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Here are older questions and answers processed by "Ask the Physicist!"
QUESTION:
suppose there were a crazy
architect who wanted to build a conical monument or perhaps working
structure on the tip instead of at the wider base, and without any
outside support beams. could the building be supported by a massive
internal gyroscope?
ANSWER:
Yes.
QUESTION:
Am I right that planets tend
to be generally spherical? (Not perfect spheres, of course, but for the
most part sphere-like). Why is this? Your explanation of how/why fire
manifests as sphere in space seems related, but wouldn't the properties
of the matter which make up a planet have an effect on the shape of the
planet? To this end, is it possible to have a hotdog-shaped planet, or
will rotational and orbital forces (and others?) always collect a
planet's matter into a sphere?
ANSWER:
The main reason is that astronomical objects
are usually formed by smaller pieces coming together. Initially, very
small stuff (dust size) sticks together by colliding and since this is
a random process, is just as likely to stick a little piece on any part
of a larger piece with which it collides; therefore the initial "seeds"
tend to be roughly spherical. Then, when a piece gets big enough to
have significant gravity, it starts "sucking up" neighboring smaller
pieces to grow but, because the gravitational force is spherically
symmetric (i.e. it is equally strong in any direction the same
distance from the object), it tends to maintain a basically
spherical shape as it grows. If it were hot dog shaped, gravity would
pull harder on the ends than the sides and it would tend back toward
spherical shape. I read a good analogy somewhere: Suppose you want to
build a building 1000 stories high, essentially a nonspherical bump on
our essentially spherical earth. You would likely fail because gravity
would cause it to collapse under its own weight. 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.
QUESTION:
If density plays a factor in
the strength of an objects gravitational field, why is it not part of
Newton's law of gravitation?
ANSWER:
The universal law of gravitation is strictly
valid only for point masses, not extended mass distributions. What you
then do is to imagine two bodies each made up of an infinite number of
infinetesmal point masses, find the force between each pair, and add
them all up to get the net force (if you know calculus, this process is
called integration). That is where the density comes into play since
the (infinetesmal) mass at a point is proportional to the density
there. Interestingly, if the density of each object is spherically
symmetric (i.e. the density depends only on how far you are from
its center) and if the shapes are spheres, you can actually treat each
as a point mass with all the mass at the center and get the right
answer.
QUESTION:
Why do clouds float? Water
vapor is heavier than air. And what determines their altitude?
ANSWER:
Actually, clouds are not water vapor
but either water droplets or ice crystals. If it were simply water
vapor, it would mix with the air and diffuse about sort of uniformly.
In fact, the amount of water vapor in the air is what determines
humidity. Clouds stay aloft for the same reason that dust motes
floating around, also heavier than air: air drafts push them around.
For more information go to the
weather network.
QUESTION:
Is Einstein's theory of general relativity correct? It's
taught in high schools (well, special relativity) and universities but
there's a web site, http://www.relativitychallenge.com/index.htm, that
claims there are mathematical errors in it. Also, are new theories
(such as the string theory) based on relativity or do they assume that
it's wrong?
ANSWER:
What does correct mean? All observations regarding
gravity are in accord with the general theory. But, there are some
important predictions of the theory, black holes and gravity waves for
example, for which we have indirect evidence but not direct evidence.
Is it possible that someday some experiment may not be in accord with
the theory? Of course! This theory cannot be the final word because
nobody has been able to reconcile it with quantum mechanics, one of the
other most important theories of nature. Special relativity is another
issue and is generally considered certainly correct since it simply
describes the nature of space and time. Regarding your website, I
certainly will not critique it because I know that the world is full of
persons with their own personal theories of the universe desperately
seeking attention for their ideas. I will acknowledge that no theory is
immune from mathematical errors which, even if present, do not
necessarily negate the validity of the theory.
QUESTION:
If you're driving a car at 100km/h on a flat, straight
road, and the passenger is flying a remote controlled plane at 100km/h
beside the car; what would happen if the plane is steered so that it
flys inside the car window? Would it appear to hover, as it's speed,
relative to the ground below is the same as the car, or smash into the
windshield, as it's speed relative to the car has increased to 200km/h?
ANSWER:
Before coming inside, the airplane is at rest relative
to the car, so if it keeps flying it would hover when it entered. But,
is that possible? How does the airplane fly? By moving through the air.
