QUESTION:
Why only magnetic moment due to spin is considered in proton nmr spectroscopy. what about magnetic moment due to orbital motion??
ANSWER:
You are right, a proton in a nucleus will have both spin and orbital
magnetic moments. But the whole idea of NMR is to put magnetic moments in a
magnetic field and they tend to align. To flip the spin requires a certain
amount of energy. So you send photons of the right energy, typically radio
frequency radiation) and they will flip the proton spin. You need not worry
about other moments because the photons are tuned only to the right
frequency for spin.
QUESTION:
In understanding that two objects (no matter their phisical properties) will accellerate at the same rate towords the same gravitational force in a perfect vacuum, do we take into consideration the gravity of the falling objects themselves? Would the larger gravitational force of the larger falling object be perfectly offset by the larger ammount of inertia or tendency to resist motion?
ANSWER:
The force F between masses M and m whose centers are
separated by a distance R is F=GMm/R^{2}. But,
the acceleration a of m is a=F/m=GM/R^{2}.
So, as you ask in your last sentence, the inertia (m in a=F/m)
cancels out the mass (m in F=GMm/R^{2}) in the
acceleration. This is actually a profound finding for the following reason.
In Newton's second law, mass is, as you say, inertia (inertial mass, m_{I});
in Newtons universal law of gravitation, mass is the ability to feel or
create gravity (gravitational mass M_{G}). The fact that
gravitational accelerations are the same implies that m_{I}=m_{G},
not something you would necessarily expect. Why these are equal was not
really understood until the general theory of relativity was proposed by
Einstein.
QUESTION:
i am a metal worker and maybe it's dumb but i have thought up an idea for renewable energy. I need to build a huge tank capable of holding 10000 of water, it will have a coned base with a hose coming off it and running back up into the tank, returning the water to it's point of origin....question is...with 10 tons of water down force, 1 meter from the ground....would a decent amount of water travel up that hose pipe and back into the tank???..
ANSWER:
It makes no difference if you have a whole ocean of water, the water at the
bottom can never push water higher than the surface of the tank, can never
cause a continual current like you envision. There is a very detailed
answer for a very similar question to yours that
you should look at.
QUESTION:
In the event horizon of the black hole, a pair production of a particle and an anti particle and the falling of one into the black hole and the releasing of the other from it gives us the Hawkins radiation, but my question is , isn’t it possible for both the particle and the anti particle to fall into the black hole. If that happens defiantly the vacuum energy of the space decreases how can the universe compensate this decrease in energy to validate the law of conservation of energy.
ANSWER:
There are two ways you can imagine the creation of the
particleantiparticle pair
(PAP):

Vacuum fluxuations where a particleantiparticle pair is spontaneously
created just outside the event horizon, violating energy conservation.
However, the uncertainty principle (UP) allows this violation but only
for a very short time. If one escapes, it adds energy to the universe so
the captured particle must have negative total energy thereby decreasing
the mass of the black hole and maintaining energy conservation. If both
particles are captured, they are required by the UP to recombine thereby
keeping the mass of the black hole constant.

When
a virtual pair is created just outside the event horizon, the intense
gravitational field of the black hole can provide the energy for the PAP
to become real rather than virtual. In this case, the energy acquired by
the pair will be lost by the black hole reducing its mass. If only one
of these is captured, the energy of the escaped particle will be added
to the universe but the energy of the captured particle will be added to
the (already reduced) energy of the black hole resulting in a net loss
of energy of the black hole equal to the energy of the escaped particle.
Energy
of the whole system is conserved.
QUESTION::
If the speed of sound is inversely proportional to the density of a material, why does sound travel faster in solids (it is the most dense).
I have read that it takes more energy for sound to travel in dense materials so it takes longer but then neighbouring molecules are closer so sound does not have to travel that far, making it faster. This doesn't make any sense because it says the more dense a medium is, sound is both faster and slower.
Also, how does bulk modulus affect the speed of sound.
ANSWER:
You are oversimplifying because the density
ρ
is not the only thing which determines the speed of sound. In general, the
speed of sound v may be written as v= √(K/ρ)
where K is a parameter which specifies the "elasticity" of the
material. (Note that v is inversely proportional to √ρ, not ρ
as you state.) For example, just qualitatively, you might say that steel is
much denser than air but it is also much more elastic. The way K is
specified for gasses is quite different than for solids. For an ideal gas,
K=γP where P is the pressure of the gas and γ=5/3 for
monatomic gasses and 7/5 for diatomic gasses. On the other hand, for solids
K=B+4G/3 where B is the bulk modulus and G is
the shear modulus. For solids K is enormously bigger (~10^{11}
N/m^{2}) than γP for gasses (~10^{5} N/m^{2}).
That answers your question. So, let's just do a rough calculation to see if
I get reasonable values for air and steel:

Taking air as mainly diatomic
γ≈ 7/5 ,
atmospheric pressure P≈10^{5} N/m^{2}, ρ≈1
kg/m^{3}, v ≈374
m/s; a little high, but close.

For
steel, B ≈1.6x10^{11
}N/m^{2}, G≈7.93x10^{10} N/m^{2},
ρ≈7900
kg/m^{3}, v≈5800 m/s; this is exactly the speed of sound
in stainless steel!
QUESTION:
The earth's radius varies depending on latitude. How many times greater is the acceleration of gravity at the poles of the earth than the equator if the radius of the earth is 99.5% smaller at the poles than the equator?
ANSWER:
To calculate exactly would be impossible because the earth is not a perfect
spheroid and its mass is not uniformly distributed. But you can get a good
idea by calculating the difference in gravitational force for spheres of the
same mass but having the two different radii; these radii are 6378 km
(equator) and 6357 km (poles), about 0.33% larger at the equator. Since the
gravitational force is inversely proportional to the square of the radius,
this would give about a 0.11% smaller g at the equator. This effect
is smaller than the effect due to the earth's rotation which results in
about a 0.34% reduction at the equator (due to centrifugal acceleration).
QUESTION:
What is the purpose of utilizing a percentage of body weight to determine how much weight to bench press/push/whatever? I know this seems like a fitness question and not a physics question, but what I am interested in is WHY weight would be used to determine how much one could (or should be able to) lift/push?
For example: A gym teacher wants to grade his students on their strength. He decides to use abililty to push a weighted sled across the floor as the measure. He wants to make the task equally difficult for every student in order to make the grading fair. So, he decides that each student will push 2x his/her body weight for 5 minutes and the grade will be based on how FAR the student is able to push.
So, Student A weighs 170 pounds and pushes 340 pounds (including the weight of the sled) for a total of 160 yards. Student B weighs 240 pounds and pushes 480 pounds (again, including the weight of the sled) for 80 yards. Student A pushed farther and gets a better grade, but Student B complains that he had to push much more weight so he should not get a worse grade.
Does Student B have a legitimate complaint or does his heavier weight contribute somehow to his ability to push that doesn't have anything to do with his strength? As in, does his weight help push the sled in some way?
Sorry, I don't know enough about physics to ask this question using proper physics terms like force, mass, etc. I hope you will still answer my question!
ANSWER:
I cannot comment on the rationale for correlating weight to strength. I can
certainly comment on the physics of your particular example of sled pushing.
I would first of all comment that this example is certainly not one solely
of strength because, since it is a timed activity, endurance as much as
strength is being tested; if one student, for example, were a heavy smoker,
he would likely become exhausted more easily. As a physicist, I would equate
"strength" with force. The specific example you give, though, seems to me to
be more related to energy (work done by the student) or power (rate of
energy delivered) than strength; purely in terms of strength, the heavier
student exerts more force. The force F which each student must exert
depends on the weight w he is pushing and the coefficient of friction
#956;
between the sled and the ground, F= μw.
The work W done in
pushing the sled a distance d is W=Fd= μwd.
The power generated if W
is delivered in a time t is P=W/t=μwd/t .
Both students have the same
μ
and
t , so W_{A}/W_{B}=P_{A}/P_{B}=d_{A}w_{A}/d_{B}w _{B}=(160x340)/(80x480)=1.42.
So student A did 42% more work, generated 42% more power, than student B.
From a physics point of view, B demonstrated more strength, A demonstrated
more power. I would judge that this is not a fair way to assign a grade. It
would be interesting to see if A (B) could move B's (A's) sled 80 (160)
yards.
QUESTION:
Say I have 2 factories producing an object, we'll be arbitrary and say bottle caps. Each produces them at a rate of 1 per second. Now ignoring things like raw materials etc...
One stays on Earth, producing its bottle caps constantly.
The second one (somehow) gets accelerated to (again being arbitrary) 90% the speed of light and remains at that velocity until completing a roundtrip 100 years long.
When the moving factory returns, 100 years later from the perspective of the factory on Earth...will there be a difference in the total number of bottle caps each has produced?
ANSWER:
This is just a variation of the twin paradox.
The number of bottle caps produced by the moving factory will be N √(10.9^{2})=0.44N
where N is the number produced on earth.
