mass.

ANSWER:
The radius of gyration r _{i} of an object of mass
m and moment of inertia I (about some axis) is defined by
I=mr _{i} ^{2} . Therefore what it means is that r _{i
} is the distance from the axis that a point mass m would be
to have the same moment of inertia. The radius of gyration is never zero
unless I= 0 or m = ∞.

QUESTION:
Why do we think the speed of light in a vacuum is constant when we have different neutrino densities in vacuum

ANSWER:
Neutrinos do not interact with photons, so even if there are
neutrinos present, it might as well be a vacuum.

QUESTION:
I asked the sunglasses vendor in his shop how I can check if their glasses are indeed polarized and they asked me to look through the glasses at a certain spot on the wall. I did so and there appeared an object (an icon in the shape of a fish in this particular example) which disappeared when I looked at this spot without my glasses on. Beats me! I was taught at school how polarized glass can block light, but I have never heard how it can possibly make things appear!

ANSWER:
You can cause reflected light to be polarized. Imagine that there is
some substance which is colorless but that you can paint on a wall and
light reflected from it will be polarized. Paint a picture of a
butterfly on the wall. You will see it only if you can distinguish
between polarized and unpolarized light.

QUESTION:
Say a spacecraft was launched such that it was fired away from the moving earth and remained "parked" in a stationary position relative to the sun (much like tying a dog to a lamppost) while the earth completed its annual orbit around the sun. After one year, the earth would again encounter this motionless craft, which would appear as if it was unavoidably hurtling towards us.
While a totally useless purpose for such a spacecraft, does this particular orbital method have a name; ie, "heliostationary" or other such label?

ANSWER:
The orbit you describe is impossible. Any object placed in the
earth's orbital path can only move around the sun once in a year, would
keep pace with us (ignoring any interactions between it and earth).
Communication satellites, which I presume you are thinking about, are in
the plane of the equator, have circular orbits, and have a period of 24
hours so that they appear to be stationary when viewed from
earth. A good place to look to see this is the
NASA
page which shows all satellites; look about 6.5 earth radii out from
the center for the geosynchronous orbits.

QUESTION:
Why the rotating things are more stable? For example, a rotating top can stand upright on the ground while a still top cannot; one can spin a basketball on his finger, but one can never keep the ball still on his finger.

ANSWER:
Angular momentum. A spinning object has angular momentum and to
change the direction of the angular momentum vector you must exert a
torque on the object. So, if the rotating top is standing straight up
there is no torque on it and so it just keeps spinning, does not fall.
If the top is not standing straight up, there is a torque on it due to
its own weight. Interestingly, though, the top does not fall but rather
undergoes a motion called precession. It is a bit complicated to explain
all this in more detail.

QUESTION:
What would happen if I could extend a straw from just above ground-level to the vacuum of space (ignoring the physics of the weight and length of the straw). I am curious what would happen inside the straw from a vacuum/pressure perspective. Would our atmosphere be sucked out into space?

ANSWER:
The pressure change from top to bottom would be exactly the same as
as the air right outside the straw.

QUESTION:
I've read that scientists haven't been able to fully decipher spectral analysis of the sun, while here I thought all the elements in the Periodic Table were already identified. Have there been any ideas put forth about what these elements might be?

ANSWER:
The sun is a very different environment than on earth. Most of the
atoms whose spectra we can see are not neutral atoms but are ionized,
that is, they are missing some of their electrons. I doubt that we have
detailed data on the spectrum of neon with only three electrons, for
example.

QUESTION:
I am a professional juggler. One of the patterns I juggle has balls thrown to two different heights. The right hand throws across to the left and vice versa. Basically I know that the balls thrown from the right hand reach their peak in the same amount of time as the balls thrown from the left take to peak and fall down again (to be caught by the right at the same level as they were thrown).
When I learned this juggling pattern I was still in school and I remember using Newton's laws to work out that the ball thrown from the right hand peaks at exactly 4 times the height of the ball thrown from the left. (If you google this, a ball dropped from 4m will fall to the ground in 9ms and a ball dropped from 1m will fall to the ground in 4.5ms)
Please tell me how I worked out the relative heights using Newton's formulas and my knowledge of the relative time the balls take in the air! (If you need to see an example, you can google it, it's a standard juggling pattern known as "triple singles" because of the triple spins thrown if you juggle it with clubs.)

ANSWER:
This requires only that you know the equations of motion for uniform
acceleration because the ball has a constant acceleration of g =-9.8
m/s^{2} when in flight (the minus sign meaning it is speeding up if
falling, slowing down if rising. These equations are y=v _{0} t- �gt ^{2}
and v=v _{0} -gt where v is the velocity at time
t and y is the height (at time t ) above where it was
thrown (at time t =0) with velocity v _{0} . So, we have
one pair of equations for the ball that goes higher (I will call it #1) and
another pair for the other ball (#2):
y _{1} =v _{01} t- �gt ^{2}
and v _{1} =v _{01} -gt and
y _{2} =v _{02} t- �gt ^{2}
and v _{2} =v _{02} -gt.
Suppose #1 goes to a height h _{1} before coming back. The
time t to reach h _{1} is found by knowing that v _{1} =0
at that time, so t=v _{01} /g. Put that time into the
equation for y _{1} and find h _{1} =�v _{01} ^{2} /g .
Now, you tell me that you want #2 to make the full trip up and down in the
same time; that means that it will make the trip up in half that time, t =�v _{01} /g.
If you now put this time into the equation for y _{2} (which
we would call h _{2} at this time), you will find that h _{2} =⅛v _{01} ^{2} /g= �h _{1} .
Of practical interest to the juggler is that you must give #1 twice the
initial velocity of #2 to achieve the desired heights.

QUESTION:
Why are atoms so small? What law dictates that atoms have to be so small? Why are there no "big" atoms? Why are their size limited?

ANSWER:
Atomic structure is determined by the constants of nature: what is
the mass of an electron, what is the mass of a nucleon, what is the
strength of the electrostatic force, what is the strength of the nuclear
force, what is Planck's constant. People who think about the problem of
the values of fundamental constants emphasize that even a slight
difference in the values of nature's constant would make our universe
completely different �stars might
not form, atoms might not be able to make molecules, etc .

QUESTION:
What is the speed of expansion of Universe?
Is this speed is increasing?

ANSWER:
As I have stated, I am not an astronomer and usually avoid
astrophysics questions. I can give you a rough idea of the numbers you
seek, though. The most distant object yet observed is about 13 billion
light years away and has a red shift of about z =10 which
translates into a speed about 98% the speed of light. Keep in mind,
though, that this was the speed of this object 13 billion years ago.
Also, the speed at which objects recede from us depends on how far away
they are. Regarding whether the speed is increasing, recent observations
indicate that it is; in fact the most recent
Physics Nobel Prize was for this discovery. As to why it is speeding
up, it is a mystery right now and is related to
dark energy which
is the term normally associated with this acceleration.

QUESTION:
You have on many occasions (most recently, several days ago) you been asked about "centripetal & centrifugal" forces. As with all classical mechanics texts I've read, you have explained that (with respect to rotating bodies) the centrifugal forces are fictitious. . As a long time reader, I sometimes scan through old answers for thought & study. In the very last archived section & first question, you were explaining the Earth's obliqueness @ the equator. You said that "The earth is actually slightly oblate, that is, fatter at the equator than the poles, because it is rotating and the centrifugal force pulls it out at the equator". What centrifugal force? Are you referring to the net gravitational force exerted by the moon, or to something else like to a set of axises whose origin is located on the spinning equator? A point on the equator is certainly being subjected to a centripetal acceleration toward the earth's center. These forces are sometimes called tidal forces?

ANSWER:
I love the little cartoon on the recent post .
If you are in an accelerated frame of reference and try to use Newton's
laws to understand how things behave, you can only succeed if fictitious
forces are introduced which simply means that in the accelerated frame
things behave as if those forces were there. In a rotating frame there
are several fictitious forces, centrifugal being one of them. Consider a
situation where you have a spinning rigid sphere and you are on the
equator. You would experience two real forces, your weight W and the
normal force N up which, let us say, is being measured by a scale. If
the scale is very accurate you would note that the two were not equal
even though you are in "equilibrium". The reason is simply that you are
not in equilibrium but moving in a circle so the net force cannot be
zero. But, if you want to have Newton's 1st law be true in your frame,
you must invent an outward "force" C equal to the negative of the sum
N +W such that N +W +C =0. If the sphere is not rigid but plastic, the force
C will cause the sphere to squish out. As the rotation
gets faster, the centrifugal force gets bigger. It has absolutely nothing to do
with the moon or tidal forces or any real force at all, it is a result
of the noninertial frame behaving differently from an inertial frame.
Another interesting fictitious force in a rotating system is the
Coriolis force which causes the circulation of weather systems. An
explanation is given in the video below.

VIDEO

QUESTION:
why it is happen that when we look outside window of moving car ,nearby objects move faster and far off objects seems to be stationary

ANSWER:

Parallax .

QUESTION:
A plane wave consists of in phase electric field and magnetic field components. There are node points in space where the fields go to zero. What happens to the wave energy at those points?

ANSWER:
If at any instant the field is zero, the energy density at that
point is zero. The energy density at any point is time varying as the
field changes. The total energy of the whole wave (summed energy density
over the whole volume of the wave) remains constant.

QUESTION:
The Michaelson/Morley experiment could not detect the ether. If you factored out the ether issue should there have been a difference in the 2 split beam lights using Newtonian physics? If so and you didn't detect any difference in speed that would have been revolutionary regardless of the ether issue. To conclude that the absence of a difference regardless of an ether means the speed of light emitted will always be the same makes Einstein's conclusion all the more remarkable. I can't get an answer anywhere. They just deal with the failure to detect an ether.

ANSWER:
19th century physics assumed that all waves necessarily travel
through a medium. Since light was known to pass through a vacuum, it was
assumed that there is some mysterious medium which we had not yet
detected and called the luminiferous
� ther.
(Sound familiar? Dark matter?) Whatever it is called, the
Michaelson-Morley experiment was attempting to detect the earth's motion
through that medium. So, what the experiment showed was that there is
no medium . Simply from that, it follows that the speed of light must
be the same regardless of the motion of the source or observer, so it is
often taught in physics courses that this was the origin of relativity
since Einstein must have realized that the negative result implied
constancy of the speed of light. I have read that Einstein once said
that he did not remember whether or not he was aware of the results but
they certainly did not play a role in his coming to the conclusion that
the speed of light is the same for all observers. The reason is easily
found in classical electromagnetism, Maxwell's equations; see an
earlier answer .

QUESTIO N:
What is the maximum speed a human body can accelerate to if sliding at a 30 degree angle for 3 miles? I am watching Alien vs Predator and the premise is there is a pyramid under the ice in antartica. the predators drilled a 30 degree angle hole three miles deep to get to the pyramid. someone fell into the hole and slid all the way down. I do not doubt that he would be going very fast probably fat enough to splatter on any surface he hits once he reaches the bottom. I am just curious at the maximum speed he would accelerate to and if he would even be able to breathe. not to mention how long it would take to slid 3 miles at that angle.

ANSWER:
If there is no friction, sliding 3 miles at a 30^{0} angle
will end with the same speed as a vertical drop of 1.5 miles. This is
about 488 mph. The time for the sliding would be about 45 seconds. Of
course, there is sliding friction which could be made to be very small
with a very slippery surface, but the air drag would become very
significant at high speeds so there would be some terminal velocity when
he would stop accelerating.

QUESTION:
Why the electron don't collapse into the nucleus?

ANSWER:
Why the earth don't collapse into the sun? An object moving under
the influence of an attractive central force can move in a stable orbit
indefinitely. The puzzle with an electron is that, because it is an
accelerating electric charge, it should radiate electromagnetic energy
and spiral into the nucleus. But, that is what happens classically and
quantum mechanically (which is what governs the physics on the atomic
scale) it does not radiate if it is in an "allowed" orbit.

QUESTION:
Would you give an example of how centripetal force (if I am correct in thinking and not confusing centrifugal force) helps keep the lacrosse ball from falling out of the net when cradling (semi circular motion) the ball. She is in 7th grade and used the example of the water bucket spinning Ina circle to explain to her friends why it works, but I would like a more thorough explanation.

ANSWER:
I will try to make this comprehensible to a bright 7th grader.
Newton's second law says that what a force does is cause an object to
accelerate. In everyday life, accelerate means speed up or slow down,
but in physics it can also mean that the velocity changes direction
without speeding up or slowing down. So, when something moves in a
circular path, even with constant speed, there has to be some force
present. Here is one example. Consider the International Space Station
which moves in a circle about 200 miles above the earth with an
approximately constant speed of about 18,000 mph. So it is accelerating
because it is moving in a circle, always changing its direction. What is
the force which causes that acceleration? It is simply the only force it
experiences, its own weight which is a force which points toward the
center of the earth. A force which causes a change in direction is
called a centripetal force and must have a magnitude of mv ^{2} /r
where m is the mass, v is the velocity, and r is
the radius of the circle. Now, a more down-to-earth example. If you have
a ball of mass m attached to a string of length r and you
wish to swing it over your head, when it passes over the top of the
circle it has two forces acting on it, its weight and the tension in the
string, both downward forces. If it is going very fast, the tension in
the string will be very big, maybe big enough that the string would
break because the string was not strong enough. As it goes slower and
slower, the required centripetal force gets smaller and smaller so the
tension also must get smaller because the weight cannot change.
Eventually you will be able to go over the top with the tension in the
string zero, that is the string is not needed at that instant. But, that
is the smallest speed you can go because if you try to go slower, the
weight will be bigger that the centripetal force needed and so the
string will go slack and the ball will fall out of its circular path. If
you had used a stick instead of a string attached to the ball, you could
go through as slowly as you wished because, unlike a string, a stick can
push. There is no such thing as a centrifugal force, it is called a
fictitious force and is added to make Newton's laws work in an
accelerating reference frame. E.g ., if you are sitting in a car
speeding up, for example, you would swear that there is a force pushing
you back but there isn't; you feel the back of the seat pushing you
forward to accelerate you and your brain interprets this as being pushed
back against the seat. See cartoon at the right.

QUESTION:
When the total energy for an atom is calculated I've seen terms for the electrical potential energy and the angular momentum of the electron around the nucleus. However, I have not seen the angular momentum component of the atomic particles about thier own center of mass. Why is this term ((moment of interia)*(angular velocity)^2) excluded?

ANSWER:
The total energy of an atom of mass m (at rest) is simply
mc ^{2} . To calculate this you have to know all the binding
energies of electrons in the atom and of protons and neutrons in the
nucleus. But, you never say that there is a contribution of
�Iω ^{2} (I think you are missing the
�) from the orbital motion of the electrons because
this is a quantum mechanical system and this is a classical expression
(incorrect). Intrinsic angular momentum (usually called spin) is simply
an intrinsic property of a particle and any effect it might have on the
energy of the particle is just part of what we measure as mass.

QUESTION:
Does a black holes violate the first law of thermodynamics? I mean isn't energy and matter the enter a black hole is lost reducing the energy of the Universe?

ANSWER:
When an object is captured by a black hole, energy is conserved. As
black holes capture more and more objects, their energy (mass) gets
bigger and bigger.

QUESTION:
I am looking for the formula or way to calculate how much thrust/energy is needed to launch something into space. It is a little more complicated than that though, I want a way where I can find the differences in the amount of energy or thrust needed to propel an object of X weight from different altitudes into space (Ex. from sea level and then from 1,000 ft.)

ANSWER:
Thrust and energy are two completely different things. Thrust is the
force you must apply and that could be anything greater than or equal to
the weight of the object. If the weight were 1000 lb, you could exert an
upward force of 1001 lb and it would eventually escape "into space" by
which I presume you mean far enough that the earth's gravity is
negligible. The energy E required can be found from knowing the
potential energy U of an object of mass m in the gravitational field of a mass
M . This is given by U=-GmM /r where G =6.67x10^{-11} ,
r is the distance m is from the center of the earth, and M
is the mass of the earth. If m and M are in kilograms and r is in
meters, U will be in Joules. The energy required to lift m
from r _{1 } to r _{2 } is E =GmM (1/r _{1} - 1/r _{2} ).
Don't forget that for sea level r _{1 } is the radius of
the earth, that is the r 's are not altitudes, they are distances
from the center of the earth. There is your formula.

QUESTION:
I've recently gotten into the hobby of amateur electronics and one thing that perplexed me was current draw. It's my understanding that a component will "only draw as much current as it needs". So for example if I have a 20milliamp LED it will only draw 20 milliamps, even if my power source is capable of supplying more.

ANSWER:
If you look into the specs of the device, you will see that it is
rated to draw 20 milliamps for a particular applied voltage . If
you apply less voltage, the bulb will draw less current and if you apply
more it will draw more (or be destroyed).

QUESTION:
If a ball is thrown straight up into the air, is the acceleration at the highest point (apex) zero or non-zero? And why

ANSWER:
The definition of acceleration is rate of change of velocity (just
like velocity is rate of change of position). Even though the ball is at
rest at this instant, its velocity is changing. You can tell this
because a very short time before (after) the ball's velocity was (will
be) nonzero.

QUESTION:
i wanna ask u that infinite specific heat mean the amount of energy you add up to the object but its temperature doesno increase at all (as i read this definition in one of the answers given by u) so where did the energy go then....and does that mean that the temperature of any object will not rise ever if infinite specific heat is supplied?

ANSWER:
The kind of thing you are asking about, although I never heard of
its being called "infinite specific heat" happens only if there is a
phase change. The best known examples are melting (or solidifying) and
vaporization (or condensation). For example, it takes energy to remove a
molecule from a liquid and have it become a gas molecule; that is where
the energy goes.

QUESTION:
''Bernoulli's principle does not explain, the lift on a JET PLANE''. This statement is true or false?

ANSWER:
Let's just say it is not the whole picture. See
earlier answer .

QUESTION:
In the two slit experiment, photons are directed through two slits and form a wave defraction pattern on a wall. If an observer tries to "measure" the photons, the resulting pattern on the wall changes. If this test were run as a double blind experimant, where one set of observers are the "measurers" and one set are the observers (with no idea of whether or not an attempt to measure is being done), do the second set of observers see the light pattern on the wall change? If so, would that suggest that one "measurer"'s action can affect the "measured"s actions for the whole universe?

ANSWER:
The first measurement puts the photon in a particular state (having
passed through a particular slit) and therefore any subsequent
measurement will get the results consistent with that. Therefore, it
makes no difference whether it is the same or different observers who
measure the photon going through a particular slit. Since the photon is
part of the "whole universe", any measurement of the photon affects the
universe.

QUESTION:
I'm a big fan of SpaceX and space in general. So as you may know the Dragon Spacecraft is performing maneuvers to dock with the station. At this moment I'm watching the Dragon orbiting about 2 miles directly below the ISS.
My question is: how can the spacecraft hold its position below the station when it should need to be going faster given that it's closer to the earth. Or is the difference so small that the effects of orbital decay are not even noticeable until many orbits later?

ANSWER:
Kepler's third law says that the ratio of the periods of two orbits
is equal to the square root of the ratio of the cubes of their radii:
T _{1} /T _{2} =(R _{1} /R _{2} )^{3/2} .
Suppose that R _{1} =R _{2} + δ ,
then
T _{1} /T _{2} =(1+( δ /R _{2} ))^{3/2} ≈1+3δ /(2 R _{2} )
where I have used the binomial expansion to approximate the root because
certainly
δ /R _{2}
is very small. Now, put in some numbers: the radius of the orbit is
about 4000 miles and
δ
is 2 miles, so T _{1} /T _{2} =1+6/8000=1.00075.
This means that if T _{2} is 90 minutes, T _{1}
is 90 minutes + 4 seconds. So, your last statement wraps it up pretty
well.

QUESTION:
i read in my previous classes that light consists of photon packets...
i want to know that what do photon packets exactly means...
if photon packets are matter..
does that mean that what we see as light is just ignited photon packets?

ANSWER:
Light is an electromagnetic wave and, like all waves, light carries
energy. It was assumed until early in the 20^{th} century that a light wave
could carry any amount of energy. However, many experiments (e.g.
photoelectric effect or Compton scattering) showed that there is a
minimum amount of energy light can carry. That minimum amount is called
the photon or also, more generally, the quantum of the electromagnetic
field. The energy E of a single photon depends on the frequency
f of the corresponding light, E=hf , where h is
Planck's constant. It does not mean that photons are matter because a
photon has no mass. I have no idea what you mean by "ignited".

QUESTION:
Can two sound waves interfere with each other when they intersect orthogonally?

ANSWER:
Yes, if you mean by "interfere" what physicists mean: the net
disturbance at any point in space is the sum of all the individual
disturbances there. This is called the superposition principle.

QUESTION:
Why is kgf considered mass even though it is the weight of a body?

ANSWER:
There is always confusion in "everyday life" between mass and
weight. In countries where metric units are used, kilograms are used as
weight, for example you might buy 2 kg of potatoes. Technically, this is
incorrect because in science, kg is a measure of mass and Newton (N) is
a measure of force and weight is a force. That is why it is a good idea
to add the f to your kgf so you realize that it is meant to be a force,
not a mass. The kgf is the weight of an object in N divided by 9.8 m/s^{2}
which is the nominal acceleration due to gravity. Therefore, 1 kgf=9.8
N.

QUESTION:
Light is scattered as it passes through translucent objects. But we use translucent paper etc. in pinhole camera to see the image from outside. Why the light coming from the pinhole forms an image on translucent paper, why it is not scattered?
How can we see the image from outside of pinhole camera so clearly, why it's not blurred.

ANSWER:
The pinhole forms an image on the paper , so if light can pass
through the paper at all, you can see the image. However, if you try to
look at a distant object through the paper, what you are looking at is
not located on the paper, so that light is scattered which fails to make
an image of what you are trying to look at. Think of putting the paper
over a printed page; what you are looking at is located at the surface
of the paper and you can read it, but if you pull the paper away you
cannot.

QUESTION:
Does sound travel differently across different mediums? Also, does gravity have an effect on sound?

ANSWER:
Yes, sound waves have different speeds and different characteristics
of absorption in different media. Sound is affected by gravity only if
gravity causes the medium to be inhomogeneous, usually just an effect on
the density being different at different heights, like the atmosphere.

QUESTION:
if a heating is rated at 2kw,200v,what does this mean?

ANSWER:
It means that if connected to a 200 volt power source, the device
will consume 2000 watts of power, that is, 2000 joules of energy per
second. If well designed, then most of this energy comes out as heat.

QUESTION:
If a body moving at the speed of light develops infinite inertia and mass, would't it then take an infinitesimal amount of energy to affect such a mass?
Why then is gravity, or any form of energy/matter able to affect light in various appreciable ways?
Could it be justifiably said that all "light affecting" matter/energy systems have an infinitesimal amount of energy and inertia as well?

ANSWER:
I think you must mean infinite, not "infinitesimal". Your question
is not really applicable to light because since it has no mass, it does
not have infinite mass like matter would. You should read my faq page
about how gravity bends light
if you want to understand this better.

QUESTION:
We need to use a recorded pheasant call on an MP3 player to replicate the call of pheasants, for a biology research project. We want it to be the same loudness as a real pheasant.
We used a sound level meter to record decibel levels of pheasants at different known distances from the microphone, at a game farm. We then turned up the volume on the MP3 player until we got virtually the same "curve" of decibel levels vs. distance as we recorded for live birds.
But even so, we cannot hear the recording from nearly as far away as we can hear a real bird calling, under similar environmental conditions. What makes the difference, and what can we do about it?

ANSWER:
Two things occur to me. First you have to worry about the "radiation
pattern" from a calling phesant and from whatever you are using as
speakers. I would guess that the real call is more like a point source,
radiates more or less equally in all directions; your speakers probably
are more directed. So, if the speaker is not pointed directly at you,
you will hear a lower intensity than if they were. You probably
calibrated along the "line of sight" of the speakers and if you had them
pointed right at you, they would probably sound loud enough. The other
possibility is that the MP3 recording does not record all frequencies
which might be present in the natural call. Hence, the readings from the
two sources might be the same but the sound your ear hears could be
different. This latter case, though, would have the recorded sound more
rather than less, I would think.

QUESTION:
If your gravity gets stronger as you go to the center of the earth because of its density, does my gravitational pull become more on the most dense part of my own body?

ANSWER:
Gravity is determined by mass, not density. Furthermore, for a
sphere, the gravitational force depends only on the mass inside where
you are. Therefore the gravitational force gets smaller as you go toward
the center.

QUESTION:
I can't grasp why it takes such a high speed to escape Earth's atmosphere.
Suppose you install a skyhook (OK, it can be purple) in space. You hang
a very long ladder down to the ground. Then a person just walks up it
(assuming some sort of air pack) to the top.

ANSWER:
This is mainly an issue of semantics. Escape velocity is the minimum
velocity you must give a projectile at the surface of a planet for it to
never come back. But, obviously, as you note, if you slowly lift
something up, adding energy as you go, there is no limit to how slowly
you could "escape".

QUESTION:
My question is about the practical business of cosmology. Given the 3-body problem (generalised to an N-body problem) for the planets and comets etc. of our solar system, as described by Poincar� most notably, how do cosmologists calculate the future positions of planets, comets etc. ? Does the accuracy of these calculations vary with the magnitude of variables like time and distance? Do they make simplifying assumptions to make calculations easier? Do they introduce additional parameters perhaps to account for other neighbouring solar systems and other influential masses in the milky way? Am I right in thinking that machine-based computational methods are necessary and that these can take a while to run?

ANSWER:
Beyond the 2-body problem, no celestial mechanics problem can be
solved in exact closed form. Most problems of interest are nearly 2-body
problems and "minor players" can be introduced approximately using
perturbation techniques. This has been done for hundreds of years, in
fact this is how the planet Neptune was discovered: irregularities in
the orbit of Uranus were deduced to be due to some other unknown planet
and calculations by Le Verrier predicted where to look for it and,
voila , there it was! In very few cases of interest are gravitational
forces from outside our solar system of any importance.

QUESTION:
In QM we learn that a wave packet is a superposition of many waves of different frequencies. From this we get the impression that waves of different frequencies do interfere. On the other hand, we study in telecom that different channels are allocated different carrier frequencies so that the waves do not interfere.
The question is, do waves of different frequencies interfere?

ANSWER:
"Interfere" simply means that at any point in space at any time the
total disturbance equals the sum of all the individual disturbances.
This is called linear superposition and is always true. In
telecommunications you have many waves of many frequencies all
interfering like crazy but you tune out everything except what you want
to receive. That is what a radio receiver does, for example.

QUESTION:
I�m trying to calculate what would be my equivalent body mass if I was propelled into a wall at 16kph. What I�m trying to establish is what would be the force of impact i.e. would it be double my weight or more. I�m trying to understand what happens to the body when it�s involved in a car crash.
I have been using the following factors.
Speed 16 kph
My weight 80 kg
Travel distance to hitting an immovable object 0.5m.

ANSWER:
For the thousandth time, there is now way to calculate what you want
with what you have. Your body mass is what it is, there is no
"equivalent body mass". 80 kg is your mass, not your weight. Travel
distance has nothing to do with how much force you feel when you hit the
wall. The average force you feel during impact is given by mv /t
where m is your mass, v is your speed and t is the time it takes you to
stop. In your case, m =80 kg, v =16 km/hr=4.44 m/s, but t
(in seconds) is unknown; so, F =71.1/t . Suppose it takes you 0.1 s to stop. The
average force you would feel would be about 700 N which is almost 10
times larger than your weight. You can find similar answers on the FAQ
page.

QUESTION:
A charged particle moving with a speed 'v' enters a uniform Magnetic field 'B',perpendicular to it, now as it enters it and starts moving in a circular path of constant radius 'r' = Mv/qB (M- mass of particle), this particle has some acceleration, but MAXWELL's ELECTROAGNETIC theory says that a charged particle if it has some acceleration must lose energy in the form of radiation, so radius of the path in the above situation should gradually reduce, but it is constant WHY? (you can take charged particle as proton.) if we balance forces in above situation it comes out to be
Mv^2/r = qvB
which reduces to r = Mv/qb
constant why?
am i missing out some force, or do we have any hidden supply of energy

ANSWER:
In almost all situations involving circular motion of charges, the
energy carried off by the radiation is so small compared to the kinetic
energy of the particle that you would have to wait years to see even a
tiny change of radius. Radiate they do, though.

QUESTION:
I was wondering if there was an equation that would tell me how much electrostatic force is required to counter the downward force of gravity on X amount of water...
I know it has to be equal, I'm more interested in the laws that govern the laws of electricity / static.

ANSWER:
If the electric field is uniform, water will feel no force because
water molecules have zero net charge. Although the water molecule has no
net charge, it does have what is called a dipole moment which means that
one end of the water molecule (where the hydrogen atoms are, see figure)
has an excess of positive charge and the other end has an equal excess
of negative charge. Therefore, the molecule will experience a net force
if the electric field changes magnitude across the molecule. The force
on one molecule is given by F=p ( ΔE /Δx )
where p is the dipole moment and
( ΔE /Δx ) is the rate
at which the electric field changes when you move from one side of the
dipole to the other. A water molecule has a dipole moment of p =6.2x10^{-30}
C∙m and a weight of W=F =2.5x10^{-25} N. Therefore
( ΔE /Δx )=4x10^{4}
(N/C)/m to levitate the molecule. This means that the electric field
must change by 4x10^{4} (N/C) as you go 1 meter, or across the
size of an atom, about 10^{-10} m, a change of 4x10^{-6}
N/C. This does not depend on how much water you have as long as
( ΔE /Δx ) is the same
everywhere.

QUESTION:
This may be simple to answer but do you generate any forward momentum on a treadmill....i say no but cant seem to prove that to a stubborn friend

ANSWER:
Momentum is a vector and so you must stipulate momentum with respect
to what. On a treadmill, your momentum relative to the ground is zero
but your momentum relative to the treadmill is the same as if you were
walking or running at the same rate on the ground. Note that at the same
rates, you actually do more work running on the ground than the
treadmill because of air drag not present on a treadmill.

QUESTION:
Are electrons in every atom equal? If that is the case, can it be that when two objects touch their electrons get mixed or interchanged during contact?

ANSWER:
All electrons are indistinguishable. Even in a single atom (with
more than one electron), you cannot label the electrons. Each electron
is really all electrons in the system. The quantum mechanical way to
insure that electrons are not in some way labelled is called
antisymmetricization, sometimes referred to as an exchange interaction.
Similarly, if two atoms interact with each other, any electron could be
in either atom.

QUESTION:
I have a gravity question in a sky scraper on an upper floor Am I feeling gravity from the floor or from the earth and if it is from the earth why don't I feel more of pull on me the higher I go

ANSWER:
Technically, both the earth and the floor exert gravitational forces
on you, but the floor has such a small amount of mass that its
gravitational force is negligible. Regarding your second question, the
nature of the gravitational force is to get smaller as you get farther
away from the source. On top of the skyscraper, however, you will not
notice an appreciably different force because you have not moved far
enough away from the earth. If the skyscraper were as tall as the radius
of the earth, the force you feel would be four times smaller.

QUESTION:
Since the free electron of copper are moving around some at high speed why do they not escape from solid altogether?

ANSWER:
It takes a certain amount of energy to remove an electron from a
metal, often called the work function. The kinetic energy of the
electrons in the metal at normal temperatures is smaller than the work
function so they do not escape. However, if you heat the metal up,
electrons will be emitted if it is hot enough; this is called thermionic
emission.

QUESTION:
Suppose we have Bob the astronaut sitting without a space suit in a spaceship full of air on a mission to Mars. Bob is very fond of balloons and is holding on to a nice, big, red helium balloon via a piece of string. Bob is sitting facing the front of the spaceship. Mission Control decides to slowly accelerate the spaceship. In which direction will the balloon move relative to Bob? Why?

ANSWER:
I do not like this question because the reason a helium balloon
floats is that the buoyant force, which floats it, arises because the
pressure in the air is bigger underneath the balloon than above it; on a
mission in empty space, the pressure everywhere in the cabin is the same
and so the balloon would not go up! Let's just have Bob ride in an
accelerating car right here on earth. There are three forces on the
balloon, its weight W , the buoyant force B ,
and the tension in the string T . B and
T are both vertical, and so, for the balloon to have an
acceleration in the direction of the acceleration a of the
car, T must have a horizontal component in the direction
of a . All this is shown in the figure to the right.
Therefore the balloon will move backward opposite the direction of the
acceleration. In the spaceship where there would be no buoyant force and
no weight, the balloon would appear to accelerate backwards until the
string was straight and "horizontal".

"ENHANCED" ANSWER:
It has been pointed out to me (again, thanks to Michael Weissman at Ask
the Van ) that my answer would be correct only if there were no
air in the spacecraft. What I failed to think about was that if there is
air in the cabin, forward acceleration will cause the pressure at the
rear of the cabin to be greater than at the front; therefore there would
be a buoyant force on the balloon from back to front. So, if the mass of
the balloon is less than the mass of an equal volume of air, the
direction the balloon would move would be forward, not rearward. If you
are familiar with the equivalence principle, it is even easier to
understand. The equivalence principle states that there is no experiment
you can perform which can distinguish between an accelerated frame and
being in an unaccelerated frame in a uniform gravitational field with
the same accelerateion (due to gravity). So, if the acceleration of the
ship were g , the balloon would have to behave just the same as it
would on earth except forward now would play the role of up on earth, so
the balloon would go forward just like the balloon on earth would go up.
(My example of the car on earth, above, would also have a "pseudo
buoyant force" in the direction of a , so both this force
and the horizontal component of the tension would be in the direction of
a meaning that the balloon would always go the opposite
direction of a here on earth.)

QUESTION:
Consider a vehicle that we want to move from point A (with a starting velocity of zero) to point B (with an ending velocity of zero) as fast as possible. The vehicle has a maximum acceleration limit and a maximum deceleration limit. It would seem that the fastest way would be maximum acceleration until your vehicle's velocity and distance to the goal indicate you are at the maximum deceleration limit. How can you calculate the maximum velocity reached, location of the transition point and the time to reach the goal?

ANSWER:
Interestingly, I have recently
answered precisely this question.
To get your transition point, call it w , you must add w=x _{1} (t _{1} )=�a _{1} t _{1} ^{2} =s /(1+(a _{1} /a _{2} )).

QUESTION:
I know that ionization energy can be looked up in tables on a per element basis.
Is ionization energy further refined by the isotope?
For example, do Hydrogen-2 and Hydrogen-5 have different first ionization energies?

ANSWER:
When you look at atomic spectra, there are very tiny splittings of
lines called hyperfine structrue. Hyperfine structure has more than one
cause, but one is the fact that the energy spectrum of an atom is
affected by the isotope. The reasons are that different isotopes have
nuclei of slightly different sizes and that different isotopes can have
different nuclear spins which couple to the spins of the electrons. So,
if the energy spectrum of an atom is affected by isotope, the ionization
potential will be affected. That said, the effect will be almost
unmeasurably small.

QUESTION:
If you are given one photon how can its velocity c be a constant without violating the uncertainty principle? If delta (momentum, velocity) is zero then the delta x seems infinite?

ANSWER:
For a photon, the velocity and momentum are not proportional. The
enegy of a photon of frequency f is E=hf and the momentum
is p=hf /c. (h is Planck's constant and c is
the speed of light.) However, there is still an issue with the
uncertainty principle. If you know the frequency exactly (hence the
momentum), you are totally ignorant of the position. Any photon for
which you know a finite region of where it is, must have an uncertainty
in frequency.

QUESTION:
I'm an eighth grade science teacher who needs a simple explanation and I am pooped trying to locate it on the Internet. I understand the basic idea of an electromagnet. I also understand the basic methods to change the strength of the electromagnet: change the amount of current, number of wire loops, etc. The problem is that one source states that the diameter of the iron core also has an impact on strength. A second source states that the DISTANCE from the core to the wire has an impact.
Question 1: Is it only the diameter of the core? Not simply length? If I use a LONGER iron nail with the same diameter as a shorter nail, will that impact the strength of the electromagnet? Why?
Question 2: We just take our students to the lab and wrap wire around a nail and hook it to a battery to make an electromagnet. What does distance from the core to the wire do?

ANSWER:
First some basics about long solenoids. The magnetic field inside is
uniform (constant all the way across the area) as long as you are not
close to the ends and has a magnitude of B= μ _{0} NI /L
where N is the number of turns, I the current, L
the length of the solenoid, and μ _{0 } is the permeability
of free space. First note that the diameter of the solenoid does not
matter, so the field inside is the same regardless of the diameter. So a
nail put inside a solenoid which wraps tightly around it is the same as
the field the nail experiences if the wire has the same number of turns
per unit length but a much bigger diameter. Therefore, since it is the
field which causes the iron to become magnetized, both situations will
give rise to equally magnetized nails. The proviso is that the ends of
the nail cannot be close to the ends of the solenoid. The fatter
solenoid will have a weaker field at the ends. If you have a nail twice
as long it will have the same strength as one with half the length as
long as there are twice as many windings; what matters is N /L .

QUESTION:
I've read that Uranium 235 has a half life of about 700 million years and Plutonium 239 has a half life of 24,100 years yet when the United States dropped "Little Boy" which contained Uranium 235 on Hiroshima and "Fat Man" which contained Plutonium 239 on Nagasaki, the radiation levels quickly disappeared within a year, nowhere near the 24,100 years let alone 700 MILLION years so my question is, how did the radiation levels disappear so quickly?

ANSWER:
It is not the radioactivity of the fuel which is the problem, it is
the radioactivity of the fission products. When the uranium or plutonium
nucleus splits (the source of the energy), there are two much lighter
nuclei produced which are far from stable nuclei and these subsequently
decay with half lives often on the order of decades. If you took the
uranium in the bomb dropped on Hiroshima and scattered it uniformly
around the city, the level of radioactivity would be so small as to be
difficult to even detect.

QUESTION:
Our son perished in an automobile accident in 2005 in which he hit a tree. Since then I have noticed many instances that when people hit other cars, walls, telephone poles, and the like, somehow survive. However if a tree is involved, the outcome is usually mortal. Is there a reason related to physics which may account for this?

ANSWER:
My condolences on your son's death. The important thing to think
about in an accident is the force to which somebody is exposed. In turn,
because of Newton's second law, F=ma , the acceleration determines
the force you experience �the bigger
the acceleration a , the bigger the hurt.
Acceleration is the rate of change in velocity, a = Δv /t
where
t is the time to stop
and Δv for coming to a dead stop is just v , the speed you
hit with.
So if v is big or t is small, the force will be big. A
large tree is essentially immovable, and so the result is that the time
of collision is very short resulting in a very large force. Of course,
if you have an airbag, that results in the person stopping more slowly
than the car, often saving a life. On the other hand, if you hit
something which can move or break, the time of collision is made longer
and so the force is made smaller.

QUESTION:
I often hear how we "Don't know where the electron is". What do these physicists mean when they don't know where the electron is? Isn't it orbiting the nucleus of an atom like all the picture I've seen in science books and so on or are they referring to something completely different?

ANSWER:
The picture of electrons going around in neat little well-defined
orbits is useful but only as a rough picture, not at all an accurate
description of an atom. I think the best way to think of an atom is that
it is distributed around the volume near the nucleus, like a cloud. This
cloud represents the probibility of finding the electron at any
location, if dense it is likely to find the electron there, if not, it
is not so likely. The electron may be thought of as being wherever the
cloud is, so you know a volume in which you will find it, but you cannot
predict exactly where it is at any time. In the picture to the right,
you can see where the "orbit" is, where the electron cloud is densest.
Even if it did move in an orbit, you would have now way of knowing
precisely where it was at any time.

QUESTION:
Heavy unstable nuclei usually decay by emitting alpha or beta particle but why not single proton or neutron?

ANSWER:
It is all a matter of energetics. If a particular decay mode does
not take you to a state of lower energy, it does not happen. There are
certainly cases of neutron decay (in very neutron-rich nucle) or proton
decay (in very proton-rich nuclei), but they are so far from naturally
occuring nuclei that they are seen only if we take a lot of trouble to
create them. Beta decay turns out to be a very efficient way of changing
a proton to a neutron or a neutron to a proton inside the nucleus if it
is slightly proton- or neutron-rich. Alpha decay happens in very heavy
nuclei because an alpha particle is very tightly bound and therefore has
a reasonable probability of forming inside a nucleus.

QUESTION:
What is difference between displacement amplitude and pressure amplitude and which has more direct influence on loudness of sound?