Before entering the car, the airplane sees a 100 km/hr "wind" going
opposite his direction of flight (assuming that the air outside the car
is still); if the plane were at rest relative to the air, e.g.
if there were a real 100 km/hr tailwind, the plane would drop to the
ground! But that is exactly what the situation would be when the plane
entered the car--it would suddenly be at rest relative to the air and
drop to the floor. If it were a helicopter instead of an airplane,
which flys by moving its wings (the rotor) through the air instead of
moving the air over the wing, it would hover inside the car.
QUESTION:
What are all of the types of particles in physics? The
ones I know of are:
- electrons
- photons
- phonons
Could you give me the complete list with short descriptions?
ANSWER:
What is a particle? How about a grain of sand? How
about an atom or a nucleus? There are more than 150 particles known in
elementary particle physics, but physicists no longer think of them as
"elementary" because they can be thought of as being built from more
elementary building blocks. One important example of those building
blocks is the quarks. The whole issue is quite complicated, more than I
can answer in a forum like this one. I recommend you do some reading.
One good web site which does a good job at disentangling the whole mess
is Wikepedia.com.
QUESTION:
If an airplane's wing shape lets it fly. (curved allows
more air
to flow over it thus air pushes up on the flat part of the wing at
least
that's what they tell me...)
How does an airplane fly upside down? Wouldn't the air
pushing the flat part of the wing force it into the ground?
ANSWER:
The standard textbook explanation of how airplanes fly
is a considerable oversimplification. In addition to the Bernoulli
effect to which you refer, the "angle of attack" is also important. I
have included details in a previously answered
question.
QUESTION:
Is it theoretically possible to make a magnet-only powered
motor capable of spinning forever?
ANSWER:
I guess that if you could make something without
friction it could spin forever. You wouldn't even need the magnet. But
in the real world, perpetual motion machines are forbidden by the
second law of thermodynamics.
QUESTION:
What would happen if a car was traveling at the speed of
light, and then turned on it's headlights? I read this question in a
magazine but didn't understand the answer too well. Can you try to
explain it please?
ANSWER:
This is easy to answer: a car cannot travel at the
speed of light.
QUESTION:
A question I have to research...... A light source is 2m
below the surface of the water in a calm pool. Find the radius of the
circle through which the light travels from the water into the air.
Take the refractive index of water as 4/3 and air as 1.
ANSWER:
Since this sounds like homework, I won't work it out
but I will tell you the idea. Light coming straight up from the source
hits the surface perpendicularly (zero angle of incidence, relative to
the normal to the surface) but light not straight up strikes at some
angle of incidence other than zero. As you go farther and farther away
from the straight up point, the angle gets bigger and bigger.
Eventually the angle becomes greater than the critical angle and no
light can escape.
QUESTION:
I have a simple question concerning electorn orbit
changes. How fast do electrons stay in an unstable orbit before
dropping to their basic orbit?
ANSWER:
There is no single answer to this question because it
depends on what atom you look at and specifically on which orbits are
involved in the transition. Typical lifetimes are on the order of
nanoseconds (billionths of a second).
QUESTION:
What is gravity - I am told it is a force that pulls
objects together - but how and why?
ANSWER:
This is a question about which whole volumes have been
written. It is one of nature's four fundamental forces and it is, by
far, the weakest force in nature. The reason why gravity is so weak is
one of the great unsolved problems of physics. In simple, classical
terms, gravity is a force which is caused and felt by objects which
possess a property called gravitational mass which, as far as
we can tell, is any material thing. You might find the statement that
it is weak to be surprising since it is the force of which we are all
most aware. But consider this: the weight of a pin is the graviational
force which the whole earth exerts on it but that force is
easily balanced by a small magnet (which uses the electromagnetic
force, another of the four). If you want to probe more deeply you need
the theory of general relativity. Here the idea is that the presence of
mass actually causes the space around it to warp and this warp of space
results in masses wanting to move toward each other. An often-used
analogy is to imagine a bowling ball placed in the center of a
trampoline which causes there to be a sag in the center; now place a
marble on the trampoline and, of course, it rolls toward the bowling
ball.