QUESTION:
Okay, so for a while I've been wondering why bass travels through certain materials that higher frequency sound doesn't. I have a theory which could be wrong but what I thought is that bass, having a larger wavelength, passes through particles that are bigger, and that shorter wavelengths can't get through. Somewhat like red light and blue light traveling through the atmosphere. Okay, so I looked it up everywhere (almost) and couldn't find a straight answer, even my physics professor didn't know surely why this happens. So, maybe you could. Why can I hear just the bass when someone blasts music in their car?
ANSWER:
The reason is that the attenuation of sound intrinsically depends on the
frequency of the sound, the higher the frequency the greater the
attenuation. The theory of attenuation is rather complicated, so I will just
sketch the results. For a
Newtonian fluid,
Stokes showed that A(d)=A_{0}e^{} ^{ αd }
where
A(d) is the sound amplitude a distance d into the
material, A_{0}=A(d=0), and
α= 8 π ^{2} η f^{2} /(3ρv^{3});
here
η
is viscosity of the fluid, f is frequency of the sound, ρ is
the density of the fluid, and v is the speed of sound in the fluid.
The important thing to notice is that the amplitude decreases like exp(βdf^{2})
for any Newtonian fluid. For example suppose you compare f_{1}=20
Hz with f_{2}=1000 Hz for a thickness of d=1 cm=0.01 m
and A_{1}(0.01)=ĽA_{0}=A_{0}e^{β·0.01·400}=A_{0}e^{4β}
which can be solved for β, β=0.35 s^{2}/m; then, for
f_{2}, A_{1}(0.01)=A_{0}e^{0.35·0.01·1,000,000}=A_{0}e^{3500} = A_{0}x 9x10^{1521}≈0.
Finally, for any material, Stokes' law of attenuation still may be written
as
A(d)=A_{0}e^{} ^{ αd }
except α=βf
^{ε }where 0<ε<2. Water and
most metals have ε≈2. But even if ε is smaller, it is always
positive which means that the attenuation is always greater for a higher
frequency.
QUESTION:
We know the relative size of the solar system measuring thru the distances of the planets circling our star.
Do we know the thickness of this rotational disc?
Is the gravity of our star concentrated only on this single orbital plane as our star rotates?
If not, why do planets not circle verticaly as well as on horizontal planes?
As far as I know we have only sent space craft thru our orbital disc towards other planets. Do you believe if directed a flight up from our orbital plane would the suns gravity lessen where our space craft could fly faster?
ANSWER:
I should first address your speculation that gravity is stronger on the disc
near which the planets orbit. In fact, gravity depends only on how far you
are from the source, so if you went 100,000,000 miles from the sun in any
direction you would find the same gravitational force. There is not really
any welldefined disc whose thickness you can measure; all the planets are
nearly all orbiting in the same plane and the inclinations of the orbits
relative to say, the plane of the earth's orbit, are all less than 3 ˝^{0}.
But that does not mean that everything is near that plane. Many of the dwarf
planets are much out of the plane of our orbit; Pluto, for example, is about
18^{0} inclined relative to our orbital plane. Outside the most
distant planets is the
Oort cloud, a collection of rocky and icy bodies which is distributed
roughly equally in all directions; comets are thought to originate in the
Oort cloud and when they enter the inner solar system they come from all
directions relative to the orbital plane. The reason that the planets are
all nearly on the same plane is that when the sun first formed from a huge
cloud of dust and rocks, it had angular momentum and as the system collapsed
under its own gravity, it spun faster and faster and the resulting
centifugal force caused it to flatten like a giant pizza dough.
QUESTION:
An image is formed
when the reflected ray meet each other at a point. But why does an image is
not formed by the intersection of incident rays? After all they are the same
light rays and it's not like there's some kind of reaction which takes place
or something.
ANSWER:
The light rays from each point on the object all diverge away from that
point. Unless there is something to change their directions, e.g. a
mirror or a lens, they never meet.
QUESTION:
If I were to throw a pencil up into the air, would it come down at the same speed as I threw it up?
ANSWER:
It would if there were no air. However, air drag takes energy away from the pencil and so it would arrive back at your hand going more slowly. For speeds you are likely to be throwing the pencil, though, this effect is small and often neglected in elementary physics classes.
QUESTION:
The demonstration wherein one pours water from a clear vessel and one shines a laser pointer (a red laser pointer in my case) through the water and it appears the laser beam bends with the water pouring out of the clear vessel;
What causes the beam to bend with the water? Would this same type of effect work with other fluids than water, such as air or glass?
ANSWER:
It is caused by
total
internal reflection. If the angle which the light hits the surface
between the water and the air is glancing enough, the light will be
reflected back into the water rather than be refracted into the air. Total
internal reflection can happen whenever light strikes the interface with a
material of smaller index of refraction; the minimum angle for which it can
occur depends on the relative indices of refraction.
QUESTION:
I was reading a introductory book on particles. I read that neutrinos only respond to weak and gravitational forces, isn't that fact that it responds to gravitational forces prove that it has mass? So why was this years Nobel Prize awarded for figuring out that neutrinos had mass?
ANSWER:
No, responding to gravity does not imply mass. Photons, which are known to
be massless, are deflected when passing near a large mass. See
FAQ page.
QUESTION:
If a train runs at 100km/h and if in this train somebody drops a rubber ball and catch it, let's say it drops down 1m and up in one second, it means that the ball has travelled 2 meters in one second for the person in the train.
If I am at the station looking at the train when it passes me by and let's say it is made of glass so I can see what is happening, I can see the ball dropping down and up but for me as the train is moving the ball will have travelled 2 meters plus the distance travelled by the train in one second so more than 2 meters.
I don't think that space time dilatation plays at this sort of speed so how is it explained?
ANSWER:
Actually, you have not calculated the distance traveled correctly. But
because the ball is speeding up on its way down and slowing down on its way
up, you have set up a problem with unnecessarily difficult mathematics to
calculate the distance seen by the person at the station. I will alter your
problem somewhat to make it easier to calculate: the ball is made to first
move down at constant speed and then to move up with the same constant
speed. If the time is still 1 s, the necessary speed is v_{y}=2
m/s. The train is moving with speed v_{x}=100 km/h=27.78 m/s
so the ball and train move a distance 27.78 m horizontally during the one
second the ball is moving. But the distance the ball moves is not
27.78+2=29.78 m but rather
2√( 13.89 ^{2}+1^{2} )=27.85
m because the ball will take a zigzag path. But the ball also has a
different speed as seen from the station, v=√ (v_{x}^{2}+v_{y}^{2})=27.85
m/s. We therefore conclude that the time it takes the ball to travel 27.85 m
is 1 s, so everything hangs together. Your error was to ignore the fact that
the ball has a horizontal component of its velocity in the frame of the
station. You are right that relativity plays no role here.
QUESTION:
Suppose I have a charge +q and there is a point P , Suppose I place a conductor between the charge +q and P . Since there are free electrons in it , Negative electrons move towards +q and equal positive charge inside the conductor near P , The conductor has charge distribution like a dipole. So If I want to calculate E field at P . I could use superposition principle to find E at P due to +q and E due to dipole. But Gauss's law says that dipole doesn't contribute anything to E field at P. Can you explain me 'intuitively' (Not in equations) why the dipole wont contribute anything to the field at P ?
ANSWER:
Gauss's law does not say that the dipole contributes nothing at P. If you
put a spherical Gaussian surface enclosing both q and point P the net
charge enclosed is q but that does not mean that the field due to the
dipole is zero everywhere on that surface. All you can say is that the net
electric flux passing through that surface is q/ ε_{0}.
Gauss's law is usually useful in determining a field only if the field can
be argued to have constant magnitude and normal to the surface everywhere.
Superposition, not Gauss's law, should be used for this problem.
QUESTION:
If I was in a spaceship
moving at half the speed of light (in normal space) and I measured the time
it took the beam of light from a laser pointer to travel from its source to
two points inside my ship, one in the direction of travel and one in the
opposite direction, wouldn't C demand that I observe different time
measurements? If the time measurements are the same, then when I pointed in
the direction of travel, the light would have to be traveling at C plus the
speed of the ship from observations made outside of the ship. If the
measurements are different, then shouldn't we be able to observe the same
effect on earth because of the direction of travel of our planet, solar
system, Galaxy, and group of galaxies?
ANSWER:
I am not sure what you are asking here. Does this
earlier answer help you? One thing is clear is that you do not really
appreciate that the speed of light in vacuum is independent of the motion of
either the observer or the source. This is one of the main postulates of
special relativity. See the FAQ page for questions related to this.
QUESTION:
Sir,I am a mechanical engineer by profession but very interested in reading physics fundamentals. Recently I went through the fundamentals of electro magnetism and I got this doubt. Consider two charges each of charge +q rigidly fixed in a train moving with a constant velocity, V. Let the train speed be negligible compared to speed of light so that we can treat the problem in nonrelativistic terms. The fixity condition ensures that it overcomes electrostatic forces and remain motionless. A traveler in train sees both of them at rest and there wont be any magnetic forces developed between the charges.