ANSWER:
Sound waves are longitudinal waves which means that the atoms in the
gas oscillate in the direction parallel to the direction the wave
travels. An average atom, if you watch it, vibrates back and forth. So
you can characterize the wave by specifying the frequency and maximum
value (amplitude) of this oscillating displacement. The loudness
(intensity) of the sound is proportional to the square of the amplitude.
The figure to the right indicates this displacement vibration with the
little red arrow. But, because all the atoms are vibrating like this,
the result is that at any time the atoms in some regions are more
tightly packed than if there were no sound and, in other regions, less
densely packed. Hence the pressure at any particular place changes with
the frequency of the sound. The intensity of the blue circles in the
figure indicate the pressure at any time. The amplitude of this pressure
oscillation is the maximum deviation from the pressure in the gas if
there were no sound. Again, the intensity is proportional to the square
of the amplitude of the pressure oscillations, but it is obviously a
different proportionality constant.

QUESTION:
I am using a drive roller to push against a round steel ring to rotate a turn table. The drive roller pushes against the plate by way of a spring so that it maintains contact with the plate. I understand that a roller that would provide more friction would more likely prevent the roller from slipping - A steel roller would slip more than say a rubber roller. My question is: If I use a soft rubber roller it will deform more than a hard rubber roller. The more it deforms the more surface area of contact I have so it slips less. If the force of my spring stays the same, does some of the force that is pressing against the ring get used to defore the softer roller? - I may get better traction even if some of the force is used to deform the roller, because I am increasing my surface contact and co-efficient of friction, but does deforming the wheel steal some amount of force away from ring?

ANSWER:
This is not a simple question because friction is seldom as simple
as presented in elementary physics courses. Some of your assumptions are
true, some are not. In particular, you are right that it is best to
choose a material with a larger coefficient of friction to minimize
slipping �rubber on steel is better
than steel on steel. But your assumption that more surface area gives
more friction is not true. The main reason that wide tires, for example,
have more traction is that the rubber used to make them has a higher
coefficient of friction than that used for narrower tires. This is an
oversimplification but is a good first approximation. If you press
harder on the deformable wheel you will get more friction but not
because the area gets larger, rather because the force pressing the surfaces
together gets bigger. See an
earlier answer for a
very detailed discussion. And you are right that it costs energy to
deform the wheel and most of this will be lost, so it takes much less
power to run a harder wheel. This "lost" energy is not really lost but
appears in the wheel heating up which might cause problems for your
application.

QUESTION:
I always hear about how certain physics works best for large objects and some other set works for very small objects. Is there a field of study for the points in between? I'm wondering if there's been a measurement of the points where the equations fail.

ANSWER:
Nanophysics, which studies materials whose sizes are on the order of
nanometers, is concerned with the issues at the interface between
quantum and classical systems. For example, when does a cluster of atoms
get big enough that its properties are the same as the bulk properties,
properties like conductivity, specific heat, bulk modulus, etc .?
A branch of nanophysics, mesophysics, is also interested in these types
of problems.

QUESTION:
We were having a discussion at work regarding calculating tension in ropes that were connected at a common point where a weight was attached. We kept getting different answers and now no one knows who is correct. There are two ropes connected at point O with a weight of 10 lbs. The rope on the left is designated AO and rope on the right designated BO. Rope AO is at an angle of 70 degrees from the horizontal and rope BO is at an angle of 30 degrees from the horizontal. Therefore rope AO is shorter than BO. I have calculated the tension in rope AO to be 8.79 lbs and tension in rope BO to be 3.48 lbs. These makes sense to me since the ropes are at different angles the tensions in each would be different. But why don't they add up to 10lbs? How could the tension of each rope added together be greater than the weight acting upon it?

ANSWER:
Since your answers are right, I will not go through the solution to
the problem. Each rope is doing two things: holding up part of the
weight and pulling on the other rope. It is the "pulling on the other
rope" part which results in your "paradox". Part of the falacy in your
reasoning is that forces are vectors and you cannot add two forces by
summing their magnitudes and expect it to mean anything. If you add your
two tensions properly, the horizontal components will be equal and
opposite and add to zero, and the vertical components will add up to 10
lb.

QUESTION:
Which string of guitar produces highest pitch?

ANSWER:
That depends on how the instrument is tuned. A stringed instrument
is tuned by adjusting the tension in the strings. The speed of waves on
a string which has a mass density
μ kilograms per meter and tension T in Newtons is v =√(T /μ )
in meters per second. Since v=fλ where f is frequency
and λ is wavelength, f =(√(T /μ ))/λ.
For a string like a guitar string, both ends are fixed so the length is
L and the fundamental pitch is that for which λ= 2L , so
the frequency is f =(√(T /μ ))/(2L ).
So, there are three ways you can change the pitch:

Change T ;
this is what you do with tuning pegs.

Change μ ;
note that the lower-pitch strings are often thicker than the
higher-pitch strings.

Change L ;
this is what you do when you press the string to the fret with your finger.

QUESTION:
when we see the blue arc of electricty, for example we take a batterie and place wires on positive and negitive ends then slowly bring them together, what speed is this blue arc of electricity traveling ? i ask because the blue arc is light ? so that must mean its travleling at 300,000,000 meters per second ? correct?

ANSWER:
The filament in a light bulb produces light we can see. Is that
filament moving with the speed of light? No, it is sitting still. And
the electrons moving in the filament which cause it to glow are moving
very slowly. The arc you see results from air molecules breaking down in
a very strong electric field between the two wires, causing electrons to
jump from atom to atom causing the atoms to give off even more electrons
and so forth, a cascade effect. Nothing in the arc moves with a speed
anywhere close to the speed of light.

QUESTION:
why can't infa-red knock off an electron?

ANSWER:
An infrared photon does not have enough energy to detach an electron
from an atom or to remove an electron from a metal. The energy of a
photon is proportional to its frequency, inversely proportional to its
wavelength. If an infrared photon could knock out an electron, so could
visible light which would mean that everything around us would be
continually being charged just by having light shined on it.

QUESTION:
Regarding power in a water stream - web/text book searches reveal the formula (neglecting losses and efficiency) to be:-
P = density of water (Kg.m^-3) x flow rate (m^3.s^-1) x water head (m) x g (m.s^-2)
Now I can fully understand the first two functions. It is intuative that power will be proportional to density and flow rate, but why the last two functions? Surely flow rate itself is a function of both head of water and gravity, so why are they incorporated again?

ANSWER:
I do not want to get into the derivation of this formula. I can tell
you that you can often tell whether something is right or not simply by
looking at the units and seeing if they come out right. Power is energy
per unit time. The unit of energy is the watt, 1 W=1 J/s=1 kg m/s^{3} .
What if you used only the first two terms of your equation? The units
would be kg/s, the rate of mass flow in the stream; this is not
dimensionally correct. If you multiply by g you convert this to
weight of water (a force) per second. If you multiply this times a
distance (head), you convert it to work per second, that is power. Note
that your equation is dimensionally correct.

QUESTION:
For a body exerting a gravitational force that is not a black hole (i.e. it has a finite gravitational force), there must be a point at which its gravitational strength stops increasing. On Mt. Everest, gravity is slightly less strong than at sea level. But at what point does the gravitation pull stop increasing? If I'm on Earth, is there an event horizon equivalent, say, if I plunged down towards the middle, or is it a point at the dead center?

ANSWER:
Let us say the body is a sphere with uniform mass density. As you
get closer the gravitational force increases until you reach the
surface. Then it starts decreasing and becomes zero at the center. The
reason is something called Gauss's law which says, essentially, that
only the mass inside where you are probing the force effects it.
Therefore, if you are a distance r inside the earth, only the
mass in a volume 4 π r ^{3} /3
exerts force on you. Since the gravitational force depends on 1/r ^{2} ,
the force inside decreases linearly with r as you go to the
center.

QUESTION:
How much faster does time pass, out in the middle of nowhere in a space that is not effected by gravity? Like how much faster will time go compared to a clock on earth if we stuck another clock out in a magical spot in space where the gravitational pull of galaxies would not effect it?

ANSWER:
The equation for gravitational time dilation is
√(1-(GM /(Rc ^{2} ))) where G is the universal
gravitational constant, M the mass of the object, R the
distance from the object, and c the speed of light. At the
surface of the earth this is approximately 1-7x10^{-10} and, for
R= ∞ it is 1. So time passes about 7x10^{-8 } % faster in
empty space.

QUESTION:
I am a writer and I am writing a science fiction novel with some events
taking place in a space station. My question is: Is it possible to have
a zone in the station that has gravity while another zone (rather
adjacent one) has zero gravity? This is important to me because although
the general plot of the story is fiction, the science part of it has to
be true to what we know so far to be possible.

ANSWER:
The only way we know to create "artificial gravity" is to be in an
accelerating frame. For example, an accelerating elevator in empty space
will appear to have a gravitational field �you
could stand on the floor, would feel your "weight". A more practical way
to achieve acceleration is to move in a circle or radius R with
speed v . You would then experience an acceleration v ^{2} /R
which, if equal to earth's gravitational acceleration (9.8 m/s^{2} ),
would feel just like the gravity you were used to. You probably saw a
similar thing on the movie 2001: A Space Odyssey , where Dave
exercized on a spinning track. You could have an adjacent part of the
station which was not rotating. Gravity which just switches on, like on
the starship Enterprise on Star Trek , is not possible with
currently known physics.

QUESTION:
how can an object reach a limiting terminal velocity while the gravity express theory says if we dig a hole through the planet we can jump in said hole and reach the other side in 42 seconds? doesn't that mean there is no terminal velocity and due to the gravitational efects of a conatant 32 ft per second will occur no matter the weight or height. don't they contridict themselves? if it takes 42 seconds to get to china thres a hole in the center of the earth won't that exceed terminal velocity? if not, how?

ANSWER:
A falling object has a terminal velocity because of air drag. I
believe the time to pass through a hole through the center of the earth
is 42 minutes, not seconds. And this number assumes that the earth has a
uniform mass distribution (that is anywhere you measure the mass of one
cubic meter, you will always get the same answer. It is also assumed
that there is no air in the hole or else, as you worry, there would be a
terminal velocity and you would never get to China. The situation is
further (if there is air in the hole) complicated by the fact that the
deeper you go, the smaller the gravitational force becomes; therefore,
the terminal velocity gets smaller as you go deeper, so you would speed
up for a while on the way to the center but then start slowing down.

QUESTION:
I know that 1u is equal to 1.6605x10^-27. Where does atomic mass come
from and why is it not the proton?

ANSWER:
When defining a unit of measurement (like u, the atomic mass unit),
one of the most important considerations is to have a readily available
standard for comparison. The atomic mass unit is defined to be 1/12 the
mass of a ^{12} C atom where ^{12} C is the isotope of
carbon with 6 protons and six neutrons. Carbon is a very abundant
element, easily obtainable. So, the atomic mass unit is 1/12 the mass of
12 g of ^{12} C (1 mole) divided by Avagadro's number:
12/6.02214x10^{23} /12=1.66054x10^{-24} g=1.66054x10^{-27}
kg. The mass of a proton is much more inconvenient to measure.

QUESTION:
How is it possible for anything at all to be composed completely of atoms? For instance, a piece of wood, or any kind of large solid object... If everything is composed of tiny atoms(not to mention electrons, quarks, etc), is it possible that we are talking about an enormous, unimaginable number of atoms? Otherwise, how? This has bothered me for a long time

ANSWER:
Indeed, there are many atoms in any macroscopic object. The number
of atoms in a typical macroscopic object, say your pencil, is on the
order of 10^{24} atoms. That is a million, million, million,
million. There are more atoms in your pencil than there are stars in the
entire universe. To get an idea of the size of an atom (really tiny) see
my FAQ page.

QUESTION:
How is it possible to send a satellite in polar orbit, considering that in Einstein's view of the universe as a "fabric", gravitational motion is caused by the bending of the "fabric" due to the mass of the earth along a particular plane? (It seems counterintuitive that any orbits other than those along the same plane as the moon is on would be possible.)

ANSWER:
As I have emphasized many times before, the
"trampoline model" illustrating
gravitational warping of spacetime is merely an illustrative cartoon.
There is not really a sheet of "fabric" which is warped; that is a two
dimensional space which is easy to visualize being warped if viewed in
three dimensions. It is next to impossible to visualize a three
dimensional space being deformed in four dimensions or, even worse,
trying to visualize the four dimensional spacetime being deformed in a
five dimensional picture! The warping is not "along a particular plane".

QUESTION:
According to E=mcc, one unit of mass is equal to 34,596,000,000 units of energy and therefore, 1 kilogram of unranium has the same amount of energy as 1 kilogram of wood. Is my understanding of the famous formula correct?

ANSWER:
If you were able to turn all the mass to energy, then any kilogram
would do. However, very seldom can you do that. One example where you
can turn all the mass into energy is matter-antimatter annihilation. For
wood, you use chemistry to burn it and the tiniest fraction of the mass
is turned to energy, maybe something like 0.0001%, way smaller than you
could ever measure. For uranium using nuclear fission, something more
like 0.1% of the mass is converted to energy, way bigger than the wood.
I do not think wood-powered power plants will be competing with nuclear
reactors any time soon.

QUESTION:
how does the bulb lights up so quickly within some seconds........does that mean that the flow of electrons is as fast as the speed of light......plz explain this in detail..................................

ANSWER:
The wires in an electrical circuit are conductors which means that
there are many electrons more or less free to move. When you turn on the
switch, you apply a voltage across the ends of the wire going to the
lightbulb and that results in an electric field which exerts a force on
the conduction electrons causing them to move. In a typical household
circuit, the conduction electrons move very slowly, maybe something like
1 cm/hour on average. However, the electric field which drives them is
set up with the speed of light and so all the electrons in the circuit
start moving almost instantaneously. I have a QUESTION for you: why do
you put big strings of periods in your ANSWER?

QUESTION:
Do you know why do golf balls have dimples? (does it make them more aerodynamic and if so why?)

ANSWER:
I am The Physicist , I know everything! Well �maybe
not everything. I do know why golf balls are dimpled and the reason is
also why tennis balls are fuzzy. Aerodynamics can be very complicated,
but I can explain it qualitatively. Air drag on a projectile is caused
mainly by the turbulent wake behind the object; the pressure is lower in
this turbulent region so there is more force on the front pushing back
than there is on the back pushing forward. As shown in the figure to the
left, if the air moves smoothly over the ball, the turbulent wake is
very broad. But, if you break up the boundary layer of air with some
kind of roughness on the surface, you get a much smaller turbulent wake
and therefore less drag, as shown in the figure to the right. It is
counterintuitive that a rough ball encounters less air drag than a
smooth one, but true.

QUESTION:
how vacuum fluctuation cause the formation of particle-antiparticle in vacuum?

ANSWER:
I read an article about quantum mechanics and it mentioned particles "popping into existence". Is the assumption that the particles are popping into existence from another invisible/unknown state of energy which exists, or is the claim that they are appearing out of "absolute nothingness" and violating the law of conservation of energy?
If the currently held belief is that the particles are actually appearing out of pure nothingness, how is it being proven that there is nothing there in the first place? I'm not a scientist obviously but I would have thought that absolute nothingness could not possibly change state into matter, and that there must be a real and invisible force already present which is temporarily changing state into those particles. Where the particles popping into existence even at absolute zero? Surely there must have been some energy somewhere !!!!

ANSWER:
The answer to the first question is that vacuum fluctuation is the
scientific term for the creation of particle-antiparticle pair. The
second question gets more into the physics of this kind of pair
production. The conservation of energy is one of the most important laws
of physics, however the Heisenberg uncertainty principle allows it to be
broken provided that it is violated for a short enough time.
ΔE Δt≈ħ, where ΔE is the amount by which
conservation is violated, Δt is the time the violation lasts, and
ħ≈ 6.6x10^{-13} keV-s where keV (kiloelectron volt) is a
unit of energy. The mass energy of an electron-positron pair is about
1000 keV, so if they are created out of nothing, they may exist for Δt≈ 6.6x10^{-16
} s and then annihilate back to nothing. A proton-antiproton pair
has 2000 times more energy and so could last for only about Δt≈ 3.3x10^{-19
} s So, there is no such thing as "pure nothingness" as you put it,
but you will never actually find anything there because anything which
"pops into existence" pops right back out in almost no time. Such
processes are called virtual pair production.

QUESTION:
Light is considered as particle because of photoelectric effect. This is only one evidence. Are there any other experiments that shows light is made up of particles?

ANSWER:
Historically, there is an equally important experiment demonstrating
that light sometimes behaves like a particle �the
Compton effect. If electromagnetic radiation (x-rays were used in the
orginal experiment) their scattering from electrons can only be
explained with particles, not waves. Also, any time atoms or nuclei
deexcite, they emit single photons.

QUESTION:
Can I say the fact that light is able to travel through vacuum shows it consists of particles (since nothing is waving in a vacuum, so it make no sense that wave can exist in vacuum)?

ANSWER:
No, any electromagnetic radiation can be either a particle or a wave
in vacuum, you find what you look for. This is called wave-particle
duality. A wave can travel through a vacuum because what is "waving" are
electric and magnetic fields which can exist just fine in a vacuum.

QUESTION:
I was wondering why the Earth's magnetic field has such a huge range and influence even though the field strength in Gauss/Tesla is so very low. Why is is it not 'overpowered' by magnetism of magnets of natural rock like lodestone?

ANSWER:
The magnet which provides the earth's field is huge, deep in the
interior of the earth. However, outside the earth itself the field is
quite small because you are far from the source. Nevertheless, because
the source is so strong, this small field persists all over the surface
of the earth and far into space. Any magnet can overwhelm the earth's
field, but ultimately is puny compared to the earth's magnet and will
"overpower" the earth's field only near where it is. For example, if you
have a strong magnet on your sailboat, you cannot reliably navigate with
a magnetic compass. Also, local geoligical conditions will influence the
earth's field which is evident in the variations in the direction of the
earth's field.

QUESTION:
mass and weight are not the same things but on earth mass is considered=weight.why? and what does the weighing machine measures...........mass or weight?if suppose we go deep inside the earth what will be the resultant effect on our weight?

ANSWER:
Because the weight of something (the force which gravity exerts on
it) is proportional to mass, all you need is an instrument (usually
called a scale) which is linear and calibrated. The simplest scale is a
spring which, if you push it down the amount of compression is
proportional to the force. So, if you put a 1 kg mass (which has a mas
of 9.8 N) on the scale and it compresses by 2 mm, if you put a 3 kg mass
on it it will compress by 6 mm. So you draw up a little scale and attach
a pointer and you measure weight but you specify it in kg which is
conventional (but a kg is not really a weight). If you go deep inside
the earth, the weight would be smaller so you would have to recalibrate
your scale. Similarly, if you went on top of Mt. Everest, the weight
would be smaller also and you would have to calibrate your scale again.
In general, since the acceleration due to gravity varies around the
earth, you should calibrate your scale wherever you are. A simpler way
to measure weight is to use a beam balance which measures weights by
comparing them to known weights. Measuring weight is the most convenient
way to measure mass. If you wanted to measure mass M directly,
you would have to measure the acceleration a for a known force F ,
M=F /a , not so convenient.

QUESTION:
If the moment of an object is its tendency to turn about an axis, why is it that the moment of an object depends on not only force, but also distance.
I mean, it's obvious to me, that an objects tendency to turn will be a function of the force applied to it, but it's not so obvious why it matter how far away you are from the axis.

ANSWER:
I
am a little confused by what you are asking. Physicists usually use the
word torque to mean the ability of a force to cause an object to have an
angular acceleration. So, I will assume moment=torque and tendency to
turn=angular acceleration in your question. First let us write the
definition of torque,
τ =Fd sin θ
where d is the length of the line drawn from the axis to the
point where F is applied and θ is the angle between F
and d . You are comfortable with a big F giving a big
torque; this not necessarily right, because if
sin θ is zero, the torque is
zero regardless of how hard you push. For example, if there is an open
door you wish to close and you push on the edge opposite the hinges but
directly toward the hinges, the door will not close no matter how hard
you push. Finally, I will address your question why it matters what d is.
Imagine the door is closed and you push perpendicular to the door but at
a point very close to the hinges; even if you push very hard, the door
will not open or open very slowly.

QUESTION:
I wrestle for a sport hobby. One element in wrestling is pinning your opponent's back to the mat. This requires one to apply a force through the opponent chest without rolling the oppenent but applying the most force possible. If you model this problem, a human chest can be considered a cylinder and the mat has some friction coefficient. What would be the maximum force vector that can be applied to the cylinder, relative to the radius and weight of the cylinder and the coefficient of mat friction? The force vector cannot run perpendicular to the mat (ground). This is impractical as this is the least stable option (e.g. Standing on a log).

ANSWER:
I am not sure this is the best model to use, but I will follow your
lead because anything more realistic will be very complicated. There are
two things you do not want to happen: the cylinder must not roll and it
must not slide. Therefore, to keep it from rolling, the force must be
directed through the point of contact with the floor. Now, the diagram
to the right shows the forces (not necessarily to scale):

your force F
which makes an angle
θ with the vertical;

the weight of your opponent W ;

the normal force N of the mat up on your opponent; and

the frictional force f , the maximum value of which is f _{max} =μN .

For
any given situation, F sinθ-f =0 and N-W -F cosθ= 0.
Therefore, the maximum frictional force for this θ , μ , and
F is f _{max} =μ (W +F cosθ ) and
so F _{max} =μW /(sinθ-μ cosθ ). That is the
answer to your question, it depends on θ , μ , W , and
F but not on the radius of the cylinder.

If θ= 90^{0} , horizontal, F _{max} =μW.

If sinθ-μ cosθ= 0, you can exert an infinite force; for
example if μ =0.2, θ _{critical} = tan^{-1} (0.2)=11.3^{0} .

For any angle less than θ _{critical } you can also exert
an infinite force without the cylinder sliding. Your reasoning for a
vertical force not being practical is not correct because "stability"
has nothing to do with it; rather the practical reason is that you
cannot exert a force greater than your own weight vertically.

For any angle greater than θ _{critical } the maximum force
will be smaller, the smallest at 90^{0} .

I
guess next you have to worry about the mechanics of how you are going to
exert this force. Your feet cannot slide on the mat and your weight will
always be vertical.

QUESTION:
I just heated a large cup of tea in my microwave for 3 minutes, which prompted me to wonder how much water could instantly be boiled by a moderate-sized nuclear bomb. My guess is: quite a lot.
So perhaps a more practical theoretical question would be, what would happen to the water if a moderately-sized nuclear bomb were detonated 1/2 mile above a body of water?

ANSWER:
I am afraid this is too vague. To even make a very rough estimate I would need the area and depth of the body of water and what you mean by moderate-size (megatons TNT?)

FOLLOWUP QUESTION:
May we suppose a 1 MT yield and Lake Kabekona, where I have vacationed for years. Lake Kabekona is 2252 acres (9 square KM) in area, is oblong in shape, and has a maximum depth of 133 ft (41 m).

ANSWER:
This is a very rough calculation just to give you order of
magnitude. A megaton of TNT is about E =4x10^{15} J. I
take the average depth of the lake to be 10 m so the volume is about 10^{8}
m^{3} and the mass, M , is about 10^{11} kg. If
you detonate a half mile above the lake, I figure that about 30% of the
energy goes to the lake. The specific heat of water, c , is about
4x10^{3} J/kg^{0} C. Now, the increase in temperature
Δ T is
given by
Δ T= Δ E /(Mc )=0.3x4x10^{15} /(10^{11} x4x10^{3} )=3^{0} C.

QUESTION:
I am considering getting a tattoo of the equation for Newton's First Law of Motion and want to make sure it is correct (for obvious reasons). I am in my forties and this would be my first because it is the first thing that really spoke to me. Basically, to keep moving and doing :)

ANSWER:
Newton's first law is not conveniently stated in an equation. Are
you committed to that?

FOLLOWUP
QUESTION:
So this what I had; does this make sense from physics standpoint? I read it as "in the absence of any external force, the acceleration is zero" which to means covers the meaning I want, which is "keep moving and you WILL keep m oving, but sit still and you WILL keep sitting still". It is a reminder for me to stay active, adventurous and to keep moving..

ANSWER:
OK, how about the equation to the right? Here is what that
equation says: the rate at which momentum (p ) changes is zero. If
this were Newton's second law there would be force F in place of
the zero. Momentum is mass times velocity, but the word momentum fits in
well with what you want to convey, I think. The equation says that if
the momentum is zero, it stays that way, if it is not, it stays that
way. Partly, Newton's first law is a special case of Newton's second law
(when there is no external force) and that aspect of it is often called
the law of inertia. The first law also has a deeper meaning, but that is
not relevant for what you want.

QUESTION:
I am not a physicist and these questions should make it pretty clear.
Could you calculate the mass of Human? If so would you then be able to
put that in to the mass-energy equivalence and figure out how much
energy I am?

ANSWER:
Sure. Suppose your weight is 200 lb, then your mass is about M =90
kg. The speed of light is about c =3x10^{8} m/s and so
E=Mc ^{2} =90x9x10^{16} ≈8x10^{18}
J. To put this into perspective, the energy consumption of the whole
world in a year is about 1.85x10^{10} megawatt hours≈6.7x10^{19}
J, so the energy could supply the whole world for a little more than one
month.

QUESTION:
I am trying to teach quantum physics for the first time. I am not sure what the link is between photoelectric effect and plancks constant or how to link it meaningfully for the students.

ANSWER:
The graph to the right shows photoelectric effect data. Plotting the
energy ΔE of the electrons
as a function of the frequency
ν
of the of the light which caused them to be ejected yields a straight
line. But, to understand why the data behave like they do, you must
assume that the energy given to the electrons is from photons, not try
to understand it in terms of waves. Now, the photon gives all its energy
to the electron and, since it takes a certain amount of energy, call it
W , to remove an electron from the metal, then the energy of the
electrons must be
ΔE=h ν -W
where h is the slope of the line. It just so happens that h
(the slope of the line) turns out to be Planck's constant, and that is
the link you seek.

QUESTION:
Why are the astronauts floating with no weight in their space shuttle in space?

ANSWER:
They are not weightless, even though everybody says so. Weight is
the force which the earth exerts on you and that does not go away. It is
the same as if you were inside a freely falling elevator �you
still have a weight but, because you are freely falling too, you feel
like you are weightless. The shuttle is like a freely falling elevator.
I think you might get a feel for this if you play with
Newton's Mountain .

QUESTION:
I am having trouble comprehending the idea of a neutrino having no mass. Wouldn't that mean that it simply does not exist? How does something exist that does not exist? Do we just not have the technology to measure it's mass?

ANSWER:
First, let's address the question of the "existence" of massless
particles. Possibly the most abundent particles in the universe,
photons, are massless. Photons are particles of light, and there is no
argument that they exist. The restriction which must be obeyed by
massless particles is that they must, in a vacuum, travel with the speed
of light. You can never find a photon at rest. So other massless
particles might exist. For many years neutrinos were thought to be the
other known massless particle in our universe. However, recent
experiments have shown that neutrinos cannot be massless, but their
masses (there three kinds of neutrinos) are known to be nonzero.
However, they are exceedingly tiny, hundreds, maybe thousands of times
smaller than the mass of an electron. Therefore most neutrinos travel
with a speed very, very close to the speed of light.

QUESTION:
If we on earth measure time the way we do according to earth rotation etc .. Is there a different measurement of time in relation to the universe or is it all to do with rotation ? As the earth slows down would time slow down?

ANSWER:
Earth's rotation has not been the standard for time for over 300
years. If earth's rotation slowed down, time would not slow down. Time
is measured relative to a particular atomic vibration (atomic clock)
which is constant (as far as we know) anywhere in the universe. The rate
at which clocks click does, however, depend on the velocity of the
clock. This is time dilation in special relativity but not really
observable in everyday life.

QUESTION :
I am a photographer and general tinker-er. I have started building
pinhole cameras and during the course of building them I came across
some equations on obtaining the optimal pinhole size. I find I un derstand
the equations well, but I can not find a solid explanation for the 2.44
constant used in the calculation for the Airy Disk (2.44 x light wave
length x focal length). Everything I have read is very vague and I want
an explanation. Why do we use this number? I read that it is like pie,
it just is, but is to what, waves? Is it a speed? What does it
represent? I want to understand.

ANSWER:
This verges on being too technical for this web site, but I always
want to encourage folks to "have an enquiring mind", so I will take a
stab at it. This question requires that you understand a little about
diffraction, interference of light waves. If we shine light through a
slit and the slit is very large compared to the wavelength of the light
we are looking at, we get an "image" of the slit on a screen. If the
rays of light hitting the screen come from very far away compared to the
size of the slit, the image of the slit will be the same size as the
slit itself. The sun coming in through a window creates an image of the
window, the same size, on the wall. So, I now want to make images of
smaller and smaller slits; what happens is that we come to a point where
as we make the slit smaller, the image starts getting bigger. The figure
above shows what the "image" of a very narrow slit looks like (below)
and a graph of the intensity (above). It is fairly straightforward to
show (see any elementary physics textbook) that a sinθ=λ
gives the angle of the first dark spot in the pattern for wavelength
λ ; the geometry showing a and θ are shown to the
right. Now, a pinhole is not a slit, but it is very similar and you
would expect its pattern to be very similar. Indeed, instead of getting
stripes you get a bullseye as shown to the left. Doing the analysis
similar to that which leads to the angle of the first minimum gives a
slightly different result,
a sinθ = 1.22λ
where a is the diameter of the hole. In essence, this is
where your factor of 2.44 comes from; where does it come from? It gets
pretty technical here! When you solve the pinhole diffraction problem,
you work in cylindrical coordinates because the problem has cylindrical
symmetry. As is often the case with problems with this symmetry, the
solution (for the intensity in this case) involves a Bessel function, a
special mathematical function which whole books have been written about.
The intensity is given by
I=I _{0} [J _{1} (½ka sinθ )/(½ka sinθ )]^{2} ,
where J _{1} is the first order Bessel function and k= 2π /λ .
Now, we are interested in when the intensity is first zero; the first
zero
of J _{1} (x ) is for x =3.832, so ½ka sinθ=(πa /λ )sinθ= 3.832,
or a sinθ =1.22λ . Next, we convert the angle to
lengths by approximating (see diagram above) sinθ≈x /R so
x= 1.22Rλ /a or 2x =2.44Rλ /a ; 2x
is the diameter of the smallest spot to which a collimated beam of light
can be focused. Since you say that you "understand the equations well"
except where 2.44 comes from, I guess my task is complete. Apparently
(from the little research I did) the optimal size is if the minimum spot
size equals the hole size, i.e . 2x =a _{opt} ,
so a _{opt} =√(2.44Rλ ). There is an alternative form
of optimum size which is based on the Rayleigh criterion for resolving
two spots (which stipulates that the central maximum of one image is on
the first minimum of the second) which has a _{opt} =√(3.66Rλ ).
(A similar problem is the structure of the
compound eye
of insects, elegantly discussed in Feynman's Lectures on Physics ,
Vol. 1.)

QUESTION:
I am confused about the units for acceleration: mathematically, how can
m/s/s and m/s^2 be used interchangeably?

ANSWER:
It is just like arithmetic or algebra: (3/4)/4=3/4^{2} , (x /y )/y =x /y ^{2} .

QUESTION:
Why do the planets remain in position in respect to their position (or
distance) from the Sun. Why doesn't the emmense gravity of the Sun
slowly pull the planets towards it?

ANSWER:
See an earlier answer . If a
planet is in a circular orbit, the immense gravity does pull on it but
the result is not to make it move toward the sun, rather to bend its
orbit.

QUESTION:
if 50 billion neutrinos pass through your body every second .... I have
read this so many times .... why do only a few show up in the detectors
below ground. I thought that neutrinos could travel through anything.

ANSWER:
What are you asking? If they are very unlikely to interact with your
body, they would also be very unlikely to interact with the underground
detectors. It is really hard to detect neutrinos and that is why the the
detectors are so huge (millions of gallons) �to
enhance the chance of "catching" one.

QUESTION:
It's my understanding that a uranium atom was split to create the
nuclear boom. Has anyone ever split a mercury atom? I am reading ancient
text and putting pieces together here.

ANSWER:
There would be no point since no mercury isotope is fissile, that is
you cannot sustain a chain reaction and if one fissioned it could not lead
to others fissioning. See an earlier answer .

QUESTION:
I am having a hard time understanding the relationship between mass
and energy. E=mc2 shows energy affects mass, but not how. I can
understand energy affects mass. But not energy is mass. A coiled up
spring with potential energy weighs more than a uncoiled spring. But I
don't understand why according to what I am reading. If have this right,
mass increases because of energy. We know this because we measure mass
by weight, and weight increases with energy. W=mg. and g is a constant
so m is increasing.

ANSWER:
Here is what E=mc ^{2} means: mass is a form of
energy. That is all you need to understand, that and if you do work on
something, you change its energy. Although we normally measure mass by
weighing something, as you suggest, that is not a fundamental way to measure
mass because the weight of something depends on where it is. The right way
to measure mass is to measure inertia using Newton's second law, M=F /a ;
measure acceleration for a known force and you have mass. All that is beside
the point for the crux of your question �does
the mass of the spring increase when it is compressed. The answer is yes
because you must do work to compress it and that added energy will appear as
mass (assuming a perfectly elastic spring). In elementary physics we can
introduce a potential energy function to account for this added energy, �kx ^{2} ,
but if you were measure the mass while it had this potential energy, you
would find that that is where the energy is "hiding". So how come nobody
ever noticed this before? Let's take a typical spring, say 200 N/m and
compressed by 10 cm. Then the added energy is 1 J. This corresponds to a
mass of M=E/c ^{2} =(1 J)/(3x10^{8} m/s)^{2} ≈10^{-17}
kg. Good luck trying to observe that! However, there are cases in nature
where potential energy showing up as mass can be easily observed. An atomic
nucleus is a bound collection of neutrons and protons. Imagine that you
pulled them out, one by one. You would have to do work, and so you would add
energy to the system when you disassemble it. You could just say that you
have added potential energy to the system, but if you measure the mass of
the nucleus and the compare it to the sum of the masses of all the proton
and neutron pieces, you would find the mass of the nucleus around 1%
lighter, an easy amount to observe.

QUESTION:
How can laser light cast a pattern resembling the double slit
experiment when shone on the tip of a needle, a structure that does not
resemble a double slit?

ANSWER:
The only way this "resembles" a double slit pattern is the
appearance of fringes. Diffraction can occur any time you have more than one
path light can take to reach the screen. Imagine light coming from right
from the two sides of the needle. Then, these two rays could interfere
either destructively or constructively depending on the viewing angle, so
the pattern could look very much like a double slit pattern.

QUESTION:
Does the transfer of energy takes place when when we push a huge rock
with all our might?where does energy go which we spend? I thought that
the transfer of energy does not take place with respect to the wall,and
muscular energy is changed into heat energy.BUT MY TECHER IS SAYS When
we push huge rock and fail to move it,the energy spent in doing so is
absorbed by the rock.This energy is converted into potential energy of
configuration of the rock which results in its deformation.Ho wever,this
deformation is not visible on account of the huge size of the rock.

ANSWER:
I always hate to say that a teacher is wrong, but he is in this
case. If this were true, then if you pushed long enough you could observe a
change in the rock. Also, the surface the rock is on is pushing up with a
force equal to the rock's weight, a force much larger than you can exert,
and that does not result in any significant deformation. What is happening
here is discussed in an earlier
answer . Essentially it is a biological answer, not a physical one. The
way muscles work makes them consume energy even if they do no external work,
and that energy would presumably end up as heat.

QUESTION:
We say that the moon has the gravitational force one sixth in
comparison to that of the earth. If the earth has held an atmosphere
with its gravitational force which is 480 kilometers high, why can�t the
moon even hold an atmosphere one sixth of the height of the atmosphere
which the earth has, that is, 80 kilometers?

ANSWER:
Holding an atmosphere of a given composition and temperature by
gravity is an "all-or-nothing" thing. To understand, you have to first
understand escape velocity. If you throw something upwards it falls back;
throw it harder and it goes higher; throw it faster and faster, and
eventually you will reach a speed where it will never come back down. This
is called the escape velocity and it depends, of course, on how strong the
gravity is. So, the escape velocity from the moon is much less than the
escape velocity from the earth. Now, we need to talk about the speeds of
molecules in a gas. At a given temperature, there is a distribution of
velocities, called the Maxell-Boltzmann distribution, shown to the right.
The average speed is determined by temperature (the hotter, the faster) and
mass of the molecules (the lighter, the faster). So, as shown by the figure,
a few are always going very fast, maybe even greater than escape velocity.
So, suppose that some of the molecules have speeds greater than escape
velocity, perhaps the
�% shown on the right in the diagram; then these will escape. But now the
distribution will not be correct, so it will readjust itself so some of the
molecules go faster; then these, in turn, escape also, and so on until all
molecules have escaped. On earth, helium and hydrogen have very high speeds
so that earth's gravity is not capable of holding helium and hydrogen in its
atmosphere and there is virtually none of these in the atmosphere. But
heavier molecules (N_{2} , O_{2} , CO_{2} , etc .)
have such a tiny fraction above escape velocity that they do not escape (at
least at any rate that can be measured). But, on the moon, all gases would
behave like hydrogen and helium do on earth and all escape.

QUESTION:
the theory of relativity states that "the speed of light is the same
in all inertial frame of refence" thus doesnt accelerate. but Einstein
said that light is bent by gravity like a ball projected at an angle.
thus, light will accelerate towards the direction of gravity. and does
accelerate.

ANSWER:
Acceleration does not have to mean change speed, it can also mean
change direction. The speed of the light is the same after being bent. If
the light moves parallel to the gravitational field, it does not speed up or
slow down, it changes wavelength.

QUESTION:
I am now in Gr 11(18 years old) and i had a project on the
conservation of moment. I did the experimant with two trollys with
different masses each time, My conclusion was that the total momentum
after the collision did not = the total momentum before the collision
because my calculations proved it. Now iv handed my project in but im
still struggling to find out what the causes are of the discrepancies
between the values in momentum before and after each of the collisions
were? (I did think about friction but is there other reasons?)

ANSWER:
Call the horizontal direction the x -axis. The linear momentum
along this axis is conserved unless there is an external force with a
component along this direction. Likely, the only significant force along
this direction is friction. I would think that this would be pretty minor
and that momentum should be conserved to an excellent approximation. I
assume that the collision lasted a very short time, so that is another good
reason to expect momentum to be conserved. I am a little disturbed by your
statement " � because
my calculations proved it".
Calculations
do not prove something, measurements do.

QUESTION:
If you stand on a shrinking planet so that in effect you get closer
to its, your weight will increase. But if you instead burrow into the
planet you get closer to its center,your weight will decrease. Explain
please

ANSWER:
On or inside a sphere, the force depends only on the mass inside
you, not on that outside you; also, the force decreases as 1/r ^{2}
where r is the distance to the center. As the planet shrinks (mass
the same but radius smaller), the mass below you stays the same but the
distance gets smaller, so you get heavier. If you burrow, the mass inside
decreases proportional to r ^{3} (the volume of a sphere is 4 π r ^{3} /3)
so, the force changes like r ^{3} /r ^{2} =r
and therefore decreases as r decreases.

QUESTION:
So if light from a flashlight is a form of energy, can it be felt?
Is it possible for a blind person to know if he is under a street light?
How bright does a light have to be for me to feel it?

ANSWER:
Indeed, light does exert a force, normally referred to as radiation
pressure. When light strikes you it is either absorbed or reflected. Light
which is reflected exerts twice as much force as light which is absorbed
because it does not lose any energy. But, the force for any reasonable
intensity is extremly small and for the force to be big enough to feel would
require an intensity where you would be burned up by the heat from the light
you absorbed. To give you an idea of the magnitude, a laser pointer exerts a
force on the order of 3x10^{-12} N; to exert a force of about 1 oz
you would need about 100 billion laser pointers.

QUESTION:
if you have a rope the length of the equator around earth, and then
you give the rope 3' feet slack, how high would the rope rise above the
surface?

ANSWER:
I hope you don't think that the rope would spontaneously rise up if
it were longer than the circumference of the earth; you would have a
slightly slack rope laying on the ground. If you are asking the difference
in radii between one circle with a circumference C and another of
circumference C+ δ , that
is not really physics. But, it is easy enough to do. If C is the
circumference of the earth, then C= 2πR _{ } where R =6.4x10^{6}
m and
C+ δ =2πR' where
R' is the radius of the circle your rope would make if δ =3
ft≈1 m. Then R'-R=δ /(2π )≈0.16 m. Notice that the answer is
independent of R if the circumference of the earth were 1 m and the
rope were 2 m long, the height of the rope above the earth would still be
0.16 m!