QUESTION:
I had the following question on my physics test and could
not figure it out. A child holds a sled weighing 77.0N at rest on a
frictionless incline at 30.0 degrees. Find a) the magnitude of the
force the child must exert on the rope, and b) the magnitude of the
force of the incline exerts on the sled. Answer a) 38.5 N b) 66.7N
ANSWER:
There are three forces on the sled, the force due to
the child (F), the weight of the sled (W),
and the force from the incline (N). Since the incline is
frictionless, N has only a component perpendicular to
the incline; F has only a component parallel to the
plane. Hence N must be equal to the magnitude of the component
of the weight perpendicular to the incline (77cos[30]=66.7) and F must
be equal to the magnitude of the component of the weight parallel to
the incline (77sin[30]=38.5).
QUESTION:
This goes all the way back to high school (as well as
college) physics. What is friction? It was always taught to me as
simply something that exists between two surfaces and the friction
between two surfaces (their relative coeficient of friction) must be
empirically determined. That is how I was taught about friction. It
seems to me with our deep understanding of physics down to sub-atomic
and maybe string level, SOMEONE has had to come up with a better theory
of friction than what I learned. It there an post-Newtonian theory of
friction and is there a way to calculate the coeficient of friction
between two surfaces?
ANSWER:
I have previously answered a question similar to
yours. Link here.
QUESTION:
I am currently reading "The Elegant Universe" by Brian
Greene. In his description of the "horizon problem" of cosmology,
Greene describes how the uniform background radiation is too uniform
according to the standard model of the big bank because light would not
have time to travel between two currently distant regions of space no
matter how close to the moment of the big bang we go. I don't
understand this because it either suggests that a.) the two regions of
space have traveled with a relative velocity of GREATER than the speed
of light or b.) that there must have been an initial displacement
between those two regions of space sufficiently large to account for
light being unable to travel between the two regions. Neither of these
seem consistent with other aspects of physics in the standard big bang
theory. What am I missing here? Am I just not understanding Greene's
explanation?
ANSWER:
from L. A. Magnani:
The horizon problem is indeed a problem for STANDARD
big bang cosmology (i.e., the version developed in the 1950's - 1980's.
The way out is to invoke a brief period of "inflation" during the early Universe when spacetime expands much more rapidly than the expansion rate we infer from galaxy redshifts today.
This inflationary explanation was proposed in the mid 1980's by Alan Guth and others and is supposed to be produced by a phase transition in the vaccum of some kind or other - the cause of the inflationary epoch is still a matter of debate.
The confusion for this person, I think, is arising from mixing up the sound speed (which is what is necessary to establish thermal equilibrium between two regions) with the speed of light. The inflation does not have to occur at greater than the speed of light. It just has to occur at a velocity greater than the sound speed of the medium to effectively thermally decouple opposite sides.
From J.-P. Caillault:
The commonly accepted solution to the horizon problem is
Inflation, which was when the early universe must have expanded
exponentially (faster than the speed of light, but this doesn't
violate relativity since it's the universe itself which was
expanding, not anything moving within it). Most
cosmologists now accept Inflation as part of the "standard big
bang theory," but this NYU person is probably thinking of the big
bang paradigm that prevailed prior to the introduction of the
Inflationary idea (by Alan Guth in the late 1970s).
from L. A. Magnani:
I think JP is right. The thermal speed may not be relevant
because
the coupling is between photons scattering off plasma, rather than
what goes on in a gas if the particles are doing the energy exchange.
But the expansion of spacetime can go on at faster than the speed of light - something that is recognized also in standard big bang cosmology.
QUESTION:
Please explain insimple terms what E=MC2 stands for.
ANSWER:
This means that mass (m), which
measures the inertia of a quantity (and to which its weight is
proportional) is just a form of energy. The amount of energy (E)
is enormous because the factor c2 is the square of
the speed of light and c=3x108 m/s=186,000
miles/second is a huge number. To give an example, suppose that you
could completely change a pound of something into energy. The amount
you would have would be about 4x1016 Watt-seconds which is
about 10 billion kilowatt-hours; this amounts, approximately, to the
total energy output of all nuclear reactors in the US in 4 days!
QUESTION:
Two twins, Bill and Ben are 22.0 years old and they leave
Earth for a distant planet 8 light years away. The twins depart at the
same time on Earth, and travel in different space ships. Bill travels
at 0.9c, while Ben travels at 0.5c. What is the difference between
their ages when Ben arrives on the new planet?
ANSWER:
This sounds suspiciously like a homework problem to
me! But I thought it was particularly interesting so I answered it
anyway! To see the solution, link
here.
QUESTION:
In an idealized case when no air resistance and
engine-fuel factors are considered would the same plane travel the
distances Vienna - Tokio and Tokio - Vienna for the same time? Why?