Now consider an observer in platform. For him, both charges are moving with a constant velocity, V equal to the train velocity. Each charge will develop magnetic field according to this observer as per BiotSavart law and there will be mutual attraction as each charge is moving under the magnetic field of other.Thus, an observer in train sees no magnetic forces whereas an observer in platform sees mutual magnetic attraction. How do you explain this?
ANSWER:
There is no rule which says that the either the electric or magnetic field
must be the same in all frames of reference, even slowly moving frames like
you specify. The real root of your problem is that electromagnetism is
intrinisically relativistic, even at slow speeds; the electric and magnetic
fields of classical electromagnetism are really both components of the
electromagnetic field which is a tensor and when you change inertial frames,
you cause a transformation of that tensor into another where both the
electric and magnetic field pieces of it are different. In your second case
you would also find that the electric fields were slightly different from
their original values but the differences would be very tiny; the magnetic
fields, though, are nonzero but small, but small is very big compared to the
original magnetic field of zero.
If you
are interested, I will give here the electric and magnetic fields for one of
the charges moving with velocity v in the +x direction.
E'=iE_{x}+ γ(jE_{y}+kE_{z})
and B'=(vxE')/c^{2
}where
γ=1/√[1(v/c)^{2}] and i, j,
k are unit vectors; the vector E is in the frame
where q is at rest and E' and B' are when
q is moving.
QUESTION:
I see the equations for computing the slowing of time, based on speed compared to light speed. Do we always use c=3^10 cm/sec? I am asking since light speed is slower, sometimes much slower, in mediums other than air or space. If we use a smaller value for c, the time dilation factor is enhanced.
ANSWER:
The significant thing in the theory of special relativity is the speed of
light in vacuum. What matters in the theory is the universal constant
c, not the speed in the medium you happen to be looking at.
QUESTION:
Is it true that a person flying in an airplane is actually living a shorter amount of time, than a person standing on the ground?
ANSWER:
You must specify " … living
a shorter amount of time … "
with respect to whom . The
person in the airplane sees time progressing at a normal rate. However, the
person on the ground will see the clock of the person on the airplane run
slowly, so he will perceive the traveling person to be " living
a shorter amount of time", i.e. she will have aged less when she
returns to earth. You should read FAQ Q&As on the
twin paradox,
how clocks run, and the
light clock.
QUESTION:
If I was travelling in a bus and the bus was travelling at around 60 mph and
I jumped in the air, how come I don't fly to the back of the bus. Like with flying, if
I was in an airplane travelling at around 500 mph, and I was floating without an aid in the cabin why do
I travel with the plane not just get smashed to the back?
ANSWER:
It is impossible for you to be " floating without an aid ",
so let's assume that you are standing on the floor of both the bus and the
plane. In both cases it is because of Newton's first law. Someone watching
you from the ground would see both you and the vehicle as moving with the
same speed. From that point of view you are in equilibrium because the only
forces on you are your weight down and an equal force up from the floor.
Newton's first law says that if you are moving in a straight line with
constant speed you will continue to do so if all forces are in equilibrium.
When you jump up, the floor no longer exerts a force on you, so you are no
longer in equilibrium vertically and you will fall back to the floor. But
you are still in equilibrium horizontally (there are no forces on you
horizontally) so you will continue, right along with the bus or plane, to
move horizontally with the same speed. Actually, I did this the hard way
because Newton's third law will still be valid in the moving bus or plane,
so when you jump up you will not move horizontally because there are no
forces which act horizontally on you to accelerate you (make you start
moving horizonatlly).
QUESTION:
If I have two soccer
balls the same size, yet one is heavier than the other, which one will go
farther with the same kicking force involved? How does momentum and Newton's
Second Law come into play here?
ANSWER:
I will assume the balls fly through the air, not roll or slide on the
ground. This question has no single answer. Before getting too deeply into
this, I need to modify your question a little. The force does not determine
what happens to the soccer balls (see FAQ
page), so let us change " same kicking force" to "same kicking
impulse"; impulse is essentially force multiplied by the time it was
applied. That is also more convenient because the change in linear momentum
(mv) is equal to the impulse (J). So, with equal impulses, the
lighter ball has a larger initial speed: v_{light}=(m_{heavy}/m_{light})v_{heavy}.
So, if both balls are launched at the same angle and if air drag can be
neglected, the lighter one will go farther because it started with a
larger speed. Maybe that is all you wanted. But, it is altogether possible
that air drag will not be negligible. The force of air drag may be roughly
approximated as F=ĽAv^{2 }for a sphere where A
is the crosssectional area of the balls. (This is true only for SI units
since Ľ is not dimensionless.) So now you see that the ball launched the
fastest experiences the larger air drag slowing it down. But the heavy ball
has even more advantage since Newton's second law says a=F/m so even if the two balls
experienced the same force, the less massive ball would have a larger
magnitude of acceleration, slow down faster. Now it becomes complicated
because you have to know the masses, the impulse, and the angle of the
launch to calculate which would go farther.
QUESTION:
Let’s say you have a very long strong tube uncapped at both ends, and laying at sea level. It would have one atmosphere pressure inside. Put a cap on one end and stick that end down to near the bottom of the Mariana Trench. Now pop the cap off. Wouldn't this create a perpetual geyser from the end at the surface? It seems to me that it would come out with enough pressure to run a generator. Would it be considered perpetual motion or just hydraulics? Why wouldn’t this setup solve the world’s clean energy needs? Please pop my bubble. I can't be the first to think of this. What am I not accounting for? Inside the tube I don’t think gravity would be a player. Neither would the pressure fading outside the tube.
ANSWER:
Once the water reached the top of the tube the column of water would be in
equilibrium, just the same as if you had put the tube down without capping
the bottom. You might get a spurt at first but then you would just be left
with a column of water in equilibrium. I guess you could analyze it roughly
by ignoring viscosity and the drag of water moving on the pipe. Suppose the
crosssectional area of the tube were A, the depth of the bottom is
D, and the density of water is
ρ .
At some time after uncapping, the distance up to the current level of the
water in the tube I will call z. The force up on the column of water
would be F_{bottom}=(P_{atm}+ρg D) A,
the weight of the column of water is W=ρgzA,
and the force down on the top is
F_{top}=P_{atm}A. The net force on the
column is F= ρgA (Dz)=ma= ρAza
where a is the
acceleration and m= ρAz
is the mass. So, a= ρgA (Dz)/ ρAz=g [(D/z)1].
Suppose we take g≈10 m/s^{2} and D≈10^{4} m;
then a≈10[(10^{4}/z)1] m/s^{2}. The graph of
the acceleration is shown to the right; note the logarithmic scale for
acceleration to show the huge variation from bottom to top. Note that it is
huge, about 10^{5} m/s^{2} for 1 m and falls by four orders
of magnitude when the column of water is halfway to the surface, finally
falling to zero when the pipe is full. Suppose we start with z=1 m
and the column at rest (where a≈10^{5} m/s^{2}).
After 10 milliseconds (10^{2} s) the column will have acquired a
speed of approximately v=at=1000 m/s and the water level would have
risen approximately to z=1+˝at^{2}=6 m. Now, the speed
is almost immediately huge and so the assumption of no frictional forces
becomes impossible. The drag of due to the water moving in the tube for the
column of water going 1000 m/s will be huge
and keep the water from speeding up so rapidly. Also, the air above the
column of water will not just get pushed up will get compressed and
therefore push down on the column with a pressure greater than atmospheric.
It is therefore clear that my simple calculation above is only a best case
scenario and the drag and added pressure will take energy away. The drag
might get big enough that the water in the tube would boil and release steam
to futher impede its progress. I believe that you will end
up with column of water slowing down as it approaches the surface of the
ocean and it will just come to rest there. In any case, when the tube was
filled, even if still moving, the column of water would have zero net force
on it. If you tried to have it do work on your turbine, it would soon loose
all its kinetic energy and stop. Alas, no clean energy solution!
ADDED
THOUGHT:
Robert M. Wood pointed out to me that the most energy you could possibly get
out of the rising water would be the amount of work you performed to get it
down there. If you neglect the weight of the tube itself (or, imagine first
lowering the tube open at both ends and then pushing the water down with a
piston), this work would be
W=˝ρgAD^{2}.
Neglecting any energy loss, this would have to be equal to the kinetic
energy of the column of rising water when it got to the top, K=˝ρDAv^{2},
so v=√(gD). For the case above, v=316 m/s, pretty
modest. And keep in mind that all the water in the tube is now in
equilibrium, so as the water goes above the surface, it is slowing down as
it gains potential energy. And, if you imagine the tube projecting high
above the surface, you can calculate how high h the water will rise
before falling back. The center of mass of the column of stopped water is
y=˝h and its mass is ρhA, so mgy=˝h^{2}ρgA=˝ρDAv^{2},
so h=v√(D/g)=D. Therefore, with no friction, the
water would rise up and oscillate back and forth for ever. But, as argued
above, friction will be extremely important so this oscillation will quickly
damp out and will never rise up to a height equal to the depth of the ocean.
See the video in the answer to the followup question below as an example for
a miniature of this.