QUESTION:
Is there a metaphor or a simplistic explanation that I can use to
explain to middle schoolers that electricity and magnetism are
manifestations of the same force? Age-appropriate textbooks just state
that this is so without explanation, or else they simply give the
example of an electromagnet, which doesn't answer the deeper question.
Some books don't even explain that they are related! Introducing the
ideas of vectors and tensors makes the explanation too long and I lose
most kids' attention. Do you have any ideas?

ANSWER:
Sure, there are lots of ways to demonstrate the connection. Easiest
is to simply refer to an electromagnet. Here we have an electric current
going around a coil of wire and it will attract iron. You can easily make
one by wrapping wire around a nail and connecting it to a battery and show
that the nail becomes a magnet. Moving electric charges (which is what the
current in the wire is) cause a magnetic field. This connection was first
discovered by
Oersted , a Danish physicist of the 19^{th} century, who noticed
that, when he hooked up an electrical circuit, a nearby compass deflected
from north. You can demonstrate the connection also by changing a magnetic
field, for example by thrusting a magnet into a coil of wire. The changing
magnetic field causes an electric current to flow in the coil. Here is a
nice animation to demonstrate this. This is the basis of how electricity
generators work.

QUESTION:
I am a senior in high school with an AP physics class. During class
we came upon a problem that stumped us. If a satellite orbiting earth
encounters air resistance that slows it, What will happen to the path of
the satellite? Will it increase it's angular velocity and presume an
orbit closer to earth? Will it spiral around and towards earth
eventually crashing?

ANSWER:
You must have done the calculation of a near-earth orbit velocity.
In order for a satellite to achieve orbit at an altitude small compared to
the radius of the earth, it must have a minimum velocity, about 18,000 mph.
If you try to launch it with a smaller velocity, it will come back to earth
before getting all the way around. You might like to play with
Newton's mountain . So, to stay in that orbit, the speed must remain
unchanged. But, if there is air drag, it will slow the satellite down. The
result is that it will lose altitude; the lower it goes, the more air it
will encounter and the faster it will lose speed and therefore altitude,
until it finally hits the ground (or water). From an energy perspective, a
satellite in a stable orbit has a constant energy; but friction takes energy
away and so the altitude will decrease.

QUESTION:
If the higgs boson composes a field that is present throughout the
universe and gives all matter mass, then why is it so hard to find.
Shouldn't the higgs boson be as easy to detect as let's say a photon,
electron, or proton.

ANSWER:
It is not the Higgs boson which imparts mass, it is the Higgs field.
The Higgs boson is the quantum of the field just as the photon is the
quantum of the electromagnetic field or mesons are the quanta of the strong
interaction. So, it is not as if there were a bunch of Higgs bosons sitting
around waiting to be observed, they exist as virtual particles and have to
be created to be observable. An analogy is the electric field � if
you have an isolated electric charge, photons are popping into and out of
existence but you cannot see them.

QUESTION:
If two cars are traveling at 50 mph towards each other what is there
speed when they pass? My boyfriend said it doubles to 100 but that makes
no sense! They do not increase speed nor does their speed differ! How
can the action passing another car moving at the same spew miraculously
double your speed?

ANSWER:
Velocity is one of those things which you must specify what it is
relative to. In your example, the speeds 50 mph are the speeds of cars
with respect to the road . As they pass, their speeds remain the same 50
mph. What your boyfriend is talking about is the speed of one car with
respect to the other. Each car sees the other one approaching at a rate of
50+50=100 mph; there are no sudden changes in speed, each sees the other
going 100 before, during, and after passing. This is called the velocity
addition theorem. If the two cars collide head-on, it is the same as if one
car were parked and the other car going 100 mph. If the two cars were
traveling in the same direction, each sees the other as being at rest,
50-50=0 mph; the minus sign signifies moving in the opposite direction.

QUESTION:
If two quantities have different dimensions, can they be multiplied
or added? please support the answer with examples because i really can't
differentiate between the dimensions and the units !!

ANSWER:
First, let's be clear on what are meant by dimensions and units.
Dimensions are fundamental notions of three quantities �mass
(M ), length (L ), and time (T ); units are the particular
operationally defined ways we measure the three dimensions�the kilogram (kg)
for M , the meter (m) for L , and the second (s) for T
are the so-called SI units, the ones scientists prefer. There are other
units for other quantities like force or energy or power, but all are
ultimately describable in terms of kg, m, and s. So, can two quantities with
different dimensions be added? Suppose we add 2 kg to 5 s; this would have
no meaning. It's the old "apples and oranges" thing, it just makes no sense.
On the other hand, we can multiply (or divide) things which have different
dimensions. For example, if a car goes 80 miles in 2 hours, its average
speed is (80 miles)/(2 hours)=40 mph; if a 2 kg object has an acceleration
of 3 m/s^{2} , the net force on it is the product 6 kg∙m/s^{2}
and the frequently occruing SI unit for force is given a name, the Newton
(N): 1 N=1 kg∙m/s^{2} . Can you add things with the same dimensions
but different units? Yes, but the answer is usually not very useful. For
example, 1 ft + 1 m is the length of a foot and a meter, but that doesn't
say much; it would be better to convert to one or the other units before
adding, such as 1 ft= 0.3048
m and so
1 ft + 1 m=1.3048 m. Also, it not a good idea to mix units when
multiplying quantities. For example, how far does a car going 50 mph go in 3
minute? The distance D=vt= (50 mi/hr)(3 min)=150 mi∙min/hr is pretty
meaningless because how far is a mi∙min/hr? Better to convert to the same
time units, for example 3 min=1/20 hr, so D =50x(1/20) mi=2.5 mi.

QUESTION:
If I move a magnet towards a solenoid and the two ends of the
solenoid are closed, there would be a temporary induced current in the
solenoid according to Faraday's law of Induction. If the two ends are
opened, there is no current but an induced voltage can be measured
between the two ends. Does the latter contradict Faraday's law since the
solenoid does not formed a close circuit?--

ANSWER:
Faraday�s law states that if there is a changing flux through some
area, then there is an induced emf around the edge of that area. No closed
circuit is required, but if you put one there the emf will cause a current
to flow in it.

FOLLOWUP QUESTION:
If the magnetic flux through a circular conducting loop changes, there is an emf induced in it. Since emf is similar to potential difference, which requires two points for measuring, where is the starting and ending points for measuring the emf in the loop?

ANSWER:
When you talk about a potential difference, you need two points.
But, the definition of potential difference is the potential energy
difference for a positive test charge divided by the charge (so volts are
joules/coulomb). But potential energy difference is the work you must do to
move the charge from one point to another. In electrostatics, this work is
path independent, it does not matter how you move it from one point in space
to another; similarly, the work done if you move it over a closed path is
zero. This is a conservative force. But, when magnetism gets involved, the
electric field no longer need be a conservative field. So, if you take a
charge and move it around some closed path, the work you do might very well
be nonzero; think of there being an electric field which curves around and
"bites its own tail" like a snake. So here you would have an EMF around a
closed path and you do not need to have two points to think about it. That
is what is going on in Faraday's Law situations and you do not need two
points and talking about potential differences really makes no sense because
the "battery voltage" is all around the circuit.

QUESTION:
Hey! I was told that the human uterus can exert up to 400 n or 100 lbf of...force(?) when doing its uterus thing and pushing out a baby. I have no frame of reference for these numbers/units, and my wikipedia research provided me only with a lot of not-worth-the-mental-effort equations and definitions.
I'm just trying to put these facts in perspective. For example, is 100 lbf comparable to pushing a couch across the room? Lifting a small car? Doing 40 pushups? 600 pushups? Are any of these examples even remotely related to the units "n" or "lbf?"
I'm not seeking any kind of technical answer about how many calories explode when you exert 400 n of superpower to move a million-joule flavored kilogram 9 meters (by now you can see my level of physics expertise), just a simple "Oh 100 lbf, that's like X, which you are very familiar with and can relate to! What a neat thing to know about the uterus, we have all gained a new appreciation for its strength!"

ANSWER:
A small woman weighs about 100 lb; 100 lb is the force you would have to exert to hold her up. A 100 lb woman having a baby would exert a force with her uterus which would be the same force she would have to exert to do a pullup.

QUESTION:
Since the Brownian movement of atoms describes temperature as the rate at which atoms move, and Einstein showed that particles can move no faster than the speed of light, doesn't this imply an upper limit to which temperature can rise?

ANSWER:
Actually, temperature should be thought of more as the average
kinetic energy of the constituents, not the speeds of the constituents.
What's the difference, you might ask? At everyday temperatures, the kinetic
energy of an atom is
�mv ^{2} , but this is no longer true if v is not very
small compared to the speed of light. While there is a limit to v ,
there is no limit to the kinetic energy and hence not to temperature.
(Actually, the temperature of anything could not correspond an energy
greater than the total energy of the rest of the universe!)

QUESTION:
If an object has a greater mass will it reach it's terminal velocity more quickly than an object with less mass?

ANSWER:
It also depends on the geometry. Suppose that two objects have the
same geometry but different masses. You should first look at an
earlier answer where I show that the velocity of a
falling object as a function of time is v =√(g /c )tanh(t √(gc ))
where g is the acceleration due to gravity (9.8 m/s^{2} ),
c =�A/m (A is the area presented to the onrushing wind and
m is the mass), and tanh is the hyperbolic tangent function.
Then the characteristic time (the time it takes to achieve about 76% of the
terminal velocity) is 1/√(gc )=√(4m /(Ag )),
so the larger m is, the longer it takes to reach terminal velocity.
(Incidentally, you must work in SI units for this to work out. Notice that
it does not look like √(4m /(Ag )) has the dimensions of
time but the number 4 has hidden units in it to make things come out in
seconds.)

QUESTION:
I know if two cars traveling 50 mph and they collide head on that the impact is 50 mph. If one is traveling at 5 mph and one is traveling at 95 mph, is the impact 95 mph?

ANSWER:
Really, you know this? First of all, what does "impact is 50 mph"
mean? If you collide with another car going 50 and you are going 50 the
other direction, it is the same as hitting a car at rest if you are going
100 mph. It is also the same if one is going 5 mph and the other 95 mph in
the opposite direction �like hitting a
stationary car when going 100.

QUESTION:
i got a good question if threw a rock downward at the same speed i threw it upward... wouldnt they meet at one point because they all go the same speed?

ANSWER:
The equation for motion is y=v _{0} t- �gt ^{2}
where v _{0} is the initial velocity, g is the acceleration due to
gravity, y
the height above (or below) where you threw it from, and t is the
time since it was thrown. I assume you are throwing the two at the same
time, so one has an equation of motion
y _{1} =v _{0} t- �gt ^{2}
and the other
y _{2} =-v _{0} t- �gt ^{2} .
When
y _{1} =y _{2} , the two are passing, so v _{0} t- �gt ^{2} = -v _{0} t- �gt ^{2} ,
and the two will pass only when
v _{0} t=-v _{0} t, only when t= 0;
this is when you threw them �they will
never pass again. They do not go the same speed, they have the same
acceleration. Their speeds are always different (except at the instant they
were thrown).

QUESTION:
if i have a set distance, an acceleration rate and a deceleration rate is there a way of calculating the maximum attainable speed in which to still stop within that distance? ie distance of 52.15m, acc 2.19m/s/s and deccelareation rate of 6.67m/s/s. what would the theoretical maximum speed be?
[ Not
home work just work! I'm an accident investigator for lancashire
police.]

ANSWER:
The equations of motion for uniform acceleration a are given
by x (t )=v _{0} t + �at ^{2}
and v (t )=v _{0} +at where
x (t ) is the distance traveled in time t ,
v (t ) is the velocity at time t , and v _{0} =v (t =0).
I will adopt the notation that

x _{1}
is the distance traveled by the accelerating object in a time t ,
v _{1} its velocity at time t , t _{1}
is the time after starting that acceleration stops, and a _{1}
is the magnitude of its acceleration; and

x _{2}
is the distance traveled by the decelerating object in a time t ,
v _{2} its velocity at time t , t _{2}
is the time after starting deceleration that it stops, and a _{2}
is the magnitude of its deceleration.

The object starts at
rest and so the distance it travels in time t is x _{1} (t )=�a _{1} t ^{2
} and its velocity is v _{1} (t )=a _{1} t.
Now the object starts decelerating (physicists do not like that word!)
beginning with a speed of v _{0} =v _{1} (t _{1} )=a _{1} t _{1
} so the distance it travels in time t is x _{2} (t )=a _{1} t _{1} t -�a _{2} t ^{2
} and its velocity is v _{2} (t )=a _{1} t _{1} -a _{2} t
(note the minus sign because it is slowing down). Now, v _{2} (t _{2} )=0=a _{1} t _{1} -a _{2} t _{2
} (it comes to rest at the end), so t _{2} =a _{1} t _{1} /a _{2} .
The total distance traveled is s =x _{1} (t _{1} )+x _{2} (t _{2} )=�a _{1} t _{1} ^{2} +a _{1} t _{1} t _{2} -�a _{2} t _{2} ^{2} =�a _{1} t _{1} ^{2} +a _{1} ^{2} t _{1} ^{2} /a _{2} -�a _{2} (a _{1} t _{1} /a _{2} )^{2} =�a _{1} t _{1} ^{2} (1+(a _{1} /a _{2} )).
Final results, using your numbers:

t _{1} =√[2s /(a _{1} (1+(a _{1} /a _{2} )))]=5.99
s,

t _{2} =a _{1} t _{1} /a _{2} =1.97
s, and

v _{max} =v _{1} (t _{1} )=a _{1} t _{1} =13.1
m/s==29.3 mph.

QUESTION:
A stable compound nucleus has a lower mass than the sum of its constituent protons and neutrons. I was told this is because the bound state has lower energy. A similar situation occurs in the atom: the stable bound state of an atom has a lower energy than its constituents, and thus the mass of the atom is slightly less than that of the electrons and protons that make it up. This leads me to believe that a bound system minimizes energy and hence mass. Why does this reasoning fail when thinking about the binding of quarks? The up and down quark masses are around 1-6 MeV, but the bound states of quarks, rather than being less than the sum of their constituents, are much larger (pion ~130 MeV, proton/neutron ~1 GeV). It seems that it is not necessary for a stable bound state to minimize energy. This seems strange to me. Do you see where my confusion is coming from?

ANSWER:
You are sort of muddled here. Here are the basics:

Any bound state is less massive than the sum of
its parts; this is easy to convince yourself of because it obviously
takes work to disassemble any bound state and therefore energy must be
added. This energy (which you put in) shows up as increased mass.

By definition, the ground state is the
"boundest" state of a system. Therefore the ground state is the least
massive.

Conversely, if a system binds, the resulting
object has less mass than what you started with and therefore energy is
released. For example, CO_{2} is less massive than C+O_{2}
and that is where the energy comes from when you burn carbon �fossil
fuels, essentially.

Because of the
way quarks bind, there is no such thing as a free quark and therefore it
is difficult to talk about its rest mass. To zeroth approximation, it
takes infinite energy to unbind a quark so it would have infinite mass.

QUESTION:
11.2 km/s is the escape speed from Earth. Is it possible to escape from Earth at half this speed? Is it possible to escape from Earth at one quarter this speed?

ANSWER:
You need to understand what is meant by escape speed. It is the
minimum speed which you give an object at the earth's surface so that it
will never come back. Of course, you could escape with any speed, say 1
cm/year, but only if you keep pushing on it to keep it going; but that is
not what escape speed means.

QUESTION:
ma+kx=0 →(A) is an equation which represents mass spring system. It is a linear equation of second order. If I want to make it non-linear I make it like ma+kx+k1(x^3)=0, i.e. I add another term k1(x^3). Why can�t I add k1(x^2) to (A) to make it non-linear?

ANSWER:
Because the force would be repulsive when x <0, that is, it
would not be a restoring force. Your equation for the spring is essentially
F=-kx. For F to be a restoring force, it must be an odd
function of x .

QUESTION:
how much "energy" or "power" or "force" would be needed to join separate hydrogen and oxygen atoms to create water?

ANSWER:
Actually, it is the other way around �you
get energy out when you combine hydrogen and oxygen to make water. It is
called burning hydrogen and you could make it happen just by putting
hydrogen gas through your gas stove and lighting it. Since heat and light
come out of burning gas, it is clear that energy is released. Some people
think that one long-term solution to our energy needs is to dissociate the
oxygen and hydrogen in water (by putting windmills in the ocean, for
example) and then burning the hydrogen which would simply return the water
to the environment.

QUESTION:
I am an audio engineer and I have a question related to sound and
waveforms. What is the physical difference between a sound wave and an
electromagnetic wave? I understand that sound waves are mechanical
waves, and need a medium through which to travel, while electromagnetic
waves can exist in a vacuum. But what is the actual difference in energy
or physical makeup between the two? If I took an analog sine wave
oscillator with a frequency of say 440hz (which would be audible), and
sped it up to let's say, 1GHz, would I be generating a Microwave? Or is
there some physical difference between a sound wave and an
electromagnetic wave?

ANSWER:
Two things could hardly be more different than sound and light. The
only thing they have in common are that both are waves. In the case of
sound, the waves are longitudinal, that is, the stuff which is "waving" is
waving in the same direction as the wave is traveling; think of a hanging
slinky spring which has compression pulses moving in it. Sound needs a
medium in which to travel; since the molecules in the medium are what
transmit the wave, if you take away the medium, you necessarily take away
the wave. In the case of light, the waves are transverse, that is, the stuff
which is "waving" is waving perpendicular to the direction the wave is
traveling. What is waving in an electromagnetic wave is electric and
magnetic fields. Since the wave is not comprised of the medium through which
it might be moving, no medium is required �the
wave can travel through empty space. You can read more detail about
electromagnetic waves in an
earlier answer .
Your 1 GHz sound wave is most certainly not a microwave, it is ultrasound.

transverse
longitudinal

QUESTION:
why dont jupiters moons get sucked into jupiter if its so massive and
has such a huge graviational pull?

ANSWER:
The sun is much bigger than Jupiter. Why don't its "moons" (AKA
planets) get sucked into it? Given the right velocity, any object can have a
stable orbit around some much more massive object.

QUESTION:
Recently, I was involved in a 4-car pileup on the freeway. The roads were dry, and I was traveling at 40-45 mph in a 2003 Honda Accord
(gross weight: 4300 lbs) when the car in front of me, at about 1 car length
away, started braking. I braked and stopped 3-5 feet behind the car in
front of me before the guy behind me rear-ended me.
The CHP officer claims that I caused the accident by making an unsafe lane
change. The guy behind me was driving a 2005 Toyota Sequoia (gross weight:
6500 lbs.) at 30-35 mph and he was roughly 1-1.5 car lengths behind me when
I got in front of him. I argued that if I could come to a complete & safe
stop going up to 10 mph faster than him, then there's no reason why he
couldn't stop in time. The CHP officer says that since his car has more
mass, he needed more room to stop. I think the difference in speed should
account for the difference in mass. So, who's right?

ANSWER:
I cannot judge on this. If you made an unsafe lane change, I think
you will not gain much by arguing. Certainly 1-1.5 car lengths is too small
for the speeds you were going. To first approximation, though, if you are
getting maximum braking (just barely not skidding) or if you are skidding,
the distance to stop does not depend on mass, just the initial velocity (and reaction
time).

FOLLOWUP QUESTION:
Would you mind explaining why distance to stop doesn't depend on mass, just velocity, if you are getting maximum braking?

ANSWER:
The force of friction if you are just about to skid on a level road
is F= μ _{s} mg
where m is mass, g is acceleration due to gravity (9.8 m/s^{2}
or 32 ft/s^{2} ), and μ _{s} is a constant (called
coefficient of static friction) determined by the surfaces (let's say rubber
on dry asphalt). But, Newton's second law says F=ma where a is
the acceleration. Therefore, ma = μ _{s} mg
and so m cancels out and the acceleration (which determines how
far you go) is independent of m : a = μ _{s} g.
It is easy to show that the distance you go before stopping is s =�v _{0} ^{2} /(μ _{s} g )
where v _{0} is the speed when you start braking. I find that,
taking μ _{s} ≈ 0.6, s ≈110 ft for 45 mph and 70 ft
for 35 mph. These are the absolute minimum (which you can approximate if you
have antilock brakes) and assuming zero reaction time (which is impossible).
Also, if the vehicle locks the wheels and skids, you go farther yet because
the physics is all the same except the coefficient of friction is smaller if
the surfaces slide.

QUESTION:
My wife filled a metal thermos with soup (she says it was warm / room temp). She closed the lid tight and put it in the fridge. Next day, I take out of fridge and have trouble unscrewing top. As soon as I get it loose the lid pops up, soup explodes all over me and the counter, and it looks as if the soup is boiling for a few seconds. Actual temp is much cooler than I would have expected it to be for just being in the fridge.

ANSWER:
Here is what I think. The soup has some air above it. When sealed
the contents of the thermos are very hot. When they get cold, the pressure
in the air gets very much lower, so you have a partial vacuum inside the
thermos �that is why you have trouble
opening it at first. But, when you finally get it cracked, outside air
rushes into the low pressure region thrusting out the soup all over the
place. I expect "�cooler than I would have expected�" is a misperception on
your part.

QUESTION:
If you have a hollow copper sphere and connect the + terminal of a
battery to the copper ball, is the sphere considered to be charged? If
so, what is the charge on the sphere?

ANSWER:
You are raising the potential of the sphere to the voltage of the
battery. Assume the negative terminal is grounded so we can think of zero
potential at infinity. Then the potential will be V=kQ /R where
V is the battery voltage, Q the charge on the sphere (which
will be uniformly distributed on its surface), R the radius of the
sphere, and k =9x10^{9} Nm^{2} /C^{2} .
Therefore, Q=VR /k .

QUESTION:
The current concensus is that while moving close to the speed of light time is slowed. At 99% the speed of light 7 days is equivalent 500 years outside the craft movingthat speed. So wouldnt the amount of fuel consumed by the shuttle only be 7 days worth of fuel. Would this open up possibilities of time travel to the future at a relatively low expense?

ANSWER:
When you get to cruising speed, you do not need much more fuel. The
problem is to get the fuel to get up to the high speed. Also, if you are on
the ship, your clock runs just like you would expect. Only to an observer
moving by you is your clock running slowly.

QUESTION:
if mass equals energy and my mass is 86.6kg then how much energy theoretically could the mass in my body produce if it was released all at once i tried to figure it out and came up with 7.729�10^18 joules (courtesy of wolfram alpha) is that right? and could you put that into some kind of perspective that i might be able to understand.

ANSWER:
Well, of course this is just Mc ^{2} =79.9x(3x10^{8} )^{2} ≈7.7x10^{18}
J. The entire energy consumption of the whole world in a year is about
1.85x10^{10} megawatt hours≈6.7x10^{19} J. So your mass could supply
the whole world with energy for about a month.

QUESTION:
I am a lineman and a intriguing question has been brought to my attention and would love some help. How high would a wooden pole weighing approximately 1500 pounds need to be dropped in order to crush a steel toed boot with a crush rating of 2500 pounds?

ANSWER:
Usually I trash such questions (see my
FAQ page). However in this case you
tell me the force which the toe can withstand (which I will call F =2500
lb) and the weight of the pole (which I will call W =1500 lb). We have
to make one approximation, that the force is constant over the time it takes
to crush it and you will also need to know how much the toe crushes (I will
call that S ). I will not bore you with the details, but I find that,
if H is the height from which it is dropped, H /S =(F /W )-1.
For your numbers, H/S=0.67; that is pretty small. For example, if the toe
crushes by S =3", dropping the pole from H =2" would be enough
to crush it. I would avoid dropping such a thing on my toes.

QUESTION:
Why is it that if you blow a spider suspended by her web she floats out but then when this pendulum swings back it stops when the web is vertical and doesn't swing back and forth? Is it due to the air friction as it comes back to equilibrium or perhaps the dynamic structure of the web strand that absorbs energy that would have made the web swing back and forth?

ANSWER:
It is caused by air drag. This is called a damped oscillator. If
there were no air, the spider would swing back and forth with constant
amplitude, just like a clock pendulum (apart from the little friction from
bending the thread she hangs from. A spider has so little mass that her
terminal velocity is very small �drop
her off the roof and she will not get hurt because she quickly comes to some
constant velocity because the air drag, which can be approximated as being
proportional to her speed, quickly becomes equal to her weight. If the air
drag is not too big, the pendulum will swing back and forth with ever
decreasing amplitude; this is called underdamped. For larger drag, as in the
case of your spider, she never crosses over the equilibrium and just slowly
approaches the bottom of her swing; this is called overdamping. There is a
third possibility called critically damped, but it is qualitatively just
like overdamping, so let's not go there. There is a neat
web site (pun
intended) where you can play with the damping for a mass on a spring, not a
pendulum, but the idea is the same. (The constants there are k is a
measure of the stiffness of the spring, m is the mass, and b
is a measure of how big the drag is.) Two examples from that applet are
shown below. The first might be a pea on a thread where it swings back and
forth a few times before stopping. The second case is your spider whose
swing just dies away.

QUESTION:
I want to know that inside C.R.O. in elecrton gun, how do electrons pass through accelerating anode, due to positive potential these should stick on the anode but it does nt happen, why? and how focusing anode works in c.r.o.

ANSWER:
The field on the other side of the anode is very small, so electrons
which have a trajectory which will allow them to pass through the small hole
will continue through.

QUESTION:
ok!
Suppose we have to masses m attached to a line of length 2r rotating around their cm. Lets further suppose that there is a mechanism of some kind that when activated lets the line extend in length for example 2 times longer.
The masses of the line and the mechanism are supposed to be very small compared to the masses m and the masses m are point masses so their individual rotations can be ignored.
Initially the masses have speed v, so:
total angular momemtum J = 2mvr
total kinetic energy K = mv^2
after the line is allowed to become 2 times longer then, because of the conservation of angular angular momentum, the velocities of the rotating masses will be v/2 (J=2mvr= 2m*v/2*2r).
But now the kinetic energy is 1/4*mv^2
So where did the energy go?

ANSWER:
Suppose we model your mechanism as a thread which connects the two
masses which are beads which can slide frictionlessly on a long rotating
rod. The two are each a distance r from the rotation axis, as you
suggest, and have a speed v . Now, cut the thread. Each bead starts
moving out the rod and both angular momentum and energy are conserved
because there are no external torques or forces doing work. When each gets
to a distance 2r from the rotation axis, it has a tangential
component of its velocity, as you suggest, of v _{t} =v /2
as required by angular momentum conservation. But, each also has a radial
velocity; you can easily find the radial velocity
v _{r} from energy conservation: E _{1} =mv ^{2} =E _{2} =2x �m ( v _{t} ^{2} +v _{r} ^{2} )=m (�v ^{2} + v _{r} ^{2} )
so v _{r} =(v /2)√3.
Finally, to be sure we reproduce the problem you are interested in, we must
stop those beads from sliding out, maybe just put a couple of little hooks
to "catch" them when they get to 2r . The little hooks had to do work
to stop the beads and that is where the energy went.

QUESTION:
The weight of a body is less when it is submerged in water, due to the upthrust.
But the weight of the water column above the body can be far more than the upthrust(weight of the displaced water).
So is it not that the weight of a body can increase when it is submerged in water?

ANSWER:
The weight of an object in a fluid is not less than its weight
anywhere else �the weight is the force
the earth exerts on something. However, there is an additional force when it
is in a fluid, the buoyant force (which you call "upthrust") which makes it
appear that the object has less weight. You do not understand the
nature of your "upthrust", however. When the object is completely submerged,
the water above the top pushes down on it as you suggest; but the water
under the object pushes up with an even greater force. It is the difference
of these two forces which is the buoyant force.

QUESTION:
if photons were assumed to have non zero mass,how will the potential between two electrons be modified

ANSWER:
If the field quantum (which is what the photon is for the
electromagnetic field) has mass, the force is short-ranged. The
electromagnetic force falls off like 1/r ^{2} , a very
long-range force because the photon is massless. The strong (nuclear)
interaction field is mediated by massive mesons and is a very short-range
force.

QUESTION:
I've been reading about special relativity and how space dilates when things travel close to the speed of light. Apparently it would be possible, at close to light speed, to travel to our nearest neighbouring galaxy in 50 years. Why, if light travels at light speed, does it take thousands of light years for light to travel to the other galaxy? Why does space not dilate for light?

ANSWER:
The 50 years you allude to is the clock on the fast-moving ship.
Somebody on earth watching you would see it taking thousands of years. You
can never win a race with light. If a photon had a clock (it does not) it
would not run, time would stand still, and space would all be at one place;
I always like to insist, though, that photons do not have a point of view,
do not carry clocks or meter sticks, so it is pointless to ask "what would
you see if you were on a photon?"

QUESTION:
[This is the last of several communications between The Physicist and
the questioner. Essentially, he was asking the
old question about how much force
does an object which fell some distance exert. I was unable to provide
an adequate answer because I did not know enough details about the
device he used to stop the falling mass. The Physicist ]
�Finally, i s there a correllation between the force that is imparted and the distance the weight falls ?

ANSWER:
Yes. But the nature of the correlation depends on how it stops.
First, if an object of mass m falls from a height h , it ends
with a speed v =√(2gh )
where g =9.8 m/s^{2} is the acceleration due to gravity. I
will do the two simple possibilities for stopping:

The mass always
stops in the same time Δt. In this case, the average force
felt over the time Δt is F _{avg} =mv /Δt=m √(2gh )/Δt ;
so the force is proportional to the square root of the height.

The mass always
stops in the same distance d . In this case, the average
force felt over the distance d is F _{avg} =�mv ^{2} /d =mgh /d ;
so the force is proportional to the height.

QUESTION:
You have explained about terminal velocity a lot, but I cannot find answers to this: the guy who wants to jump from an altitude of 120,000 feet to break the previous record, will supposedly break the sound barrier on the way down. Please explain how this will happen. Also, then when he gets closer to earth, he will slow down, as the air resistence increases?
Why will this guy not burn up, like comets going through the atmosphere? And why does it seem that comets are going so fast, they should be hitting terminal velocity and actually not be going any faster then 120mph?

ANSWER:
I have addressed this
problem before in much detail; it is a very complicated problem and
difficult to make quantitative measurements. He will not burn up because
meteors enter the atmosphere with a velocity far larger than he will ever
have.

QUESTION:
two massive particles m1 and m2 are released from a large distance.what are the speeds of particles when distance between them is 'r'?

ANSWER:
The total energy of the system is zero and the total linear momentum
is zero. When they are a distance r apart, energy conservation and
momentum conservation give the equations
�m _{1} v _{1} ^{2} +�m _{2} v _{2} ^{2}
-(Gm _{1} m _{2} /r )=0 and m _{1} v _{1} -m _{2} v _{2} =0.
If you solve these you will find that v _{1} =m _{2} √[G /(r (m _{1} +m _{2} ))]
and v _{2} =m _{1} √[G /(r (m _{1} +m _{2} ))]

QUESTION:
if a planet were suddenly stopped in its circular orbit ,show that it would fail into the sun in time (T into sq.rt. of 2 divided by 8) where T is time period of revolution.

ANSWER:
Kepler's third law states that the square of the period of an orbit
T is proportional to the cube of the length of the semimajor axis
A , T ^{2} /A ^{3} =constant. For a circle, the
semimajor axis is the radius. For a straight line, the semimajor axis is
half the length. So, T ^{2} /A ^{3} =T' ^{2} /(A /2)^{3}
or T' =T /√8.

QUESTION:
i have read that gasous molecules have no effect of gravitaion whereas gasous with high molecular weight and hydrogen flies in the air, regarding this concept i have an other query that hydrogen stop flying after thermosphere and no gas is yet discovered in the space.

ANSWER:
I do not understand. All molecules in the atmosphere experience
gravitation. It is gravity which holds our atmosphere to the earth, gravity
which accounts for atmospheric pressure, gravity which results in the air
getting less dense as you go up in altitude. At temperatures on the earth's
surface, many helium atoms and hydrogen molecules have speeds greater than
escape velocity and therefore all of those gasses leak out into space.
Interstellar space does contain both helium and hydrogen.

QUESTION:
When two identical out of phase waves going in opposite directions meet (deconstructive interference), there is a moment where the waves cancel out each other. At this moment the rope is flat. Where did the energy go at this specific moment? did it disappear ?

ANSWER:
Why would you think the energy is gone? Even though the rope is
straight at this instant, it is moving so it has a bunch of kinetic energy.

QUESTION:
Hi, I had a question about airships (zeppelins, blimps, etc.). We all know that they work by filling the balloon with a gas that is lighter than the surrounding air (hydrogen, helium, hot air, etc.). My question is, what would happen if you filled the balloon with nothing (i.e. a vacuum)? I understand that in blimps and balloons you need the air to give the balloon structure, but what about in the case of a zeppelin or other rigid-structure airship that has a solid frame that provides structure? Nothing, having no mass, is certainly lighter than air, so would that cause lift, just like with helium?

ANSWER:
Yes. Archimedes' principle simply says that the buoyant force equals
the weight of the displaced air, so the volume is all that matters. If the
buoyant force is larger than the total weight of the ship, up it
goes.

QUESTION:
Although the Kinetic Theory of Matter states that the particles that make up all matter are in perpetual motion, we know that this isn�t really correct because solids don�t move.

ANSWER:
Actually, the atoms in the solid are constantly moving around, they
are not stationary. Think of them as being connected to their nearest
neighbors by tiny springs, constantly bouncing around.

QUESTION:
my 8th grade class is studying sound. We are trying to find out why sound is so much louder when it is cold out? We have eliminated the snow as a reason since it should act as an insulator. We know that sound moves slower in colder temperatures.

ANSWER:
I have actually never noticed this, but I think I can explain it. If
sound is "louder" it must mean that the air is absorbing more of the sound
(like if you filled the space between you and the source with foam rubber
which would absorb the sound). It turns out that the temperature really is
not of much importance. What matters is how much water vapor there is in the
air, that is, what the humidity is. As you probably know, cold air in the
winter is much drier than warm air in the summer which is why we humidify
houses in the winter and dehumidify them in the summer. There is a
nice website
where you can calculate the absorption of sound as a function of frequency,
temperature, and humidity. For example, I find that for a frequency of
10,000 Hz, the attenuation is about 16 dB/km for 0^{0} C, 0% humidity
(cold, dry day) and 76 dB/km for 30^{0} C, 100% humidity (hot muggy
day).

QUESTION:
why does the object you throw in the air while you are walking move with you so that you catch it? I'm a third grader doing a demonstration for the science fair. I looked up the laws of motion and could'nt find an answer.Can you please explain to me why this happens and suggest ways i could get more information on the internet.

ANSWER:
Suppose you are not walking and throw a ball straight up. You are
giving it a velocity which is entirely in the vertical direction. But, if
you are walking (or running, or in an airplane going 600 miles per hour),
when you throw it "straight up" it really has a horizontal velocity equal to
your forward velocity in addition to the vertical velocity. Since there are
no horizontal forces acting on the object, when you release it it continues
moving forward with exactly the same velocity you have and, when it comes
back down, you are there to meet it.

QUESTION:
how can time be measured using speed (time=distance over speed) when speed and time are so closely related. That's like the rule when writing a dictionary: never use the word your trying to define in the sentence which defines it, it creates paradoxes.
My point is, of course, that time is in fact a measure of distance also. When you look at the roots of how time was invented by mankind, you'll see time is related to how far the sun travels across our sky in a day, that being able to be split into 12 and we have ourselves hours. A day, of course, was just a measurement of whether or not the sun had come up, still a measurement of distance because the sun was no longer on this side of the earth but on the other or w/e you get the point. Time seems to be a very man-man concept yet it obviously exists. Is it just me or is time mathematically a very incomplete subject?

ANSWER:
Here is a brief tutorial on how time and length are defined in the
SI system of units. No longer are astronomical cycles used to define time.
The second is based on an atomic clock which very reliably runs at a
constant rate. The second is defined the be
" � 9,192,631,770 periods
of the radiation corresponding to the transition between the two hyperfine
levels of the ground state of the^{ 133} Cs
atom�"
(ref.: Wikepedia ).
The meter is defined in terms of the second and the speed of light. A
" � metre
is the length of the path travelled by light in vacuum during a time
interval of 1 ⁄ 299,792,458 of
a second �"
(ref.: Wikepedia ).
You should not be bothered by this because the speed of light is a universal
constant, no matter who measures it, the speed is always the same.

QUESTION:
There is an urban myth, that if you put a can of liquid in a warm
woolly sock, soak the sock in water, and hang the sock in the summer
sun, the liquid gets colder. Is this true?

ANSWER:
This is no urban myth, this is a simple scientific fact. The sock
need not be warm. To understand this, take a wet towel and spin it rapidly
around for a few seconds. You will find that it gets very cold. That is
because it takes energy to evaporate a liquid and when the water evaporates
it extracts that energy from the water not yet evaporated. Most canteens are
covered with a fabric; the reason is that the fabric can be soaked and it
will keep the water inside cool even when the temperature is very hot. In
the desert water is sometimes carried in fabric bags which never dry out
until the bag is empty because the water slowly seeps out and evaporates.
There are also evaporative coolers which are very efficient air conditioners
if humidity is not an issue.

QUESTION:
i know that when we turned on a light switch, it takes like just 1-2 seconds for the light to come out. the light comes out due to the electron flow in the circuit but the process for the electron to flow usually takes a long time. i am curious here to know that do we wait for the electrons in the wires to reach the light bulb before it light up and also when the current is flowing how long it takes for the bulb to begin emitting visible light?

ANSWER:
1-2 seconds? The light comes on almost instantaneously (unless it is
a fluorescent light). It is true that the electrons move very slowly but
when you turn on the switch, an electric field is set up in the wire with
the speed of light so all the electrons in the wire begin moving almost
immediately.

QUESTION:
How large or small is light year relative to our everyday life?

ANSWER:
A light year is the distance that light travels in one year. It is
almost incomprehensibly large compared to any lengths in your everyday life.
A light year is about 6x10^{12} miles. A spaceship traveling the speed of the
shuttle (about 20,000 mph) would require 34 thousand years to go one light
year.

QUESTION:
the concept of electrostatic potential says that potential of a point in an electric field is the workdone in bringing a unit positive charge from infinity to that point without acceleration and against the electrostatic force. my question is if there is no acceleration in the unit positive charge it means the net force on it is zero because f=ma, if net a=0 then net f must be zero. and workdone is the product of force and the displacement in the direction of force. if f=0 then work should also be zero in every case and hence the potential of every point in electric field should be zero . but it is not so. how ?

ANSWER:
F=ma =0 refers to the net force on the charge. The work which is
associated with the potential energy is the work done by the force of you
pushing it in which is not zero. It is actually more elegant and correct to
say the change in potential energy is equal to the negative of the work done
by the field which gives the same result.

QUESTION:
what is the force that keeps the stars in the galaxy in motion

ANSWER:
Newton's first law says that nothing is required to keep something
in motion. What forces do is change motion. In the case of a galaxy, there
is general motion around the center of mass of the galaxy and the force
which causes the stars to revolve around this point is the gravity they feel
from the presence of all the rest of the galaxy. It has been observed, in
recent years, that their motions are somewhat different from what we would
expect from all the other mass in the galaxy; such observations have led to
the "invention" of some invisible additional source of gravity, called "dark
matter".

QUESTION:
Are there any conditions under which a fan could actually cool a closed, empty (of people) room.
Background: I understand that fans cool people not air. But could, for example, a fan blowing air onto a wet sheet cool a room by forcing more water to evaporate and thereby use up heat? In this or any other case, if heat is being turned into some other form of energy, what form is that?

ANSWER:
If the fan is inside the room and the room is isolated (perfectly
insolated), there is no way that the net effect in the room is to cool down.
Think about it: electrical energy is entering the room to keep the fan
going. It is possible to use some of this energy to evaporate some water as
you suggest, but some of the energy will certainly end up heating the air.
Eventually, when all your water is evaporated, all the energy will go into
heating the room. If you put your fan outside the room and blow air into the
room through your wet sheet, it can cool the room. This is the principle of
evaporative coolers which are useful coolers in dry climates where removing
humidity is not an essential feature of air conditioning.

QUESTION:
if we bring a magnet near an electronic device like e safe or a battery powered device will it eventually absorbs all the electric charge store i it.

ANSWER:
If you put an electric charge in a magnetic field, nothing happens.
Therefore, the magnet will certainly not "absorb" all the electric charge in
a battery. In fact, a battery does not "store" charge, it stores energy.

QUESTION:
If the Earth's mass suddenly doubled, what would immediately happen to the gravitational force between the sun and the Earth and what about the centripetal force?

ANSWER:
The gravitational force would double. The centripetal force would
double. The orbit would be unchanged.

QUESTION:
I have been watching a lot of basketballgames lately and I have started to wonder:
Why does the net go up when scoring?
The ball goes down so I simply cant figure out why the net goes up. Can you help me?