ANSWER:
I'm not sure what you are getting at here. For
starters, for no air resistance the plane could not fly! The most
important issue in time differences in long distance flights is
head/tail winds, but without air resistance, we would ignore these.
Let's assume, rather than no air resistance, equal resistance for any
direction (perfectly still air). Then, if there are identical airspeeds
in any direction there would be identical groundspeeds, so the answer
to your question would be that the times would be equal.
QUESTION:
Relativistic theory says mass increases to infinity as
speed of light is approached. Yet accelerators routinely accelerate
particles to near light speed ( 99 percent in some cases) without the
particles ever getting anywhere near infinite mass. Why? And if there
is some mass increase, what is the largest ever recorded and at what
"speed"?
ANSWER:
You should look at masses relative to the rest mass.
The mass of a particle traveling with a speed of .9999c is
about 76.6 times its rest mass; this is about how fast an electron with
kinetic energy 6 GeV at the CEBAF accelerator at Jefferson Lab
travels. If you were to make the energy 1000 times larger, the
speed would only increase by about .01%. Records of largest mass are
not kept. If you find the highest speeds recorded for a mass m0,
then the mass will be given by m=m0[1-v2/c2]-1/2.
QUESTION:
If temperature is defined as the average kinetic energy of
molecules in a mass, then why is there not a universal molar specific
heat for all substances in all states?
ANSWER:
In fact, the molar specific heat for most solids at
temperatures near room temperature is nearly constant as you suggest.
For a gas, however, if it can change volume it can therefore do work
and so only part of the heat (energy) added increases the temperature
and part comes out of the system as work.
QUESTION:
Suppose you have a one inch perfectly square bar of length
L resting on a flat perfectly rigid surface. You have a roller of
weight W resting at some point on the bar. The roller is a cylinder so
its contact with the bar is a line perpendicular to the length
direction and parallel to the surface. What is the vertical force at
each point along the bar due to the weight? (The point is the end of a
line across the bottom of the bar that is perpendicular to the length
and the question is about the force on that line.)
A practical application for this is the determination of the weight of
a roller required to compress bonding tape to stick a bar to a sheet of
metal. A particular pressure is required to cause the adhesive to
adhere properly.
ANSWER:
For the application you have in mind, your model
(contact being a line) is too simple to yield a meaningful answer. The
force which the cylinder exerts on the bar is W and that force
is (ideally) distributed evenly across the line. However, pressure is
defined as force/area and since the area of a line is zero, the
pressure is infinite! That should give you no problem sticking the
tape! To compute the pressure you need to let the line have some width w;
then, L being the width of the rod (and length of your line),
the area of contact will be wL and the (average) pressure would
be W/(wL). This is still a simplification since the
force is likely to vary across the distance w, but that is
maybe more detail than you want (or that I could probably deduce!)
QUESTION:
I would like to know what cold is. I would like to know
what is happening on the atomic level that makes (we'll say) air, cold
and pass on that coldness.
I know that heat is a form of radiation and that the molecules get
excited and knock eachother around, but what do the molecules of cold
do and how do they do it? Is it atomic spin, do they crowd other atoms
and force them to slow down?
ANSWER:
Cold is not a noun, it is an adjective. What you want
to know is what is the difference between something cold and something
hot; and you want to understand on a microscopic (atomic) level. Let's
talk about the air in the room. As you probably know, all the molecules
in the air move around and their velocities have a distribution, some
fast, some slow, some moderate, etc. But you could, in
principle, measure all the speeds and take the average. Now, the higher
the average speed is, the hotter the gas is. That is what hot
means--speedy atoms, on average. If you want to get more technical, you
can talk about the average kinetic energy of the molecules in the room,
kinetic energy being mv2/2 where m is the
mass of the molecule and v is its speed. Now, temperature (that
number we usually use to measure hot or cold) is merely a measure of
the average kinetic energy of the molecules in the room.
QUESTION:
why does "fire" behave the way it does in space? why
does it not "go up" like fire on earth, but instead look like a sphere
or something? also, he says that he's heard it's more "dangerous"
[sic] [unpredictable?] in space and never goes out? is any of
that latter claim true?