FOLLOWUP
QUESTION:
Thank you for finding interest in my question and for being able to let the nuts and bolts slide. The calculations you created from my questions were like a body punch to my self esteem. I didn't understand a lick of it. And now I'm feeling like the Dickens character asking for more gruel. Your response knocked me for a loop. You pointed out two things I wasn't accounting for. The weight of the water in the tube and the air resistance. The air resistance I've resolved/incorporated.
The equilibrium (if it occurs) from the weight of the water in the tube is a stopper and I hate it. Because it seems so common sense. Then I thought of this,(1000 ft.depth= 30 atmospheres = 420 psi.~), in your model if you bring the tube up at a 70 degree angle instead of the assumed 90, because of the longer tubing, there wouldn't be enough pressure to fill the tube. Also, if you put an uncapped tube from the surface to depth at a 70 degree angle as soon as you reached depth the tube would have too much water in it for the pressure and water would have to be expelled from the bottom until it reached the length of the 90 degree tube length. (I can't even do the baby geometry needed to figure out how much longer the tube would be.) Neither of these seems right to me. What do you think?
What I picture happening is the moment the cap is removed at depth the whole ocean rushes to fill the void at the top and the water pressure at the outlet would be the same as at the intake. The entire tube would become an extension of the 420 psi zone at 1000 ft. If inside the tube was a vacuum and capped at the top wouldn't this be the case? Wouldn't the pressure be 420 psi at the cap?
Here is my main reason for not being able to let this go. (this is my first version of this train of thought) Same long strong tube only this time it is assembled and capped in the vacuum of space 500 miles above earth and the capped end brought down to the surface. When the cap is removed I expect the atmosphere at that point to start rushing up the tube, through it, and out the end in a very long term geyser. And even though the tube is 500 miles long equilibrium would never be reached.
Aren't both examples using the same principal? It seems to me that it is the same principal and that the water at depth would view this one atmosphere pressure at the surface as the next best thing to a vacuum and continue trying to fill it. Just as the tube from space would leak away our air.
Doesn't pressure equal stored energy?
This thought is like a song stuck in my head. I'm tired of thinking about it. It's been months now. Validation would be best for all but, at this point I would settle for a kill shot. Can you help (either way)?
ANSWER:
You have so many misconceptions that I am not even going to try to
convince you with the physics arguments. A couple of comments: The reason
that the pressure is so high deep in the ocean is that the water deep down
must hold up the weight of all the water above it; to to try to ignore
weight of the water in the tube misses the whole point. Suppose you had a
tube 10 km long full of water (not underwater). The pressure at the bottom
would equal to the weight of the tube of water divided by the area of the
tube (+P_{atm}). Also, I have
previously answered your
question about the air in a tube in the atmosphere—it would not suck out all
the air in the atmosphere for the identical reason your water tube would not
create a perpetual geyser. Since you do not know any physics, let me
convince you another way; remember physics, at its heart, is an experimental
science, so let's do an experiment. Your idea should work equally well if
the ocean were only 5000 m deep, maybe not just so robust a geyser, right?
That is, your idea should scale. So, try the following experiment. Take a
straw and put a little cork to plug the bottom and push it down into a glass
of water; pop that cork out some way and see what happens. Does the water
come up the straw and continually squirt out the top (tiny geyser) or does
it come up to the level of the water surface and stop? Maybe it will rise a
bit above the surface but it will settle back to the level of the water in
the glass.
QUESTION:
Suppose we have a horseshoe magnet. Now we bend it in such a way that it becomes doughnut shaped and poles remain in contact with each other. In this situation what will happen to the magnet? Will it behave as a magnet? Where would be its poles? What will happen to domains inside the magnet?
ANSWER:
A horseshoe magnet is just a bent bar magnet, so let's start there. I have
shown a short stubby bar magnet, but you can imagine that as it gets longer
relative to its width the field inside will get more and more uniform. Now,
bend it around into a circle and it will look just like a toroid. The
magnetization inside will remain pretty much the same as before you bent it
and there will just be field inside the torus, not outside. The domains will
all stay pretty much the same.
QUESTION:
the Coriolis effect..
on a freely falling body... is to east...?
motion of earth is towards east right.? so we must feel that falling body deflects to the west isn't it..???
ANSWER:
I cannot give the full derivation of the motion of a particle in a rotating
coordinate system, it is much too involved. I can give you the results,
though. The coordinate system we will use is the coordinate system (x',y',z')
shown to the right. The Coriolis force is given by 2mv'x ω
where v' is
the velocity of m and
ω is the angular
velocity of the earth ( ω≈ 7.3x10^{5}
s^{1}). Now, for a body dropped from some height h, the
direction of the velocity is in the negative z' direction, so the
direction of
v'x ω
is east. This is not what
you would intuitively suspect (as you note), but it is true. An expression
for how far eastward it would drift before hitting the ground is x'=[( ω·cosλ)/3]√(8h^{3}/g)
where λ is the latitude. For example, at the equator where λ=90^{0},
and you drop it from 100 m, the deflection would be x'=2.2x10^{2}
m=2.2 cm.
QUESTION:
What is impulse? My text book explains this as force acting for a short interval of time. Can you please explain this concept?
ANSWER:
What is the effect of applying a constant force F to a mass m?
Newton's second law tells you that you cause the particle to accelerate,
that is if you exert the force over some time t the velocity will
change from v_{1} to v_{2}. Writing Newton's
second law, F=ma=m(v_{2}v_{1})/t.
Rearranging this equation, Ft=m(v_{2}v_{1});
so you can see that the amount which the product mv (which is called
the linear momentum p) changes is numerically equal to the product
Ft which is called the linear impulse J. If the force is
constant, the time need not be short; t could be an hour and J
would still tell you the change in p. I hope you can appriciate that
J=p_{2}p_{1} is simply an alternative
way to write Newton's second law. If F is not constant, you can still
calculate J but it is just harder. In essense what you do is
calculate Ft for many vanishingly small values of t all along
the path the particle takes and sum them to get the total J. (If you
know integral calculus, this is just integrating Fdt over the
time interval.) Of course, Newton's second law is a vector equation, so J
and p are vector quantities, J=p_{2}p_{1}.
If you have a system of particles which interact only with each other, the
net force on the whole system is zero because of Newton's third law, so
J=0 and p does not change as the system moves
around; this is called momentum conservation. Here p means the
momentum of the whole system, the vector sum of all the momenta of all the
particles.
QUESTION:
I've started reading an elementary particle textbook and it doesn't explain why a moving charged particle radiates an EM field. I was hoping you could help with that.
ANSWER:
The semantics of "radiates an EM field" is somewhat ambiguous. A charged
particle at rest creates a static electric field. A charged particle
moving with constant velocity (constant speed in a straight line) creates
both an electric field and a magnetic field and both change with time. But
neither of these situations are said to radiate electromagnetic
fields. Electromagnetic radiation propogates through space as waves; visible
light, for instance, is an electromagnetic wave. Electromagnetic fields can
be created when a charged particle accelerates. E.g., an antenna radiates
radio waves when electrons in the antenna are made to oscillate back and
forth (accelerating). The derivation of how accelerating charges radiate is
a topic in an intermediate E&M course and beyond the scope of this site.
QUESTION:
In the famous book by H.G Wells  "The Invisible Man"  the invisible man explains that his invisibility is due to light getting passed through a membrane (his skin/tissue) clearly , i.e not reflecting or refracting. Is this true? Can a object me made invisible by this method? Further, has there been some actual research in this topic and any advancements?
ANSWER:
First, it is not "true" —this
is a book of fiction! More pertinently, is it possible? I would say, given
what we know today, it is not possible because for the light to not be
refracted the speed of light in the body would have to be the speed of light
in a vacuum, and there is no material for which this is true. Research
regarding invisibility is focused on "invisibility
cloaks".
QUESTION:
A charged parcticle when accelerated radiates electromagnetic radiation or light.
All atoms on earth are accelerated therefore the electrons and protons which are charged are also accelerated so why isnt light radiated?
ANSWER:
First of all, atoms are neutral, have zero net charge. Second, the
accelerations involved because of the earth's motion are very small.
QUESTION:
Why does centripetal force do what it does?
It is a force so why does the object start moving in a circular path and not simply in the direction of the force?
Also it should provide acceleration...so even if some how the object move in circle why doesnt it spiral in??
ANSWER:
If an object is at rest and you exert a force on it, that force will not
cause the object to "start moving in a circular path". You cannot simply
label something a centripetal force. If an object is moving in a circle with
constant speed, it is accelerating because the velocity is constantly
changing. The velocity vector can change either by changing its magnitude
(speed) or its direction. An object moving in a circle therefore requires
some force to accelerate it, for example the sun's gravitational force on
the earth, which points toward the center of the circle. If you stopped the
earth in its orbit, the gravitational force would indeed cause the earth to
crash into the sun. You need to simply look up a derivation of centripetal
acceleration in any elementary physics text you would understand this
better.
FOLLOWUP QUESTION:
unhuh...
how can it change direction without changing its magnitude??
therefore in time dt
the velocity in downward direction becomes adt
where v' is equal to the vector sum ie v'=((v)^2+(adt)^2)^1/2,
so how is the velocity supposed to be constant when the object is not
decelerating tangentially?