ANSWER:
The unstretched net is a little smaller than the ball, so the ball
pulls it down when it goes through. But the net is attached to the rim and
so the net pulls the rim down. But, the rim is made of steel and therefore
acts like a spring; when the rim bounces back up it pulls the net up.

QUESTION:
What makes particles, atoms, planets, galaxies and everything else spin?

ANSWER:
A better question might be "why do we see almost nothing which is
not spinning?" Imagine a huge cloud of dust and gas. What are the chances
that its total angular momentum is zero? Because it is big, even tiny
velocities of its constitutents far from the center will have substantial
angular momentum. As the cloud collapses under its gravitational attraction
(I am thinking of a star forming, but it could apply to anything which forms
by getting smaller), it will keep the same angular momentum but will spin
much faster. This is angular momentum conservation which is true if the
system you are looking at is only minimally interacting with anything else.
Notice how the spinning dumb bell to the left speeds up when it gets
smaller.

QUESTION:
what is the spin orbit force?

ANSWER:
An elementary particle may have two different kinds of angular
momentum, spin and orbital. For example, if you think of an electron of mass
m in an orbit of radius r and speed v around the
nucleus, it has orbital angular momentum which classically is mvr and quantum mechanically is
ħ √[ℓ(ℓ +1)] where
ℓ is an integer. But, it also has a spin angular momentum of
ħ √[�(�+1)]. Now, if
there is some way these two separate angular momenta can interact with each
other, the result will be that the energy of the electron will depend on the
relative orientations of the angular momentum vectors, either parallel or
antiparallel. This is called spin-orbit splitting or fine structure in
atomic physics. In atomic physics it is pretty easy to understand why the
angular momenta couple. From the electron's point of view, it sees the
nucleus orbiting it and therefore sees a magnetic field due to that current;
the electron has an intrinsic magnetic moment (it looks like a tiny bar
magnet) antiparallel to its spin which wants to align with that field which
means that it takes work to move the electron to the unaligned orientation
meaning aligned is at a lower energy. The spin-orbit interaction is a very
much bigger effect in the structure of the nucleus but its origin is not as
easy to understand as for electons in atoms.

QUESTION:
Does a battery weigh more when full than discharged?
If not, then why is E= mc2 not relevant - assuming that discharging does some sort of work like emit light, heat or cause motion?
I googled this for an hour: I found about an equal # of YES/NO responses.

ANSWER:
I'll weigh in on this one (pun intended)! Yes, if you add energy to
anything you increase its mass. Let's get an idea of how much. The energy
stored in a 3.6 V lithium battery is about 10^{4} J, so the
corresponding mass would be m=E /c ^{2} =10^{4} /(3x10^{8} )^{2} ≈10^{-13}
kg. This is about 100,000 times smaller than the mass of a single dust
particle. Lots of luck trying to measure this mass change!

QUESTION:
Can you settle an agrugment here Please!!!!! If you have two cars and they are driving towards each other can a accurrate speed be determinded. My son thinks that physics would apply here.

ANSWER:
It is not really clear what you are asking. If you want to know if
one car can deduce how fast the other is going (relative to the road) then
the answer is yes if the the first car knows what his speed is (relative to
the road). For example, if the first car is going 50 mph north and measures
the second car to be approaching him from the south at a speed of 120 mph,
the speed of the second car (relative to the road) is 70 mph. Did you get a
speeding ticket from a cop in a car approaching you and with a radar gun? He
can do it!

QUESTION:
what is the difference between "inertial mass" and "gravitational mass"?

ANSWER:
Inertial mass is the property an object has which quantifies what
its resistance to being accelerated is. Gravitational mass is the property
an object has which allows it to feel and cause gravitational forces. It
turns out that, as shown in the theory of general relativity, they are
actually the same thing and there is only one kind of mass.

QUESTION:
Why or what makes molecules move? I understand the relation between the RMS velocity and temperature but why do molecules move in the first place. Or to put the question another way if there was an isolated molecule (of hydrogen, say) at absolute zero floating in space (no other particles can impart any energy to it) and you started to irradiate it would it spontaneously shoot off? and if so why - and why doesn't that violate Newton's 1st and 3rd laws of motion?

ANSWER:
If you ignore the internal structure of the atoms or molecules, the
only way a gas can accept added energy is by adding kinetic energy to the
atoms/molecules. Your second question really does not make sense because
temperature is really not defined for a single particle, it is a statistical
concept. Besides why would it be at absolute zero anyway? And, if you
"irradiated" it (by which you mean add energy to it), it would just go
faster.

QUESTION:
planets do not loose energy when they orbit the sun due to interplay of centripetal and centrifugal force. Then why is this system not able to explain that electrons do not loose energy while orbiting the nucleus?

ANSWER:
Your premise that energy is conserved because of an "interplay
of centripetal and centrifugal force" is utter nonsense. The reason that the
energy remains constant is because the only force present is a conservative
central force for which a potential energy may be introduced and there are
therefore no external forces doing work on the system. In the case of an
electron moving in an orbit, we expect it to lose its energy because
accelerating electric charges radiate energy (which is how a transmitting
antenna works, for example). Technically, an accelerating mass is expected
to lose energy by radiating gravitational waves but no direct evidence for
gravitational waves has ever been seen.

QUESTION:
how long does it take a 1 kg mass with an area of 1 square meter to get to terminal velocity when dropped. actually any size will work but i need the mass and area.

ANSWER:
Technically, it never reaches the terminal velocity but approaches
it. Also, you can only approximately do the calculation using an approximate
expression for the air drag force which I often use, F ≈�Av ^{2
} where A is the area presented to the onrushing air and v
is the speed (and this is only true in SI units). I cannot tell from your
question whether you want a general solution for any mass (m ) and
area or just for this specific problem. So, when I solve the problem for the
velocity as a function of time I get the solution v =√(g /c )tanh(t √(gc ))
where g is the acceleration due to gravity (9.8 m/s^{2} ),
c =�A/m , and tanh is the hyperbolic tangent function. For
your specific problem, v =6.26tanh(1.57t ), so the terminal
velocity is 6.26 m/s because tanh(∞)=1. The graph of this is shown to the
right; as you can see, the terminal velocity is reached, for all intents and
purposes, by about 1.5-2 seconds. Sorry if this is more technical than you
wanted; oh, and the tanh is defined as tanh(x)=[(e^{x} -e^{-x} )/(e^{x} +e^{-x} )].

QUESTION:
I have a question regarding electron degeneracy pressure. I understand that this pressure arises from the exclusion principle which states that no two electrons can be in the same quantum state. How then is this pressure overcome (say in the collapse of a white dwarf to a neutron star)? I take it the exclusion principle is not violated at this point?

ANSWER:
This is the effect which makes solids and liquids difficult to
compress. However, if the pressure becomes large enough, as in the
gravitational collapse of a burned out star, the electrons are forced into
the protons to form neutrons, and neutrinos are released, a different quantum state so there
is no violation of the Pauli principle per se . As often happens, I
learned something new researching this question. I have always
thought that the electrostatic
repulsion between electrons on the surfaces of "touching" objects was what
caused them from passing through each other. Indeed, Freeman Dyson showed
that it is instead the electron degeneracy pressure which is responsible.
Learn something new every day!

QUESTION:
I'm curious about why when you kick a stone it hurts more than if you kick a sponge of the same mass?
Newton's third law suggests that if you kick them with the same force, then they should exert the same reaction force back on you. So why do hard objects like stones hurt more than soft objects like sponges?

ANSWER:
What matters is how long the impact lasts. Suppose your foot has a
mass m and a speed v , then the momentum of your foot is mv .
Suppose the time it takes your foot to stop is t . The average force
your foot experiences if it comes to a dead stop, is given by F=mv /t .
Your foot stops much more quickly when you hit the stone than the sponge,
t _{sponge} >t _{stone} , so F _{stone} >F _{sponge} .

QUESTION:
I was wondering that, if you jump off a building on a chair, lets assume that the building is 20 meters high, and at the last second before you hit the ground, you jump off the chair, how less damage could the impact give to you?

ANSWER:
I answered a similar question in an
old answer , jumping in
a falling elevator. There is no way that jumping can save you because the
amount of upward speed you could give yourself is tiny compared to your
downward speed. In your case (standing on a chair), you would have even less
chance because, because the chair has a smaller mass than you, the main
effect of jumping would be to make the chair go down faster.

QUESTION:
According to newtons 3rd law of motion, a rock sitting on the ground pushes against the groung, and the ground pushed back against the rock with force. explain why this force doesnt cause the rock to accelerate upward from the ground according to newtons 2nd law.

ANSWER:
You need to look at all the forces on the rock. Let us say that it
has a weight of 5 lb. Then the earth pulls down on the rock with a force of
5 lb (that is what weight means). But Newton's first law says that if the
rock is in equilibrium (which it clearly is), the net force on it
must be zero. Since the only other thing which can possibly exert a force on
the rock is the ground, it must exert a force up on the rock which is equal
in magnitude to 5 lb making the net force zero. Nothing needs to be said
about the third law. If you want to invoke the third law, then the fact that
the ground pushes up on the rock with a force of 5 lb, the rock must push
down on the ground with a force of 5 lb. You could also say that if the
earth pulls down on the rock with a force of 5 lb, the rock must pull up on
the earth with a force of 5 lb.

QUESTION:
My nephew asked me why snow is white.

ANSWER:
There is a good explanation at
HowStuffWorks.com .

QUESTION:
My son wanted to do a science project on time dilation, we read that it can only be proved with thought expts., like twin paradox. I also read that some expt. Can be made with two parallel mirrors[when light is passed through the mirror it reflects and bounces making a v_shape pattern]., which proves speed of light differs in different frames. Is this a proper expt. To prove time dilation or is not a proper proof?, pls let me know, my son is 8 yrs old he is very much interested onthis topic.

ANSWER:
What does "proved" mean to you? I think you want, for an 8-year old,
to make it plausible and demonstrable, not proven. Proven, I guess, would
include a derivation of the Lorentz transformation from which time dilation
follows naturally, and that is surely far beyond your son. The first thing
you need to do is to get him to appreciate the one thing you must accept in
order to understand relativity, that the speed of light is the same no
matter who measures it. So, if you shine a light forward from your rapidly
moving spaceship, you will measure its speed to be 186,000 miles/second and,
even if your speed is 93,000 miles/second, somebody you are going by will
still measure the speed of the light to be 186,000! You will find this
discussed on my FAQ page and will
probably need to paraphrase those discussions for your son. Now, how can you
make time dilation plausible, given the constancy of the speed of light?
That is where the
light clock you allude to comes in. But this experiment does not "prove[s]
speed of light differs in different frames", quite the contrary: it assumes
that the speed of light does not differ and shows that it follows
that moving clocks run slower than stationary clocks (time dilation). But,
is it all academic, we can only "observe" it for "thought experiments"
(traditionally called by the German name gedanken )? No, the theory of
special relativity is one of the best verified in all of physics, never has
there been any indication that any of it is wrong. But to easily see time
dilation in action requires very high speeds. For example, consider a pi
meson, an elementary particle which has a half life of 18 nanoseconds (10^{-9}
s). Suppose we make one in the lab with a speed of 80% the speed of light.
Since the speed of light is 3x10^{8} m/s, we expect it to go (3x10^{8} )(18x10^{-9} )=5.4
m. But when we actually observe the pi meson, it goes 9 m. Why? Because this
particle is like a clock that ticks once and dies; that clock is moving very
fast and therefore ticks much more slowly than if it were just sitting here,
so it can go a lot farther than we expect it to. Another good example is the
satellites used in GPS systems. Their speeds are quite small compared to
light speed, but it is extremely important to measure time very accurately
for the systems to be accurate; if corrections for time dilation were not
applied, you would never be able to have GPS with near the resolution we all
take for granted.

QUESTION:
A steel railing on my outside deck connects to the house near our bedroom. When it is very cold outside the railing vibrates - a very low harmonic hum, almost like a train in the far distance. The sound keeps us awake and can cause a headache. I have solved it by strapping a heavy piece of carpet over the railing which diminishes the vibration and noise. This only occurs in the cold with a wind. Never the rest of the year. Why?

ANSWER:
Wind can cause things to vibrate. If the object is free to vibrate
and the wind causes vibrations of one or more of the resonant frequencies of
the object, the result can be quite dramatic. The classic example, of
course, is the collapse of the
Tacoma Narrows bridge .
Since most materials expand when heated, it would appear that something is
contracting which allows the railing to vibrate when it is cold. I do not
know how it is attached to whatever it is attached to, but I would check to
see if the attachments could be tightened up when it is cold. That might
clamp the vibrations. If not, your carpet (which is called damping) may be
your only option.

QUESTION:
I've been wondering if I had a telescope fixed on a star that has gone super nova, but the light of the explosion had not yet reached earth, if I mounted said telescope to a rocket traveling close to light speed and flew towards the star would I see the evolution of the star to it's final fiery demise being accelerated due to:
a. The fact that I am traveling towards the light source b. The effects of time dilation
Also, would it begin to glow with a blueish tint due to the doppler effect?

ANSWER:
This is another question which asks "how fast do clocks appear
to run?" You will indeed see the supernova you are moving toward as
happening sooner and faster than if you watched from earth. It also will be
blue-shifted because of your motion. To understand this better, consider the
return trip of an
earlier answer.

QUESTION:
How electrons & protons combine to form neutrons in a neutron star?

QUESTION:
What should happen when electrons are squeezed into the nucleus?

ANSWER:
What happens is the inverse of beta decay, a proton and electron go
to a neutron and a neutrino: p+e� >n+ ν _{e} .

QUESTION:
for almost five years since we were taught, we've known that current is a scalar quantity having direction that does not obey vector laws, but, obeys Kirchoff's laws. However, I'm recently studying electrodynamics from Griffiths and it says that current is vector quantity. Its very confusing. If current is vector, it has magnitude and direction that obeys vector laws and also directions with respect to Kirchoff's laws. My feeling is that it is vector at microscopic scale and scalar at macroscopic scale. But, how come?

ANSWER:
Current density J is a vector. The current through
some area A is defined as _{A} ∫ J ∙ dA
over the area. So, current can be either positive or negative
depending on the sign of the dot product. It is not a vector, but its sign
matters, particularly when applying Kirchoff's laws as you have found.

QUESTION:
If a rocket is in space and you add thrust to the rocket to make it move forward for 10 seconds, and then 2 minutes later you add the same amount of thrust to move the rocket forward again. Since you are already going X MPH would the rocket increase speed based on the additional thrust or would it stay the same speed since you are already going X MPH? Being that there is no friction (very little) in space, would each thrust forward add to the forward motion and increase the speed? Also meaning that if you continued to add thrust without stopping, even in small amounts, you would continuously go faster and faster until you ran out of thrust? I undertsand that the faster an object goes, the more mass it has, but I do not understand why that would need additional thrust if there is no friction in space.

ANSWER:
All this turning on and turning off "thrust" just complicates
things. Let's focus on a simpler situation where we simply apply a constant
force to the rocket continuously. In classical physics, this would just
result in the rocket going faster and faster without bound at a constant
rate. However, when the speed becomes comparable to the speed of light, we
are constrained to not exceed the speed of light and acceleration ceases to
be constant even though the force is constant. For a detailed discussion of
this problem, see an
earlier answer .

QUESTION:
If Earth is moving through Space at an incredible rate, why do
relatively tiny changes in speed effect humans so strongly on earth?
If Earth is moving through the Universe, within the galaxy, within whatever
larger system, at astronomical speeds, how does something like a roller
coaster or a plane ride - which changes the speed from not moving on Earth
so incredibly fractionally compared to the overall speed we are travelling
through the universe - have such a great effect on the Human Body.

ANSWER:
The way that the earth moves subjects you to very small accelerations and
acceleration is what you feel.

FOLLOWUP QUESTION:
So we know for a fact that the earth isnt accelerating or decelerating through space any fast than a fast car pulling away from a stop light? In fact, significantly less? That seems rather improbable to me. How in the world could it be that we can easily achieve acceleration noticeable enough to feel on the ultra-small scale of earth - such as in falling off a cliff, or skydyving - while the enormous earth in an enormous solar system, in an enormous galaxy, in an enormous universe that's expanding at an unbelievable rate somehow reflects an acceleration less than this - unnoticeable to us. To put it another way, if the universe is expanding, how could it possible be that the de/acceleration of the universe is less than the meager capabilities of a motorbike.

ANSWER:
You confuse speed with acceleration. Just because something is going
very fast, it is not necessarily accelerating. What matters is the rate at
which a velocity vector is changing. The greatest acceleration due to the
external sources you list is due to the earth's rotation. The speed at the
equator is about 465 m/s which is about 1040 mph, pretty fast compared to
your motor bike. But, what is the acceleration related to this motion? It is
about 0.034 m/s^{2} (from v ^{2} /R ), about 300
times smaller than the acceleration due to gravity. Another example is the
acceleration due to earth's orbit around the sun. The speed here is much
bigger, about 30,000 m/s, but the acceleration is about 0.006 m/s^{2} .

QUESTION:
I would be grateful if you might help me with a difficult problem in mechanics that I have tried to solve for a long time without success. The problem is a part of a bigger problem that I am trying to solve and is therefore important to me. Here is the problem:
How much power P [W] is needed to force a satellite (modeled as a point particle) to move in a perfect circular orbit with radius r [m]? The mass of the satellite is m [kg] and its tangential speed v [m/s] is constant.
Important conditions:

It is assumed that NO GRAVITY affects the satellite!

The mass loss due to fuel consumption is neglected.

The power is assumed to act under ideal conditions, i.e. there is no energy loss due to friction or the choice of engine.

It is also assumed that v << c, i.e. relativistic effects are neglected.

I have received many answers saying that P = 0 as a consequence of the definition of mechanical work. But this cannot be true since no gravity affects the satellite. Energy is needed to change a linear path in space and a circular orbit is one such motion. In absence of gravity no particle with mass will loop in a circle without any energy.

ANSWER:
What you have been told before is correct. No work is done by the force causing the satellite to move in orbit. Since no work is done, we can ask whether the energy is conserved. Since there is no gravity, the only energy
to think about is the kinetic energy of the satellite and since the speed is the same always, so is the kinetic energy. Or, look at it this way: suppose we are in otherwise empty space but that there is a fixed point to which we can tie a rope, the other end of which we tie to the satellite which we give a shove so that it moves in a circle around the center. Does the string need to supply any energy?
Do you need to connect a machine of some sort to the rope? Your basic
premise, "Energy is needed to change a linear
path in space", is simply wrong; a force is needed, but not energy. This is
not a "difficult problem".

QUESTION:
Have there been any theories presented by rational and peer-accepted scientists, trying to pair the combining force of Gravity with the separating force of Dark Energy into a single equation or system of equations?

ANSWER:
This already exists, has existed ever since the origin of general
relativity. At the time Einstein developed general relativity to explain
gravity, it was believed that the universe was static. He therefore included
a term in the general relativistic equations called the cosmological
constant. Later, when Hubble found that the universe was expanding, he
removed the constant calling it his "greatest blunder". If now it is
reinserted, dark energy is simulated. However, this is far from a complete
theory or explanation, seems to be pretty empirical.

QUESTION:
A friend claims that the inside of an LP record (or CD for that matter) spins faster than the outside. I say this is impossible because if it were true, the record/CD would tear itself apart. Who is right?

ANSWER:
LPs and CDs spin differently and therefore must be discussed
separately. First we need to clarify what "spin faster" means. Since a disk
is a rigid object, each point has the same angular velocity,
revolutions per minute (rpm) for example. Each point on any disk rotating at
100 rpm goes around 100 times per minute. However, if you look at the speed
of any point on the disk, the farther out you go the faster the point goes;
a point 4 inches from the axis will move with a speed twice as large as a
point 2 inches from the axis. An LP is sometimes referred to as a "33"
because it always spins with an angular velocity 33 1/3 rpm. The groove
moves past the needle slower and slower as the needle moves in during the
playing. So, if you were to microscopically look at what the groove looked
like for some note, that would look different depending where on the record
that note was recorded. CDs, on the other hand, are designed so that the
speed with which the data on the disk pass the laser reader is always the
same, so as the CD plays, the angular velocity is constantly adjusted to
keep the linear velocity of the data being read the same. The CD, unlike the
LP, reads from the middle out and as the laser moves out the angular
velocity decreases. So, clearly neither of you is right: inside and outside
spin the same, outside moves faster, CDs and LPs have different motions, and
nothing tears itself apart.

QUESTION:
Kind of a simple question, I think, but just curious. If I check
in at an airport with my luggage, someone said if I lay my bag down it will
weigh less than verticle. Obviously the thought may not get charged for over
50lbs or whatever if I lay it down. If you're married you probably
understand where I'm coming from:)

ANSWER:
The pressure the bag exerts will be smaller, but the weight is unchanged.
The scale they use at check in measures weight, not pressure.
No comment on the marriage statement.

QUESTION:
First, take a large round, flat-bottomed container and pour in 3 inches or more of water. Sprinkle in about a quarter to a half teaspoon of sand. First stir the water at random and notice how the sand distributes itself in a random manner. Then, swirl the water around in the container with a spoon or other utensil in a continuous clock-wise or counter clock-wise motion for several seconds. Remove the stirrer. Why do the particles of sand aggregate in the center of the vortex? Since they are heavier than water (they sink), why aren't they spun to the sides of the container due to the negative centripetal force? Is this behavior similar for air and, for example, saw dust in industrial dust collectors?

ANSWER:
Seems it is not a trivial question. There is a ton of stuff
here , all of which I have not waded through.

FOLLOWUP:
I followed up on the reference to "Einstein's Tea Leaves" and found
a very good
article .

QUESTION:
At what velocity and angle would one have to launch a paintball from the surface of the moon for it to achieve and maintain an orbit?
Paintballs weigh approx 4 grams and are approx .685" in diam.

ANSWER:
First of all, the mass and size of the paintball is irrelevant. The
orbital speed for any particular orbit is the same for any mass (small
compared to the mass of the moon). And, you cannot launch from the surface
if you want it to not eventually hit the surface again, you must launch from
above the surface. But let's say that we launch horizontally a few feet
above the surface (and neglect any mountains it might hit). The minimum
launch speed would be about 1680 m/s ≈3760
mph.

QUESTION:
While using an elliptical machine for exercise, as one increases the resistance from the machine for a period of time, does one actually exert more force with his legs? Or is the maximum force a person can exert just the force from his body weight (gravity) which doesn't change as he exercises? Can one press harder with one's legs if he doesn't have something to brace against to increase that force?

ANSWER:
I do not know the details of an elliptical machine, have never used
one. But, I can tell you that you can exert a force up on your body with
your legs which is greater than your weight. This is what you do when you
jump. The fact is that if you do so, your torso accelerates up. For example,
if you are in a squat and stand up quickly your legs are exerting a force
greater than your weight; the quicker you stand up, the bigger the
force. I do not know what it means to "increase the resistance" of the
machine, but if it means go faster then it would be sort of like doing knee
bends faster.

QUESTION:
While airing up a heavy truck tire, my friend said if I lifted the tire off the ground (relieved the load) it would air faster.
This seems wrong to me. 110# is 110# (internal pressure) regardless of the external load.
Who is right, and more importantly, why?

ANSWER:
Take the extreme situation as an illustrative example: when sitting
on the ground the unfilled tire will be flattened down almost to the rim.
This might be the case if the wheel is mounted on the truck. Now, when
pumping it up, the increased air must do two things, increase the pressure
and increase the volume. Increasing the volume means work is done which
means it takes energy. If the tire starts out round, adding air only
increases the pressure and does not do the work to increase the volume. So,
I would guess that the wheel which starts out round would fill faster than
one you need to expand. If the tire is already round when you start there
would be no need to lift it. Nevertheless, the compressor is probably strong
enough that the difference will be pretty small. Should be an easy enough
thing to actually try it. Physics is an experimental science, after all.

QUESTION:
When a balloon collects a charge it gains electrons and loses the positive charge.why is it that the balloon will neve have a positive charge, everywhere I look says it will always be negative, but doesn't have an explanation.

ANSWER:
Something is charged negatively by adding electrons and positively
by removing electrons. Balloons are made of rubber, an insulator. Insulators
are difficult to remove electrons from because all the electrons in the
atoms are tightly bound. There is no reason why electrons could not be added
to an insulator, though.

QUESTION:
If I am in a rocket ship that is moving away from Earth at very near the speed of light and my friend is on another rocket ship also headed away from Earth at very near the speed of light but in the opposite direction. Do we see each other as moving away from each other at greater than the speed of light?

ANSWER:
No, because velocities do not add the way you expect them to in
Galilean relativity which is v'=u+v where u and v are
the speeds of the two ships and v' is the speed that one sees the
other moving. This turns out to be incorrect if u and v are
not small compared to the speed of light c . The relativistically
correct addition formula is
v' =(u +v )/[1+(uv /c ^{2} )]. So, if
u=v =0.9c,
v' =1.8 c /(1.81)=0.994 c .
If you had looked at my FAQ page you would have found this discussed; the
link would get you to
here .

QUESTION:
I'm trying to program a robot to shoot a basketball into a
basket. I've been trying to simplify doing this with only projectile motions
equations (The ball weighs 11.2 oz and probably is effected by drag). The
robot would be able to find the height and the distance. I was wondering if
it is possible to find the initial velocity and the angle I would need, just
from the height and distance?

ANSWER:
Possible, yes. Easy, no. It is a pretty complicated problem which does not
have a simple formula you can use. I would try the simple model first and
see how it works.

FOLLOWUP QUESTION:
So are you telling me to just use the projectile motion formulas? When I do them I end up with 3 variabled, time, initial velocity and theta

ANSWER:
I thought your concern was the air drag. So, when you do the
kinematic equations, there are two:

One is the horizontal position which must end
up at x=R where R is how far horizontally the basket is
from the launch point; R=v _{0} t cos θ
where v _{0} is the initial velocity and t is
the time of flight.

The other is the
horizontal position which must end up at y=h where h
is how far vertically the basket is above the launch point; h =v _{0} t sin θ- �gt ^{2
} where g is the acceleration due to gravity (9.8 m/s^{2} ).

So, you have two
equations with three unknowns,
v _{0} , θ , and
t . The first thing to do is reduce it to one equation with two
unknowns by solving the R equation for t and substituting that
into the h equation. The result is h=R tanθ- �gR ^{2} /( v _{0} cosθ )^{2} .
Now, think about it�you cannot mathematically know both θ and
v _{0} because you have only one equation; but you
shouldn't expect to from a physical point of view either. If you change v _{0
} you will have to change
θ too to hit the target. So, there are an infinite number of
right answers here, not one unique answer. Nevertheless, if you program your
robot to always shoot with the same velocity, you can solve for the angle.
So, you need to solve the above equation for θ and that will tell you
how to aim. It is not trivial algebra, but not real hard either and I do not
intend to present the general solution because it is complicated. Be aware
that there will be two solutions for every situation which you can easily
understand, I think. Think of a rifle shooting at a target where you have to
aim a little above the target to hit it; if you also shot way up in the air,
there would be some large angle where it would come back down through the
target. I will leave the general solution to you; I will do one simple
special case to illustrate, R =4 m, h =2 m, v _{0} =10
m/s, and I will approximate g as 10 m/s^{2} : 2=4sinθ /cosθ- 0.8/cos^{2} θ.
Rearranging and simplifying (and using sinθ= √(1-cos^{2} θ )),
0.4+cos^{2} θ= 2cosθ √(1-cos^{2} θ ). Now
square this equation and rearrange: 5cos^{4} θ- 3.2cos^{2} θ+ 0.16=0.
This is a quadratic equation in cos^{2} θ. Thus, cos^{2} θ= (0.585
or 0.055) and cosθ= (0.765 or 0.234) and θ=( 40.1^{0} or
76.5^{0} ). Of course, you could also fix θ and solve for v _{0} .
This is easier. For example, if you fixed θ at 40.1^{0} , 2=4tan40.1^{0} -80/( v _{0} cos40.1^{0} )^{2}
which leads to, guess what, v _{0} =10 m/s. However, you will
run into trouble here because since you are playing basketball, the ball
must be coming down when it gets to the basket, and that will not always be
the case for a randomly chosen set of R , h , and θ. For
example, in the example I did, 40.1^{0} would not work because the
ball would be going up when it got to the basket (v _{y} =1.2
m/s); for the other solution, 76.5^{0} , the ball would be going down
(v _{y} =-7.4 m/s) like you need to make the basket. I got
these by solving for the times from the x equation and then using the
velocity equation (v _{y} =v _{0} sinθ-gt ).
You have your work cut out for you!

QUESTION:
Here is a question I have long pondered: If a man is riding in the back of a pickup truck, and the truck is traveling forward at 500 feet per second, and he fires a pistol toward the rear of the truck, and the pistol fires the bullet at 500 feet per second, will the bullet leave the barrel?

ANSWER:
When you say "the pistol fires the
bullet at 500 feet per second", this is the muzzle velocity, the velocity of
the bullet with respect to the barrel . So, yes, the bullet will
definitely leave the barrel. And if you are the man in the truck, you see
the bullet go backwards with a speed of 500 fps. But, you see the rest of
the world going backwards with a speed of 500 fps also. So, somebody
standing on the ground will see a bullet drop straight to the ground! (This
assumes we can neglect air friction.)

QUESTION:
I learned the basic premise that things moving fast can have negative effects on other things when I was in kindergarden. In elementary school I came up with a question. If light travels so fast, why doesn't it completely obliterate anything it hits, by smashing it to pieces? I had my sister ask her physics teacher. The answer he sent back to me was that light was too unfocused, and that objects with zero rest mass do not have that kind of effects on things which do have mass. (Great answer to give to a kid in elementary school.)
This question remained with me as I got older, and inquiring my own physics teachers I only learned enough to become further confused about the subject. I know when something with positive rest mass, like a bullet or a baseball hits something it transfers momentum and kinetic energy. I was told that the effect light has on objects heating them etc, is an effect of kinetic energy. But I am still having trouble really understanding the concept. With learning the difference between energy, and momentum only helping slightly.
If I throw a baseball with high velocity and it hits something, like a block of metal the impact of the ball will cause the block to move. If I shine a laser at the block and just keeping adding more and more power would it ever be possible to accelerate the block with the laser? (Assuming it could avoid being melted)

ANSWER:
You are interested in force so you need to ask about linear momentum
(which we usually think of as p=mv , but I will modify this a little
below) because, in a collision, each object experiences a force which is
proportional to the change in its momentum. When a ball strikes another
ball, during the time they are in contact each exerts a force on the other
(which are equal and opposite, Newton's third law). The struck ball,
initially at rest, rebounds forward because of this force and the striking
ball slows down because of the equal/opposite force it feels. If the
striking ball moves faster, it has more momentum and therefore exerts a
larger average force during the collision. In fact, the force is precisely
equal to the time rate of change of momentum; this is Newton's second law.
Now, to your question. Light can exert a force but not because it goes fast
but because it carries momentum. Even though photons have no mass, they
still do have momentum. So, when light strikes something, it does exert a
force on it; that force, for everyday situations, is just too small to have
much of an effect because normal intensities of light just do not have all
that much momentum. I will give you a couple of examples of so called light
pressure. GPS satellites, to work accurately, must have their orbits
calculated extremely accurately and corrections for radiation pressure from
light from the sun must be applied. It has been proposed that future
spaceships in the solar system could have gigantic reflecting sails and use
radiation pressure as a means to be accelerated away from the sun; each
photon comes in with a momentum p=E /c where E is the energy of the
photon and c is the speed of light. Since the photons are reflected,
their change in momentum is 2E /c which is also the momentum
transferred to the spaceship.

QUESTION:
gallile said that the force of gravity is the same on everything but the force of rubbing with the air molecule that change the speed of falling,so shouldnt 2 balls made from the same materiel but one is bigger than the other falling from the same heigth,the smaller touch the ground first cz her volum is smaller than the force of rubbing should be smaller than the big ball,but that is not the fact i dnt understand?

ANSWER:
First of all, Galileo never said the force of gravity is the same on
everything because the notions of force and gravity were not even conceived
before he was dead! Anyhow, that is not even true; the force of gravity is
greater for greater masses. What is the same is the acceleration due to
gravity, which Galileo supposedly demonstrated by showing that two objects
of different masses hit the ground simultaneously. This is only
approximately true because there is a force due to the air resistance as the
balls fall. Galileo did not take air resistance into account and balls of
different masses and sizes will not hit simultaneously. If you look at my
FAQ page you will find lots of discussion of problems involving air drag and
you will find that the force is proportional to velocity times the area
presented to the oncoming wind (approximately). If the two balls are made of
the same material, their masses are in the ratio (R _{1} /R _{2} )^{3
} and their cross sectional areas are in the ratio (v _{1} R _{1} /v _{2} R _{2} )^{2} .
As each ball falls, it initially is accelerating down but this acceleration
gets smaller as the speed increases because of the increasing air drag. Each
ball eventually falls fast enough that the air drag is equal to the weight
and falls with a constant speed called the terminal velocity. So, the weight
(mg ) and the drag ( �πR ^{2} v ^{2} )
are equal. Taking the ratio of the equations for the two balls,
(R _{1} /R _{2} )^{3} =(v _{1} R _{1} /v _{2} R _{2} )^{2}
so v _{1} =v _{2} √ (R _{1} /R _{2} ).
Suppose that #1 is the smaller ball and R _{2} =4R _{1} ;
then v _{1} = � v _{2} .
Since the smaller ball has the smaller terminal velocity, the larger (more
massive) ball will reach the ground first. So, in fact, your intuition was
wrong.

QUESTION:
Given the same frontal area, mass, surface area, etc. Why is the drag coefficient less for a hemisphere than a sphere? A sphere seems more streamlined but has a larger drag coefficient.

ANSWER:
It is true that drag is approximately proportional to cross
sectional area and therefore you would expect the two to have the same drag.
However, the key word is approximately and computation of drag
coefficients is a complicated thing and depends on aerodynamic details which
can really only be determined numerically. The nature of the wake and the
turbulence are crucial and often not intuitive. My favorite example is the
use of tailgate nets in pickup trucks. It is obvious that opening the
tailgate on a pickup truck will make the truck more streamlined, right? That
is the idea behind those nets which are marketed to save gas. However,
opening the tailgate considerably increases drag; the tailgate traps a
"bubble of air" which moves with the truck.

QUESTION:
Often times they refer to the flipping of a coin as having an outcome that is random. If this be the case, I am wondering how then a particular side of a flipped disc had a person applied pressure to one portion of a disc and after having applied all necessary forces, i.e. wind resistance, gravity, etc., and had observed the disc up on a flat surface calculated a particular image to be heads or tails at random. I understand that it has to be either or and there is "probability" of an imagine over another. How can one be more likely than the other if you flick it the exact same way every time so that one side would never end up showing. How can probability be so disconnected from physics if even conceptually?

ANSWER:
Since flipping the coin is done by a person, it is simply impossible
to "flick it the exact same way every time".
Also, some of the forces, air drag in particular, have aspects which involve
turbulence which is chaotic and cannot be predicted precisely nor
replilcated each throw. Perhaps in a vacuum one could, by using a machine to
flip the coin and having it "stick" when it hits the floor, create
a situation where there would be at least a bias for heads or tails.

QUESTION:
Is there any relationship between Lennard-Jones potential and Van der Waals force ?

ANSWER:
Yes. The Van der Waals force is the long range attraction between
neutral atoms or molecules. The Lennard-Jones potential represents an
empirical force which has both a short range repulsion (r ^{-12} )
and long range attraction (-r ^{-6} ); the attraction part could be called Van
der Waals.

QUESTION:
I am trying to find out how much energy was produced. When I struck. Headonatree head on
The car weighted 2945 pounds travelig at 75the miles per hour.
I weighed 285 pounds and was nor wearing a seat belt

ANSWER:
No energy was "produced"; the kinetic energy of your car was
converted into heat, sound, and the work necessary to crumple up your car.
The total kinetic energy of you and your car is
�mv ^{2} where m is the mass (in English system
units m =(285+2945)/32=101 lb s^{2} /ft) and v is your
speed (v =75x3600/5280=51 ft/s). So the total energy is �x101x51^{2} =131,000
ft lb. That is about 0.067 kilowatt-hours; if you could take that amount of
energy and spread it out over an hour, you could run a 67 watt light bulb.

QUESTION:
We learned in class that centripetal acceleration is equal to v^2/r, when velocity is constant.
we also learned that acceleration is the rate of chance of velocity. Since magnitudly, velocity is not changing, wouldn't a=0?
What does the value of centripetal acceleration actually mean if the magnitude of velocity is not changing? If the acceleration is 15m/s^2, but velocity is constant.. then the body is not actually accelerating at 15m/s^2.
So basically what does the magnitude of centripetal acceleration actually tell you? What does it mean?

ANSWER:
Velocity is a vector, that is it has both a magnitude (usually
called speed) and a direction. Acceleration is the time rate of change of
velocity and velocity can change by either changing its magnitude or its
direction (or both). An object in uniform circular motion is accelerating
because the direction of the velocity, which has constant speed, is
changing. Acceleration is also a vector and its direction is toward the
center of the circle for uniform circular motion. The problem is that in
everyday life we usually think of acceleration as something speeding up but
that is not the general meaning in physics.

QUESTION:
I have learned that every subatomic particle behaves in a wave like particle that can described using the wave equations, however, what exactly does it mean when they say waves? I know photons travel in waves, but how do protons, electrons, and other particles exhibit wave behavior.

ANSWER:
It is one of the tenets of quantum mechanics that any particle is
also a wave. Look for a particle, you will find particle-like properties,
look for a wave, you will find wave-like properties. This was first proposed
by Louis de Broglie in 1924 and he won the Nobel prize for his hypothesis in
1929. It has subsequently been verified innumerable times in experiments.
For one example of diffraction of protons by a sphere, see an
earlier answer .

QUESTION:
I am a Stuntman for TV and film and we have created a device called a ratchet. It is a pneumatic piston that very quickly pulls a line (rope, cable etc.) to "yank" someone or something a certain distance. (explosions, etc...) So, if I had a 200 lb man at the end of a line, what dynamic force is generated on the line at 6 ft per second. (no ground dragging friction, it generally would pull you up in a 45 degree angle with the aid of rigging and pulleys). I have heard that this can multiply by 4 or 5 creating an dynamic load of 800 or 1000lb on the line for that instant. is that accurate, more, less? or is there a formula. please, if possible, answer in layman terms.

ANSWER:
What matters is not the speed, but how quickly you get there. Once
you get to constant speed, the tension in the line will always be less than
or equal to the weight. It is easiest to understand if we keep it simple and
lift vertically. Just to hold the 200 lb man vertically obviously means the
tension will be 200 lb. But, suppose we want to accelerate him to 6 ft/s in
0.1 s; his acceleration is 60 ft/s^{2} . Newton's second law says
that there must be a net force upward which is equal to his mass times his
acceleration. This is a little tricky using English units (ft, lb, s), but
without any details, just accept that his mass is 200/32. Therefore the net
force needed is (200/32)x60=375 lb. However, there is already a 200 lb force
down (his weight), so the tension in the line would have to be 375+200=575
lb, almost three times his weight. Once you stop accelerating him, it makes
no difference what his speed is, the tension in the line will be 200 lb.
Pulling horizontally (ignoring friction) the tension would only be 375 lb,
so pulling at 45^{0} would be somewhere between. You could replicate
this calculation for any acceleration you wanted; 0.1 s is pretty fast, but
the man could take more since 60 m/s^{2 } is only about 2g's of
acceleration.

QUESTION:
What is the minimum mass that an object can be in order to bend the path of a photon? Does the bending of the path result in a decrease in the speed of the photon?

ANSWER:
Any mass creates a gravitational field and any gravitational field
will result in a photon being deflected. However, for the deflection to be
big enough to observe the mass has to be really huge. The speed of a
deflected photon is unchanged.

QUESTION:
I recently helped my 11 year old son design and test a science project.
The project hoped to measue a quantifiable difference between a "corked"
baseball bat and a non-corked bat.
I can't help but thnk I did not consider all the necesarry factors.
We have only limited resources so we tried to make it as simple as possible.
We built a gantry and mounted it upon a rectagualr base. From the gantry the bat swings from a vertical position about 150 degrees above the contact point of bat on ball, hitting the ball which has been placed on a tee.
It is
hit straight out from the contact point onto a long sheet of waxed paper (which records the spot where the ball hit), The spots where the balls hit are measured and recorded.
In an attempt to create an experiment that uses a constant bat speed, we decided to use gravity as the "force generator" since it would be the same every time.
One of the "beliefs" about the benefit of corking a bat is that the reduced weight of the bat allows the bat to be swung faster/harder (at least as I understand it).
I don't know if that is in fact the case but, (here's my question):
Did I err by making gravity the "constant" in my experiment?
Galileo showed that two objects of different weight arrive at the ground at the same time. (haha-I think) So the two bats we are testing will fall through their arc at the same rate of speed, won't they? One wil not fall faster because it is lighter will it?
I can't see how this differs from a baseball player swinging a bat horizontaly and having it move faster becasue it is lighter.