ANSWER:
This is a good question and you can find it discussed
many places on the web. Two are
Wonderquest and
Scientific American. I will add my two cents' worth: How could a
flame go up if there is no direction identifiable as up? If all
directions are equivalent, then a physical phenomenon like a flame
would of necessity be spherically symmetric like the space. Regarding
the "never goes out" part, a flame is just a chemical reaction and if
either the fuel or the oxygen runs out, the flame will stop. Regarding
"dangerous", on earth the "updraft" of the flame allows oxygen to be
"sucked in" at the point of combustion making for more efficient
burning, so that would likely be a more dangerous situation. I believe
that something burns more quickly in gravity for this reason.
QUESTION:
While driving on an interstate highway the other day, I
was idly day-dreaming about driving my vehicle up onto the ramps of a
trailer that was being towed directly ahead of me. I was traveling at
70 MPH relative to the ground. The trailer ahead of me was moving at a
speed slightly slower than me. As I approached, I thought - my wheels
are turning at a rate that moves me along at 70MPH relative to the
highway pavement, but if I start to drive up onto the ramps of the
trailer ahead of me (which was traveling at - say- 68MPH relative to
the pavement), once my turning wheels hit the ramps, I would be
instantly accelerated to 138MPH - the sum of the trailer's speed
relative to the ground plus my speed relative to the ground - clearly
faster than I had been traveling an instant before I went onto the
trailer.
Now - what if I was traveling at the speed of light, and the vehicle in
front of me was traveling at just under the speed of light? Why
wouldn't my speed suddenly exceed the speed of light as soon as I drove
up onto the ramps?
ANSWER:
Where did you get the idea that you would be
accelerated to 138 mph? Your speed with respect to the road would
remain 70 or close to it. Your front wheels would have to make a little
adjustment because it would be like the road suddenly started moving
backward at 2 mph so your wheels would have to start spinning as if you
were going 72, but that would probably be no big deal (but your tires
might give a short squeal like when you brake or accelerate quickly).
Now, anything which you can understand using everyday examples like the
one you give is not applicable to objects which are traveling near the
speed of light; that is one of the important truths about the theory of
special relativity, that our intuition is not a good measure of what is
reasonable if we are in a regime (high speeds) where we have no
experience.
QUESTION:
Time moves slower for someone who is moving very fast than
for someone else at rest.' How can this be said if motion is
relative? Example: Two people are floating out in space, and person A
sees person B zip by him very fast. B of course sees A zip by him. If
it is correct to say that A is moving and B is at rest, and correct to
say B is moving and A is at rest, how can you decide who is moving
slower through time, and thus ageing slower than the other?
ANSWER:
To each observer the other's clock running slow. They
do not appear to be running slow, they are running
slow. This is very puzzling as you note in your question, but it is
also very true. The trouble is that you have the standard intuitive
feeling for time, that there is some ablsolute and correct clock with
respect to which all others can be compared. In fact, time is such that
there is no problem with A seeing B's clock being slow and vice
versa. There is no way that they can sit down in a room together
and look at the two clocks to find out who is right because they are
not in the same reference frame. To do that, one or the other would
have to accelerate and the theory of special relativity is not
applicable to accelerating frames. You can understand time dilation
without talking about acceleration by studying the "twin
paradox" which I have discussed in answer to a previous question.
QUESTION:
Can an object have positive acceleration and be slowing
down?
ANSWER:
Yes, of course. Acceleration, like velocity, is a
vector quantity and so it has a sign in a particular coordinate system.
(I will restrict myself to one dimensional examples for clarity.)
Suppose that an object has a velocity v1=-2 m/s and
then, five seconds later, has a velocity v2=-1 m/s;
this object is moving in the negative direction (direction of
decreasing coordinate) and is "slowing down". Now let's compute its
average acceleration, the change in velocity over the elapsed time: (v2-v1)/t=(-1-(-2))/5=+0.2
m/s2, a positive acceleration! The best known example of
this is an object thrown straight up into the air: the object slows
down on the way up and speeds up on the way down, but the acceleration
is always the same.
QUESTION:
What is the rate of speed at the point where the two
blades of a scissor meet as they close?
ANSWER:
That is determined entirely by how fast you "snip",
how big the scissors are, and how far from the pivot point the mesh
point is. For example, suppose the scissors are initially open 30
degrees (p/6 radians) and
close in 1/10 of a second; the average angular velocity of one blade
relative to the other would be w=p/6/(1/10)=5.24 radians/s. If the
blades extended from 1 to 10 cm from the axis, then the relative speed
of one blade with respect to the other would be calculated as v=Rw=5.24 to 52.4 cm/s where R is the
distance from the pivot.