ANSWER:
Like I said in my original answer, look this up in an elementary physics
text — I
am not a tutor. The reason your argument is invalid is that dt is
infinitesimal so if adt is perpendicular to the
velocity vector v, the result is to change direction. (Limit(as
dt —> 0){ √[v^{2} +(adt)^{2}]}=v.) Just use
your common sense: a car going in a circular path has a constant magnitude
of velocity, right? And nonconstant direction or velocity, right?
QUESTION:
Two bodies A and B are at rest and in contact with each other. Now if some arrangements made body A exerts a pressure of 10 NEWTON on body B then according to NEWTON'S 3rd law of motion body B will also exert an equal force of 10 NEWTON in opposite direction so the resultant force should be zero, while practically we see that there will be a resultant force of 10 NEWTON acting on body b and if mass and other conditions of body B are such that it moves by applying 10 NEWTON force on it. Then it will start moving. HOW?
ANSWER:
You have this all confused. But, you are not alone! When doing this kind of
problem, you must choose a body to focus on; only forces on that body
affect its motion. So, if you are interested in what body B will do, you
look only at that body. The force which B exerts on A is not a force on B.
So, if B has a mass of, say, 2 kg and there are no other forces on B, it
will have an acceleration of a_{B}=10/2=5 m/s^{2}.
Since the 10 N force continues, A and B remain in contact, so a_{A}
must also be 5 m/s^{2}. One force on A is the 10 N force from B
which is in the opposite direction as the acceleration. But the net force on
A must surely be in the direction of the acceleration, so there must be some
other force (probably due to you pushing on A) on A which is bigger than 10
N and points in the direction of the acceleration. For example, if the mass
of A is 4 kg, F10=4x5=20 N, so F=30 N. Finally, you could
look at A and B together as the body. In that case the forces they exert on
each other do cancel, the acceleration is 5 m/s^{2}, and the total
mass is 6 N; therefore, there must be some external force (you again)
causing this 6 kg to have an acceleration of 5 m/s^{2}: F=ma=6x5=30
N. All three ways of looking at this problem are consistent with each other.
QUESTION:
Can I make a simple bike reflector using 20 thru 40 degree angles?
Everything I have read uses 45 and 90 degree angles.
ANSWER:
I do not know how 45^{0} would work. The point is for the incident
light to be reflected in exactly the opposite direction from the direction
it came in. The only way I know to achieve this is with
corner reflector
which is three mirrors connected as in the inside corner of a cube.
QUESTION:
Why simple harmonic motion is called simple?
ANSWER:
Take a look at the figure above. The red curves are all pure sine waves, just
having different frequencies. These are all called simple harmonic motion
because they may be expressed as a single sine wave. Now, take all four of
these and add them together. The result is the black curve. This curve is
still harmonic (which means that it is periodic, it repeats itself after
some time, called the period, has elapsed) but it is not "simple" because to
describe it you must use four different sine waves.
QUESTION:
I am a Martial Arts instructor who also happens to have a BS in mathematics (well, statistics, sorta like math). I like to explain to my students the science of the art in order to help ground them in reality from a lot of the BS involved in Eastern Mysticism, such as board breaking.
My question is this, wooden weapons, such as nunchaku are usually either round/cylindrical or octagonal. A common technique is to place the weapon lengthwise down the arm and block an incoming attack with the wood. Traditional tonfa are octagonal, but modern police batons (with the side handle) are round. Which would protect the blocking arm from a baseball bat strike (for example) to a greater degree? Would an octagonal shape disperse the incoming force more than a round shape? I believe the octagon would spread the applied energy to a greater degree than a cylindrical/round shape and allow less energy to travel through the wood and into the arm, but I cannot find the math to back this up.
ANSWER:
The only reason that I would say that the octagonal shape would be better is
that it has a larger area of contact with the arm and therefore would
distribute the applied force over a larger area. For example, you would not
just want a point in contact with the arm since all the force would be
applied at a single place. Even better in this regard would be a rectangular
shape. This is probably a negligibly small effect, though. I cannot see how
the shape would have any effect as you suggest to " … allow
less energy to travel through the wood …"
QUESTION:
How much energy is required to move the electron sufficiently far away from the proton such that it does not experience the proton's electric field. I am kind of confused by this question. The effect of electric field will never end as far as we take the electron from the proton. And if we use the relation V=E.L then if E=0, then
L will be infinity?
ANSWER:
Technically, yes the field will be exactly zero only at L= ∞.
But,
that is only if your proton and your electron are the only things in the
universe! You can calculate the ionization energy, the energy necessary to
move an electron from the ground state of hydrogen to infinity; it is 13.6
eV=2.18x10^{18} J. You can also calculate the energy necessary to
move it to any other distance. For example, the energy to move it to about
0.5 μ=5x10^{7} m (which is about 100 times the radius of the
hydrogen atom) is 13.599 eV, essentially the same as to infinity. So you can
see that most of the energy is supplied in close to the proton because that
is where the field (and therefore the force) is strongest.
QUESTION:
So I've been trying to figure this for a while, if a positively charged and a negatively charged quark orbit each other or actually come in contact, also could they form a functional "atom" and be able to have electrons interact with them?
ANSWER:
Twoquark particles are called mesons (one quark plus one antiquark). If you
had a positively and negativelycharged quark you would have an uncharged
meson and so it would not bind an electron and you could not have a
meson+electron atom. However, if you had a positively charged pion, it could
bind an electron. For example, a positive pion
π^{+} is
composed of an up quark (charge +2e/3) and a down antiquark (charge +e/3)
and would have a charge +e and be able to form an "atom" which you might
call pionium.
ADDED
NOTE:
Whoops! Pionium is an "atom" composed of one
π^{+} and one
π^{}. I cannot find any reference to this pion+electron
system, so I guess you can call it whatever you like.
QUESTION:
i love physics and this question i asked to my teacher and principal but they couldn't answer it so my question is about third law of motion "every reaction has an equal and opposite reaction" so when a truck moving with constant speed hits a stationary car so according to newton's 3rd law of motion truck shoud be stopped after collision because car applies equal force on truck which it have during collision.
ANSWER:
Why should it be stopped? Certainly the truck experiences the force which
the car exerts on it, but every force does not have the effect of stopping
the object which experiences a force. If you are running toward a fly which
is hovering at rest in your path, you feel the force of the fly but it does not
stop you.
QUESTION:
Hi ...here is a link to Felix Baumgartner's freefall jump from space 
https://www.youtube.com/watch?v=vvbNcWe0A0
135,890 feet  or, 41.42 km (25.74 mi)
In the video you see speed at which he is supposedly travelling, 729 mph (1173km/hr), 46 seconds after he jumped. The footage is cut then seconds later speed strangely falls to 629 mph...anyway after 4 mins and 18 seconds of free falling he releases parachute. My questions are What speed would he be going at a 4 mins and 18 seconds into freefall? Also how do you explain the deacceleration when it was at 729 mph then down to 629 mph?
ANSWER:
(You might be interested in an
earlier answer.) There is a better video at
https://www.youtube.com/watch?v=raiFrxbHxV0; this video shows data
acquired by instruments on Baumgartner. Here some clips from that video:
The first three are the times also
showing speeds; these are roughly the speeds you refer to in your question.
The second three show speeds and altitudes. The third graph shows the whole
history of altitude, speed, and mach speed. The speeds are all larger than
indicated in your video. The speed at 4:20, 113 mph, answers your first
question. The speed at 0:50 is about the maximum, 847 mph or mach1.25. The
speed at 1:01 has dropped to 732 (about the same change as your video); I
suspect this deceleration is due to two factors: there is getting to be much
more air resulting in higher drag and, probably after he broke the record he
oriented his body to increase drag and slow down. The official maximum speed
after everything was calibrated was 843.6 mph.
QUESTION:
Will the drag coefficient be the same in air and water?
ANSWER:
There is no simple answer to this question. The drag coefficient C_{D
}is a constant characterized by the geometry of an object used to
calculate the drag force F_{D} if it is proportional to the
square of the velocity v: F_{D}= ˝ C_{D} ρAv ^{ 2 }
where A is the area
the object presents to the fluid flow and
ρ
is the density of the fluid.
Whether or not this equation is true depends on a quantity called the
Reynolds number Re=L ρv / η
where L is a length along the direction of flow and
η
is the viscosity of the fluid. It turns out that only if Re>1000 is
the velocity dependence approximately quadratic. If Re<1 the drag
force is approximately proportional to v. Anywhere between these
extremes the drag force is a combination,
F_{D} ≈˝ C_{D} ρAv ^{2}+kv.