ANSWER:
The main problem here is that you are not replicating Galileo's
dropping two balls experiment. Your situation is one of rotational motion,
not simple falling, and the effect of gravity on the two bats is going to be
different. Since your son is only 11, I will try to keep it nonmathematical.
The time it takes the bat to drop (and the speed it will have at the bottom)
is not determined by what the mass is but it does depend on how that mass is
distributed. Think about the center of mass* of each bat; since the corked
bat is lighter at the extreme end of the barrel, its center of mass is
closer to the knob than the uncorked bat. So, as a first approximation,
think of each as a simple pendulum, a tiny point mass on a string, where the
point mass is at the center of mass and the string is thought of to run
between the hinge and the center of mass. A simple pendulum with a short
string has a shorter period than one with a longer string, so the corked bat
should actually be moving faster at the bottom than the uncorked bat because
its center of mass is closer to the hinge. And, if you had a video
camera and could examine its movie frame by frame, you could determine the
speed of each bat at the time it hit the ball.

Again, this is a rough approximation because neither is really a simple
pendulum as I have used to try to make it 11-year old friendly (you can
easily make simple pendula and show him that the shorter gets to the bottom
quicker). In real life, these are called physical pendula and you have to
worry about how the mass is distributed in detail (moment of inertia if you
have ever studied physics), not just where the center of mass is, but I
think my explanation should give a good qualitative overview.

So, the answer to your final question is that one bat does fall faster but
not because it is lighter but because its mass is differently distributed.

*You can easily find where the center of mass of each bat is by finding
where each balances.

QUESTION:
My Son & I were are having a discussion about how far a golfball would go on the moon vs. the Earth, if hit @ a 45 degree angle. The ball, club, strike force and any other possible variable would be identical with the exception of gravity and atmospheric resistance. The accepted story is the golfball would travel the equivalent of the difference between the Earth and Moons' gravitational pull as the multiplier or divisor. I contend the lack of atmospheric drag on the moon would cause the golfball to to go measurably farther. What say you?

ANSWER:
I say that the lack of air on the moon gives the moon ball more
advantage than gravity alone. You are right.

QUESTION:
Suppose a spherical ball is tied wiht a string , is moved in a horizontally circular path with other end as centre . Now if we apply more force to the end of string at centre to move the ball faster , then what force is responsible for speeding up the ball? However all the forces are perpendicular to the tangential velocity.

ANSWER:
If you apply more force to the end of the string, the ball will be
pulled into a path of smaller radius. So, the velocity of the ball will have
a component in the direction of the force and so work is done to speed up
the ball.

QUESTION:
if the speed of a car doubles, why does the force of impact quadruple

ANSWER:
Here is that question that keeps coming up again and again: when one
thing going some speed hits some other thing at rest, how much force is
exerted? THERE IS NO WAY TO KNOW FROM THAT INFORMATION. See my FAQ
page, first two questions. If two cars crash into an immovable wall, each
loses all its kinetic energy. Kinetic energy is
�mv ^{2} , so the energy converted into heat and sound is 4
times bigger if the speed is 2 times bigger. If each stops in the same
distance, the average forces will differ by a factor of 4, but you would
expect the faster car to "crunch up" farther, not the same, so the average
force would probably be less than 4 times bigger.

QUESTION:
I am planning on using a strain gauge to measure force (N) exerted over time for a shark captured on a long line. Is there any way I can convert these values to joules of energy used given that it is a static pull i.e. no distance travelled?

ANSWER:
If I understand, you are not hauling this shark in, he is just
pulling trying to get away but failing. He is consuming lots of energy, but
your line could just be tied to the pier and therefore does no work.
Therefore, the force will not give you any information about energy
because, as you note there is no distance. It takes no energy to hold the
shark. The shark herself is doing work against the water and consuming
energy and you cannot measure that by measuring how hard she is pulling.

QUESTION:
if a plane is flying 240 m/s in the air will the pilot hear the doppler effect?

ANSWER:
Yes, but it is not the usual. Usually we think of the Doppler effect
as being due to either the source or the observer moving. In this case, the
source (engine) and observer (pilot) are at rest relative to each other.
However, the medium through which the sound moves is moving at a rate of 240
m/s, not so far from the speed of sound itself in still air.

QUESTION:
When an atomic explosion takes place, matter converted turned
into energy. Is the reduction in mass accounted for by a quantified
reduction in the total number of subatomic particles?

ANSWER:
No, the change in mass can occur even if the total number of particles
remains constant.

FOLLOWUP QUESTION:
This sounds very strange and interesting! If what you say is true, then I am wondering where is the mass coming from? Do the particles themselves reduce mass?

ANSWER:
This is just E=mc ^{2} . Any nucleus weighs less than
the sum of its parts. Here is how you understand that. Does it take energy
to pull one proton or neutron out of a nucleus? Of course it does because it
is bound in there and you must do work to get it out, thereby increasing the
energy of the system. So it takes a whole bunch of work to disassemble a
nucleus into its constituent protons and neutrons. Where does that energy
go? Into mass, the mass of all the constituents is greater by an amount W /c ^{2}
where W is the work you did. So, if a heavy nucleus splits (fission,
like in a conventional bomb) or if very light nuclei fuse (fusion, like in a
star or in an H bomb), the mass after the fission or fusion is less than
before since energy is released. It is subtle to understand why, but I have
explained it in an
earlier answer .

QUESTION:
This question came up in class and we went over it, but I still could not understand it. Assume the earth had sufficient nutrients and adequate water, no toxins present, uninterrupted growth, division every 20 minutes, and had e coli bacterium. How long would it take for this mass of bacterium to equal the weight of the earth? Is it possible?

ANSWER:
I think your teacher was trying to illustrate exponential growth. If
you have not had any calculus, you will not understand my answer. Anything
whose population changes in a way that is proportional to the population
itself gives rise to exponential growth or decay. So, dN /dt=cN
where N is the number at some time t , dN /dt is
the rate at which N is changing, and c is a constant of
proportionality. If c >0, N is increasing (growth); if c <0,
N is decreasing (decay). This should make sense to you�if
you double the number, you double the rate at which the number is growing.
The solution to this equation is N=N _{0} e^{ct}
where N _{0} is the number at t =0. In our case we will
choose N _{0} =1 (start with a single bacterium). What we know
is that if t= 20 min, N =2, so 2=e^{20c} . Solving
this, c =20/ln2 where ln is the natural logarithm. Now, denoting the
mass of one bacterium as m , we know that the mass M at any
time must be M=Nm =m e^{20ct} and if we let
M=M _{earth} , and solve for t we find t =20 ln(M _{earth} /m )/ln2.
I will let you do the math and look up the constants. I get a little less
than 2 days.

QUESTION:
I am attempting to build a 20 foot tall pole for use in a play.
It will have minimal weight at the top of it, maybe 1 lb (a small flag). I
am planning on using 2 10 foot long sections of Schedule 40 PVC pipe with a
coupling to make the pole. I'm curious how large in diameter and how heavy
my base should be. I was considering making a form 4 feet in diameter out
of plywood and 2x4s and filling with about 200 lb of concrete. I want to be
sure it is safe so if there are any drafts at the top or if someone happens
to bump into it at the base that it won't fall over and hurt anyone. I
believe the 4 inch Schedule 40 PVC pipes are approximately 10 lbs each, so
the whole pole should only be about 20 lbs. Any recommendations on the
base design or where I could find more information would be appreciated. I
would like to minimize the base as much as possible without sacrificing the
safety of anyone on stage.

ANSWER:
With a 4' diameter base, if somebody pushed on the pole at a
distance of 4' above the ground, I figure that to push it just to the point
of starting to tip it over would require a force of F=W /2 where W is the
weight of pole plus base. The way I got this was to sum torques about the
bottom of the pole, Fh-Ns= 0, where h is the distance above the
ground where F is applied, N is the normal force up from the
floor (equal to the weight W ), and s is the distance from the center
to the edge where N is assumed to act if it is about to tip. So, for
your design, a 110 lb force would be required and would have to be exerted
until it had reached about 11^{0} of tilt before it actually fell
over. It seems very unlikely that somebody bumping into it would tip it. To
be safe though, couldn't you find a way to attach the base to the floor?

QUESTION:
This question has been bothering me lately and hopefully you can answer. If you're in a car and the car is moving forward and if you would throw an item in the air why doesn't it go flying backwards considering it is not attached to the car or the people in the car?

ANSWER:
Actually, it does not keep perfect pace with the car �it
gets pushed back by the wind which you can feel if you stick your hand out
the window. The effect depends on what you drop. If you drop a small, heavy
marble, a passenger in the car would see it drop straight down,
approximately. If you drop a cotton ball, I think you would agree that it
would not appear to drop straight down.

QUESTION:N:
I was wondering what happens to massless particles - are they subject to gravity/able to to be acted upon by forces? If so, do they travel at a diffent velocity to particles with mass?

ANSWER:
The only known massless particles are photons and they, like any
other massless particles that might exist, travel with the speed of light
(they are light). We used to think that neutrinos were massless, but they
are now known to have a very tiny mass, much smaller than other elementary
particles. Regarding the effect of gravity on photons, yes, photons are bent
by gravity even though they have no mass. See my
FAQ page.

QUESTION:
What would happen if
one split a radium molecule? This question is from my ten year old daughter's who has been reading about/admiring Marie Curie for the past two years.

ANSWER:
I think you must mean split a radium nucleus. Indeed, if you were to
induce a fission of radium, energy would be released just like it is from
any other fission of a heavy nucleus (like ^{235} U or ^{239} Pu),
but you could not make a reactor with radium as a fuel because it is not
fissile which means that it can be induced to fission by causing it to
absorb slow neutrons and to therefore sustain a chain reaction. The
energetics are such that if it were split into two fragments, energy would
certainly be released. You can tell your daughter that I had the pleasure of
working with Marie Curie's grandaughter Mme. H�l�ne Langevin-Joliot, also a
nuclear physicist, at Saclay National
Laboratory outside Paris in the 1990s.

QUESTION:
Why does torque depend on lever-arm distance?
I don't really understand the logic behind it. Please don't use ideas of work,energy,moment of inertia etc to explain this. I just want a logical explanation for this using ideas from linear dynamics.

ANSWER:
In translational physics, force is a measure of how effective a push
or pull is at getting an object moving (accelerating it). In rotational
physics, torque is a measure at how effective a push or pull is at getting
an object rotating (angular acceleration). So, think of opening a door.
Suppose that you push the door out where the door knob is; your lever arm is
large and this push is very effective in opening the door. Now, suppose you
push with the identical force one centimeter from the hinges; the same force
with a small lever arm is much less effective in opening the door.

QUESTION:
What should happen in the following experiment ... If you keep the height and length of a ramp the same and only vary the mass of the car by adding washers to add mass, should the car travel a lesser or greater difference? Some sources I've read say that the acceleration should decrease with an increase in mass, which means the distance traveled should also decrease. However, other sources say that the momentum will increase with the added mass causing it to travel further.

ANSWER:
Simple physics says there should be no difference. However, usually
the heavier of otherwise identical cars wins. I have discussed this
extensively in earlier answers which are linked to in my
FAQ page.

QUESTION:
Does a theory eventually turn into a law or are they entirely separate? I have talked to many people about this subject and they have given me different answers.

ANSWER:
Usage by physicists of the words law and theory are
ambiguous and vague at best. Newton's second law in its usual form, F =Ma ,
for example, is untrue for speeds not small compared to the speed of light;
the theory of special relativity has never failed to predict the behavior of
nature. See an earlier answer .

QUESTION:
My 13 year old daughter is studying physics and has asked a question I have no idea how to answer so I am hoping you can help. She has been told at school the energy of a wave is proportional to the square of it's amplitude and therefore independent of frequency. However she told me she has also read that energy of a wave is given by E=hf (E is energy f is frequency h is Plank's constant) and so is proportional to frequency! She is utterly confused by this and asks which one is correct?

ANSWER:
Your daughter has encountered one of the most important innovations
of 20th century science �quantum
mechanics. Light is both a wave and a particle (called a photon). If
you devise an experiment to prove that light is a wave (e.g. Young's
double slit
experiment), you will succeed. If you devise an experiment to prove that
light is a particle (e.g. the
photoelectric
effect ), you will succeed! Both energy statements in your question are
correct. But they are not really contradictory. The intensity of light is
the energy per unit time through an area, Joules/second/square
meter=Watts/meter^{2} . If you look at this as waves, this is
proportional to the amplitude squared. If you look at it as particles, this
is proportional to the number of photons passing through the area.
Therefore, red light of some intensity has more photons than blue light of
the same intensity because each "red photon" has less energy than each "blue
photon".

QUESTION:
What is the wavelength range of the radiation from the sun that actually hits the ground?

ANSWER:

QUESTION:
if 2 lines start on the same point. And they are separated by a 1 degree difference in their heading. How far apart will they be extended out 365 inches. What I'm trying to illustrate is 2 people walking together in marriage - barely different - how far will they be apart in 1 year (365 days) because of 1 degree of separation.

ANSWER:
To most easily do this, convert 1^{0} to radian measure, 1^{0} = π /180=0.0175
radians. Since the angle is 0.0175=s /r (see figure at the right),
s =365x0.0175=6.37 inches.

QUESTION:
I'm attempting to expand my understanding of some fundamental principles of physics, such as the laws of motion and energy. I've run into a bit of a puzzle dealing with the conversion and conservation of energy in reference to moving objects.
I understand the conversion of energy in, say, a moving car: chemical (potential) energy of the gas is converted to kinetic energy, and this in turn is converted to heat energy when the breaks are applied.
Where I get hung up is when energy is used to slow a moving object. Using the car as an example, rather than breaking one might downshift, thus using the engine to slow the car. This is where my understanding breaks down. It seems *more* chemical energy is being converted and put into the car to slow it down. What is the sum of this kinetic energy being converted to?

ANSWER:
Lots of the energy your engine generates is lost to friction. If you
drive down a straight, level road, you are constantly consuming the chemical
energy in the gasoline just to keep going. This energy all ends up as heat
or sound in compliance with energy conservation. If you take your foot off
the gas, you are not providing enough energy to keep going and you therefore
must slow down, again because of energy conservation. If you downshift and
take your foot off the gas, your engine goes faster but the gas you are
providing is the same so friction in the engine is increased while gas
consumption is not, so you slow down faster still.

QUESTION:
A better example may be a rocket in space. Let's assume it's not acted upon by gravity. The rocket engines fire, converting chemical energy into kinetic energy. Now, to come to a stop again, the rocket must turn around and again fire its rockets - and this is where I get confused. Though it "naturally" makes sense that this would "decelerate" the rocket, I'm confused about the principle of conservation of energy. It seems that in both instances of firing the rockets, kinetic energy has been put *into* the rocket. What is this kinetic energy being converted to when it is decelerated?

ANSWER:
This is not really the same question at all, so I will answer it
separately. It is conceptually easier because there is no energy being lost
to friction. The way a rocket works is that chemical energy is used to
propel mass out the back (or front) and the rest of the rocket recoils; it
is like the recoil of a rifle. Here is my simple model of a rocket: it is a
2 kg rocket with 1 kg of fuel. Now, the fuel is ejected so that the rocket
recoils with a speed of 1 m/s; what is the speed of the ejected fuel? This
is determined by momentum conservation. Linear momentum is the mass times
the velocity, so before the rocket "burns", the momentum is zero and
afterward it is (2 kg)x(1 m/s)-(1 kg)V where V is the speed of
the ejected "fuel"; the minus sign is because they move in opposite
directions. Conserving momentum (equating old with new), we see that V =2
m/s. Now, what is the chemical energy consumed? It is the change in kinetic
energy
�mv ^{2} before and after; originally, kinetic energy was
zero and afterwards it was �x2x1^{2} +�x1x2^{2} =3 J which is
positive, energy was consumed. Now, consider braking. Suppose the 3 kg
rocket+fuel has a speed of 1 m/s; so, its momentum is 3 kg m/s and its
kinetic energy is 1.5 J. Now, the fuel is burned so that the 2 kg rocket
comes to rest and the ejected fuel has a speed V . Again, momentum
conservation gives us V ; in this case V =3 m/s. So now the
energy of the system is �x1x3^{2} =4.5 J. So the change in energy of
the system is again positive and equal to 3 J. The consumed chemical energy
all went into kinetic energy of the fuel whereas in the acceleration
situation, part went into the rocket and part to the fuel.

QUESTION:
What kind of force would it take to get a 1 or 5 or 10 gram object out of earth's orbit? What amount of force would be necessary to guarantee that the object leave the solar system? Or the Galaxy?

ANSWER:
To lift an object, you need to exert a force just a tiny amount
bigger than the object's own weight. So, if you had a 10 lb object and
wanted to move it away from the surface of the earth, first exert a force
just greater than 10 lb and it will start moving upward. Once you have it
moving, you can reduce the force to 10 lb and it will continue moving up
with a speed of whatever you had given it. But, as you get farther and
farther away from the earth, its weight gets smaller (see the answer right
after this one), so it will start speeding up. So, now keep adjusting the
force to keep it going with constant speed. So you see that the force
required to move it far away is very small.

QUESTION:
Does every particle inside an object have a
gravitational pull form all the objects in the universe. So each
individual particle would have a pull of 9.81 ms-1 or sorry acceleration
caused by some sort of mystical pull (any help appreciated).

ANSWER:
Two objects of mass m _{1} and m _{2}
separated by a distance r exert a force on each other of F= 6.67x10^{-11} xm _{1} m _{2} /r ^{2} .
For an object of mass M at the surface of the earth this just happens
to be F =9.81M . For that same object at an altitude equal to the
radius of the earth above the surface, that force would be four times
smaller. In principle, all objects in the universe exert forces on each
other, but gravity is only significant for very large masses and/or
relatively small distances.

QUESTION:
I have read somewhere that hot water freezes quicker than cold water.
if it is so, then please explain me the reason behind this.

ANSWER:
See an earlier answer .

QUESTION:
I know that freshwater can not be heated above 212 at sea level. What about freezing? Can freshwater attain a temperature below 32 degrees? If so, is there a theoretical minimum?

ANSWER:
At very high pressures, liquid water can exist at about -25 ^{0} C.
See the water P-T diagram
on Wikepedia. This happens at a pressure of about 3000 atmospheres.

QUESTION:
I just learned about the conservation of momentum in collisions in school. I wanted to check this in real life so I took some high speed footage of a friend hitting a tennis ball and after checking the frames, I found out that the momentum of the system before the racket hit the ball and after is not the same. It differs by about 50% and it occurs in all of the trials I had (3 in total). Is momentum not really conserved in real life?

ANSWER:
First of all, you need to understand under what conditions linear
momentum is conserved. Total momentum of an isolated system is always
conserved*.
Now, if you just look at the tennis ball, the racquet exerts an
average force on the ball over some time t so that the momentum
should change by an amount Ft . So, if you measure the momentum of the
ball before and after it is hit, it changes by an amount equal to Ft .
But, do not forget that momentum is a vector and the ball changes direction,
so if it has mass m and comes in with speed v _{1} and
goes out with speed v _{2} , the change in momentum is mv _{1} +mv _{2} .
So, do not expect the momentum of the ball to be conserved. Suppose you look
at the ball plus the racquet as the system; here momentum is not conserved
either because your friend's arm is pushing on the racquet and therefore
delivering an impulse during the collision. To see momentum conservation,
imagine the racquet in the middle of space moving to the right with some
speed and the ball moving to the left with some speed; after they collide
the linear momentum will be the same as before because this is an isolated
system. Regarding your final question, real life is seldom as simple as
introductory physics would have it, but real life can often be understood to
a good approximation.

* It
can actually be a little broader than that since, to change the momentum of
a system, an external force must deliver an impulse which is, essentially,
some external force times the time over which it acts [I do not know your
level, but impulse is
∫F (t )dt ].

QUESTION:
If a clock traveling at high speed runs slower than one fixed on earth does the same apply for computers processing information. Say two computers at the same rate are writing a copy of war and peace, one at a high velocity one on earth, would the one on earth be farther ahead. And if so could I, being on earth then tell the person at high velocity what will be printing out soon via radio signals ahead of the computer on their aircraft.

ANSWER:
Once again, I make reference to the diagram I use to explain the
twin paradox. For a full explanation of this diagram, see earlier answers (1
&
2 ). In a nutshell, this shows the stay-at-home twin (computer in your
question) moving up the time axis and sending out a light pulse once a year
and the traveling computer traveling out to a distant star and back, also
sending out a light pulse once a year. Think of each light pulse as being
sent after turning 20 pages of War and Peace (pretty slow computers).
Then, your idea is seen to fail because it takes time for the information to
travel between the two computers. For example, the second pulse (40 pages
read) sent out from the earth-based computer does not reach the traveling
computer until it is turning its 120^{th} page. In your question,
the traveling computer never comes back. But, if it does come back, the
traveling computer will begin getting advance information after he has
received 8 signals (at which time he has also sent out 8). When the
round-trip journey has finished, the earth-bound computer has read 400 pages
while the traveling computer has read only 240.

QUESTION:
If I creat a huge (I mean massive) float system with a pully generator that will create energy from the rise and fall of the daily tide, will I eventually disrupt the tidal pattern?
I've had the concept for years but am afraid of the outcome. Then I weigh in the damage we do daily other ways, and it would be as almost abundant as the sun.
I am only worried about the tide. I realize I'd disrupt the waves.

ANSWER:
Tidal power is already used. You can get a pretty good overview from
Wikepedia . Nothing
you build will seriously disrupt the tides but it can have local
environmental effects like riverbank erosion, disruption of ecosystems,
etc .

QUESTION:
My son has already totaled one car in a curve, but he wasn't hurt. He thinks rigid suspension will solve problem, but he had rigid suspension in the car he totaled. I heard somewhere that when you reach a certain speed car can't handle curve. I guess the bank of the curve is involved. (I watch car races) I have limited math skills. But he would understand the math, and I would use it to convince him to slow down.

ANSWER:
When you round a curve a force is required because a car is
accelerating when it turns even if its speed is constant. That is because
the direction of your velocity is changing, just as truly an acceleration as
when the magnitude (speed) of your velocity is changing. The force required
to keep you moving with a speed v in a circular path of radius R
is mv ^{2} /R where m is your mass. On a level
road the only possible source of such a force is the frictional force
between your wheels and the road. You can see that this is the case because
on an icy road, which can provide almost no friction, it is almost
impossible to turn at any speed. So, you can see that the faster you go the
more frictional force you need from the tires to be able to negotiate the
curve. In fact, since it depends on the square of the speed, doubling the
speed quadruples the needed force. Now, the nature of static (not sliding)
friction is that there is a maximum amount you can get which depends on the
nature of the surfaces in contact (rubber and asphalt for this case) and how
hard the surfaces are pressed together. In the case of a car with mass m ,
the weight is mg (where g is about 9.8 m/s^{2} =32 ft/s^{2} )
and that is how hard the two are pressed together. The maximum force you can
get is then F _{max} = μmg= mv _{max} ^{2} /R
where μ is a constant determined by the nature of the surfaces (e.g.
rubber on ice has a much smaller μ than rubber on asphalt). So, the
fastest speed you can negotiate a curve on a flat road is v _{max} =√(μgR ).
For example, rubber on dry asphalt has μ ≈0.6 so for a curve of radius
40 m, v _{max} ≈√(0.6x9.8x40)=15.3 m/s=33.6 mph. If the road
were wet, you would skid out at an even lower speed. If the road is banked
the math is a bit more complicated and, you would find, you can go faster.
Nevertheless, the suspension has nothing to do with the speed where you will
lose traction with the ground. See also an
earlier answer .

QUESTION:
Has the quantum concept "entanglement" been experimentally proven? Where might I find a layman's discussion of it?

ANSWER:
There have been numerous experimental verifications, the easiest to
understand probably being entangled photons and their respective
polarization states. There is a very nonmathematical discussion in Roger
Penrose's book The Emperor's New Mind . While not mathematical, it is still pretty tricky to
understand.

QUESTION:
Suppose we have tu push a large block on a very rough surface. It is assumed that the surface is so rough that we are unable to push it. We will say that the friction is acting in the opposite direction to our force and our force is not strong enough to overcome the force of friction. But when we are not pushing the block, then also the block remains stationary. There is no net force acting on the block. How can we say that the friction is acting only as long as we are pushing the block and ceases to act as soon as we withdraw our own force.

ANSWER:
If an object is at rest, the sum of all forces on it must add to
zero. If static forces is one force then it will adjust to whatever the
other forces are to maintain that zero sum. If your block is sitting on an
incline, it does not go down to zero when you stop pushing on it; rather, it
adjusts to equal the component of the objects weight which is trying to push
it down the incline.

QUESTION:
If you are in a car and inside the car is a floating Helium balloon, if the car makes a hard turn, what direction does the balloon go?

ANSWER:
The opposite direction the car is turning. However, it might not be
too pronounced because the air drag as it starts to move would restrain it.
Of course, what it is really trying to do is keep moving in the direction
the car was originally moving as the car turns, it just seems to you that it
is being pushed out. A similar question may be seen in a
previous answer .

QUESTION:
So if me and a friend are picking up a weight (30lbs) from either side lets say its flat like a board, we're each doing work to pick up the weight yes related to the angle and other factors but is it possible for each of us to be holding up 30lbs? Or is the weight always divided between the two of us, equaling less than 30lbs for each?

ANSWER:
If the board has its weight uniformly distributed and each of you is
holding up one end, you each are exerting an upward force of 15 lb. If one
of you is holding on closer to the middle than the other, he holds more
than the guy at the end. But the sum of your two forces will always be 30
lb.

QUESTION:
While discussing atoms with some students we "briefly" discussed atoms and atomic structure being mostly empty space. The idea of solid matter being made of atoms is difficult to grasp for middle school. More so the idea that atoms do not touch. So of atoms never touch, why do I make sounds when I clap? Or create heat and friction when rubbing my hands? And how does fire affect items if atoms never touch? Even I couldn't answer these students questions!

ANSWER:
The key, I think, is that objects do not have to touch to exert
forces on each other. Think of dropping a stone. Althought it is not
touching anything it speeds up as it approaches the ground. Something is
causing the stone to gain energy as it falls. Atoms (small size, about 10^{-10}
m) are positively charged nuclei (much smaller size, about 10^{-15}
m) with electrons on the outer parts of the atom. The electrons have a mass
about 2000 times smaller than the nucleus, so you can see why atoms are
often referred to as being mostly empty space. But, when two atoms come
close to each other the negatively charged electrons in on atom see the
electrons of the other atom and are repelled, that is they feel a force.
This leads to friction. Fire is just rapidly moving atoms in a gas which
then exert forces on atoms they are close to causing chemical reactions to
occur (i.e. burning). Finally, I guess I do not like the statement
that atoms do not "touch"; the electrons in one atom often overlap those of
neighboring atoms.

QUESTION:
Hello, today i watched a programme about the universe. It stated that if a person in a space craft were to travel at incredibly fast speeds, by the time he got back to earth, being in space for a week, a huge ammount of time would have passed. I thought this was a facinating deduction, so i thought about it for a second and thought of this question: if the person in space could if possible, communicate to people on earth like on a webcam, or phone, what would happen? Would the people on earth speed up? Or what would happen?

ANSWER:
See an
earlier answer .

QUESTION:
I have only ever seen depictions of the solar system with the same design: a series of planets in a line. But I've been wondering, all of these graphs basically depict the planets as orbiting more or less in the same plane (presumably with some wobble).
So, is this true? Or is that for easy of orbit-distance-comparison? And if they are all in the same plane... why?

ANSWER:
All the planets orbit very nearly in the same plane. The reason is that the early solar system condensed from a big cloud of gas and dust which was spinning. As it got smaller it spun faster (conservation of angular momentum) and flattened out like a pancake. The planets formed by clumping of the matter in this pancake.

QUESTION:
Since the acceleration due to gravity of a planet like Earth can be given by the equation g=(GM)/r^2, where G is the universal gravitation constant, M is the mass of the body, and r is the distance from its center, what would happen to g very close to its center? The limit of that equation (assuming M is the mass of Earth for example) as r approaches 0 is infinity. How can you have an infinte amount of acceleration? What exactly would happen?

ANSWER:
The equation you quote is true only for point masses or outside
of spherically symmetric mass distributions. It may also be applied inside
of a spherically symmetric mass distribution with one important revision �M
means only the mass inside of r . Therefore, the force at the
center of the earth is zero. You might be interested in one of my
FAQ s.

QUESTION:
Assuming Einstien's theory, that matter is comprised of energy, that in fact, energy and matter are just two forms of each other, is correct; where does the energy in our bodies go, as we move throughout the day? how can we lose energy, without losing mass?

ANSWER:
If energy is used, mass is lost. But, think about it. The amount of
energy in a mass is huge, so the amount of energy you use (which comes from
chemical reactions) is so small that you could never hope to measure it. For
example, the energy you would burn to climb a 9 m rope would be about 9000
J. This would correspond to a change of mass of about m =9x10^{3} /(3x10^{8} )^{2} =10^{-13}
kg.

QUESTION:
Is the net observation of either of two parties traveling toward one another at half of the speed of light equivalent to either one witnessing the other as traveling at the full speed of light, in that either one can be justified as "at rest" relative to the other one?

ANSWER:
You would think that, wouldn't you? Well, for very high speeds the
intuitive ideas of velocity addition don't work. You should read an
earlier answer for the details, but in the case you are imagining, each
party will see the other approaching with a speed of 80% the speed of light.

QUESTION:
To get something off the ground you need to push the object with enough force against the ground (and atmosphere) to get it to overcome gravity and 'fly', but what would a theoretical space craft have to push against to get it to move in the vaccuum of space? I've always seen shows and movies pushing space crafts around with puffs of air and can't understand what the air is pushing against to make it move.

QUESTION:
How does a spaceship travel in space if the engines have nothing to push against?

ANSWER:
Both these questions illustrate a common misconception that the way
a rocket works is by the exhaust pushing on something. The way a rocket
works is that the rocket recoils when it "throws out" the exhaust. Imagnine
being in empty space with a rifle; when you fire the rifle you and the rifle
recoil. That is how a rocket works.

QUESTION:
What formula do you use to prove that a car can not jump abridge without a launch pad there to lift it off the ground? I know it can't happen but I don't know how to prove it!

ANSWER:
You don't need no stinkin' formula. You just need a simple
qualitative argument. When an object is in free fall (meaning it is not in
contact with anything, so after it has left the ramp), it can only do one
thing �fall. (I am assuming no effect
from the air; an airplane obviously does not fall.) What "fall" means here
depends on how it is moving:

If it is
initially moving upwards (like off a launch pad), it will slow down.

If it is
initially moving downwards, it will speed up.

If it is
initially moving horizontally, it will speed up, downwards.

Therefore, if it is
not originally moving upwards, it can never reach a point either level with
or higher than where it was launched from.

QUESTION:
NASCAR "drafting" energy: If "drafting" behind another vehicle is lucrative enough in energy savings to merit the inherent danger it entails -- then recouping some of that energy by the vehicle generating it should be lucrative as well.
In general, how much energy is available from a vehicle's "wake pressure gradient"? (my term) Could some of that pressure gradient potential be "tapped" with say a literal pipe-tap from the vehicle into the low pressure area behind it's own bumper? Could the resulting pressure differential be used mechanically in some way to result in a net energy savings for the vehicle itself?
Or is the "drafting energy" only available to a disconnected/uncoupled vehicle?

ANSWER:
Aerodynamics is a very complex topic, usually best done with
powerful computers and extensive wind tunnel measurments. Drafting is very
interesting. It turns out that it is advantageous to both cars, but the
trailing car gets the lion's share of the advantage. The reason has to do
with the change in geometry of the two cars compared to the individual cars.
So, it is not so simple as there being a source of energy in the wake of the
leading car which the second car taps. You can be sure that the first car
may not somehow extract energy from the turbulent wake behind it; there is
no free lunch in the world of physics.

QUESTION:
i'm just wondering when we see the sky diving shows, why do the sky divers tilt downwards a lot?Does this have a effect on the forces acting on him? how?

ANSWER:
The air drag force is proportional to the cross sectional area
presented to the onrushing air. If the sky diver orients himself in the
direction she is moving, she goes faster; if she spreads out perpendicular
to the direction she is moving, she slows down.

QUESTION:
a body is moving with a velocity 'v' with respect to (w.r.t.) a frame of reference s1.It bumps into a very heavy object and comes to rest instantaneously,its kinetic energy(1/2*m*v^2) as seen from the frame s1 is completely converted to thermal energy.Now a man moving with a uniform velocity 'V' (in the direction of the body) w.r.t. s1 observes the body , he notes that its initial kinetic energy of the body is 1/2*m*(v+V)^2 and that after it rams into the heavy body as 1/2*m*V^2 and concludes that the thermal energy produced is m*v*V+1/2*m*V^2.Which of the two answers is correct?

ANSWER:
So, to answer your question we just need to find out how much the
kinetic energy changed as seen by both observers, right? Where you have gone
wrong is that you have not taken the energy of the wall into account and you
assume the wall has infinite mass. Of course, no wall has infinite mass but
if yours did, it would have, for the moving observer, infinite kinetic
energy both before and after the collision but those two infinities would
not be the same! What you need to do is assume the wall has a mass M
and the body has a mass m . So, your first observer sees a speed after
the collision (conserving momentum) of u=mv /(M+m ) and a change
in kinetic energy of
ΔE =�(M+m )u ^{2} -�mv ^{2} =�mMv ^{2} /(M+m )
which is the energy converted into thermal energy. In your second scenario,
u=V + mv /(M+m )
and
ΔE =�(M+m )u ^{2} -(�mv ^{2} +�MV ^{2} )=�mMv ^{2} /(M+m ),
exactly the same. You can also note that as M - � ∞,
ΔE - � �mv ^{2} .

QUESTION:
Hey I've been reading some General Theory of Relativity and i understand that gravity warps space and time yet on earth what does gravity do does it warp anything.

ANSWER:
The warping of space-time is why you fall out of a tree. The warping
of space-time is why relativistic corrections must be made to clocks in
satellites so that GPS will work. It is not what "gravity does" it is what
the mass of the earth does and it is what gravity is.

QUESTION:
Nuclear fusion reactions are kind of straightforward with the elecrotmagnetic force accelerating the two daugheter nuclei doing the large amount of work. Is the high speed of the neutron formed at nuclear fusion reactions due to an acceleration by the strong force, or is that reasoning too simplistic?

ANSWER:
You have it wrong, the electromagnetic force is not the source of
fusion energy. If you imagine using an accelerator to bring two light nuclei
together, that is just the "match" to ignite the fusion. The source of the
energy is the fact that the mass of the fused nuclei is less than the mass
of the unfused nuclei; mass is converted into energy. In fission, the source
of the energy of that speedy neutron you are looking at is, again, from mass
being converted into energy. When heavy nuclei split apart, the mass of the
products is less than the mass of the original nucleus and thus energy is
released. E=mc ^{2} . You might like to get more detail from an
earlier answer .

QUESTION:
why does not moon escape toward the sun due to more powerful gravity of sun compared with earth?

ANSWER:
Is the moon not orbiting the sun? Just because it is in orbit around
the earth does not mean that it is not also in orbit around the sun. The
sun's gravity does not cause it to escape, just to follow the earth around
the sun.

QUESTION:
We talked about this in my physics class but I was still confused. What are all the forces acting upon an elevator as it moves from rest upward and stops at its floor? To me the passengers would have something to do with the force, or am I wrong?

ANSWER:
You are right, the passengers have something to do with the force,
but if you say the weight of the passengers is a force on the elevator, you
are dead wrong. The weight of the passengers is a force on the passengers.
The passengers exert a force down on the elevator, the magnitude of which
depends on what the acceleration of the elevator is.

QUESTION:
If atoms are mostly empty space and if gold leaf can be as thin as one or two atoms thick, why can't we see straight through it? I understand in thicker objects there are atoms behind the atoms so you can't see straight through except that isn't the case for transparent objects like glass and diamonds. How can we see through them?

ANSWER:
First, gold leaf is nowhere near 1-2 atoms thick. It can be made a
little less than 10^{-7 } m which corresponds to about 1000 atoms thick.
And, in fact you can see through gold leaf and the light passing through is
bluish green. Gold is a metal and metals have conduction electrons which are
more or less free to move around; the entering light has electric fields to
which the electrons respond and absorb their energy. Transparent things like
glass are insulators and therefore do not absorb much energy from the light.
The detailed description of light transmission is very complicated. You
might be interested in
this site for
information about colors of gold.

QUESTION:
OK, I'm a screenwriter endeavoring to get things right in my work. So, here goes:

I am confused by what seems to me to be the mixing of Newtonian descriptive language with Einstein's whenever I hear someone talking about gravity. For example, the use of terms like "attraction" of gravity is really a Newtonian throwback, is it not? And, actually, isn't it incorrect as it is the deformation of space (G. Relativity) that causes the effect we refer to as"gravity"? Or, conversely (and my question), is it correct to purge Newton from my dialogue about gravity? If so, and I'm left with space, doesn't that mean, strictly speaking...(question 2):

Is space gravity?

ANSWER:
This is a little tricky, because it involves semantics as well as
physics. Newton was the first to appreciate the nature of gravity, its
"action at a distance" nature. He was able to determine, by showing how the
gravitational force of the sun could explain the motion of the planets
(which was known from astronomical observations made by Brahe and
systematized by Kepler), how gravity depended on distance r between
attracting bodies (1/r ^{2} ), and their masses (M _{1} M _{2} ),
and how strong the force was. This was an amazing achievement, but, it must
be noted, it was purely empirical. It never addressed why or how
the masses caused the "attraction". There is that word (attraction) that you
suggest might be archaic. But, I would argue that regardless of whether we
understand the theory behind the attraction or not, it is still clear to
anybody that two bodies with mass attract each other. Einstein's amazing
achievement was the theory of general relativity where the attraction was
shown to be due to the warping of space-time, the why/how holy grail for
gravity. It is just another step in the progression of science: data are
gathered, the data are interpreted, an empirical theory for the interpreted
data is proposed, and finally the reasons behind the empirical theory are
proposed. I see no reason to "purge" Newtonian gravity terminology from a
discussion of gravity, it is true and useful in many ways. I am quite sure
that NASA does not use general relativity and the warping of space-time to
calculate trajectories of space probes on their way to outlying planets. Any
discussion of gravity should, in my opinion, not "purge" Newton (nor Kepler
nor Brahe nor Copernicus, for that matter). I might add that general
relativity is not regarded as the last word on gravity; there is no theory
of gravity for very small distances, a theory of quantum gravity is sought
but elusive. Hence, gravity stands alone and physicists seek a way of
uniting it with the rest of physics. Finally, although I am not so
mainstream here, I believe that the recent observations which are attributed
to dark matter and dark energy may simply be indications that
general relativity is not as complete and correct as is generally believed.
The answer to your second question is that gravity is the warping of
space-time by mass. I would urge you to have a look at my
FAQ page for some earlier answers
about general relativity.

QUESTION:
I've started doing year 12 work for next year and one of the topics is circular motion. One of the examples of a centripetal force in the text book is when a car turns along a curved track, and it says the centripetal force is friction. But i dont understand this, as i thought that friction is the resistance to motion and the wheels arent facing the 'centre' of the curved track, and thus how is the centripetal force friction?
This also lead me to wonder how a car moves in physics terms, coz i also have heard that a car moves due to friction with the ground, which i also do not understand when friction is the force opposing motion. So if you could, can you also explain why cars move once the motor has provided a torque on the wheels, because im battling with this concept of friction especially.

ANSWER:
It is a mistake to think that friction always takes energy away. For
example, if there is a box sitting on an incline and not sliding, what is
keeping it there? It is the static friction between the box and the incline
and it is not changing the energy at all; if there were no freiction, the
box would slide down. The car going around a curve stays on its circular
pathe because of friction. If there were no friction, it would be impossible
to negotiate a curve, you would slide off (like when the road is icy). The
wheels are not slipping, so, again, it is static friction because at any
instant the tires are at rest relative to the road surface. The only
acceleration of a car going around a curve with constant speed is toward the
center of the circle, so that has to be the way friction acts. For a car to
start moving, there must be friction (again static friction); if there were
no friction, the engine would cause the wheel to turn but it would slip and
the car would go nowhere. The friction must exert a force backward on the
wheel in order for it to not slip. Therefore, the road exerts an equal and
opposite (forward) force on the wheel. The friction force on the car by the
road is the force which causes the car to have a forward acceleration. Once
you are moving, there are lots of other kinds of friction which come into
play �air drag, friction from the wheel
bearings, frictional drag from the engine and transmission�which all cause
the car to want to slow down, and the forward force of the road on the
wheels balances these forces when keeping the car going at constant speed.