QUESTION:
how can we prove that earth can be considered as an
inertial frame of reference to a good approximation?
ANSWER:
The definition of an inertial frame is one in which Newton's
first law is true. That is, if an object has zero net force on it, it
will move with constant velocity (i.e. move with constant speed
in a straight line or be at rest). The degree to which this is true
determines the degree to which it is an inertial frame. A rough measure
would be to ask what is the maximum acceleration of the surface of the
earth; to calculate this I will ignore accelerations due the motion of
the earth around the sun and the motion of the sun around the center of
the galaxy since these are demonstrably smaller than the acceleration
due to the earth's rotation. Consider an object with mass 1 kg at the
equator. It is sitting "at rest" on a scale which reads, of course, 9.8
N. But, does it really? The weight of the object is 9.8 N but the scale
reads the normal force between the mass and the scale, and these cannot
be exactly equal because the object is accelerating because it is
moving in a circle of radius RE=6.4x106 m
with a speed of v=2pRE/T=2p(6.4x106 m)/(1 day)=465
m/s; so the acceleration is a=v2/RE=0.034
m/s2. So the scale would read a "weight" of 9.77 N for an
object whose weight was actually 9.8 N. If you can tolerate errors on
the order of 0.3%, you may consider the earth to be an inertial frame
of reference.
QUESTION:
What would happen if the two slit experiment were done
with electrically neutral particles like neutrinos or neutrons?
ANSWER:
The interference has nothing to do with electrical charge,
only with the wave-like properties of the interfering particles. In
order to be able to observe interference, the slit spacing should not
be large compared to the deBroglie wavelength of the particles. The
wavelength is l=h/p where h
is Planck's constant and p is the linear momentum of the
particle. Once you know the wavelength and the slit spacing d,
the analysis proceeds just as if you were doing the Young's double-slit
problem for light, dsinq=nl.
n=0,1,2... for maxima.
QUESTION:
I am having a VERY difficult time with this question and I
don't know why. I have gotten 4 different answers! I am
just not sure which formula to use, or how to start anymore.
Please help. The question:
Part 1: Suppose you throw a 3 kg ball straight up at 40
m/s. Using energy conservation, calculate how high the ball would
go if there was no air resistance.
Part 2: Suppose that the ball actually reached a maximum height
of only 75 m. How much energy was lost due to friction with the
air on the way up?
ANSWER:
Part 1:
E1=mv2/2=2400 J
E2=mgy=3x9.8xy=29.4y
but, E1=E2, so y=81.6
m.
Part 2:
E3=mgy'=3x9.8x75=2205 J
therefore, DE=E3-E1=-195
J.
195 joules of energy was lost.
QUESTION:
Since the earth is being continuously bombarded by cosmic
rays, why can't we develop some method of harnessing the energy from
these particles?
ANSWER:
Many of these are either trapped in the Van Allen radiation
belts by the magnetic field of the earth or lose much of their energy
interacting with the atmosphere. Although I am not an expert, I believe
that the total energy content would not be large and efforts would be
much more productively directed at harnessing energy reaching us from
the sun.
QUESTION:
Why does an object gain mass the faster it travels? Where
does this mass come from? Do scientists believe there is any way to get
around this and somehow travel at or past the speed of light?
ANSWER:
Mass is inertia, that is resistance to acceleration. When we
say that the mass increases, this simply means that it gets harder and
harder to accelerate a particular object as it speeds up. So, think of
mass as inertia, not "stuff" and then you won't have to worry about
where that "stuff" came from. Another way to think about the problem is
to ask how much energy it takes to accelerate something up to some
particular speed; it would take an infinite amount of energy to
accelerate anything all the way up to the speed of light. Special
relativity, on which all this is based, is very well verified by many
experiments, so no scientists believe that there is any way to
accelerate something beyond or even to the speed of light. There is
some discussion of tachyons which are particles traveling faster than
the speed of light, and they would behave in predictable ways. The
problem is, that they must have always been there since you can't get
there from here! Nobody has ever observed a tachyon and they (in my
opinion) likely do not exist.
QUESTION:
What will happen if the centripetal, gravitational force of Earth exerted on the moon becomes stronger ?