C_{D} ≈constant
only for the case Re>1000; the drag coefficient would then be the
same for identically shaped objects. It is interesting, though, to get a
feeling for what the relative speeds in air and water corresponding to Re=1000
are. The necessary data at room temperature are
ρ _{air}≈1
kg/m^{3},
ρ _{water } ≈1000
kg/m^{3},
η _{air}≈1.8x10^{5}
Pa·s, and
η _{water}≈1.0x10^{3}
Pa·s. Then v_{water}/v_{air}≈(1.0x10^{3}/1000)/(1.8x10^{5}/1)=0.056
(which is true for any Re). Taking a sphere of diameter 0.1 m as a
specific example for the critical Re=1000, C_{D}=0.47,
A=7.9x10^{3} m^{2}, L=0.1 m, v_{air}=1000x1.8x10^{5}/(0.1x1)=0.18
m/s and v_{water}=1000x1.0x10^{3}/(0.1x1000)=0.01
m/s. Since the speeds are relatively low, you can conclude that for many
cases of interest the quadratic drag equation holds for both water and air.
Therefore, for many situations you can approximate that the drag coefficient
in air and water are about the same. For a specific case you should check
the Reynolds numbers to be sure that they are >1000. The bottom line is that
if the drag force depends quadratically on velocity, the drag coefficient
approximately depends only on geometry, not the properties of the fluid.
Keep in mind that all calculations of fluid drag are only approximations.
ADDED
COMMENT:
As I said above, for 1<Re<1000, F_{D} ≈av+bv^{2
}where a and b depend on Re; e.g., a≈0 for
Re>1000 and b≈0 for Re<1. So, for Re<1, a
is just a constant and F_{D}≈av≡˝ C_{D} ρAv ^{2}
so C_{D}=2a/( ρAv )=2aL/(A ηRe )≡c/Re
where c is a constant. If F_{D} is proportional to
v, C_{D }is inversely proportional to Re; this is
called Stokes' law. As we saw above, Re>1000 leads to C_{D}=constant.
Ferguson and Church have derived an analytical expression which very
neatly reproduces data for the transition from Stokes' law (1/Re) to
constant C_{D}. Calculations for a golf ball are shown at the
left (constant C_{D}, the green line, is called turbulent in
the legend). Note, as we have stated, that the transition occurs in the
region
1<Re<1000. Note the sudden drop in the data around Re=500,000.
I believe that this must be due to the
dimples on the golf ball which
reduce drag.
QUESTION:
In induced magnets, why
does the end nearer the magnet have the opposite polarity to that of the
magnet?
ANSWER:
Think about a piece of iron. It is like a whole bunch of tiny bar magnets,
each of which has a N and S pole; but the iron is usually not a magnet
macroscopically because all the tiny magnets are oriented in random
directions. But, if you bring the north pole of a permanent magnet up close
to the iron, the south poles of all the tiny magnets turn to point toward
the north pole of the permanent magnet.
QUESTION:
I think if an object is turning, it has more gravity than an object which is not turning
ANSWER:
You are right. However, it really has nothing to do with the turning, per
se. When something is turning it has rotational kinetic energy and
therefore a spinning planet has more energy than an otherwise identical
planet not spinning. In general relativity you usually hear about gravity
being caused by mass warping spacetime. However, mass is just the most
obvious source of gravity and what it is which really warps spacetime
is energy density and mass has a lot of energy (E=mc^{2}). The energy due to the turning
is infinitesmal compared to the mass energy of the object so you would never
be able to distinguish the difference due to the turning by looking at the
gravity.
QUESTION:
I've heard that a bullet falls to the ground
at the same speed [in the same time] no matter the charge pushing it forward. More powder and it moves faster but vertical speed is the same. I imagine two equal weight gliders of 20/1 and 30/1 lift/drag ratios would also express the same vertical speed?
My question is, wouldn't this gravity effect also hold true underwater? An underwater glider expressing 30/1 L/D would travel 50% further, faster than a 20/1 L/D in a given vertical?
ANSWER:
I will assume that we are talking about horizontally fired bullets. All
bullets do not hit the ground with the same speed, rather they all take
(approximately) the same time to get to the ground; I have corrected your
question because I believe that is what you meant. Your comparison to
gliders with different lift/drag ratios is not appropriate, because the
thing which results in the times of vertical fall being the same is the
assumption that there is no lift at all. In other words, the vertical speed
of a bullet will always be much less than the terminal velocity so there is
no appreciable component of the drag in the vertical direction. (This is not
true if the gun is fired from a very high altitude since there will be
enough time for the verticle component of the velocity to approach the
terminal velocity.) Underwater (I am just talking about bullets, not
gliders) the bullet will not behave the same because the terminal velocity
is much smaller than in air, in other words the drag is much larger.
QUESTION:
Quantum mechanics has popularized the idea of very small particles existing in two or more points in space during the same instant in time. Is this to be taken at face value as a fundamental aspect of quantum particles, or is this simultaneous existence just a concept used to express our inability to observe these particles without inadvertently altering their speed or location?
ANSWER:
See an earlier answer.
QUESTION:
I read from my history book that when nukes were first created we didn't know that the radiation their explosions produce is harmful to one's health, even to one's life maybe. So my question is, how come neither physicists or medical personnel ever figured out that radiation can be dangerous until we had first victims of radiation?
ANSWER:
Ever hear the expression "hindsight
is 2020"? It is unrealistic to expect that when something new is
discovered that we should somehow know all the effects that might have on
anything else. At the time radioactivity was discovered, nobody even knew
what atoms were composed of or what their structure was. And it was found to
be very tiny bits of matter (e.g. what we know as electrons today)
and who would have thought that getting hit by something trillions of times
lighter than a speck of dust could be harmful? It took experience before it
was appreciated how dangerous it could be. One of the best known examples of
such experience is the case of
radium watch dials.
Radium, mixed with a phosphor, was painted by women onto watch dials and it
would glow in the dark. The workers were encouraged to point their brushes
with their lips to make the fine lines required. Subsequently many became
ill with cancer and other radiation sickness. It seems stupid now, but the
dangers were not known until they were discovered. Also many of the health
effects were longterm effects, the effects not appearing until years or
decades after exposure. One example I know about from personal experience is
the effect of radiation treatment for acne which was popular in the 1950s. I
very much wanted to have this done but my father forbade it, making me
furious with him. Years later people who had had this treatment started
coming down with cancer —thanks,
Dad! Yet another example is the shoefitting xray machine. When I was a kid
it was really fun to buy new shoes because you could look down at an xray
of all the bones in your foot and how well the shoe fitted you; again, when
more was learned about radiation, these machines were all sent to the scrap
heap. When it began to be appreciated that there were dangers, studies began
to be done to try to set allowable dosage levels. But, it is unethical to do
such experiments on people, so lab animals had to be used which always
presents problems with scaling and other variables. Much of what we know
today was learned afterthefact by doing longterm statistical analyses
over decades of medical records.
QUESTION:
Magnetic flux according to my book is total no. of magnetic field lines passing through a given area in magnetic field. ok but why there are not infinite no. of magnetic field lines, because magnetic field line are defined as the path that a magnetic north monopole would take if left in north part of magnet, so if i take a monopole and leave one atom away from the previous position then it should take a slightly different path and that path should be considered as magnetic field line,so in this way i can draw millions of line.
ANSWER:
Magnetic flux is well defined:
Φ_{M}≡∫∫B·dA,
the area integral of the magnetic field. If the integral is over a closed
surface, the flux is zero; this is the famous situation which tells you that
there are no point sources (called magnetic monopoles) of magnetic fields
like there are for electric fields. If you like, you may interpret this as a
number, but that is not really fundamental. If you talk about uniform
magnetic fields which are perpendicular to a plane surface, Φ_{M}=BA
, is the flux through an area A. Then, if you interpret
Φ_{M}
as a number, you would simply say, for example, there were 10 lines through
an area of 1 m^{2} if the magnetic field is 10 T. You could ask the
same question about electric flux,
Φ_{E}≡∫∫E·dA .
This is perhaps a little easier to understand because you can have point
charges. For example, if you have a 1 C point charge, the electric field 1 m
from it is E=Q/(4 πε_{0}r^{2})=1/(4x3.14x8.85x10^{12}x1^{2})=9.27x10^{9}
N/C. The flux at a distance of 1 m from the charge would therefore be Φ_{E} =EA= 9.27x10^{9}x4x3.14x 1^{2}=1.17x10^{11}
Nm^{2}/C. Incidentally, this is the flux regardless of where you
measure it because the area of the sphere
(4 πr^{2})
surrounding the point charge appears both in the denominator of the field
and in the numerator of the flux and cancels. Hence, you could say that
there were 1.17x10^{11} lines of
electric field emanating from a 1 C point charge. You can see this from
Gauss's law,
Φ_{E}=∫∫E·dA=Q/ε_{0}
if the integration is over a closed surface enclosing the charge Q.
QUESTION:
Why degree of monochromaticity is always non zero?
ANSWER:
The practical reason is that you cannot make a laser which is perfectly free
from electronic instabilities and noise. But even if you were able to make a
perfect laser, you could not know its frequency perfectly. The more
fundamental reason is that monochromatic light means light with a single
frequency. But, the linear momentum of the photons is proportional to the
frequency of the light. Because of the Heisenberg uncertainty principle, the
only way you can know momentum of the particle perfectly is to be completely
ignorant of its position; you need an infinitely long wave to know its
frequency perfectly.