QUESTION:
If a black hole pulls material in doesn't it gain mass and therefore slow down?

ANSWER:
Yes, it gains mass. But, it gains the mass by interacting with
something else. So, to find out what the black hole is doing after the
collision, you have to conserve momentum of the event when it gained that
mass. It is pretty much like the collsion of two putty balls �you
start with two and end with one. For example, suppose the black hole is at
rest and the to-be-captured object is far away and has a velocity v
directly toward the black hole and a mass m . So the momentum of the
system is mv. After the collision, the mass must still be mv (that is
what momentum conservation means). So, if the mass of the black hole were
999m , the speed after the collision would be V=mv /(1000m)=v /1000.
You could go through the whole thing for other initial conditions, for
example if the black hole were moving with a speed v and the object
were at rest, the speed after the collision would be V= 999mv /(1001m)=0.998v ,
slower, as you suggested, but not really for the reason you suggested. (I
have done this nonrelativistically, but it conveys the idea.)

QUESTION:
How much energy does CERN use to smash 2 protons together to try & see the Higgs particle?
ie in laymans terms, would it be equal to lifting a chevy off the ground...or a train engine?

ANSWER:
There is a
link from CERN which fully discusses the energy content of the LHC
beams. The number they give for the beam at full intensity is 362 MJ. They
show that this is the energy of a 3200 kg Subaru going about 1000 mph. It is
the energy which could lift that car about 10,000 m, or about 30,000 feet.
But maybe you are asking about just the energy of the two protons? That is
much smaller, because the beam contains about 3x10^{14} protons and
so the energy of each proton would be about 10^{-6} J, not enough to
notice macroscopically.

QUESTION:
How would a helium balloon behave in the zero-gravity environment of the International Space Station, or any other orbiting spacecraft?

ANSWER:
The reason the balloon goes up is that the buoyant force is greater
than its weight. The buoyant force results from there being a greater
pressure on the bottom of the balloon than the top. There is a greater
pressure on the bottom than the top because the pressure is the result of
all the weight of the air above some point, and there is more air above the
bottom than the top. So, if you have the international space station in
empty space the balloon would neither rise nor fall because it has no weight
and there is no buoyant force. In orbit, it is not really a zero-gravity
environment but rather a free-fall environment. Still, the pressure
difference, if any, inside the orbiting spacecraft would be probalby be too
small to cause any buoyancy.

QUESTION:
I am a 4th grade teacher and one of my students asked me a great question....Why don't clouds move when a jet flies through them? I could not answer him. Can you help me?

ANSWER:
Well, the cloud certainly does move. The plane pushes the cloud out
of the way to make room for itself, it creates currents and eddies around
its surfaces which cause those parts of the cloud to violently move around.
Think of similar situations: Does the water move when you dive into it? Does
a watermelon move when you plunge a knife into it?

QUESTION:
What prevents velocity in a known direction from being measured by firing a photon in that direction and timing precisely how long it takes to travel a known distance? If the speed of light is in fact constant regardless of frame of reference, it would be possible to find the velocity by (known speed of light) - (observed speed of photon fired)?

ANSWER:
First of all, the speed of the photon is always the same, you would
have to measure its momentum or energy before and after. But, the whole
point of the example of bouncing a photon from something is to demonstrate
the futility of trying make a precise measurement of the position or
momentum of some particle because the act of measurement affects that which
you are trying to measure. The heart of the matter, though, is that it is a
fact of nature that some things are simply unknowable. You cannot know both
the position and speed of a particle to arbitrary precision.

QUESTION:
Consider a water tank sitting on its stand and a pipe with an inner diameter of 6 inches was used to deliver water from the tank to the ground below it.
The vertical height from where the pipe is joined to the tank and the ground below is 20 feet. However, rather than coming straight down, the pipe comes down at an angle of 45 degrees and is then joined at the bottom so that there is a 3 feet length of pipe running horizontally (180 degrees) along the flat ground. Neglecting friction loss in the the pipe and in the bends of the pipe, what will be the flow rate of the water coming out at the end of the pipe? How much energy will the water have? Is there a formula that can be used to work out the final flow rate and energy?
Sorry if the scenario sounds silly. I can provide a sketch of the scenario if it helps.

ANSWER:
I need two more pieces of information: how deep is the water in the tank and what is the cross sectional area of the tank (if it is much bigger than the cross sectional area of the pipe, it can be neglected with little error.) The shape of the down pipe makes little difference, just how far it opens from the top surface of the tank. You need Bernoulli's equation which says
that
�ρv ^{2} +ρgy +P =constant where ρ is
the density of water, P is the pressure, v is the speed of the
fluid, g is the acceleration due to gravity, y is the height.
In your case the pressure at both the top of the tank and at the outlet of
the pipe is the same (atmospheric), v at the top surface of the tank
is approximately zero, and we can choose y =0 at the ground, so v =√(2gy ). For example, if the distance from the
outlet of the pipe vertically up to the surface of the water were 30 ft, the
speed of water coming out the pipe would be v =√(2x32x30)=44 ft/s. The
cross sectional area is π x0.25^{2} =0.2 ft^{2 } and so
the flow rate R would be R= 44x0.2=8.8 ft^{3} /s=66
gal/s. A general expression can be written for your case as R =12√(20+h )
gal/s where h is the height of the surface of the water in the tank
above the top of the pipe. For this to work, the speed of the drop in the
tank must be small. You cannot say the water has energy, you can say each
gallon of water has an energy E =�mv ^{2 } where m
is the mass of a gallon of water. When you figure that out, the energy per
second is P=ER which is power; if you do that calculation in SI
rather than English units, it will come out in watts. For my example above,
R =66 gal/s=0.25 m^{3} /s, v =44 ft/s=13.4 m/s, E =�x1000x13.4^{2} =89,800
J/m^{3} (1000 kg is the mass of 1 m^{3} of water), so P =22.4
kW. That's a little surprising to me, but a 6 inch pipe is pretty big. And,
of course, you would not be able to extract all this to use.

QUESTION:
How small would the human body be if you took out all the space between the atoms???

ANSWER:
See an earlier answer .

QUESTION:
Ok, so I got into a big discussion today with my physics Honors teacher and he didn't even say who was right or wrong. The problem was that there is what you could consider a clothesline with a cable (for purpose of the problem the weight of the cable does not matter) and a weight is suspended exactly in the middle of the cable. The weight of the object was 25 N. The angle at which the cable meets the object/block is 30 degrees. What is the tension in the cable?
My theory behind this was that when you find the force tension it is for only half of the whole cable so you need to double the force tension and my teacher argued that y ou do not. So with this problem he said that the answer was 25 N of force tension and I said that it is 50 N of force tension. Who is right??

ANSWER:
I am sure that you will not be too surprised to find that your
teacher is right. Maybe this would be easiest for you if you do not think of
the tension in the cable but rather the tensions in the left and right
parts of the cable. If the angles of the two parts of the cable were
different, the tensions would be different. So, right away, the notion of
doubling the tension to account for the two halves is seen to be faulty
reasoning. If you call the two tensions T _{L} and T _{R} ,
you can see that the horizontal components of the two tensions must be the
same: T _{L} cos30^{0} -T _{R} cos30^{0} =0,
so T _{L} =T _{R} =T. Similarly, the
vertical components of the two tensions must hold up the weight: T _{L} sin30^{0} +T _{R} sin30^{0} -25=0=T /2+T /2-25=T -25,
so T =25 N. This means that the tension in each side of the cable is
25 N and you could say, in this case, the tension in the cable is 25 N
everywhere.

QUESTION:
So I was doing my physics homework the other day and we had to figure out what the the horizontal and vertical velocity of a long jumper and the answer for the vertical was less than the acceleration from gravity. Anyway this got me wandering if you tried to launch an object upward with a speed lower than the acceleration due to gravity would it actually go anywhere?

ANSWER:
When you say "speed lower than the acceleration due to gravity", it
makes no sense because speed and acceleration are different things. It is
sort of like saying "a temperature greater than the color of a carrot".

QUESTION:
Many people place sandbags in the rear of their vehicles to increase winter traction. Many say only in rear wheel drive vehicles, generally over the rear axle or behind it (for more lever arm). Thusly weighted pick up trucks are said to benefit from better weight distribution correcting "inherent oversteer."
I could find only a few naysayers who say that there's little or no advantage gained or that the sandbag can become a dangerous flying object. It seems to me that if there's enough weight to affect the handling of the vehicle then if that weight shifted, as one could expect it might during reactive emergency driving, then the handling would be, at best, momentarily unpredictable.
So, specifically regarding a new rear wheel drive small pickup (e.g. a 2011 2WD Ford Ranger) with the all season fairly aggressive tires it came with, what is your analysis of the physics of adding weight to a pickup for winter traction? Aside from being perhaps unsafe does it really help in starting, stopping, cornering and/or hills?

ANSWER:
The force which drives your car forward, which allows you to stop,
which allows you to turn and maneuver is friction. The friction you are able
to get without your wheels slipping is proportional to the weight of your
vehicle. So, it makes sense that you should increase the weight of your
vehicle, right? Well, the force which you need to accelerate, brake, turn or
maneuver is also proportional to the weight of your vehicle. So, according
to elementary physics, there is no advantage to additional weight. However,
since a vehicle is not a point mass and has four wheels, changing the
relative loads on the rear and front wheels may make a difference. It can be
a complicated problem in the real world, but based on elementary physics, I
would guess that any gain in carrying sandbags is minimal.

QUESTION:
Basically what im trying to determin is the amount of restraint required to hold an item on an incline. In my occupaton im required to restrain various items for predetermined forces. Ie 3Gs forward, 2Gs vertical. However in this case I am not given the amount. Normally I would take the weight of the object, multiply it by the force it might incur then divide it by the rated value of the restraint being used.
What I want to know is, if an object is on a 10 degree ramp, how much force is required to hold it. The way I figure is if the ramp were at 90 degrees it would be 1G and if the ramp was at 0 degrees it would be zero. So if the ramp was at a 45 degree angle would it not be 0.5 Gs. So 10 degrees is .111
20,000 lbs vehicle multiplied by .111Gs is 2220 divided by 5000lbs (rated strap strength) = 0.444 thus requiring only 1 strap
I know that there are other factors being ignored like the coefficient of friction between the ramp and the pneumatic tires of the vehicle. (in this case 0.030 )
Also the strength of the strap being affected by the angle that it is applied. (we will say it is getting 100 percent of its rated strength)

ANSWER:
You are on the right track, but the force is not linear from 0^{0}
to 90^{0} as you are assuming. Ignoring frictional forces, the
component of the weight parallel to the incline is the weight times the sine
of the angle, so sin45^{0} =0.71 and sin10^{0} =0.17. So, your
20,000 lb vehicle would require a strap capable of holding 20,000x0.17=3400
lb on a 10^{0} incline.

QUESTION:
As I uderstand it, Carbon 14 is formed by neutrons comming from cosmic rays witch interract with Nitrogen(7 protons + 7 neutrons). Nitrogen lose a proton and win a neutron wich become carbon 14 (6 protons + 8 neutrons), right ?
Then, the part that I do not get is how Carbon 14 decay to Nitrogen. Wikipedia says that Carbon 14 loses eletrons and electrons antineutrinos to become Nitrogen again, but I don't understand because Nitrogen is 7 protons. Is Carbon gaining a proton ? Can you make this more clear for me ?

ANSWER:
The ^{14} C decay is what is called beta decay. Inside the
^{14} C nucleus one of the neutrons spontaneously turns into a
proton, an electron, and an antineutrino. The electron and the antineutrino
leave and the net effect has been to change ^{14} C to ^{14} N.

QUESTION:
Why is that if you push a book slowly off the edge of a table it will rotate, but it you give it a good, hard shove, it just flies forward and doesn't rotate much?

ANSWER:
As soon as the center of gravity of the book is beyond the edge of
the table, there is a net torque which will cause an angular acceleration
around the axis along the edge of the table. If you push the book slowly,
there is a long time for the acceleration to act so it will acquire an
appreciable angular velocity. The hard shove gives the torque very little
time to act giving only a very small angular velocity. Once the book is in
the air, there is no torque on it.

QUESTION:
how does lightning conductor works during a thunder storm ?

ANSWER:
The basic idea is to provide a conducting path to take the current
from the lightning and send it to the ground. However, it also "attracts"
the lightening so that it hits the rod rather than the house. When a charged
cloud passes over, the objects under it become charged by induction. The rod
has a sharp point which means when the rod becomes charged, the electric
field near the tip becomes very strong and often strong enough to cause
electrons to stream into the air; this is called corona discharge (or St.
Elmo's fire if on the top of the mast of a ship). This charge density and
strong electric field then has the effect of directing a strike at the rod.

QUESTION:
Do protons generate magnetic field around them ?

ANSWER:
Any electric charge which is moving creates a magnetic field. Since
a proton has charage, it will generate a magnetic field when moving.
However, a proton also has a magnetic moment, that is it looks like a tiny
bar magnet. Therefore a proton has a magnetic field even if it is sitting
still.

QUESTION:
Our state is considering the allowance for an increase in tractor-trailer weights from approximately 80 to 100 thousand pounds. The idea is that consumers will benefit because a given truck can haul more goods at a lower fuel cost. Law enforcement officials are against the measure because of the increased dangers due to increased stopping distances of the heavier vehicles. I'm curious about the relationship between increased momentum and stopping distances of these big rigs but don't have a good sense of how to get at it with a calculation. Would this comparison be better studied by actual measurements?

ANSWER:
According to the first approximation, which works very well, the
weight of a vehicle has no effect on its stopping distance. See a
recent answer . The reason is that the
frictional force between the tires and the road is proportional to the
weight of the vehicle and acceleration, which determines the stopping
distance, is inversely proportional to the weight; the weight does not
matter. However, the braking system can also be a factor because the brakes,
ideally, just keep the wheels from skidding (what antilock brakes do
automatically) and if they are overburdened (for example, overheat) they can
fail. Yes, physics is an experimental science and doing actual measurements
would be the best way to determine the answer.

QUESTION:
Is a golf ball going fastest after it leaves
the club head striking it, or is there still some accelertion as it
overcomes inertia?

ANSWER:
Nothing "overcomes inertia" on its own. Only a force causes an
acceleration and acceleration is what is meant by overcoming inertia. The
ball has its largest speed as it leaves the club because that force ends
then. When the ball is in flight, only two forces act on it �air
drag which always slows it down and gravity which slows it down on the way
up and speeds it up on the way down. But gravity can never speed the ball up
more on the way down than it slows it down on the way up.

FOLLOWUP
QUESTION:
Sorry I didn't do a better job asking the
question. My family and I are not looking to be told when peak
acceleration happens, we want to prove it mathematically. We have put
together a derivation to show that at its peak velocity a golf ball is
still in contact with the club head. Starting out with the length of the
drive we use the formula for projectile motion to find the peak velocity
of the ball. We work backward from there to derive all of our unknowns
but we run into trouble when we use the formula for uniform motion with
constant acceleration to calculate how far the ball and the club travel
during the time to accelerate. Because the distance the club travels is
expressed as S = u * t and the distance the ball travels is express as s
= u * t + a * t^2 / 2, there is always a few millimeters distance
between the ball and club when the ball reaches peak velocity.

ANSWER:
There is a difference between peak velocity and peak acceleration.
For example, a pendulum has its maximum velocity at the bottom when the
acceleration is zero. What you are doing is wrong because the acceleration
is not uniform as you assume. Let us, in accordance with my original answer,
focus only on the time during which the club is in contact with the ball;
the time after it leaves the club will always have a smaller velocity than
when it left the club. The graph to the right shows what the force which the
club exerts on the ball is likely to look like. The club first touches the
ball at t _{1}
and the ball leaves the club at t _{2} . The force over the
time of acceleration is not constant because the ball behaves like a spring.
When the force is biggest, the acceleration is biggest because of Newton's
second law, F=ma . However, the ball is speeding up over the whole
time interval and will therefore be going fastest at t _{2} .
The area under the force curve is called the impulse and is equal to the
change in momentum of the ball which, because the ball starts at rest, is
mv where m is the mass and v is the speed at t _{2}
of the ball. If you call F _{avg } the average force on the
ball, then mv=F _{avg} (t _{2} -t _{1} ).
Putting in some numbers I found on the web, m= 45 g, F _{avg} =3000
lb, v =250 ft/s, I estimated that the time of contact is less than
half a millisecond.

QUESTION:
During a discussion about basic physics I was having with my daughter, she asked an interesting question. She saw a story about helicopters that mentioned the helicopter blade tips could exceed the speed of sound creating the characteristic
"whop, whop" sound, or small sonic booms. She wondered if it would be possible for the blade tips to ever exceed the speed of light. I was intrigued by this so I worked for a few hours to come up with an equation to answer her question. I am not a math expert, but using simple algebra I came up with the formula: r=(60V) / (2 π R) where r = radius of the blades in meters, V = velocity in meters per second, and R = blade RPM. From this I calculated that the blade tips would exceed the speed of light if the blades had a radius of 286.28 kilometers and were spun at 10,000 RPM. This of course assumes the materials used could withstand the forces involved. My question: is this formula accurate and given the radius and RPM mentioned, what is the formula to calculate the forces along the blades starting from the center moving to the tips (centrifugal, centripetal?)?

ANSWER:
Nothing can exceed the speed of light. The qualitative way to
explain why for this case is that a mass increases as the speed increases.
As you go farther out on the blade, the mass increases more and more until
you are, at the crucial length, requiring an infinite force to keep the mass
moving in a circle. The speed of something a distance r from the end
would be v=r ω =9.55Rr
where R is rpm as per your notation and ω is the angular
velocity in radians/s; it looks like your equation got 60/(2 π )
where you should have had
( 2 π )/60.
So, I get the length would be just 3.14 km if it could work. Suppose the
blade had a linear mass density λ kg/m and a length L . So, the
mass of the whole blade would be Lλ. If you calculate the force
classically , ignoring relativity, you would get F (x )=(L-x )λxω ^{2}
where x is the distance out to the point where you are calculating
the force necessary to keep the mass beyond from flying off. But,
relativistically, λ depends on x , λ =λ _{0} /√[1-(ω ^{2} x ^{2} /c ^{2} )]
where λ _{0} is what the mass density is if the blade is at
rest. So, you see, if ωx=v=c , the mass of the whole blade becomes
infinite meaning the force needed anywhere is infinite.

QUESTION:
If like 3 water tanks are connected by pipes so water can flow to equilibrium, will the water levels still equalize even if the tanks are sitting on springs?
To me it seems yes but mathematically proving is difficult.

ANSWER:
Yes. The reason is easier than you think. The springs do not exert
forces on the water, they exert forces on tanks. Therefore, when considering
the forces on the water the springs do not enter. The springs would compress
or uncompress as the amount of water in each tank changes, but it would be
no different from what would happen if you pushed the tanks up and
down.

QUESTION:
if you had an object that was the size of a grain of sand but its "mass" (might not be the best term) was the same as the the suns, would it be possible for this grain of sand to have its own planets revolve around it?

ANSWER:
Yes. What matters is the force which the "sun" exerts on the objects
orbiting it and that is determined by the mass of the "sun", the mass of the
objects, and how far they are from the center of the "sun".

QUESTION:
My car travels at 55 mph. Are the outer portions of my tires moving at a faster speed than my car? Im thinking it does, but have no formula for it.

ANSWER:
The point of your tire which is touching the ground is at rest. The
axel of that wheel is going forward with a speed of 55 mph. The top of the
tire is going forward with a speed of 110 mph.

QUESTION:
A vertical narrow tube has a photon emitter at one end, so that photons travel through the tube and are emitted along the verical axis of the tube (all in a frame moving left to right near the speed of light). Do the photons emitted from the end of the tube leave at an angle to the vertical, as seen by a non relativistic frame? If so, do the photons hit or move toward the wall of the tube on their way inside the tube?

ANSWER:
First, view from the frame of the tube. The photons travel straight
up the center of the tube and never hit it. Therefore, all observers have to
agree that the photons do not hit the tube. Certainly the photons, as seen
by the stationary observer, do not move vertically. Rather, they have a
component of their velocity which is v along the direction of the
motion of the tube but their total velocity is still c . The angle
they make with the vertical is therefore
θ= sin^{-1} (v /c ). What you are describing is
very similar to the light
clock.

FOLLOWUP QUESTION:
you said that both the moving observer and the stationary observer must agree on whether the photon hits the wall of the tube or not. You said the moving observer see the photon traveling straight up the tube.
but you also said the stationary observer sees the photon traveling at an angle to the tube wall. you didn't say whether that angled path resulted in the photon hitting the wall and how this discrepancy is resolved.

ANSWER:
There is no mystery here �the
tube has the same horizontal velocity as the photon does, v . So,
every time t that the photon moves a distance vt , the tube
also moves a distance vt and so the photon stays in the middle of the
tube.

QUESTION:
If we are having two balloons suspended with thread. When we blow air between them why they attract each otherhence having like charges in both the balloon?

ANSWER:
Bernouli's equation says that in a fluid the pressure gets smaller
as the velocity gets bigger. So, the pressure between the two balloons
becomes smaller than atmospheric pressure (which is pressing on the outer
part) and they go together. This is the same kind of force which helps
airplanes fly and causes spinning balls to curve.

QUESTION:
if you fall 10 feet from a hayloft, what part
of your body would hit the ground first?

ANSWER:
There is no way to answer this question. The orientation of the
falling body at any point is determined by the initial conditions. For
example, just jumping would result in feet first, diving head first would
result in head hitting first. It is also possible to change the answer by
what you do as you fall. For example, a diver can make his body spin faster
by doubling into a ball and slow the spinning by straightening back up; and
a cat can land on its feet.

QUESTION:
If my car weighs 5600 lbs and I'm traveling 35 mph. How long will it take to stop under normal road conditions?

ANSWER:
The weight of the car does not make any difference. The coefficient
of friction between rubber and dry asphalt can be anywhere from 0.4 to 0.8,
from which I calculate stopping times of 2-4 s and corresponding stopping
distances of 50-100 ft.

QUESTION:
What is the diffrences between atomic bomb and nuclear bomb. Can both weapons be consider nuclear weapons

ANSWER:
Atomic bomb is an archaic name. It is what nuclear bombs were
initially called the public was not so well informed about physics. Really,
a conventional bomb should be called an atomic bomb since it gets its energy
from chemistry which is energy from atomic forces. Nuclear bombs (fission
and fusion are the two kinds) get their energy from the nucleus. But, atomic
bomb is a name which survives and is just an old-fashioned name for nuclear
weapons.

QUESTION:
Does the amount mass of an object really not effect the gain of speed withing a field of gravity? FE: DO a large meteor and a small metal BB fall at the same speed exact speed, If the factor of wind resistance and friction are taken out? And do you have a formula for this? please keep in mind that this is not for homework but to clarify my curiosity.

ANSWER:
Constant acceleration due to gravity is an idealization which
assumes a uniform graviational field which means that the force is always
straight down, the earth is perfectly flat, the force is independent of how
far you are from the earth, the earth is not rotating, and that the earth
has infinite mass. These are all literally false, but for most everyday
problems they are an excellent approximation. If, as you stipulate, there is
no air drag, then you would be hard pressed to measure any difference in the
accelerations of a 10 ton meteor and a BB. If the altitude is very small
compared to the radius of the earth and if the mass is very small compared
to the mass of the earth, acceleration is constant.

QUESTION:
What would the transit time between Earth and Neptune be (pick your own planetary positions) assuming an acceleration/deceleration curve that maintained 1 g of force on the spacecraft for the entire journey? In other words if our hypothetical spaceship wanted to simulate gravity by means of constant acceleration as well as deceleration (for orbital capture) how long would it take to get to Neptune. I realize that one would have to take into account the trajectory of the spacecraft since getting from one planet to another is never a straight line, but any rough answer would be much appreciated.

ANSWER:
A constant acceleration of g ≈10
m/s^{2} to the halfway point will take the same time as the
deceleration the rest of the way, so we can just calculate the time for the
first half of the trip and double it. The distance from earth to Neptune is
on the order of 4x10^{12} m. Starting from rest, the position is
x= 5t ^{2} =2x10^{12} , so t =6.3x10^{5}
s. This calculation is correct only if the maximum speed is much smaller
than the speed of light; the biggest speed is v =10t =6.3x10^{6}
m/s which is still about 50 times smaller than the speed of light, 3x10^{8}
m/s. So, the time would be about 1.3x10^{6} s≈15 days. Pretty
speedy!

QUESTION:
If you are holding a 3 pound, 2'x2' sign in a 5 mph wind how many
pounds of resistance would you be holding?

ANSWER:
The weight makes no difference. The force on the sign can be roughly
approximated as
F =�Av ^{2 } where A is the area and v is the wind
velocity, but these must be expressed in SI units, A in m^{2}
and v in m/s and then F is in Newtons. Doing the conversions
for 4 ft^{2} and 5 mph and then converting Newtons to pounds, I find that the
net force from the wind is about 0.1 pounds. 5 mph is a pretty gentle
breeze.

QUESTION:
During beta decay, a neutron splits into a proton and an electron inside the nucleus. The electron is released from the atom which is the beta particle. I have studied that the atomic number of an atom increases by 1 after a beta particle is emitted. But the daughter atom has 1 proton in extra to that of the number of electron. So why does not the atom carries a positive charge of 1 unit?

ANSWER:
Immediately after the decay, the daughter atom is missing one
electron. However, the electron which escaped ensured that charge was
conserved. What will happen if there is a positive ion is that it will
rather quickly find an electron to capture and become neutral, maybe even
the beta particle that came out.

QUESTION:
Is quantum physics a theory or proven science? I have asserted that it is a theory, in debate with a friend, who demands that it is a complete proven science. With the exception of some experimentation, I can find no fqctual evidence with finished proof.

ANSWER:
It would be my opinion that there is no such thing as a "complete
proven science" and no way you can provide a "finished proof" for any
theory. It is really a question of semantics and I do not know how
philosophers of science define theory, so I am not speaking authoritatively
here. However, it seems arrogant to me to claim that any theory is incapable
of being found incorrect in some limit, just as Newtonian physics, amazingly
successful in its own realm for hundreds of years, was later to be found to
be only an excellent approximation of the theory of special relativity at
low speeds. To answer the question specifically about quantum mechanics, I
heard a statement on a Nova show the other night that there has never been
an experiment which was not described by quantum mechanics; that's about as
close as it gets to "proven science". Who knows what future experiments
might hold, though?

QUESTION:
I wonder, if we had a possibillity to accellerate a manned vehicle constantly up to, let's say, mach 5 or faster how slow/fast would the accelleration have to be to keep the human from dying? Would the pure speed itself kill the human?
And would there be a theoretical way to construct a chamber that completely isolates it's contents from the force that sudden acceleration or something like dropping it would produce?

ANSWER:
Manned spacecraft have speeds far larger than mach 5. The maximum
amount of
acceleration which a person can endure depends on the duration, the
person, training, etc . If you travel with constant velocity, you
feel exactly the same as if you were at rest no matter how big the velocity
is. Finally, there is no way to construct your hypothetical chamber.

QUESTION:
I have struggled to get the simple explanation (non-mathematical form) for my troubling question from books. Still no luck.
Whts is force ? (as a concept)
What is the relationship between energy and force? I am not a physicist so could you please enlighten me and explain the concept in layman's term for me? Or with some analogy?

ANSWER:
A force is a push or a pull. It is one of those things you need to
have an intuitive feel for to begin doing physics. It is that which
causes an object to accelerate. In physics accelerate can mean two different
things, either change speed (speed up or slow down) or change direction.
Energy is a more subtle concept. To understand what energy is you need to
understand the concept of work. In physics, work is done by a force only if
the force is exerted over some distance. For example, if you push with a
force of 1000 pounds against an unmoveable wall, you do no work, but if you
push a box across the floor with a force of 10 pounds, you do work. When you
do work on an object you increase its energy. So, when you push the box
across the floor, it acquires what we call kinetic energy, energy by virtue
of motion, because it started at rest and ended up moving. If you use a
force to lift an object so that it starts out and ends up at rest, you have
still done work but now the energy you have given it is called potential
energy, energy by virtue of position. Different kinds of energy can
transform to others. For example, if you drop something its potential energy
is converted into kinetic energy as it falls.

QUESTION:
My question is, the moment of inertia for a ring and a hollow cylinder about d same given axis is the same i.e. MR^2. its conceivable mentally, that being a rigid body, i only need to rotate one out of many rings that totally make up a cylinder, so or the stretch rule, as stated in wikipedia http://en.wikipedia.org/wiki/Stretch_rule, bt what is the mathematical proof for the rule???

ANSWER:
I never heard of the "stretch rule". Start with an object with
infinitesmal thickness dz and moment of inertia dI _{1} = [∫∫ r ^{2} ρ (r,φ,z )r dr dφ ] dz = [∫∫ r ^{3} ρ (r,φ,z )dr dφ ] dz
about some axis. Now, add another "slice" with the same shape and the same
r , φ
dependence of
dI _{2} = [∫∫ r ^{3} ρ (r,φ,z )dr dφ ] dz .
Keep adding new layers to build up the whole object. For every slice, the
integral over r , φ
is just some function of z, f (z ).
So, finally, I = ∫ f (z )dz.

QUESTION:
enrico fermi once pointed out that a standerd lecture period [50 mint] is close to one micro centuery. how long is a micro centuery in minutes. and what is the percentage difference from fermi claim?

ANSWER:
Well, this isn't physics, but 1
μCentury=(100 yr)(365.25 days/yr)(24 hr/day)(60 min/hr)x10^{-6} =52.6
min. Fermi was a real smart guy, particularly famous for "back of the
envelope" estimates.

QUESTION:
I remember from an old schoolboy science book that if one were to remove all the 'space' from the atoms in a human being, the collective mass would sit on a pin-head. And, if it were done for the whole human race, we would be nothing more than a super dense sugar-cube. I also read that if this were done for the Earth, and all the 'space' was removed from it's atoms, it would be about the size of a rugby stadium (about 500m across). Would you concur with this? And contniuing this theme, what would be the size of the Sun?

ANSWER:
Basically, you are asking what is the mass density of the nucleus
because the rest of an atom has negligible mass and the electrons can be
taken as having no volume. This number is about 2x10^{17} kg/m^{3} .
So, take a 90 kg man with a density of about 1000 kg/m^{3} and a
volume of 90/1000=9x10^{-2} m^{3} . His new volume would be
about V=90/2x10^{17} =4.5x10^{-16} m^{3} ; so his size
would be on the order of the cube root of this number, 8x10^{-6}
m=8x10^{-3} mm, about 10 microns. I am not going to do any more, you
do any others you are interested in.

QUESTION:
I know that a particle gains mass as it approaches the speed of light, but why? What actually happens to it? Can you explain in laymen�s terms how this is possible? I can imagine a large object like a spacecraft, streaking across the universe approaching the speed of light, impacting matter and virtual particles that is in its way, maybe slowing it down � the faster it goes, the more matter in the interstellar medium it encounters, keeping it from obtaining �c�. Does this have something to do with it?

ANSWER:
Nothing "happens" to the mass; if you were riding on it, you would
see no change. That is one reason why I prefer to not think of mass getting
bigger, I prefer to think of
momentum getting redefined. But, if you want to think about mass
increasing, as interpreted by someone with respect to whom the mass is
moving, here is how you can think about it: assuming that you accept that no
mass can go faster than light, as you approach c it gets harder and
harder to push it faster and you interpret this as the mass acquiring more
inertia, another word for mass. So, at low speeds, a 1 N force on a 1 kg
mass would result in an acceleration of 1 m/s^{2} , but at a very
high speed it might require 10^{6} N to achieve the same acceleration
which you would interpret as an increase of mass. Your ideas of trying to
mechanically make sense of the mass increase, while creative, are off the
mark totally.

QUESTION:
You know how The particles that make up a rock are constantly in motion. However , a rock does not vidibly vibrate. Why do you think this is .

ANSWER:
The amplitude of vibration of each atom is on the order of the
spacing between atoms, 10^{-10} m, way smaller than you would expect
to see.

QUESTION:
(Going straight up, like a rocket launch) If I impart a force of 89kN to a 6818kg mass and determine it will accelerate at 13.05 m/s^2, how fast does that mean it will accelerate over time (what is its average acceleration) assuming the same force stays applied to it, and what speed will it ultimately reach assuming a standard atmosphere?

ANSWER:
First of all, you have it wrong. The forces on the rocket are 8.9x10^{4}
N up and its own weight 6818x9.8=6.68x10^{4} N down. So, the net
force is 2.22x10^{4} N up. So the acceleration is 3.25 m/s^{2}
up. But, this is at launch, and you do not have enough information to know
anything later. First, the rocket is presumably burning fuel and so its mass
is changing so its acceleration will be increasing. Second, when it starts
moving fast air drag starts to become important (usually proportional to the
square of the speed of the object) which is an additional force down. Third,
if it goes high enough the density of the air gets less which reduces the
air drag at higher altitudes.

QUESTION:
if i have a 5 kg bird in a contained glass ball and when the bird is setting the total weight is 10 kg what is the weight if the bird flyes inside the contained glass ball does it measure 5kg or still measures 10kg best regards !

ANSWER:
I have answered this
and similar questions before.

QUESTION:
How does the weight of an object influence the angle at which it begins to slide down an inclined surface? I can find lots of info on how weight effects acceleration but not the angle needed to start to move down a slope?

ANSWER:
What allows an object to sit at rest on an incline is static
friction. However, there is a maximum amount of static friction which you
can get: the most static friction you can get is f _{max} = μ _{s} N
where N is the force pushing the surfaces together and μ _{s
} is called the coefficient of static friction and is determined by the
surfaces. This is only an approximation, not some law of physics, but it
works pretty well for everyday situations. When on an incline, N=mg cosθ
and the force trying to push the object down the plane is F=mg sinθ ;
here θ is the angle of the incline, m is the mass, and g
is the acceleration due to gravity. If it is not sliding, the frictional
force is just equal to F . But as the angle is increased, eventually
the friction is no longer able to hold the object, that is F=f _{max} ,
mg sinθ _{max} =μ _{s} mg cosθ _{max} ,
so tanθ _{max} =μ _{s} . So, you see, the answer
is that the weight has no influence.

QUESTION:
This may be a random question and It might be too obvious, but I was wondering whether gravity violates the law of conservation of energy.
Gravity seems to me, a constant force that is constantly acting on all objects but has no input energy to contribute to it.

ANSWER:
So, consider a mass, say the earth. It has a gravitational field
which has some energy density. Similarly, it has mass so it has mass energy.
That takes care of all the energy associated with that mass. Now, you can
say the same things for some other nearby mass. If that nearby mass is a
baseball, we have found it useful to introduce a potential energy function.
But I always tell my students that this may be thought of as a bookkeeping
device which automatically keeps track of the work which the earth's
gravitational field does on that baseball. Before we introduce potential
energy in elementary physics, all we have is that the change in kinetic
energy equals the work done by all forces external to that object. So,
before we "invent" potential energy, the work done by the baseball's own
weight causes its energy to change. We do not worry about nonconservation of
energy here because energy conservation only applies to isolated systems,
systems which have no external forces on them, and the earth is most
certainly an external force on the baseball. Adding potential energy to the
system simply "internalizes" the field, it is no longer considered an
external force. Now, you are worried about where the earth gets the energy
to give to the ball. The answer is all in relativity. Suppose that you take
a ball and move it out to infinity from the earth's surface. Do you do work?
Of course! But, what happens to the energy you put into that ball-earth
system? It shows up as mass! You know, E=mc ^{2} . This kind of
argument is made all the time in nuclear physics where the mass changes are
really obvious because the forces are so much stronger than gravity, but the
same argument can be made for any attractive force. This mass change is
where energy of chemistry comes from �e.g .
when you burn coal you add an O_{2} to a C and get CO_{2}
and a little less mass. When you split a U nucleus you lose a really
significant amount of mass. A ball on the ground has less mass than a ball
halfway to the moon. Strange, but true!

QUESTION:
why does a swimming pool that is 30 degrees celcius have more thermal energy than a cup of soup that has 60 degrees celcius

ANSWER:
The energy content of a cup of water is proportional to the absolute
temperature. So a cup of water at 60^{0} C=332 K has energy E _{hot}
and a cup of water at 30^{0} C=302 K has energy E _{cold} ;
so, E _{cold} /E _{hot} =302/332=0.91, the cold
water has 9% less energy than the hot water. However, if you have two cups
of cold water, the cold water has twice the energy of a single cup and so
now E _{cold} /E _{hot} =604/332=1.82 so the cold
water has 82% more energy than the hot water. And, so on �

BETTER ANSWER: (Thanks
to Michael Weissman at Ask
the Van )
" This sort of comparison of thermal E s assumes you have some defined zero, presumably
E at T =0. (We needn't worry yet about at what p or V .) So thermal
E (T ) is whatever had to be added from there to reach T . That isn't even remotely close to proportional to T. These aren't ideal monatomic gases far above the boiling point. It's largely dominated by the latent heat of melting, an almost constant energy density throughout the liquid range."
This clearly shows that my quantitative comparison was not correct; be
assured, however, that the the soup has much less energy than the pool."

QUESTION:
A few days ago my friend & I were debating about how we could find the force with which a bullet of mass m and UNIFORM velocity v would hit a wall x m away. I said that the total energy would at the moment of impact would be equal to the kinetic energy of the bullet. So the total work done would be equal to the kinetic energy right at the moment of impact. So we could easily determine the kinetic energy by dividing the product of the mass and velocity by two. Then the result would be equal to the work done by the bullet. We could find the force by the formula W=fx. Please tell me if I am right.

ANSWER:
You are not right. You are right that the kinetic energy of the
bullet is lost. However, x which you define as the distance from the
wall to the gun, has nothing to do with it. Work is the force times the
distance over which it acts and the force is not acting all the way
across the room. What you might say is that if the bullet penetrates a distance
x into the wall then the average force may be written as F =( �mv ^{2} )/x.
But, that assumes that all the work done by the force results in
stopping the bullet, but some, much maybe, of that kinetic energy is
converted into heat and sound energy. (Incidentally, uniform velocity has
nothing to do with it, all that matters is how fast it is going when it hits
the wall.)

QUESTION:
Are sound waves simple harmonic motion? Then are all waves simple harmonic motion? Which are and which aren't? How could you know the difference?

ANSWER:
Simple harmonic motion is defined to be a pure sinusoidal motion.
That is, whatever (x ) is vibrating must be expressible as x =sin( ωt ).
Only pure, single frequency sound waves are of this form. So almost no sound
is simple harmonic. All sounds, however, may be expressed as sums of simple
harmonic sounds of different frequencies; this is called Fourier analysis of
a wave.

QUESTION:
In strongman contests, the contestants pick up a
telephone pole at one end, push it upright and flip it over. The other end
of the telephone pole stays planted on the ground. Typically the telephone
poles weigh 300 lbs and are 30 feet long.

How would you calculate the actual weight of the pole being flipped at
different points during the lift? For example, in the course of one flip we
could say the pole travels in a 180 degree arc. The person is physically
lifting for the first 90 degrees or less when gravity takes over. When the
person picks it up off the ground it might be at 5 degrees, once it is over
their head it could be at 30 degrees, once they take 2 steps forward it may
be at 45 degrees, etc. - how do you calculate the actual weight of the pole
along that arc at different points?

As the person walks forward to flip the pole over their arms remain over
their head (i.e. in this example once they're at or past the 45 degree mark
in the arc), but it seems the weight would lessen along a curve as the pole
gets higher and gravity begins to assist? That is - the person (fulcrum?)
might be moving in - say - 3 ft. increments as they walk forward to gain
leverage to flip the pole over. How is the weight being lifted adjusted to
account for that variable?