ANSWER:
Suppose that the moon is in a roughly circular orbit with the
current gravitational force. If the force were to suddenly increase,
the orbit would become elliptical and the moon would come much closer
to the earth. This is shown to the right where the the moon's current
circular orbit is shown and then if the force increased at the point
where the ellipse is tangent to the circle, the new orbit would be the
elliptical (egg-shaped) orbit. The earth is not drawn to scale here.
QUESTION:
In electrostatics we learn
that the answers we get when computing voltages are independent of the
definition of ground potential. What I mean by this is that the
equations apparently "don’t care" if the ground we define to be Zero
Volts is actually at Zero, or -pi volts, or + one gadzillion megavolts.
Only voltage differences matter. So, is there a mathematical property
embedded or encapsulated within the equations of electrostatics which
ensures the physics will be invariant upon changes in the definition of
the zero voltage? If so, what is this property called? I ask this
because I am familiar with the idea that the equations of physics (for
instance, Maxwell's equations) "don't care" what the velocity of the
laboratory is in which those equations are derived, and in a certain
sense this gives rise to special relativity. Thus my curiosity about
this "don't care" condition involving voltages, and what the more
fundamental principles involved might be.
ANSWER:
The two examples you allude to are different kinds of
"relative". Special relativity, which ensures that Maxwell's equations
are the same in any reference frame, has one frame moving relative
to another. Voltages are defined by considering one point in space relative
to another. Voltage has the property you refer to (only the difference
counts) because of the way it is defined. Voltage is related to energy
(and work) and energy also has this property that the total energy is
not relevant (because potential energy may be arbitrarily zeroed), only
the change in energy. Consider an electric charge q and it
takes an amount of work Wab to move it from point a
to point b. Then the potential energy difference is Ub-Ua=-Wab
(which is a definition); because this is the definition, you
see that what U itself is is not well defined (e.g. if
I add an arbitrary constant, say 5,000 J to both Ub and
Ua, the equation is still true). Finally, the electric
potential V is defined to be U/q, so Vb-Va
is just a measure of the work necessary to move a one Coulomb charge
from one point in space to another but Vb and Va
themselves have no meaning.
QUESTION:
I am a high school junior
and I got stuck with this question while doing a prep test. And this is
the one -A braking system for a roller coaster is designed to stop it
over a distance of 15 m when the coaster enters the stopping area with
a speed of 40 km/h. What acceleration must the braking system provide?
ANSWER:
First, convert km/hr to m/s: 40 (km/hr) x (1000 m/ 1 km) x (1 hr/ 3600 s) = 11.11 m/s. Now,
write the equations of motion for constant acceleration:
x=x0+v0t+at2/2
v=v0+at
Now, taking the origin to be
where the braking begins, x0=0 and v0=11.11
m/s. Now, let x=15 m and v=0 m/s:
15=11.11t+at2/2
0=11.11+at
Solving these two equations yields a=-4.11 m/s2
and t=2.7 s.
QUESTION:
The faster you move through
space, the slower you move through time. I heard someone say on a
television program, that this astronaut, out of all the astronauts in
the world has been in space the longest. They said that because he has
spent so much time circling the earth at such high speeds, he has
traveled into the future a tiny little bit. My question is this: If we
are always moving trough space because of the earth's many movements,
isn't what they said false? Moving with the earth does count when
concidering your actual speed through space, doesn't it?
ANSWER:
What matters here is relative speed (relativity refers to how
things are in one system relative to another). So, he ages less rapidly
than you because he is moving relative to you.
QUESTION:
My 7th grade Honors Math
teacher want my class and I to find out what is the space shuttle
acceleration launch speed? My math teacher said that it woud be less
then 7 mph.
ANSWER:
Well, there is a little problem here: "acceleration launch
speed" doesn't mean anything in physics. The acceleration is one thing
and the speed is another. Speed (velocity) tells how fast the height is
changing and acceleration tells how fast the speed is changing. Maybe
your teacher means "what is the speed of the shuttle at the instant
when launch begins?" If that is so, then the answer is zero. The
average acceleration during launch is about 33 m/s/s which is about 74
mi/hr/s which means that, on average, the speed increases by 74 mi/hr
each second over the 8.5 minutes it takes to reach orbital speed. Of
course, the acceleration is smaller at the start of the launch. At the
end of the first second, for example, suppose that the whole thing has
lifted one foot up; the average acceleration would be about 1.36
mi/hr/s and the speed would be 1.36 mi/hr.