QUESTION:
This question is about relativity. If person leaves earth and heads toward a star that is 10 light years away, and he travels at near the speed of light, time will slow down for him. The round trip to the star may take him only one year, but when he returns, maybe 100 years have passed on earth. When he started out to the star, he knew it was
50 [I think you meant 10,
right? The Physicist] light years away. It seems that for him, he will have traveled 20 light years in one year. That is faster than the speed of light. What don't I understand?
ANSWER:
First of all, time for the traveler will not slow down; rather, her clock
will run slower as observed by an observer on the earth. To her, time will
run perfectly normally. However, she will observe the distance to the
destination shortened because of length contraction. You have rounded things
to what they would be if she were going the speed of light, but let's do the
whole problem as if she were going with a speed v=0.999c. The
earth observer would see an elapsed time of 100x0.999=99.9 years (your 100
years). The traveler will see the distance shrunk to D'= 10 √(1.999^{2})=0.447
light years, so the time of travel is t=2x0.447/0.999=0.895 years
(your 1 year). You should read the earlier answer on the
twin
paradox.
QUESTION:
Supposing a weightless container is filled with water. I am sure the pressure at the bottom of liquid, P1 = atmospheric pressure + height of liquid x density of liquid x g = Patm + hdg, where Patm is atmospheric pressure and d is density of liquid. We can calculated this pressure as if liquid in region A and Region C does not exist.
But how about the the pressure at the base of the container, that is P2. Is P2 same as P1? For P2, do we need to consider the whole weight of the liquid, that is inclusive of the weight of water in region A and C?
ANSWER:
It depends on what the force on the bottom is. If the container is in
equilibrium, imagine it sitting on a table. The table would exert an upward
force equal to the weight W of all the water, so P_{2}
would be W/A_{bottom} where A_{bottom}
is the area of the bottom of the container. This assumes that atmospheric
pressure is the same everywhere in the vicinity of the container; in other
words, I have ignored the buoyant force due to the air on the whole
container because it will surely be much smaller than W.
FOLLOWUP
QUESTION:
Indeed the container is resting on the table. For P1, I use h x d x g. But for P2 you use W / Abottom, So can I say P1 not equal to P2 ?
ANSWER:
Yes, but I have to admit that my answer was misleading in that I gave you
the gauge pressure, the pressure above atmospheric. So I should have said
that P_{2}=P_{atm}+W/A_{bottom}.
There is no problem that P_{2} ≠ P_{1}
because the force which the container exerts on the table is not P_{1}A_{bottom}.
Think about it —the
sides of your container exert a downward force on the bottom of the
container.
QUESTION:
Think of words (incapital letters) that can be read properly both with a mirror and without a mirror.
What are these words?
ANSWER:
Well, this is not physics and I usually do not answer such questions.
However, it is a cute puzzle. First, any letter in the word which is
symmetric upon reflection about a vertical axis will look the same in the
mirror: A, H, I, M, O, T, U, V, W, X, Y. Second, any such word must be a
palindrome composed of the allowed letters, e.g. AHA, OTTO, WOW,
YAY, MOM, TIT, TUT, … .
But, wait, it is somewhat more difficult than that! Look at the pictures of
the Camel cigarette package and its reflection, in particular, the word
CHOICE on the side. So words composed of letters with symmetry upon
reflection about a horizontal axis will also reflect the same: B, C, D, E,
H, I, K, O, X will have a reflection which is the same as the original, but
only if turned upside down. There is no restriction for these words to be
palindromes; besides CHOICE, some others would be BED, HEX, BOX, BIKE, HIKE,
CHIDE, …
QUESTION:
What is meant by virtual image?
ANSWER:
It is an image formed by light which appears to be coming from
somewhere but is not actually coming from there. When you look into a
mirror, it appears that the light from the image of what you are looking at
is coming from behind the mirror, but it is actually coming from the mirror
itself.
QUESTION:
Supposing that initial
velocity is 0 Is a displacement vs. time (squared) equivalent to a Velocity
vs. time graph? I tried to find the slope of displacementtime^2: d/t^2 =
V/t which equals acceleration But what I read everywhere is that: d/t^2 =
a/2 according to the equation d= (initial V)(time)+ (1/2)(at^2) I don't what
is wrong with my calculations?
ANSWER:
If you plot d vs. t^{2}, the slope of the line
will be ˝a.
You know that d=˝at^{2}, so suppose that I call t^{2}=u.
Then clearly the slope of the line d=˝au is ˝a. Your
equation d/t^{2}=V/t is clearly
wrong: d/t^{2}= ˝at^{2} /t^{2}= ˝a
and ˝a≠v/t.
QUESTION:
Two of us disagree on part of a sol'n given by two people with Physics background, and I want to know if I am correct, or if I am missing something in the analysis of the problem...in case I have to explain it to a student.
Question concerning Forces/impulse..... 50kg person falling @15m/s is caught by superhero , and final velocity up is 10m/s.
Find change in velocity.
Find change in momentum .
It takes 0.1 sec to catch them.....ave Force is ?
answers are: vel = 25 m/s change in mom.. 1250 kg*m/s, and ave Force = 12,500 N.
Here's where we disagree: Person B says that 12,500 N is equiv. to 25 g ?????
They try to explain that 250 m/s^2 accel. corresponds to 25g.... I said it makes No sense at all, [ I know the accel. is 250, but that doesn't in any way imply a 25 g "equivalence" to me ]. They then went further to "prove" their point........Here is their argument...
500 N/g = 12500N / ( )g ..... I agree the ( ) = 25, but say there is No justification for the 500 N / g in the first place...... any ideas where it comes from , or how to justify that value ?
BTW I teach physics on and off at the HS level.... person B is an Engineer , I think
ANSWER:
Person B is wrong but has the right idea. (As you and your friend have
apparently done, I will approximate g ≈10
m/s^{2}.)
We can agree that the acceleration is a=250 m/s^{2} and that
is undoubtedly 25g. Now, we need to write Newton's second law for the
person, mg+F=ma=500+F=12,500, so F=13,000 N. This is
the average force by the superhero on the person as she is stopped, so the answer that the
average force is 12,500 N is wrong. When one expresses a force as "gs of
force", this is a comparison of the force F to the weight of the
object mg, F(in gs)=F(in N)/mg=13,000/500=26
gs; this simply means that the force on the object is 26 times the
object's weight. So neither of you is completely right, but if there is any
money riding on this, your friend should be the winner because the only
error he made was to forget about the contribution of the weight to the
calculation of the force. I am hoping that superman knows enough physics to
make the time be at least 0.3 s so that Lois does not get badly hurt!
QUESTION:
It is said gravitons are the
expected particles to exchange the gravitational force. Do these particles
exist in real?
ANSWER:
There is no successful theory of quantum gravity, so gravitons are "expected
particles" but I would not call them "real" at this stage; hypothetical
would be a better word.
QUESTION:
Why does it take more time to go to
Dubai from Mumbai than to come from Dubai to Mumbai?
ANSWER:
There are winds at high altitudes called
jet streams. The
prevailing direction of these winds is easterly and their speeds are as
large as 100 mph. Therefore when you travel east (as from Dubai to Mumbai)
you have a tail wind, and when you travel west (as from Mumbai to Dubai) you
have a head wind. Since the greatest speed an airplane can fly, say 600 mph
for commercial jets, is airspeed, that is relative to the air, the ground
speed with a 100 mph wind would be 700 mph going east and 500 mph going
west.
QUESTION:
Why is the direction of angular displacement along the rotational
axes?
ANSWER:
There are two ways in which you can rotate something about some axis, either
clockwise or counterclockwise (as viewed from one side or the other). So, an
angular displacement has two possible directions only. It is like a
displacement in one dimension, like a bead on a long straight wire. You then
call one direction on the wire plus and the opposite direction minus. The
only way to get a similar directional assignment for a rotation is to note
that the rotation axis is just like a onedimensional axis, one way being
assigned to be positive, the other negative. Usually this is done with the
righthand rule such that counterclockwise is a positive vector on the
axis.
QUESTION:
Recently I came across such an article that Gravity isn’t actually a Force, It's the bending of spacetime that cause objects to move on curve paths.
If two objects of equal shape and masses are there they will bend the spacetime in same way by same amount.
Then according to that there shouldn't be any kind of force or something acting between them which isn't true i guess. Gravity forces will act b/w those two. Could you please
explain me this contradiction.
ANSWER:
I do not understand your logic that because they both have the same effect
on spacetime, they will feel no force. Although it is just a cartoon to help
you qualitatively understand warping of spacetime, in this case it is
illustrative. If you place a bowling ball on a tranpoline, what happens? The
surface of the trampoline is warped. Now, suppose that you place two bowling
balls a few inches apart from each other. What happens when you let go of
them? They will behave as if there were a force between them and roll toward
each other but what they are really doing is responding to the shape of the
trampoline.
QUESTION:
I have a question about gravity, or more specifically how to calculate it for the purpose of a scifi fanfiction I am making.
So Earth has the density of 5.5 grams per cubic centimeter, total mass of 5.97219×10^{24} kilograms and radius of 6371 kilometers. So how do I put these together in a formula to get the 9.81m/s gravity?