ANSWER:
First, some terminiology: the weight of the pole is 300 lb, that is
its actual weight . What you want, I believe, is the force you need to
exert to hold it at any given angle. This is a problem involving torques �you
say that, in order to hold it at some angle, the torque you exert is equal
to the torque the weight exerts, a bit more to move it. (Torques are
calculated here about the point where the pole touches the ground.) The
torque due to the weight is easy, simply 300x15xcosθ= 4500cosθ
ft-lb where θ is the angle the pole makes with the ground. The torque
you exert depends on where you push and the direction you push, an infinity
of possibilities. I have calculated for the two extreme possibilities: you
push vertically up or you push horizontally. For the vertical pushes I
assume that your hands are 7' above the ground; for the horizontal pushes I
assume that your hands are 5' above the ground. The graph above shows
vertical and horizontal forces you must exert vs . angle. Clearly, you
want to start your lift by exerting a vertical, not horizontal, force. When
you first lift it, you need to exert an upward force of 150 lb and when you
get to 13.5^{0 } your hands are at 7' and you can start moving
forward. When you get to a little more than 50^{0} , it starts to
become more advantageous to push horizontally, becoming much more
advantageous as you near the end. If you go real slowly, you will have to
exert a maximum force of about 500 lb, but when the force you need to exert
is small at the beginning you should push harder, thereby giving it some
added speed which will help it over the middle angles where the force is
maybe bigger than you can exert. Also, you will, through practice, learn to
shift smoothly from vertical to horizontal .

QUESTION:
My question is in the nuclear equation (proton + electron-> neutron). when this necular equations occurs, an x-ray is emitted. Why is an x-ray rmitted and not a phroton of a lower energy? Would the reason be?, because the photon is movind so fast it is observed as a form of light? or that gama phrotons have no mass or electrical charge, and they are pure electromagnetic energy. Or I'm I way off base?

ANSWER:
You must be thinking about electron capture, a beta decay process
which competes with
β ^{+} decay. Here the nucleus "grabs" one of the atomic
electrons and combines it with a proton to make a neutron and a neutrino
(you didn't include the neutrino). Now, think about it: there is a hole
where an electon used to be in the atom, usually the K shell. An outer
electron falls into that hole and makes an x-ray which follows the electron
capture.

QUESTION:
is central force and centripetal force the same?

ANSWER:
No. A central force refers to a force which depends only on how far
it is from some force center. For example, to a good approximation, the
earth's gravitational force is a central force. Centripetal force is the
force which is responsible for a centripetal acceleration and may or may not
be a central force.

QUESTION:
I understand according to einstiens thought experiment - if you travel near light speeds from earth and come back - you would be thousands of years in the future.
So as the subatomic particles move at near light speed are they not affected by this relativity theory - I mean are they moving in to future.

ANSWER:
Indeed, clocks on moving subatomic particles do run slower by the
usual factor
γ= 1/√[1-(v /c )^{2} ]. Particles in
an accelerator, usually protons or electrons, will not change over time, so
you would not notice it since they do not carry little clocks along with
them. However, if you make an unstable particle like a pi meson (pion) which
has a half life of about 2x10^{-8} s when it is at rest, will last
much longer if it is moving at a high speed. For example, if v /c =0.99,
99% the speed of light, then you would expect it to go a distance 0.99x3x10^{8} x2x10^{-8} ≈6
m if its clock ran the same as yours. But, when you observe such speedy
pions, they will go much farther because their internal clocks are slow by a
factor
γ= 1/√[1-(0.99)^{2} ]=7.1; they
will go, on average, about 42 m.

QUESTION:
My question is regarding an elevator in free fall.
I had a debate with my uncle about what would happen to a person in a free falling elevator. I asked him if the person would float. My uncle said no. He pointed out that a ball would not float if put on the floor of the elevator and the cable was cut, so why would a person?
To see what would happen to a ball in a free falling elevator, I put a ball on a folder (the folder representing the floor of an elevator), and I dropped the folder with the ball. The ball stayed on top of the folder as they both fell to the ground. This indicated that what my uncle said about a ball staying on the floor of a
free falling elevator was likely correct.
However, when I researched the subject on the internet, I came across many articles that said that a person would float in a free falling elevator. So I'm a bit confused. Why would a person float but a ball wouldn't?
So my question is:
1) Would a person stand on the floor of a free falling elevator or would he float above the floor in a free falling elevator (i.e. weightlessness)?
2) If the person would stand on the floor of a free falling elevator, please explain why. If the person would float above the floor, please explain why.

ANSWER:
First, let's ignore the effect of air friction or any other kind of
friction on the falling elevator. You and the elevator are both in free fall
with the same acceleration. You would feel weightless and you would
have no motion relative to the elevator which you did not have before it
started dropping. If you were standing on the floor, you would remain
standing on the floor but you would not feel the floor pushing up on you
like you normally would, so it is perhaps a misnomer to say you are standing
on the floor � you
and the floor happen to be at rest relative to each other and touching. If
you were holding a ball and released it, it would float right where you
released it. If you gave yourself a tiny push up with your toes, you would
slowly drift toward the ceiling.

Now let's talk about the more realistic case of including air friction. The
elevator would feel a drag force (like a parachute) which got bigger the
faster it went and eventually would fall with a constant velocity called the
terminal velocity. You, however, are at rest relative to the air you are in,
so you would at first feel weightless but feel weightier and weightier and
then, when terminal velocity was reached, feel normal, all the while
standing on the floor. But, I don't think the second scenario is what you
were thinking about because it is not what would be called free fall.

QUESTION:
what would be the apparent speed of light,emmited vertically from a train travelling at a velocity v,to an observer who is standing in the platform

ANSWER:
The speed of light is always the same to all observers, so the
answer is c =3x10^{8} m/s. I suspect you are trying to use a
trick to make the light go faster than the speed of light. You will not
observe the light to go vertically but rather than at an angle forward in
the direction of the train. When you apply the correct velocity
transformatons, the vertical component will get smaller than c in
just the right way that the vector sum of the horizontal and vertical
components will be exactly c .

QUESTION:
My musical instrument has been broken during an air travel, but its container, a hard case with thick foam padding inside and outside is absolutely intact! how is this possible ?

ANSWER:
Suppose that it is dropped from a height of 3 m, about 10 ft. It
would hit the floor with a speed of about 8 m/s. Suppose the thickness of
the foam is 5 cm, about 2 in. Then the instrument would stop in a time of
about 0.013 s and would experience an acceleration of about 640 m/s^{2
} which is about 64g where g is the acceleration due
to gravity. What this means is that, to stop your sitar in this short
distance would require an force of about 64 times bigger than its own
weight. Even dropping from 1 m would result in an acceleration of about 20g .
If the padding were well designed so that this force was distributed over
the whole surface of the instrument, it might survive, but probably one part
will take more than its share of the impact. So, the padding in your case
should be thought of as protecting the instrument from normal bumping and
jarring, not major drops.

QUESTION:
There are a few aircraft like the X-15 (A) that are released from other aircraft (B), which obviously then fly at a higher speed than the 'carrier' (A). In my mind, if another aircraft (C) could then be fired from B, then it would be quicker still. In theory, could C release D, D release E and so on until an insanely high speed is reached. If it were possible to build the system I have described, is there any limit to the speed it could reach? Could it not even exceed the speed of light?

ANSWER:
Your idea has already been invented and is called a multiple-stage
rocket. Don't get any fancy ideas about exceeding the speed of light,
though! Ain't gonna happen.

QUESTION:
I was watching a program about the universe and the narrator said something along the lines of "this is what a neutron star sounds like", and then went on to show a sound. How is this possible if sound is a compression wave, and therfore requires a medium of some sort? I thought that space was a vaccuum, and therefore sound waves can't travel through it so, how can we hear what a star sounds like?

ANSWER:
These stars do not make sound, they radiate electromagnetic waves in
pulses with the frequency of the rotation. So, if you take these pulses, put
them through an audio amplifier and feed them to a speaker, they make sound
with the frequencies of the pulses. To hear some,
look
here .

QUESTION:
My girlfriend and I were wondering whether an
(1) airplane travels faster over the ground as it flies east to west due to the earth's rotating underneath it, leaving out tailwind or headwind. She thinks it does not travel fast because it is in the atmosphere and she equates an example she recalled where if someone in a bus going 80 mph
(2) throws a ball straight up in the air, the ball doesn't go backwards. That just sparked an argument which I'm losing. I said it does go backward, you just can't notice because it's not in the air long enough to see a difference. It seems to me that if the bus was
tall enough and you could throw it straight up high enough, the ball
would come down behind where it was thrown because it is slowing down
while in the air.
(3) After a bit of searching, it seems that
the consensus is that the ball will come down in the "exact same" place
due to what some call conservation, others preservation of momentum. I
understand that the objects inside the bus are going 80 mph just as the
bus is - and the bus is a container. I don't understand why the ball
thrown up on the bus does not start to decelerate AT ALL after leaving
the person's hand if it has not contact with anything else beside the
air in the bus. It seems to me that it must decelerate some. I know it
would not rapidly decelerate as it would if thrown out of the window
into oncoming air; and as compared to a ball thrown 80 mph in a
ballpark, I could see if the deceleration on the bus is less rapid
because the air on the bus is going 80 and maybe it pushes the ball
forward some. But it seems to me that the air on the bus shouldn't have
much more effect than an 80 mph wind would have on a ball thrown up by a
person standing outdoors. And maybe the plane does not get an increase
in ground speed exactly equal to the speed of the earth's opposite
rotation due to whatever reasons, but it seems it should get some
increase.
(4) The atmosphere is not as fixed to the
earth as a tree, is it? Doc, (5) I'd really
rather you say something that can be used against my girlfriend here,
but any help will be appreciated. I've thoroughly enjoyed the site and
now wish I had taken physics.

ANSWER:
Whoa! Too many things going on in this question �airplanes,
balls, buses, girlfriends�I have taken the liberty to try to organize your
question by numbering parts of it. There are two issues here. First, you
always have to specify a velocity with respect to something else. What
someone on the ground measures and someone on the bus measures are
different. Second, you have to decide whether air friction is important or
not to discuss the velocity of something which moves through the air.

The airplane
flies relative to the air and if the air is at rest relative to the
ground, it makes no difference which direction the plane flies, it
travels the same ground distance in a given time.

If the air in the
bus is at rest relative to the bus and if the bus is traveling with
constant speed in a straight line, the ball will go straight up and
straight back down as seen by somebody on the bus. Somebody on the
ground will see the ball travel in a parabolic path. These statements
are if air friction is negligibly small.

The reason the
ball comes down to exactly the same place in the bus is determined by
what forces act on it. If you neglect air friction, the only force is
gravity which acts straight down causing it to slow down going up and
speed up going down. If air friction is not negligible, it is a force
which always acts in the direction opposite the velocity vector, that is
it never has a component which is horizontal which is what you need to
make it start moving (relative to the bus) horizontally. If air friction
is important, like if you throw it really fast and really high, it will
still move vertically but it will take different times to go up and
down. Unless there is a horizontal force (a wind, for example),
something moving in the vertical direction will move only in the
vertical direction.

Since you
stipulated no head wind or tail wind effects, yes, the atmosphere is
just as rigidly attached as a tree.

Sorry, your
girlfriend has a much better grasp on this than you have!

One final proviso:
remember where I said "traveling with constant speed in a straight line"?
Well, this is never rigorously true because the earth is rotating and
therefore Newton's laws are only approximately true. The effects of this
(centrifugal and coriolis forces) are so tiny that any ball throwing will be
unmeasurably small. However, for very long-range problems they can be
important. These "forces" are what are result in the circulation of weather
systems.

QUESTION:
It is well known to us that the energy is conserved everywhere in nature, infact it is the basics law in physics.But how far it is correct practically? mean to say if suppose we consume 100j of energy(inform of our food),can the net output work be of 100j or less than that or more than that?

ANSWER:
When you say you "consume 100 J of energy", you are really talking
qualitatively, not quantitatively. Your body is a machine with
inefficiencies. Nevertheless, the law of conservation of energy is certainly
true. It states that the total energy of an isolated system remains
constant, so any energy in one form always ends up in another form.
Biological systems are more difficult to do experiments with because the
isolation bit is hard to achieve. There is one little
proviso on energy conservation, but it is
not really what you are asking about.

QUESTION:
Do black holes "consume" dark matter?

ANSWER:
If dark matter exists (see my
reservations about this), then it interacts only via gravity and would
therefore be "consumed" by black holes.

QUESTION:
I know scientists have created small amounts of antimatter,but is all antimatter created equally or are there different elements ie: anti-hydrogen , anti-carbon etc.?

ANSWER:
Antihydrogen has been experimentally created. It consists of an
antiproton orbited by a positron (antielectron). To the best of my
knowledge, no other antiatoms have been created but, in principle, could be.
They would consist of nuclei with antiprotons and antineutrons orbited by
positrons. Really hard to pull it off, though.

QUESTION:
How much Electro Magnetic Radiation (in milligauss) would come off of your typical 365 kV electrical power line? Knowing distance is a factor could you give me levels at 50, 100,150, and 200 yards?

ANSWER:
I think you want the magnetic field strength which is not the same
thing as electromagnetic radiation; certainly, this is what a milligauss
measures. It is not the voltage which determines the field, it is the
current which is flowing. Since power is current times voltage, the bigger
the voltage, the smaller the current (and therefore the magnetic field) for
a given power transmission. Suppose the power is 1 megawatt. Then I figure
the current would be around I ≈3
A. The magnetic field a distance R from a current I is given
by B=μ _{0} I /(2πR )=0.06/R (measured in
gauss, R in meters). So, for R=100 m, B =0.6 milligauss; you
can do other calculations if you want. For comparison, this is about 1000
times smaller than the earth's magnetic field, so if you are worried about
health issues, don't.

QUESTION:
Theoretically, if you were traveling through space in an object
e.g., a spaceship and going at an incredible speed in one direction what
would happen to you, inside, if it made an immediate 90 degree change in
direction? wouldyou smash against the inside of the craft?

ANSWER:
First of all, you cannot make
"an immediate 90 degree change in
direction". It would require an infinite force to stop your forward motion
simultaneous with an infinite force to start your sideways motion. So, you
have to say you make the turn through a curved path. As an example, I will
choose "an incredible speed" to be 10^{4} m/s ≈22,000
mph, just a little faster than the shuttle goes. Now, at that speed let's
make a 90^{0} turn
around a curve of radius 1 km=10^{3} m. Then the spaceship and
all its contents experience a centripetal acceleration of a _{c} =v ^{2} /R =(10^{4} )^{2} /10^{3} =10^{5}
m/s^{2} . This is ten thousand times the acceleration due to gravity
which means that it would take a force of 10,000 times your weight to move
you in this circle. I would hate to see you after this maneuver, certainly
you would not be recognizable.

QUESTION:
can you help with a discussion about how it is possible for heavy aeroplanes to take off and fly. We are aware of two principles - the aerodynamics principle in which the wing (and sometimes the fuselage itself) provides the lift, and the old principle of "speed defies gravity". were early pioneers such as the wright bros. aware of BOTH principles, or was the "speed" principle used alone to begin with, then the aerodynamics principle discovered and used to improve? Also, do helicopters use both principles or the aerodynamic principle only?
There must be other principles as well, for example when thrust alone is used for a vertical take off craft, but we are mainly interested in the use and "time scale use" of the two i have asked about.

ANSWER:
"Speed defies gravity"? What is that supposed to be? I have
discussed how an airplane works in some detail in an
older answer . There are two
important factors in providing lift for an airplane, Bernoulli's principle
(which I guess you call the aerodynamics principle) and Newton's third law
as you will see in the earlier answer. When you refer to speed defying
gravity, are you suggesting that the thrust of the engines lifts the
airplane? This thrust is much less than the weight of the airplane (trying
to lift the airplane off vertically would be futile) so it is a small factor
in lifting the weight (unlike a big rocket which lifts the load this way).
The Wright brothers were the first to design and build wind tunnels and
therefore experiementally designed the shape of the wings to provide lift.
Helicopters are much like airplanes except they move the wings through the
air rather than move the air over the wings. That is why airplanes are often
called fixed-wing aircraft ; helicopter rotors are actually wings.

QUESTION:
I want to ask about weight shift in vehicle. Weight shift is a very common terminology among car enthusiast (but most of them not physicist or engineer, some might not even study science well). Magazine always comment an accessory will help in weight shift of the car, etc.
The common qualitative statements made by them in the use of weight shift are: (i) When a car turn, the weight shifts to outside wheels more; (ii) When a car accelerates, the weight shifts to the rear; (iii) when a car brakes, the weight shifts to the front.
The above is just some statements to clarify what I meant by the term "weight shift".

Here is my question:
Is "weight shift" a misnomer, or an ill concept of physics (vehicle dynamics to be specify)? The weight of an object is determined by the gravitational pull, regardless of what motion the object is doing, so how can the weight of different sections of the car change during maneuver? I think if one was to say force acting on different sections of the car during maneuver is different, then it is correct; but to say the weight shifted during maneuver, I can't get it.

ANSWER:
Technically, weight shift is a misnomer, but it is not too bad
because it is a shift in apparent weight. Normally, we think of
weight as acting at the center of gravity of something, but if that something
is accelerating, it may appear to move elsewhere. I will try to explain. In
all three of your examples, the qualitative observations are essentially the
same. For the accelerating car, the "apparent weight" shifts toward the rear wheels, opposite the direction of the acceleration . For the braking
car, it is just the same as the accelerating car except that the direction
of the acceleration vector is toward the back, the "apparent weight" shifts
toward the front wheels, again opposite the direction of the
acceleration. For the car turning there is an acceleration also, this
time toward the center of the circle (called the centripetal acceleration)
and here the "apparent weight" shifts toward the outside wheels, again
opposite the direction of the acceleration . So, if I understand one of
these I will understand them all. In every case, the force which causes the
acceleration is the static friction between the tires and the road.
The
figure to the right shows a generic case. The center of gravity is at the
point marked x . The forces labeled are the weight W ,
the force of the road up on the left tire N _{1} , the
force of the road up on the right tire N _{2} , and the
static friction force of the road on the tire f . If the car is
at rest or moving with constant speed, f= 0, W=N _{1} +N _{2} ,
and N _{1} =N _{2} ; this assumes the center of
gravity is halfway between the wheels, but if the center of gravity were
moved to the left, we would have found that N _{1} >N _{2} .
The three scenarios you mention can all be represented by this figure. If
the car is accelerating, it is moving to the right and speeding up and the
left wheel is the rear wheel; here f=ma where m is the mass of
the car and a is its acceleration. If the car is braking, it is
moving to the left and slowing down and the left wheel is the front wheel;
again, it is the friction which causes the car to accelerate (slow down). If
the car is rounding a curve, the center of the curve is to the right, and
the left wheel is the outside wheel. We still have W=N _{1} +N _{2} ,
but now the two forces N _{1} and N _{2} are not
equal. Since the car is not rotating about its center of gravity, if we sum
the torques about x it must sum to zero:(N _{1} L /2)-(N _{2} L /2)-fh =0,
so N _{1} =N _{2} + 2fh /L . So,
N _{1} >N _{2} , it appears as if the
weight (center of gravity) has shifted toward the left. So, weight shift is
a pretty good name for it, I would say; apparent weight shift would have
sounded a bit pedantic for car buffs, don't you think? (By the way, I drew
the friction only on the left wheel but if there were friction on the right
wheel instead or in addition, the conclusions would all be the same because
the torques from the right wheel friction would have been the same sign as
the left.)

QUESTION:
i know what mass is in equations, and i could define it that way but in actuality what is mass?

ANSWER:
There are two ways you can think about mass. One is inertial mass,
the property an object has which causes it to resist being accelerated; the
more mass something has, the harder it is to accelerate. The other is
gravitational mass, the property an object has which allows it to feel or
cause gravitational forces. It turns out that they are the same thing, which
is sort of the basis of the theory of general relativity, our best theory of
gravity to date. If you want to know why objects have mass, that is a
different issue; it is thought that the thing which is responsible for
endowing particles with mass is called the Higgs boson, currently being
anxiously searched for at particle accelerators.

QUESTION:
While reading " A brief history of
time by stephen Hawkings, i found that he explained something about spin
0,1,2. I wanna know that how is spin 0,1,2 different from Spin 1/2 and
how does pauli exclusion principle explains it?

ANSWER:
All these numbers are spin quantum numbers. If the spin quantum
number of a particle is s , then the intrinsic angular momentum of
that particle is S = {√[s (s +1)]}h /(2π ).
Intrinsic means that this is not due to some orbital motion, but is
associated with even a free particle. It is sort of like the earth whose
orbital motion around the sun and rotational motion about its axis are
different things; particle spin is sort of like the earth's rotational
motion about its axis. Particles with half odd-integer spin quantum numbers
(1/2, 3/2, etc .) are called fermions and obey the Pauli exclusion
principle; examples are electrons, protons, and neutrons (all with s =�).
Particles with integer spin quantum numbers are called bosons and do not
obey the Pauli exclusion principle; examples are the pi meson (s =0),
photon (s =1), and the graviton (if it exists, expected to have s =2).

QUESTION:
Assuming an object is in motion and traveling through space under its own inertia, let�s assume you stop it. In order to stop it you apply a force which results in a negative acceleration of the object causing it to stop. But since a force is met by an equal and opposite force, what is the force of the moving object?

ANSWER:
Newton's third law says that if you exert a force on the object, the
object exerts an equal and opposite force on you. But that force is on you,
not the object, and so it has no effect on the object.

QUESTION:
I have a shipping container that weighs 8000lbs. I am needing to lift it up to do some digging underneath it (to level it out).
I have a 2 1/4 ton jack (4500 lbs) that I am wanting to use to lift the container from one end at a time. My question is how much weight am I actually lifting (from the one end) as opposed to the full 8000 lbs?

ANSWER:
The crucial consideration is if the weight is uniformly distributed
in the container. In other words, could you balance it on a point in the
middle? If the answer is yes, you should be ok as long as your jack lifts
straight up vertically. It will hold about 4000 lb. If the weight is not
uniformly distributed, then when you lift the end where more of the weight
is, you may exceed 4500 lb.

QUESTION:
I have often heard of Newton's first law being referred to as the "law of inertia" but wouldn't it be much more appropriate to call the second law the "law of inertia?" The fact that a body with no forces acting on it maintains a constant velocity really just seems to establish the concept of an inertial reference frame and is true for objects of any mass. The concept of inertia seems to be represented by the second law by describing the resistance of matter to acceleration.

ANSWER:
Think about what inertia means in everyday life. If we say I have a
lot of inertia, it means that I am unlikely to change my mind if not pushed.
Anyhow, it is just semantics, so why worry about it? I agree with you that
one interpretation is that the first law is important in that it defines an
inertial frame of reference, not as a special case of the second.

QUESTION:
How do I know when a scenario is an example of Newton's first or second law? For example, when a bicycle loses speed as it coasts uphill. Is that straight line motion but the other forces acting on the bike are friction and gravity, so it doesn't remain in straight line motion? Or is it that the net force, including friction and gravity, results in acceleration in the same direction?

ANSWER:
Newton's first law is appropriate for an object moving in a straight
line with constant speed . The bike losing speed is moving in a
straight line, but not with constant speed, so Newton's second law is
appropriate. Newton's first law means that if the velocity (magnitude and
direction) is constant, the sum of all forces on the object is zero. If the
velocity is changing (acceleration), the sum of all forces (taken
vectorially) equals the mass times the acceleration.

QUESTION:
I am using "Shockwatch" indicators to measure the impact of baseballs on a helmet. I am comparing the impact of a baseball with and without a helmet. These are impact indcators from 10G to 100G. How do I make the data meaningful converting the G's to _________. Would it be newtons of force? I don't know. I am a 7th grade life science teacher helping one of my students.

ANSWER:
The "shockwatch" is likely just an accelerometer which measures
average acceleration of something over the impact time. The "G" refers to
the acceleration of gravity which is 9.8 m/s^{2} =32 ft/s^{2} ;
this means that a freely falling object speeds up by 9.8 m/s as each second
ticks by. So, if the baseball stops in 0.01 s and is initally going with a
speed of 60 mph=27 m/s, the average acceleration is 27/0.01=2700 m/s^{2} =2700/9.8=275.5
G. To convert this to a force, you need to know the mass of the object. The
mass of a baseball is about 0.14 kg, so the force over the impact time is
F=ma =0.14x2700=378 N=85 lb. You probably have your shockwatch attached
to the object you are hitting, so you need to know its mass if you want to
know the force on it. So, if you measure an "impact" of 50 G, and the mass
of the head is 4.5 kg (10 lb), then the average force is 4.5x50x9.8=2200
N=495 lb.

Important conversions: 1 N (Newton)=0.225 lb (pound), 1 kg (kilogram) weighs
2.2 lb

QUESTION:
If you took a waterproof bathroom scale that was tared correctly to register zero on land and dropped it into 30m water, would the weight of the water register on the scale?

ANSWER:
A scale is essentially a spring which indicates weight by the amount
of compression. So, think of a spring in a flexible box. If you took it down
to some depth there would forces on all sides. But, the top force would be
down and equal to the weight of water above it and the force on the bottom
would be up and equal to the weight of the water above it. So the spring
would compress just as if a column of water of cross section equal the the
area of the scale were being weighed. (I have neglected the thickness of the
scale; the forces on the top and bottom are not quite equal being at
slightly different depths.)

QUESTION:
at one point in Nova telecast "What is space?" Brian Greene explains that gravity can, as a result of Einstein's insights, be viewed as a consequence of spacetime geometry rather than as a force. Why then do physicists devote so much effort in treating gravity as a force; moreover, as force that must be accounted for in any successful "theory of everything"?

ANSWER:
I have previously
answered this question.

QUESTION:
I was wondering how it is possible to see some of the first stars and galaxies in the young universe? I understand that we are seeing light from them because the light has taken billions of years to get here but I don't understand how the atoms that make up our bodies and our galaxy got here from ancient stars who's light is just now getting here. That would seem to suggest that matter traveled faster than light.

ANSWER:
The heavy atoms from which you are made were, indeed, manufactured
inside stars long dead. However, those stars were not among the most distant
in the universe, they were stars in our own neighborhood, our own Milky Way
galaxy.

QUESTION:
Why is it difficult to calculate the terminal velocity for a cat falling from a high roof top?

ANSWER:
I do not know what you mean "difficult to calculate". We can
estimate it pretty easily, but certainly not do it precisely. First of all,
any calculation having to do with air friction is going to have
approximations and assumptions. For something like a cat, roughly 2 kg (4.4
lb), falling, it is a very good approximation to say that the drag force is
proportional to the square of the velocity. It turns out that a fairly good
approximation for the force is F= �Av ^{2}
where A is the area the falling object presents to the onrushing wind
and v is the velocity (this is only for SI units). Since it depends
on A , it depends on how the cat orients itself: if in a ball he will
fall much faster than if all spread out. Suppose we take the area of a
falling cat to be about 20 cm x 40 cm=0.08 m^{2} . Then the force
will be about 0.02v ^{2} . Now, the cat's weight is about mg =2x9.8≈20
N. When the force of air friction is equal to the weight force down, the cat
will fall with a constant velocity called the terminal velocity: 0.02v _{t} ^{2} =20,
so v _{t} =√(20/0.02)≈30 m/s=67 mph. If you google "terminal
velocity of a cat" you will find the number 60 mph, so my approximations
were evidently reasonable. There, now, that wasn't so difficult, was it?

RELATED QUESTION:
I was asked what the terminal velocity of an unladen sparrow is. I read that the average weight of a field sparrow is .5 ounces. Approximately 5 inches in size with a wingspan of 7.9inches. Lets say falling from 50ft?

ANSWER:
Look at the answer above. If the sparrow falls straight down with
wings stretched out, I would estimate his area to be
�x5x7.9=20 in^{2} =0.013 m^{2} . Following the same as for
the cat, but using m =0.5 oz=0.014 kg, I find v _{t} ≈6.6
m/s=15 mph=22 ft/s.

QUESTION:
Hello, could you tell me about the relative speed of a plane traveling let's say 400 mph. How fast would it appear to be moving from the ground at different altitudes 3000 ft compared to 30000 ft.

ANSWER:
How fast something "appears to be" is meaningful only in comparing
it with something else. To compare apparent speed at 3000 and 30,000 ft I
would compare their angular velocities, that is how long does it take to
traverse some angle, in particular compare the time for each to move from
one point in the sky to another. The distance the higher plane would have to
travel to have the same angular displacement is ten times as far, so it
would appear to be going ten times slower.

QUESTION:
According to the molecular biology book I am reading, when a chemical bond breaks, the energy in the bond is released in the form of heat.
which is to say, when a chemical bond breaks the entire atom's speed increases (the atom gains 'heat energy').
But what is happening at the electron/proton level to make this happen?
I know that when some bonds break they are 'exothermic' (give off heat-- ie,make the atoms breaking apart move faster) and when other bonds break they are 'endothermic' (absorb heat-- ie, make the atoms breaking apart move slower) but what is actually physically happening?

ANSWER:
You are talking about chemistry here, not physics, but chemistry is
really just atomic/molecular physics so I will comment. You have some
misconceptions here which need to get cleared up. Suppose you have a
molecule, say O_{2} . These two oxygen atoms are bound together and I
guess a chemist says there is a "bond" between them. If you simply break
this bond, you will never get energy (exothermic), you will always have to
add energy (endothermic). If you did not have to add energy to make O_{2}
be O+O, why would the molecule stay together on its own? This is true for
any stable molecule �it is going to cost
you energy to break it apart. You need to create a bond, not break one, to
get energy out. So, if you start out with O_{2} and are able to
cause a carbon atom to attach to it, you get energy released: O_{2} +C�>CO_{2} +energy.
You have discovered fire! More complicated molecules might have the property
that if you break a bond you make the others stronger and so that could
release energy; it will always cost energy to break a bond, but if the bonds
in the new molecule have increased by more than your cost, there is a net
gain.

QUESTION:
If you give a object at rest a push up a ramp, does it take longer, the same, or less time to slide back down to the same position? Assuming there is sliding friction. I would think the same due to conservation, but then I started thinking about the influence of gravity and friction and am now, I'm just confused.

ANSWER:
Energy is not conserved here because the friction is taking energy
away. (Actually, that energy ends up in heat, which is why the object and
ramp will be slightly warmer afterward.) To answer your question, think
about forces. There are two forces which affect the motion along the ramp,
the frictional force which is always opposite the direction the object is
moving, and the component along the ramp of the object's weight which is
always down the incline. So, on the way up these two forces both point down
the incline and on the way down the weight points down and the friction points up.
Therefore the net force slowing the object down when going up is greater
than the net force which is speeding the object up when it is going down.
Therefore the time going down is longer than the time going up. Of course,
if the friction is big enough it well never go back down and, certainly,
infinite time is longer than however long it took to go up.

QUESTION:
Who are one of the five con-men who believe or proclaim they
have found ways around the law of conservation of energy? Just would like
to know one of them and what they believed to work. Thanks physicist!

ANSWER:
What are you talking about?

FOLLOWUP QUESTION:
Was it aristotle , socrates, maxwell, or someone else who believed that they found a way around the law of conservation of energy ?

ANSWER:
Well, the notion of energy was not proposed until long after
Aristotle or Socrates were around, so we can rule them out. I never heard of
Maxwell thinking energy conservation might not be true. Niels Bohr proposed
that perhaps energy was not conserved in nuclear beta decay, but then the
neutrino was proposed by Wolfgang Pauli, and so energy conservation was
saved. I guess we should state what energy conservation is: the total energy
of a system remains constant as long as no external forces do work; or, put
more simply, the total energy of an isolated system never changes. However,
the Heisenberg uncertainty principle allows for energy to be not conserved
as long as it is only for a very short time; the total energy E of an
isolated system may change by an amount
ΔE but for no longer than a time Δt such that ΔE Δt<h/2π ,
where h is Planck's constant, h = 6.6�10^{-34} m^{2} kg/s.

QUESTION:
Galaxies recede from us at a velocity proportional to their distance. Why then is Andromeda on a collision course with our beloved Milky Way?

ANSWER:
The overall average behavior is that the universe is expanding.
However, at distances much smaller than the size of the universe there is no
fundmental reason why everything has to be moving apart. For example, the
sun is not moving away from us. And, there are attractive gravitational
forces which can be important between nearby galaxies. Astronomers have
observed many examples of galaxies colliding. The Milky Way and Andromeda
galaxies will collide in about 5 billion years. You can see a simulation of
that collision at
this link .

QUESTION:
I understand that the power that a stream of moving water can generate or develop is proportional to the velocity of the water current, cubed. Can you explain where this comes from?

ANSWER:
I never saw this before, but I think I can figure it out. The power
P generated by a force F may be written as P=Fv where
v is the speed of the agent applying the force (the water here). Now,
if an object moves through a fluid, it experiences a drag force which, for
most cases of interest, is proportional to the square of the velocity,
F=Cv ^{2} . If the fluid exerts that force on the object moving
through it, then if it is the fluid that is moving instead of the object,
the force must be the same. Therefore, P=Cv ^{3} .

QUESTION:
what is the upward force in pounds on the bottom of an empty
basin caused by a groundwater depth of 8 feet above the tank bottom? The
basin has a base of 300 square feet. What is the formula? This
is in regards to not emptying a tank fully for repair as the upward
force may displace it from its resting place.

ANSWER:
The key here is to find the pressure at a depth d (8 ft in
your case) below the surface of a liquid of density
ρ. This is given by P =ρgd where g is the
acceleration due to gravity; when you know the pressure, you can find the
force F on the bottom by multiplying the pressure times the area A
(300 ft^{2} in your case), F=PA =ρgdA . So, there
is your formula. You are probably wanting to do this in English units, but
physicists hate to work in English units. I did it in SI units where ρ= 10^{3}
kg/m^{3} , g =9.8 m/s^{2} , d =8 ft=2.44 m; so I found P =2.39x10^{4} N/m^{2} and converted
to P =499 lb/ft^{2} . So, F =499x300=149,700 lb. Actually, I
found that it is pretty easy to do this in English units: the weight
density of water is ρg= 62.4 lb/ft^{3} and so F =62.4x8x300=149,760
lb.

QUESTION:
If the LHC accelerates particles almost to the speed of light, why don't they get huge and approach infinite mass?

ANSWER:
They do approach infinite mass but really have a long way to go yet.
See an earlier answer .

QUESTION:
Lets say its 25,000 light years to the center of our galaxy. Meaning, light taking off from earth today would take 25,000 years to get to the center of the galaxy, or vice versa, from the center of the galaxy to us.
Let's say we had a ship that could travel at 99% the speed of light, and we board that ship and leave earth travelling at 99% light speed.
From earth, as we watched the ship travel, we would see that it took almost 25000 years to reach its destination.
But what about the perspective of those on the ship? How long would the journey seem to take for the travellers, moving at 99% the speed of light? How much time would have elapsed by their watches?

ANSWER:
The important factor in relativity is
γ= 1/√[1-(v /c )^{2} ]. In your case, γ= 1/√[1-(0.99)^{2} ]=7.1;
so the traveler sees the distance d he has to travel to be shortened,
d' =25,000/γ= 3520 ly. So, the time it takes is t'=d' /v =3520/0.99≈3560
yr. You could also get this from the time dilation formula, t'=t /γ= (25,000/0.99)/γ= 3560.

QUESTION:
Hey dude,
With the endless amount of stuff floating around in space, how come none of it hits earth?
And to be more specific, how come nothing so massive and unstoppable by earths defensive capabilities ever comes towards earth and destroys it?

ANSWER:
Hey Dude back at ya! It happens all the time— what
do you think shooting stars are? And, tons of space dust hit the earth every
year. In the early solar system, there were a lot more pieces of stuff
"floating around" and there were many more collisions with bigger objects
than there are today. One collision with an asteroid-sized object is thought
to have caused a major extinction of life millions of years ago. But most
remaining objects are in reasonbly stable orbits now. And, you just have no
idea how mostly empty space is—just statistically it is immensely improbable
that collisions will occur. Imagine two guys with rifles 5 miles apart
firing their rifles; what is the likelihood that two of their bullets will
collide? There was one major collision in 1994 when Comet Shoemaker-Levy hit
Jupiter. Many asteroids have highly variable orbits which could possibly
result in a collision with earth, but it is a very low probability event.

QUESTION:
If you take the same exact object and drop it from different heights like 1,000 ft... 5,000 feet... 10,000 feet... and 100,000 feet, will it hit the ground harder from a higher height or does it not matter how high you are once the object reaches terminal velocity. My husband says if you drop a rock from 10,000 feet it will do more damage than if you dropped that same rock from 3,000 feet. Is that true?

ANSWER:
If it reaches the terminal velocity before it reaches the ground,
then it makes no difference what height it was dropped from. Just as a rough
example, consider an object with a cross sectional area of about 0.04 m^{2}
(like a 9" diameter ball) and a mass of 1 kg (about 2 lb). It would have a
terminal speed of about 30 m/s (about 70 mph) and a characteristic time of
about
τ =3 s.
What the characteristic time means is that if the rock falls for a time much
greater than the characteristic time, it will have achieved terminal
velocity; e.g., if it falls for 3 τ= 9
s it will have achieved 99.5% of the terminal velocity. From 3000 ft it
would fall about 14 s if there were no air
and from 10,000 ft it would fall about 25 s. So both will certainly fall
much longer than the characteristic time and therefore both would hit the
ground with a speed of 30 m/s and have exactly the same effect.

QUESTION:
Please, can you explain me the process of how an electron can get accelerated with a magnetic field (like in a particle accelerator like CERN, by instance) ? In particular, from where do electrons get the increase of momentum: is the absorption of photons involved (since the EM fields work on the interchange of photons)?

ANSWER:
The accelerator (LHC) at CERN does not accelerate electrons, it
accelerates protons. There are many electron accelerators, though. But, here
is an important point: a magnetic field
cannot accelerate an
electric charge. Accelerators always use electric fields to speed the
charged particles up. The fact is that a magnetic field cannot do work
because the force it exerts on a charged particle is always perpendicular to
the particle's path so it can change the direction of the particle's
velocity, but not its speed. The reason you are thinking that magnets are
responsible is that there always so many magnets in an accelerator
laboratory. These are used to steer the beams of charged particles, like in
the LHC the many magnets keep the beam circulating in a big circle.

QUESTION:
How many Newtons are exerted when a 300 lb. man falls 3 ft.?

ANSWER:
Read faq page.
This question has no answer.

QUESTION:
We are working to produce a safety harness and the strap material we are using has a maximum Newton rating - we were trying to get an idea of what Newton rating would be needed to support a 300 lb. man if he fell 3 ft. Being hunters (tree stand safety harness) - perhaps we are wording the question incorrectly. Can you clarify your response?

ANSWER:
What matters is how long it takes the falling guy to stop. The mass
of a 300 lb guy is about 130 kg, the acceleration of gravity is 10 m/s^{2} ,
and so the weight of the guy is about 1300 N. You need that strong a strap
just to hang him there at rest. If he falls 3 ft (about 1 m) he will be
going about 4.5 m/s. So, let's call F the force needed to stop him
and t the time it takes him to stop; I reckon that F ≈130(10+(4.5/t )).
For example, if he takes � s to stop, F ≈3600 N to stop him. The
straps are probably pretty unstretchy, so your best bet would be to make the
harness out of a stretchy material because, don't forget, the bigger F
is the more it is going to hurt during the stop.

QUESTION:
is proton diffraction is possible?

ANSWER:
Any particle, including a proton, will be behave like a wave if you
look for it and diffraction is possible for all waves. In fact, I can give
you an example of proton diffraction which I myself measured. What is shown
in the figure to the right is the differential cross section for 800 MeV
protons elastically scattered from the nucleus ^{90} Zr plotted as a
function of the angle (in degrees) where the protons were observed.
Differential cross section is, essentially, the probability that the proton
(wave) will scatter (diffract) to some angle. This is very much like the
diffraction pattern you would observe for visible light striking a sphere �diffraction
maxima and minima. In fact, you can qualitatively understand this
diffraction pattern if you calculate the wavelength of the protons, λ=h /p ≈10^{-15}
m, and approximate the positions of maxima by the double slit relation
nλ=d sinθ where we take d to be the diameter of the
nucleus. Taking the two consecutive maxima at 8^{0} and 13^{0} ,
λ=d (sin13^{0} -sin8^{0} )=0.08d =10^{-15} .
So, the diameter of the ^{90} Zr nucleus would be about 12.5x10^{-15}
m=12.5 fm. As a check, the diameter of a nucleus with atomic weight A
is well approximated as d ≈2x1.25xA ^{1/3} fm, so for
A =90, d ≈11.2 fm, pretty good agreement for such a rough
calculation. The experiment was done at the Los Alamos Meson Physics
Facility (LAMPF).

QUESTION:
I had an argument with my roommate I was hoping to resolve. Basically, can a dart bounce off of the metal rims of a dart board past the point from where it was thrown, say 8 feet. My point was that if the force is strong enough (enough force to go 50 feet w/o obstructions) when it hit the board there would be 42 feet (I dont know if this is correct) of force left over to push back. This would be enough to push it back further than the 8 feet. He seemed to believe that due to the angle of the collision, it would bounce off and not make it.
I gave an example that from 1 feet it would surely bounce back far enough, but he didn't seem to agree. Is there a physics law or example that can explain.
Note: Dart bars are curved, so a dart could come down on an angle and hit the top curve of the metal bar and get a different angle. But we were also arguing that even if the wall was flat, it could still happen.