ANSWER:
The density is irrelevant. You need to use Newton's universal law of
gravitation for two point masses. The force F which each feels is
F=GMm/r^{2} where G is the universal gravitational
constant, G=6.67x10^{11} N · m^{2}/kg^{2},
M and m are the masses of the two masses (take M as the
mass of the earth), and r is the separation between M and m.
This also works if the masses are spherically symmetric. But, of course, you
also know in your case that F=mg. So, if you solve for g, g=MG/r^{2} ≈ 6.7x10^{11}x6.0x10^{24}/(6.4x10^{6} )^{2}=9.8
m/s^{2}.
QUESTION:
Why do electrons fall back to the ground state after jumping into a higher orbital?
ANSWER:
The simple answer is that a system will always seek a way to move to a lower
energy, like a ball rolling down a hill and not up it or sitting still. The
more complicated answer addresses whether there is, indeed, "a way" to
achieve this. For atoms, the radiation of photons in the transition is
usually electric dipole radiation and there is a corresponding
quantummechanical operator, let's call it O_{E1}. Then, you
have to look at what is called the transition matrix element for this
operator, < Ψ O_{E1} Ψ'>=∫Ψ· O_{E1} ·Ψ'dτ
which is an integral over all space and Ψ' and Ψ are the
excited and ground states, respectively. If this is nonzero, then the decay
will decay with some half life determined by the value of the matrix
element. If it is zero, the state will still probably decay to the ground
state but via a different kind of transition.
QUESTION:
We all know about second law of motion.
The example is about the coin which is placed above the cardboard and cardboard is placed above the beaker... if the cardboard is quickly pulled then the coin falls in the beaker but if the cardboard is slowly pulled then why does the coin come with the cardboard?
Why does this happen?
ANSWER:
This is usually used to demonstrate inertia, Newton's first law; since you
are interested in why it does not work if you move the cardboard slowly, the
second law also must be used. To move the cardboard from rest it must
experience some acceleration. If the coin is to experience an acceleration,
that is move with the cardboard, some force must be exerted on it. The only
force on the coin horizontally is the static friction and the greatest it
can be is μmg
(where μ is the coefficient of static friction) so the greatest
acceleration the coin could have is μg; for example, if μ=0.3,
and you caused the acceleration of the cardboard to be any greater than
about 3 m/s^{2}, the cardboard would slip under the coin. But, if
you gave the cardboard an acceleration of 0.2 m/s^{2} the coin would
move with it.
QUESTION:
How can it be said that the universe is only 13.7 billion years old? This seems ridiculous to me that anyone can make this assumption. Taking in the fact that there are black holes almost as old as the universe itself how is this possible? Stars live billions of years so the first group of black holes would be billions of years after the "big bang". Also due to the fact that the universe is expanding way faster and not slowing down like predicted it would seem to me the current model is wildly inaccurate. Can you please explain this to me? My understanding is that stars form and die out all the time so to be able to accurately determine the age of the universe another method would need to be devised. It is my understanding that galaxies form with the help of super massive black holes.(Which aren't even holes at all but super compacted spheres of matter so dense light cannot escape created by countless other bodies of mass.) Which attracts matter into tight clusters of stars we call galaxies. These galaxies would have to take billions of years to form after the creation of the first stars ever created and after the first black holes were created from some of those stars. Does the current model explain this?
ANSWER:
As clearly stated on the site, I do not do astronomy/astrophysics/cosmology.
I can at least say a little about your question, though. The age of the universe is not an "assumption", it is based on careful analyses of many measurements. The stars in the
early universe were much more massive than the sun; this
resulted from the fact that the early universe was essentially all hydrogen and a bit of helium.
Stars began forming only 100200 million years after the big bang. Very large stars burn much hotter than smaller stars and therefore have much shorter lifetimes, a few million years rather than
a few billion years for less massive stars (e.g. the sun with an
expected lifetime of about 10 billion years) which would explain why many
black holes are very old. The rate of expansion of the universe is not
inconsistent with the model of a big bang about 14 billion years ago (in
fact, the rate is one of the measurements used to estimate the age of the
universe) even though we do not understand all the details (e.g. dark
energy and accelerating expansion).
QUESTION:
I would like to ask this question because I do not agree to this but it is written in my school handout. It says that work needs only two things to be calculated which is weight and height of displacement.
Now , there is this question that says that two persons who have the same mass went to the top floor of eiffel tower. One used the stairs and the other one used an elevator. The thing is that our handouts states that they both exerted the same amount of work done because they have the same weight and they both have the same height distance covered.
I do not agree simply because the man using the stairs carried his own weight up the distance while the the other one was being carried by an elevator and I immediately thought that the elevator was the one doing the work and not the person which is why they cannot have the same work done.
If I am wrong then I would humbly accept it. However , if our physics teacher is wrong, I would need a legit source stating about the problem and saying that it was wrong or similar problem like the one given with an answer so I can correct my grade. I would really appreciate anyone's help right now.
ANSWER:
The point is that the net work done on each person was the same.
Where the energy came from is different. On the stairs, the energy is
supplied by what the guy ate for lunch. In the elevator, the energy is
supplied by the motor lifting the elevator. Regardless where it came from,
mgh of energy must be somehow supplied.
QUESTION:
How surface tension overcome insect weight and help it to stay on liquid surface, as surface tension is along liquid surface?
ANSWER:
The weight of the insect causes the water to be depressed where the legs
touch the surface, so the surface tension has a vertical component.
QUESTION:
Can we make something like antigravity on Earth, or something similar to that? And i was wondering if we could make something like antigravity with magnets, to actually reduce the weight of some object. I was thinking to make a chamber with a magnet at the bottom, and take magnetized objects and put them on the top, so it will levitate.
ANSWER:
Your device will not work. The magnet on the bottom can be made to exert a
repulsive force on a magnet at the top which tends to lift your chamber.
However, Newton's third law requires that the top magnet exerts an equal and
opposite force on the bottom magnet, so the net force on the chamber is
zero.
QUESTION:
Is rotation a reason for earth being round?
ANSWER:
No. The reason that all large astronomical objects are nearly spherical is
because the force which holds them together is spherically symmetric. If you
have a point mass, the gravitational force another mass feels depends only
on how far apart they are, not on the direction. If you want a lot more
detail, see an earlier
answer about
gravity of a cylinder which shows how the tendency is toward a spherical
shape.
QUESTION:
Using a spring balance, weigh (separately) a piece of wood and a container with water in it. Then weigh them again, but with the wood floating in the water. Does the reading on the balance change? There will be an upthrust on the wood (Archimedes' Principle), causing it to weigh less. I guess there will be an equal downforce exerted by the wood, resulting in the same weight being shown in both cases. This seems a fairly easy question, but I can't find the answer clearly stated anywhere.
ANSWER:
The measured weight will be the sum of the individual weights. The buoyant
force is the force which the water exerts on the wood; Newton's third law
requires that the wood exert an equal and opposite force on the water. As
you correctly "guess", these two forces cancel out.
QUESTION:
Good afternoon. I started learning quantum mechanics and I have a question about Heinsenberg's principle. I was thinking about a photon created by an electronpozitron annihilation. As I understand (and I hope I am not wrong here) all observers, no matter their state, can specify the point (and all will specify the same point) where the photon appears, and I think in a Wilson chamber we can do this thing. Anyway, we can specify its position with an error less than infinity. But at the same time we know it's speed with absolute precision, as, assuming we measure it in vacuum, it's speed is invariably c. So the product between (delta)x and (delta)p will be 0, as (delta)x is not infinity and (delta)p is 0. In other words we can predict it's position with a certain accuracy knowing at the same time it's speed (I am not sure, but i think that the (delta)p is not a vector but a scalar quantity). What is wrong about this as it seems the Heinsenbrg's principle is violated?
ANSWER:
The mass of a photon is zero but it has momentum. Therefore your notion that
linear momentum p is mass m times velocity v must be
wrong; the relativistically
correct expression for momentum is p=mv/ √[1(v/c)^{2}]
where c is the speed of light. For a photon, this is a little tricky
because p=0/0, but you can also write that E^{2}=p^{2}c^{2}+(mc^{2})^{2}
where E is the energy of the a particle of mass m, so p=E/c
if m=0; since E=hf for a photon, where h is Planck's
constant and f is the frequency, the linear momentum of a photon is
p=hf/c . Therefore the frequency of the photon must be
uncertain according to the uncertainty principle. Since the energy of a photon is
hf, there will be an uncertainty in the energy of the photon.
QUESTION:
The units of the Hall coefficient is m^3/C that's, cubic meter per coulomb. How so we get this?
ANSWER:
You just need to know the
definition of the
Hall coefficient R_{H}, R_{H} ≡V_{H}t/(IB)
where V_{H} is the Hall voltage, t is the thickness of
the sample, I is the current, and B is the magnetic field. The
Hall voltage may be
shown to be V_{H}=IB/(nte) where n
is the charge carrier density (units [m^{3}]) and e is the
charge (units [C]]) of a charge carrier. Therefore R_{H}=1/(ne)
so the units of R_{H} are m^{3}/C.