ANSWER:
A very light object (dart) can bounce off a very heavy object (board
attached to wall) with a speed about equal to the speed it came in with if
the collision is elastic. You may know that the maximum range for a
projectile is achieved (neglecting air friction) if it starts out at a 45^{0}
angle relative to the horizontal. Suppose the dart is thrown with an angle
of say 10^{0} , so that it comes back to the same level it was thrown
from when it gets to the dart board. Then, if it was thrown at that same
speed but at 45^{0} , it would clearly go farther (assuming the board
and wall weren't there), right? So, if it bounces elastically off the
"bar" at an angle of 45^{0} it would go back farther than it came in
from.

QUESTION:
Can you speed up the orbital speed of a electron around the nucleus?

ANSWER:
Technically, an electron does not have an orbital speed. The Bohr
model of the atom, the "planetary model" is only a rough approximation as to
what is actually going on. The best way to think about the electrons is a
cloud around the nucleus. What you can do, equivalent to speeding up the
orbital speed, is add energy to an atom which then places the electron in a
higher energy state. This can only be done in discrete steps, not in any old
amount you might wish to add.

QUESTION:
We're having this physics discussion at work (in the medical field). One of my co-workers insists that if you took a pound of anything and spread it out evenly on a scale, it would weigh less than if you piled it up in the middle of the scale. He said that the density of the mass of whatever you piled on the scale, would increase the gravitational pull moreso than if the object being weighed was spread out evenly, like say mashed potatoes, for example. What do you say?

ANSWER:
The stock answer to this question is that weight is the force that
the earth exerts on a mass and that force is independent of how the mass is
distributed. But, that is not quite right and your question is clearly a
"hair-splitting" sort of question, so I will explain. The gravitational
force between two objects depends on how far apart they are �the
farther apart, the smaller the force. The weight of something is determined
how far it "is" from the center of the earth. If you compare the mashed
potatoes spread out with the mashed potatoes heaped up, those heaped up are,
on average, farther away from the center of the earth by a few centimeters,
and therefore weigh less, not more. But, don't get too excited about
this�the difference would be about one millionth of 1%, less than you could
ever hope to measure. You would get a much bigger difference if you weighed
the same thing upstairs and downstairs in your home, again something you
know from experience is, for all intents and purposes, the same. [Rereading
the question, let me address the density argument of your colleague. It is
true (again to a really small degree) that the weight of the potatoes on the
top of the heap will press down on those on the bottom and make the potatoes
slightly more dense. However, a scale does not measure density, it measures
mass and therefore two objects with different densities but equal masses
would weigh the same.]

QUESTION:
Why do electron microscopes get better pictures than regular light microscopes?

ANSWER:
The problem with optical microscopes is that you cannot look at
anything which is comparable to or smaller than the wavelength of the light.
Visible light has wavelengths of a few hundred nanometers, ~6x10^{-7}
m, so what you are looking at has to be bigger than that, a few microns
maybe. The reason is diffraction, light does not just travel in nice
straight lines when obstructions or aperatures get to be on the order of the
wavelength, e.g . light can be bent around a corner. If light does not
go in nice straight lines, geometrical optics doesn't work. On the other
hand, we know that particles like electrons can behave like waves and their
wavelengths are inversely proportional to their momentum. So, if we make an
electron go fast enough it will have a wavelength much less than light and
therefore let us "see" much smaller things.

QUESTION:
A fly is hovering in a car. the car is going 45 MPH with the windows up. the car hits a solid object and suddenly stops. Does the fly hit the front window, the back window, or nothing at all?

ANSWER:
This is tricky. The stock answer would be that the fly would smash
into the front windshield just like you would if you weren't wearing your
seat belt. But, the fly is actually hovering with respect to the air, and so
whatever happens to the air will be what happens to the fly. The air does
not all smash into the windshield but more or less stays just where it is
relative to the car. The reason for this is that air is a collection of
molecules most of which are already going much faster relative to the car
than the car is relative to the ground. An average air molecule is going
about 1200 mph. Think of half of them going (relative to the road) toward
the back with speed like 1155 mph and half going toward the front with speed
like 1245 mph, so their average is still 1200. When the car stops, the fly
sees no significant change and just continues hovering where he is. He has
such a small mass that he does not have enough inertia to overcome the air
drag he would be experiencing and hit the windshield.

QUESTION:
Has there ever been a hypothesis that dark matter and or dark energy are forms of gravity? Does anything in string theory allow for this?

ANSWER:
Read my FAQ on dark
matter/energy. You will see that my own feeling is that so-called dark
matter may simply be symptomatic of the possibility that we do not
understand gravity as well as we think we do; I seldom see astrophysicists
or cosmologists adopt this view, but it is an open possibility until
somebody actually directly observes "dark matter particles". Dark energy can
be "understood" in terms of general relativity; Google cosmological
constant . String theory allows anything and predicts nothing.

QUESTION:
how does ac currrent move fromone place to another, if is goes back and front.

ANSWER:
It does not "move from one place to another".
For that matter, dc current almost does not either. The electrons move so slowly that it might take a day for electrons to move from
battery to light bulb. The important thing is that all the electrons in the wire are moving almost instantaneously when you close the switch,
ac or dc.

QUESTION:
Picture a perfectly aligned row of pool balls all touching each other that was 200,000 miles long. If the first ball was struck with enough force at the exact center of the ball, would the movement of the last ball occur before a laser beam could move from the beginning to the end of the balls?

ANSWER:
It would occur much later than the arrival of the light. The
disturbance down the row of billiard balls travels with about the speed of
sound. And, an important law of physics is that nothing can travel faster
than the speed of light. See FAQ page.

QUESTION:
The sun produces enough gravity to hold the earth in orbit. So why dont space shuttles get attrcted towards the sun?

ANSWER:
It does. But the gravitational force from the earth is much bigger
because, although earth's mass is much smaller than the sun, the shuttle is
much closer to the earth. So the motion of the shuttle is much more
determined by the earth than the sun.

QUESTION:
What is the "power" of gravity? i.e. how much force per second does it apply?
Power = F x V, = m x g x V, but for V = 0, what power is required to make an object hover?
A practical question might be if I have an electromagnet, how much power does it need to hover 1kg of metal?

ANSWER:
Force per second is not power, energy per second is power. To levitate something you must exert a force up on it equal to its own weight. To levitate a 1 lb object you must exert, somehow, a 1 lb force up for as long as you want it to levitate. No energy is consumed because the force does no work. Of course, if you are using an electromagnet, you must use power, but that is a feature of the design of the electromagnet, that work essentially goes into heating the coils of the magnet. If you had an appropriate permanent magnet, it could levitate something forever with no expended energy.

QUESTION:
I am doing an experiment with dropping coffee filters. By changing the
mass (adding more filters) and keeping the distance the same (4 meters)
and recording the time it takes for the stack to hit the ground. By
doing this I find the velocity. Wouldn't there be an asymptote at 9.8,
for the velocity of the dropped object?

ANSWER:
Technically, you are not measuring the velocity, you are measuring
the average velocity. But, since coffee filters have such small mass, you
can probably assume that terminal velocity is reached almost immediately
after dropping. What you are studying is the fall of objects for which air
resistance is important. I do not know what you mean by the asymptote at
9.8, but it is certainly wrong because 9.8=g is the acceleration due
to gravity, it is not even a velocity. What you want is to find out how
velocity depends on mass. This will depend how the terminal velocity depends
on mass. Now, the force of air drag will depend on two things, the
geometrical size of the falling object (but all will be the same, so you
cannot study that) and the speed the thing is moving. Usually this force can
be parameterized by F =Cv ^{2 } where C is some
constant determined by the fluid (air) and the geometry, but not the mass.
So, the faster it falls the bigger this force is and when it gets as big as
the weight of the object it stops accelerating and falls with a constant
speed called the terminal velocity which you hope to be determining. So,
drag=weight when mg =Cv ^{2 } where m is the mass
and g =9.8 m/s^{2} . Solving, v = √(mg /C );
so, if you make a graph of v vs. √m you should
get a straight line if indeed the force is a quadratic function of the
speed. And, from the slope, you could determine the constant C . If
you do not get a straight line from your data, try using
F =Cv ; then if you plot v vs. m you will find a
straight line.

QUESTION:
If I could detonate a firework inside a contained vacuum void of any other mass objects (simulating the big bag) and watched (over time) would I see the particles eventually attract to each other? Would the results of this sort of experiment prove gravity exists in every particle ever created?

ANSWER:
No. The gravitational attraction is very, very small between
the little pieces of your firecracker. So the speeds the pieces had when it blew
up would have exceeded the escape velocities from their neighbors and they
would continue moving apart forever. Escape velocity is the speed
something must have to escape from the gravity of something
else; e.g. the escape velocity from the surface of the earth is about 7 miles per
second, but this is for a much stronger gravitational force. Just to give
you another example where the mass involved is still much larger than your
masses but much smaller than the mass of the earth, the escape velocity from
the surface of a baseball is about 8 cm/hr.

QUESTION:
What is overpowering gravity and causing the outward acceleration of the universe?

ANSWER:
If I could answer that question, I would be next in line for the
Nobel Prize in physics. This year's prize went to the astrophysicists who
discovered this. The glib answer would have been dark energy , because that is
what astrophysicists call what is responsible for the acceleration. However,
this is just putting a name on something we do not understand.

QUESTION:
My question has to do with traction and the movement of a wheel(a wheel alone). Traction is essential for its movement both linear and circular. But if we throw a wheel forward it rolls some meters and then it stops(and falls). Which force is responsible for the decrease in its velocity? Cause if traction is parallel to the ground facing backwards then linear movement 's negative accelleration is explained but not angular negative accelleration. If traction is parallel to the ground facing forward then angular negative accelleration is explained but not linear. If traction is zero then which force decreases both velocities linear and angular?

ANSWER:
One of the reasons I love doing Ask the Physicist is because
I often learn things I did not know or had never thought about. You would
think that a guy who has been teaching introductory physics courses for
nearly 50 years would find this question simple. But, indeed I was puzzled
by it because, as I have found by thinking about it and talking to some
friends, I wasn't thinking beyond the friction force (which questioner calls
traction) being simply the only force in the horizontal direction and
obviously stopping the forward motion after some distance. I never addressed
the angular acceleration of the wheel before. This answer will be
long-winded because that is what I do when I have learned something which
pleases me! What frictional forces are important to understand the rolling
of a wheel? Most introductory physics classes talk only about the contact
forces of static friction and kinetic friction. Kinetic friction is not
applicable to this problem because the wheel is not slipping on the ground,
and static friction might
be important, but not necessarily. If we have a round wheel rolling on a
flat horizontal surface (don't look at the figure yet!), there are three
possible forces �the weight which must
be vertical, pass through the center of mass, and (assuming it is a uniform
wheel) pass through the point of contact; the friction, which must be
parallel to the surface and pass through the point of contact; and the
normal force which must be perpendicular to the surface and pass through the
point of contact. If you now sum torques about the point of contact (as
noted by the questioner), there are none! So, there can be no angular
acceleration; if we have stipulated that the wheel does not slip, then there
can be no linear acceleration either and the wheel will roll forever and no
friction is required. But we all know better! A real wheel will eventually
slow down. The key is that there is no such thing as a perfectly round wheel
or a perfectly flat surface, one or both must be deformed. In that case, we
have to think about a new kind of friction called rolling friction, the
friction the wheel has because of the rolling. This is different from the
static friction, and static friction may still be present still to keep the
wheel from slipping.
A perfectly round wheel cannot have rolling friction as
I showed above, it must deform which means that there is no longer a "point
(or line) of contact" but now an area of contact. Since the normal force is
only constrained to act somewhere where the two are in contact, it is now
possible (in fact inevitable) that this force will not act through the
center of mass of the wheel. That is the whole key to answering this
question. So, finally, the answer: refer to the figure where I have drawn
the forces mg , N , and f . The
weight is still constrained to be vertically down and pass through the
center of mass (blue cross). The normal force is constrained to be vertical
and act somewhere where the wheel and ground are in contact, drawn a
distance d to the left. The frictional force (which now includes both static
and rolling friction) is constrained to act at the surface and parallel to
it. I choose a coordinate system with x to the left and y up;
the axis (red cross) about which I will sum torques is at the ground
directly under the center of mass and positive torque results in an angular
acceleration which is positive when acceleration of the center of mass is
positive (counterclockwise around the axis). All is now straightforward: ΣF _{x} =-f=ma ,
ΣF _{y} =N-mg =0, Στ _{x} =-Nd=Iα=Ia /L
where I is the moment of inertia about x
and L is the distance from x to
x . Finally, N=mg , a=-f/m ,
and d =fI /(Lm^{2} g ).

Finally, a couple of real-world provisos. Of
course, N is really distributed over the whole area, but the
dynamics can be done by assuming it effectively acts all at one point just
like we assume the weight acts all at the center of mass. And, the rolling
friction might not really act at the surface of contact since it arises from
the deformation of the wheel and it might not be purely horizontal since it
is not directly a force due to the contact with the floor. So, there are
still some idealizations in my analysis, but there are always idealizations
when dealing with friction. And, the problem could have been equally well
done assuming the ground, not the wheel was being deformed. One could also
have done the analysis by summing torques about the center of mass and using
the parallel axis theorem I _{cm} =I-mL ^{2} .

I would like to acknowledge a very useful
discussion over pizza with friends and colleagues Edwards, Love, Meltzer,
and Anderson.

QUESTION:
This came to mind this morning as I was heating up my work coffee after putting milk into it. Here goes:
Given (1) a set amount of coffee at a set hot temperature; (2) a set amount of milk at a set cold temperature; (3) a set amount of heat energy to be applied to the final product, in the form of microwaving for a set amount of time, say 30 seconds; and (4) a mug at room temperature:
Which will lead to highest temperature of the final coffee/milk mixture:
A. Heating the milk first then adding the coffee to it, or B. Heating the milk+coffee mixture together?
Or, will the final temperature be the same regardless?

ANSWER:
I believe the final temperature should be larger for the milk mixed
with the coffee and heated together. The reason is that the microwave oven
will have a certain energy density inside and a larger volume should be able
to absorb more energy in a set time than a smaller volume. If you had simply
said add a certain amount of heat to the milk first or add that same amount
of heat to the milk+coffee instead, the final temperatures would be the
same.

QUESTION:
Why is the speed of sound slower in lead (25 degrees C) than fresh water when generally sound travels faster in solids than in liquids?

ANSWER:
The speed of sound is determined by two things, the elasticity of
the medium and the density
ρ
of the medium. How the elasticity is measured differs for liquids and
solids, for solids we use Young's modulus Y and for liquids the bulk
modulus B . So, the standard predictions are v _{water} = √(B /ρ )=√(2.2x10^{9} /1000)=1483
m/s and
v _{lead} = √(Y /ρ )=√(16x10^{9} /11.3x10^{3} )=1190
m/s. The experimental numbers are 1493 m/s and 1158 m/s, respectively, in
pretty remarkable agreement with the predicted values. So, let's get
qualitative. Lead is not at all elastic. Can you imagine how a bell made out
of lead would sound? How well would a lead ball bounce off a hard floor?
Also lead is very dense, so that also drags down the speed down. So, we
should not really be surprised that the speed of sound is lower than in many
other solids.

QUESTION:
What happens to ice-7 when it is exposed to conditions similar to the pressures and temperature of the Earths inner core?

ANSWER:
I don't really know what the pressures and temperatures are in
there. But, since you are interested, you can get all that info and then
look at the water P-T
phase diagram .

QUESTION:
I read that if two oppositely charged bodies of same size,mass etc. are stuck together and a third charge is kept far away from them no net attraction or repulsion takes place. But if the third one comes close enough attraction takes place. It is because they can "see inside" one another and rearrange their charges resulting to a very strong interaction. What is this "seeing inside" and "rearrangement of charges"? It would be better if a diagram related to answer is given.

ANSWER:
So, I am imagining two balls, one positive and one negative,
touching at their surface. This is referred to as a dipole charge
distribution; it has zero net charge but not zero charge distribution.
Obviously, if you get far away from this dipole, the field will be much
weaker than if there had been some charge there. However, your statement
that there is "no net attraction or
repulsion" is wrong: there if you are on the + side of the dipole and far
away a negative charge sees an attractive force, but it falls off like 1/r ^{3}
unlike the force for a net charge which would fall off like 1/r ^{2} .
As you get closer, the force gets stronger and whether it is attractive or
repulsive for a positive test charge depends on where you are. The electric
field for an electric dipole is shown in the picture.

QUESTION:
I am falling from a plane at terminal velocity. I have a ball in my hand. I throw it towards the ground. What happens? Does the ball accelerate away from me, or does it simply fall beside me? What about if it was a very heavy ball such as a cannon ball compared say to a baseball or a plastic 'WalMart' pool ball?

ANSWER:
Terminal velocity is determined by the geometry of the object, the
density of the fluid (air), and the mass of the object. The terminal
velocity in air for an object of mass m and cross sectional area A
can be roughly approximated by v _{t} ≈√(4mg /A )
(only in SI units).
You and the ball have some terminal velocity together. When you depart
from the ball you both have different terminal velocities since the masses
and geometries have changed. Suppose it was a cannon ball. Then your net
mass gets smaller and your geometry does not change much, so you will have a
lower terminal velocity and slow down; the ball has a considerably smaller
cross sectional area than it did when it was "part of" you and therefore a
larger terminal velocity, so it will accelerate down until it reaches its
new terminal velocity. Suppose it was a styrofoam ball. Then neither your
net mass nor your geometry change much, so you will have about the same
terminal velocity and continue falling at about the same rate; the ball has
a considerably smaller mass than it did and therefore a smaller terminal
velocity, so it will experience an upward acceleration and slow down until
it reaches its new terminal velocity, you will overtake it and it will
appear to go up (but is actually just going down more slowly).

QUESTION:
I have two questions regarding rotational kinetic energy. I know that
rotational kinetic energy is defined as:
KE = 1/2 * I * w^2
Where KE is the rotational kinetic energy [in units of: joules;
kg*m^2/s^2], I is the Moment of inertial [in units of: kg* m^2], and w is
angular velocity [in units of: radians/sec^2 or 1/s^2] w^2 denotes angular velocity squared
so that the units check.
If (Electron or Nuclear) Spin is defined as J, with values equal to a half
integer * h/2pi, where h is Plank's constant [in units of: Joule *seconds
or kg*m^2/s],
#1, Is the rotational kinetic energy associated with Spin:
KE = 1/2 *J *w ? (so the units check).
Since V = w * R, where V is velocity, and R is the radius (or the Electron or
Nucleus), so w = V/R, and since the fastest possible speed is the speed of
light, c,
#2, Is the greatest possible rotational kinetic energy associated with Spin:
KE = 1/2 * J * c/R ? (so the units check).

ANSWER:
You are really out in left field here. You are making the mistake of trying to understand quantum phenomena using classical ideas. Also, your quantum statements are basically in incorrect. For example, if the angular momentum quantum number is
J , the angular momentum is [√(J (J+1 ))]*h /(2 π ), not
J *h /(2 π ). Furthermore, if you try to look at electron spin classically by imagining a tiny rotating sphere you run into trouble first because you do not know the size or mass distribution, but even if you put in some reasonable guesses, you get absurd results for angular velocity or moment of inertia. Sometimes in nuclei which have rotational energy spectra it is useful to think about the moment of inertia of the nucleus, but only as a parameterization, not as a classical concept.
Also noteworthy is that a rotational nucleus is found to behave more like a
rotating fluid than a rotating rigid object. And, like any energy spectrum of a bound-state system, it is discrete, not continuous as in classical mechanics.

QUESTION:
If a motorcyclist is riding at 80 mph, is he facing or creating 80 mph winds behind him? My boyfriend insists he's facing 80 mph winds because his bike is moving at and 'hitting' air mass at 80 mph. I dont understand how his motion of mass would contribute to air mass in creating 'wind' in such precise and equal measurement?

ANSWER:
The concept here is called Galilean velocity addition. If you are in
a car going 60 mph north and there is another car going 60 mph south, you
see that other car approaching you with a speed of 120 mph. In this case,
think of the air as the other car and it is at rest relative to the ground,
that is it is a calm day; then if your velocity is 80 mph north, you see the
air approaching you with a speed of 80 mph going south. Or, try this: if you
look down at the ground, you see it going backwards, that is south, with a
velocity of 80 mph, and since the air and ground are indisputably at rest
with respect to each other, the air is also going 80 mph south to you.

QUESTION:
Why are only mass, length, and time changed as an object approaches the speed of light? Why not other "fundamental" properties of matter e.g., charge? Does this imply that charge is not fundamental or is somehow very different from mass, length, and time?

ANSWER:
The elegant way to put it is that some quantities are invariant
under a Lorentz transformation and some are not. This means that some
quantities are the same whoever measures them, some are not. Electric charge
is invariant and so is the total energy of a particle. It would be my
inclination to say that invariant quantities are more "fundamental" than
those that are not invariant.

QUESTION:
when a hand is in contact with the wall the distance between the hand and wall tends to zero then why not the gravitational force is infinity between the hand and the palm as F=GMm/r2

ANSWER:
Let us simplify the picture: you have two hard spheres which you
bring into contact with each other. The distance between their centers of
mass is still the sum of their radii, not zero.

QUESTION:
I took two cups of unequal size (one can hold a higher volume than the other), filled the largest one with water almost to the top, and left the other one empty. Then I took a twisted piece of paper towel and put one end in each cup. Capillary action caused water from one cup to go to the other and the other began to fill. My question is why did the empty cup only fill up until both cups were of equal height, rather than of equal volume.

SIMILAR QUESTION:
I've noticed that when steeping a cup of tea, the fluid will follow the string up and over the teacup lip and create a puddle around the cup, if I leave it steeping too long. What is this principle known as? I vaguely recall it from Jr. High, but am too old to remember it clearly and I can't manage to word the question properly to get a "Google" response to it.

ANSWER:
Capillary action uses the meniscus force to lift the water to the
highest point of your paper towel (string). Once it gets there, it can just flow
downhill. So, in essence, your piece of paper (string) just becomes a siphon.

QUESTION:
my question is
why is it that a projectile fired from a long-barrelled gun is subjected to less acceleration than a one fired from a short-barrelled gun? Thanks very much in advance.

ANSWER:
The average acceleration is proportional to the average force. When
the gun is fired, there is an explosion behind the bullet which increases
the pressure. This pressure pushes the bullet forward. But as the bullet
goes down the barrel the volume gets larger and so the force gets smaller.
So, the average force for a long barrel is smaller. Be sure to note that
this does not necessarily mean the shorter-barrel gun will have a faster
bullet; the longer barrel also has the property that the bullet is exposed
to the force for a longer time.

QUESTION:
is the charge of a proton calculated to be exactly equal in magnitude with an electron? and what does it mean by saying equal magnitude?

ANSWER:
No, they are measured to be equal in magnitude to incredible
accuracy, no different than 10^{-13} %. Magnitude simply means that we discard
the signs of the charges, the electron charge is -e , the magnitude of
which is e .

QUESTION:
My husband when riding his vintage motorcycle had an accident when negotiating a bend (200m radius) in the wet. it was found that the road surface only had friction coefficient of 0.3. What formula and other known parameters (like weight of bike, lean angle etc) can I use to determine what speed he could have safely negotiated the bend?
He was actually only doing about 40-43mph.

ANSWER:
I do not know where you got your coefficient of static friction
μ _{s} =0.3, but it can be only a rough estimate at best. I
find that rubber on wet concrete can have μ _{s} over the
range 0.45-0.75 and on wet asphalt over the range of 0.25-0.75. But, often
there is oil which was in the road which floats up when it rains and makes
it slipperier yet. I am assuming, since you didn't mention it, that the road
is not banked. In that case, the weight of the bike+rider and lean angle are
irrelevant. It is easy to calculate the minimum possible speed from Newton's
second law, μ _{s} mg=mv ^{2} /R , so v =√(μ _{s} gR ).
Putting in your numbers I find v =24.3 m/s=54 mph. Evidently the
coefficient of friction was less than you assumed. "Discretion is the
better part of valor" (Shakespeare, Henry The Fourth , Part 1 Act 5,
scene 4); in other words, be very careful on wet roads!

FOLLOWUP QUESTION:
The friction coefficient of 0.3 (equivalent to wet snow on tarmac) was obtained by pendulum testing in the wet to reflect the conditions at the time of the accident. You are correct in your assumption that the road was not banked. In fact the road comprises a slight adverse camber.
Given the low friction coefficient I am having great difficulty in understanding how the calculation can show that a motorcycle (at normal lean angle) can negotiate the bend at 54mph. I would not wish to attempt to negotiate that bend on the equivalent of wet snow at 54mph even in a car!

ANSWER:
This is a very standard calculation. I figure that at 54 mph it would take 13 s to traverse a 90 degree turn for this radius circle which does not seem unreasonable to be possible. Are you pretty sure of the 200 m number?
The key to this may well be the banking of the curve; if the road is banked wrong by 10^{0} ,
I would calculate
v =15.1 m/s=34 mph, quite a significant difference.

QUESTION:
why don't electrons in an atom collide with the nucleus? Doesn't the nucleus of an atom have a positive charge? if so, then it should make sense that the electrons would be attracted to it, and not just go around it in orbitals. So yeah, I'd like to know why electrons don't fly into the nucleus, and stay in their orbitals. At the quantum level, I doubt gravity has any role in it, and even if it does, the mass of the nucleus would far succeed the mass of any individual electrons.

QUESTION:
What keeps an atom from getting smaller and smaller if an electron can lose energy and be attracted to the nucleus? What keeps electrons from moving closer and closer to the nucleus?

ANSWER:
The electrons are attracted to the nucleus, that is the force that
holds them in their orbits. Your question is akin to "why don't the planets
in the solar system collide with the sun?" Quantum mechanics shows that
there is a lowest energy a system can have, called the ground state, and for
atoms this is always above zero and if an electron "popped into" a nucleus
and just sat there this would be a state with lower than the minimum allowed
energy. On the other hand, the idea of orbits for atoms is handy but not
really very good. The electrons should be thought of as a cloud and the
density of the cloud at any location is representative of the probability of finding the
electron there. This cloud extends all the way to the center and so there is
a small but nonzero chance of finding the electron inside the nucleus. The
picture at the right shows this cloud for the ground state of a hydrogen
atom. The densest part is where the Bohr orbit would be.

QUESTION:
Just like a ball when hits a wall it bounces back, why dont people bounce back in the same way, when opposite force is exerted on them by the wall (as in case of ball)??

ANSWER:
It is because of elasticity. When two objects collide, how they move
after the collision is determined by what happens to energy after the
collision. If you drop a superball on the floor it rebounds almost as high
as from where it was dropped; if you drop a putty ball, it does not. When
the superball hits the floor, it compresses but it compresses like a spring
which "stores" the energy. So the energy of the moving ball is moved into
the energy of the spring. The spring now decompresses and gives the energy
back to the motion of the ball and it flies back into the air. It obviously
did not give all the ball's energy back since the ball did not quite
rebound to where it was dropped from; the little bit of energy lost shows up
as heat and sound. The putty ball also compresses when it hits but it bears
no resemblance to a spring because after you compress it (which takes
energy), it does not decompress and give energy back. Here, the energy of the
ball's motion is used to compress the putty ball and that energy shows up as
heat and sound entirely. The superball (neglecting the little amount of
energy lost) is referred to as an elastic collision and the putty ball is a
perfectly inelastic collision.

QUESTION:
Do you weigh more at the bottom or top of a large skyscraper?
Does the mass over your head count for more than the distance?
As an example, the tallest skyscraper currently is 828m tall and wieghs about 500,000 tons if it's empty, in Abu Dhabi.
Assuming that you weigh more at the bottom, how dense or large would the mass above your head have to be to make you weigh less?

ANSWER:
There are two effects you need to think about:

Because you are a different distance from the
center of the earth when you are at the top, your weight is less. This
turns out to be a 0.03% difference.

The gravitational force due to the mass of the
skyscraper is really small. I just took all the mass to be at the center
to check this. Then, if you are at bottom, the gravitational force
from the skyscraper mass would about 0.000003% smaller than your weight
at the top which would increase your (0.03% smaller) weight at the
top and decrease your weight by that amount at the bottom. So, the
effects of the skyscraper are negligible compared to other effects.

QUESTION:
When we heat something, the particles in it move faster and faster till they reach the speed of light. But what would happen if we keep heating? The particles can move faster than the light, or there's a maximum value for the temperature (like a minimum: 0 Kelvin), or what?

ANSWER:
No, anything with mass can never reach the speed of light, but
neither can it go any faster. I have
answer ed this
question before and the answer is that there is no upper limit. I have
actually been corrected since there is an upper limit imposed by the total
amount of energy in the universe, but that begs the question. If you have
something, there is no limit to the amount of energy you can put into it,
assuming you can get that energy somewhere. There is a minimum temperature
possible, as you note, but that cannot actually ever be reached because of
the laws of quantum mechanics. So the speed of particles in an object are
constrained by 0<v<c corresponding to temperatures
of 0<T < ∞ and the extremes are
unreachable.

QUESTION:
Why electrons are used over proton for the charging of Body...??

ANSWER:
Because electrons are easily available because it takes only a little energy to remove them from atoms but protons are bound in the nuclei and require much energy to extract.

QUESTION:
Please explain why the energy of photons increase with
frequency (not how by quoting E=hf) - do you have an analogy to explain
the phenomenon?

ANSWER:
Maybe the easiest way to approach this is that, in quantum mechanics,
momentum is associated with wavelength via p~ 1/ λ~f . Now, in relativity,
E=pc so it follows that E~f. (Here, ~
denotes proportional to.)

QUESTION:
how much time will it take to travel down a 1000 ft , 30 degree ilcline with a 225 lb payload on a 200 lb soap box race car and if pay load was lighter would the time be shorter with lighter load?

ANSWER:
I have dealt with this kind of question
before , but maybe it is time to
revisit the whole thing with one answer. First, consider the ideal situation
where there is no friction of any kind. Physicists do not like English
units, so I am going to convert everything to SI units: 1000 ft=305 m, 425
lb=193 kg. I assume that you do not want all the details of my calculations,
just the pertinent results. The time does not depend at all on what the mass
is (if friction plays no role). There is an acceleration down the incline
which is a= �g. I find
that the time to the bottom is about 11.2 s and the speed at the bottom is
about 54.9 m/s=123 mph. I assume you are not crazy enough to be in a soap
box car going that speed, so friction must play a role. There are two kinds
of friction you have to consider:

Friction due to
the moving parts, like bearings, like the wheels rolling on the ground,
like wheels moving on axels, etc . Empirically, we find that this
kind of friction increases proportionally with how hard the moving parts
are pressed together and this, of course, is proportional to the total
weight of car plus rider. But, the force impelling the car down the hill
is also proportional to the total weight, so once again, the speed at
the bottom is independent of the load. There is something called the
coefficient of friction which tells you how much friction force there is
for a given force pressing the surfaces together. For example, if this
coefficient is 0.3 for a 100 lb box moving on a horizontal floor, you
would have to push with a 30 lb force to keep it moving with a constant
speed. If I take 0.3 to be your coefficient of friction, your speed at
the bottom would be about 26.8 m/s=60 mph. Since the acceleration is
about half what it was, the time is about twice as long, 22 s.

For objects which
are going faster than a few mph, air drag becomes important. Air
friction depends on two things, the shape of the object and the speed it
is going. It does not depend on the mass of the object. Actually a
pretty good approximation to the magnitude of the force of air friction
is F≈ �Av ^{2 } where A is the area presented
to the wind (this works only if F, v, and A are in SI
units). So, this force, which points up the incline just like the moving
parts friction, causes the car to slow down more. But, Newton's second
law says that a=F/m so that if F does not depend on
m , the acceleration (amount of slowing down due to this force)
gets smaller as m gets bigger. It gets a little complicated to
actually calculate the time and speed for this case, but the important
part is that this is the only place I can find where the weight of the
whole car makes a significant difference.

So, the bottom line
is that if you go fast enough for air drag to be important (and I suspect
you do), the heavier of two otherwise identical cars should win.

QUESTION:
If there is a yacht on a lake and it drops its anchor overboad. What happens to the water level in the lake. Apparently it falls. I have no idea why. Shouldn't it stay the same using common sense.

ANSWER:
This is a little complicated, so bear with me. First, you have to
understand Archimedes' principle which states that if you have an object in
a fluid, the fluid exerts an upward force on it and the magnitude of that
force (called the buoyant force) is equal to the weight of the fluid which
the object displaced. So, the boat plus anchor floating on the water have
displaced a volume of water whose weight is the same as the weight of the
boat plus anchor. Now, remove the anchor. Now the water is holding up only
the weight of the boat, so the water will have fallen equivalent to removing
water whose weight equals that of the anchor; but, the anchor is denser than
water, so the volume of the water of equal weight is bigger than the volume
of the anchor. Now, putting the anchor into the water, the water level will
go up equivalent to adding a volume of water equal to the volume of the
anchor. So the net effect has been a fall followed by a smaller rise, a net
fall.

QUESTION:
I understand that atoms are mostly empty space and that what we understand as matter "touching" is not really touching at all but the magnetic force of electrons repelling each other. My question is: if all this is true, how do we get bitten by mosquitoes or ticks or anything for that matter? How can something penetrate us if this "touching" sensation is actually magnetic repulsion?

ANSWER:
You have it wrong. It is not a magnetic force but the repulsive
electrostatic force between the electrons in "touching" atoms. But, the
force is not infinite and if you push hard enough, particularly if what you
are pushing has an edge or a point which can "pry apart atoms" at the
surface, you can penetrate.

QUESTION:
What is the magnetic field of a permenant bar magnet made of? I have researched this, but cannot find a clear answer. Leading to the question...How does a bar magnets strength dissapate over time? How long will my fridge magnet stay attached to my fridge? Where does the energy come from and go to?

ANSWER:
Made of? A field is a representation of a force which would be
experienced by, for example, another magnet. But, it is not really just a
mathematical construct, as often taught in introductory courses, but the
field exists in some "real" sense. Electromagnetic fields can have energy,
momentum, or angular momentum, so maybe you would like to think of just the
magnetic field being "made of energy". In prinicple, a magnet should never
lose its strength because the strength because it is composed of many
electron magnets which are indestructible. However, they may not all stay
aligned and external fields can cause them to get unaligned. However, I
would expect that your refrigerator magnets will not fall off in your
lifetime. Finally, it takes no energy to hold your magnets to the fridge.
Energy requires that work be done which means a force acts over a distance
and the magnet is not moving and so no work is done; zero energy is used
holding your magnets to the fridge.

QUESTION:
When a ball is thrown up in the air, does it always form a parabola? Is there any possible way that the ball could go to the top and not fall back down in a symmetrical parabola shape? ( Air resistance/ wind or something could affect this I suppose, but what about in a vaccum?)

ANSWER:
The parabolic path is only an approximation. You have to assume the
gravitational force on the ball is always constant and straight down and all
the "straight downs" are parallel. Well, the earth is not flat and the
gravitational force decreases as you go to higher and higher altitudes. But,
for usual thrown balls, etc., the approximation is excellent. Much
more important is air resistance which can have a very profound effect,
particularly for high speeds. Anyone who has played outfield in baseball can
tell you that fly balls tend to fall more or less straight down even though
they were hit at a relatively small angle. But, in a vacuum, as you suggest,
the non-flat earth is all that would cause a change from parabolic. Here is
an extreme example: if there were no air and you could throw a ball
horizonatally at a speed of about 18,000 mph, the path would be a big
circle, an orbit around the earth. See
Newton's mountain .

QUESTION:
If a light bulb doesn�t operate at a maximum efficiency, then is the law of conservation of energy correct?

ANSWER:
What does "maximum efficiency" mean? If it means that 100% of the energy
put into the bulb comes out as light, then no light bulb is maximally
efficient. But if only some smaller percentage of light that is emitted, the
"lost" energy shows up somewhere, mostly in the form of heat. Energy is
conserved for any truly isolated system.

QUESTION:
WHAT IS MATTER?
I understand that matter is composed of atoms, and that atoms are made of subatomic particles. This is the answer that most physics textbooks and google searches yield. It seems like physicists are afraid to answer this question. To say that matter is made of wave-particles is no answer at all. If it's a wave then a wave of what? If a particle, a particle of what? PLEASE I MUST KNOW!! I have lost more than a few hours of sleep over this.

ANSWER:
People lose sleep over the strangest things! I am a physicist and
certainly not afraid to answer this question! My definition would be that
matter is something which can be at rest in some inertial frame and which
has inertia, resistance to acceleration when a force is applied.

QUESTION:
How is it that my laptop is able charge as soon as I plug it in with out it shutting off? Does it work like a car where first it runs off the battery then a relay switches it to gas?

ANSWER:
The charger essentially looks like a dc power source of the same voltage
as the battery. When the charger is plugged in, it is in parallel with the
battery, so the charger both charges the battery and supplies power to the
computer.

QUESTION:
I've been told that when a gas is cooled, it takes up less volume, and therefore its density increases.
I've also been told that a gas always fills its volume (i.e. a container).
These two ideas, however, seem to contradict one another. On the one hand, cooling a gas causes it to contract and decrease in volume, which would increase its density. However, on the other hand, a gas will always fill its volume, which means its volume doesn't change when cooled. Therefore, if the volume doesn't change when cooled since a gas will always fill its volume, then its density would not increase.
In other words, if a gas will always fill its volume, then cooling it would not increase its density since the gas always fills its volume and therefore remains at the same volume. In a closed container, if a gas expands to fill the volume of the container, then how could cooling the gas increase its density? How could a gas fill its container/volume on the one hand, but then on the other hand, when its cooled, decrease in volume and therefore increase in density? If a gas fills its container/volume, it wouldn't increase in density when cooled since it FILLS its volume. Therefore, since it fills its volume, the volume of the gas would remain constant. Therefore, since the volume of the gas would remain constant, its density would not increase when cooled.
So my question is, how could a gas increase in density when cooled if that same gas always fills the volume of the container it's in?

ANSWER:
T hat's a pretty long question which requires only a short answer. If the
amount of gas remains constant, the relation between pressure, volume, and
temperature is PV /T= constant. The situation you describe has a
constraint that volume remains a constant, so P /T =constant
which means that as the temperature goes down the pressure goes down; and,
the volume stays the same. So, if you cool a gas in a rigid container, the
density remains the same.

QUESTION:
I would like confirmation about the velocity of objects moving in a circle. If an object moves in a full circle at a constant speed, is the average velocity zero? One of my friends stated that it wouldn't be zero, and that I was confusing velocity with displacement. But since velocity has direction, I thought that if I were to divide the circle into, say, ten pieces, then the velocities of each independant piece of the circle would cancel each other out because the two segments of the circle that are opposite each other would have opposite signs because they are going in opposite directions. Can you help us settle this?

ANSWER:
The average velocity is a vector and is the displacement vector divided
by the elapsed time. Since the displacement is zero, the average velocity is
zero. The average speed is not zero but is the circumference of the circle
divided by the elapsed time.

QUESTION:
Does cold radiates to your body if you open the door of a freezer? If so, how?

ANSWER:
Everything radiates heat. Nothing "radiates cold". The rate at which
energy is radiated is determined by the temperature of the object. Your body
radiates heat faster than the freezer does, so the freezer absorbs more of
your radiation than you do its radiation and you feel colder because heat
has left your body.

QUESTION:
Skydiver free falls from plane with a loaded gun, and shoots gun while free falling. Does the bullet being fired travel downward any quicker then say a bullet travelling just being dropped?
Well, during a heated discussion with co-workers, I was trying to explain how I thought the bullet fired downwards would not go any faster than a bullet just dropped by someone skydiving

ANSWER:
OK, let's say that two skydivers are side by side each falling with the same
speed. Because of air drag, both bullets will eventually both reach
"terminal velocity" which will be the same for both (assuming enough time
passes before the ground is reached). But the shot bullet will have a big
head start over the dropped bullet. The shot bullet will begin with a speed
above the terminal velocity and slow down until it reaches it. The dropped
bullet will begin with a speed below the terminal velocity and speed up
until it reaches it. Certainly, the shot bullet will reach the ground first.

QUESTION:
if i where riding a bike with a flashlight on the front end and i was going 10mph and knowing that the speed of light is constant wouldnt that mean that the light had to slow down 10 mph ?

ANSWER:
One of the basic foundations of the theory of special relativity is that
the speed of light is constant regardless of the motions of either the
source or the observers. You will find several questions on the
FAQ page which address your question.

To see
questions and answers from longer ago, link
here .