QUESTION:
Many years ago I worked for
eight months in a medical radiation laboratory that serviced medical
linear accelerators. The output from the "linac" was 4Mev and with this
being the case I used the equation E=hf to determine that the photons
being emitted must lie in the gamma radiation section of the
electromagnetic spectrum.. I was however informed that it was x-rays
the "linac" was emitting and was also told gama radiation and x-rays
are the same anyway. I have not dealt with these subjects for
many many years so can you tell me..
- Was I right to use the equation E=hf to determine the frequency of these photons and does it work out the gamma radiation is the correct output?
- Are gamma and x-rays the same thing?
ANSWER:
- You would have been right if each proton (I presume it was a proton linac with the protons having 4 MeV kinetic energy) had stopped and all its kinetic energy were carried off by a single photon. In fact, the proton beam is smashed into a piece of metal where it interacts many time with atoms, giving each atom a little of its kinetic energy. Then each atom deexcites producing an x-ray.
- They are not absolutely defined, so they can overlap a bit, i.e. a low energy gamma ray and a high energy x-ray might have the same energy. However, physicists often make the identification by where the photon comes from: gamma rays come from nuclei and x-rays come from atoms.
QUESTION:
Why do air bags in cars
reduce the chaces of injury in accidents?
ANSWER:
It is all Newton's second law. Your momentum is proportional
to your speed. To stop you, that is to take away your momentum, a force
must be exerted. The more quickly you change your momentum, the greater
the force which is required. Imagine jumping out of four-story
building. When you hit the ground your momentum disappears in an
instant and, for that to happen, you are subject to an enormous upward
force by the ground; it is this force which hurts or kills you.
However, if there is a big soft pillow that you land on, the effect is
that it takes longer to stop you and so you will experience a much
smaller force over the stopping time; therefore you are much less
likely to be injured. I think that you can make the translation of this
example to the airbag.
QUESTION:
Was looking at a program on
discovery channel that depicted a spacecraft propelled by use of a
solar sail, the interesting piece was that it was propelled by a
powerfull laser on a space station. The question is If a laser powfull
enough is fired in a direction and is powerfull enough to propel the
space craft why would'nt it (the laser that is) be pushed in the
opposite direction to the way in which the light is travelling.
ANSWER:
It would, but if you think of the space station as being
attached to the earth, the whole earth/space station system would
recoil but with negligible velocity because the net mass is so huge.
QUESTION:
Relating to free-falling
objects, if a object is thrown upwards at a velocity (v), when that
object reaches it maximun height, the velocity of the object = 0.
My question is, if you could instantaneously observe the object
while at max height (duration when V=0) then, wouldn't the acceleration
of that object be 0? I understand that g is a constant and never
changes, but wouldn't the acceleration of that object change?
ANSWER:
Acceleration has nothing to do with velocity. Rather it
measures the rate of change of velocity. The acceleration is
proportional to the force on an object and if the force is the object's
weight only, it will have always a constant acceleration which is
pointing down (acceleration is a vector). Acceleration essentially
tells you what the velocity will be a short time later, so if v=0 now,
v will be a small velocity downward a short time later. If the
acceleration were zero, the velocity would not change and the object
would stay at rest forever.
QUESTION:
If you skimmed a one
molecule thick layer of water off the surface of the earth's oceans how
much water would you have?
ANSWER:
The surface area of the earth is about 200 million square
miles and about 70% is water, so 140 million square miles; that is
about 3.6x1014 m2. The diameter of a water
molecule is about 3 angstroms=3x10-10 m. So the volume is
11x104 m3 which is about 30 million gallons. It
would fit in a cube of about 50 yards on a side.
QUESTION:
The textbooks of
physics state that 1 coulomb is a charge equal to 6.242x1018 electronic charges, and that the charge
of one electron is 1.602x10^–19 C. My question is: How did
the number 6.242x1018 come into
existence? What is its history? Did this number originate from a
measured quantity, that is, experimentally, or is it dirived
mathematically?
ANSWER:
What you are actually asking here is: "How is a Coulomb
defined and how can the charge, in Coulombs, of an electron be
measured?" (not to put words in your mouth, or anything!) It is
somewhat circuitous since the thing which is defined is the unit of
current, the Ampere (A), and the Coulomb (C) is defined in terms of the
Ampere. If you have two very long parallel wires each carrying equal
current I and separated by 1 m, the force per unit length (N/m,
newtons per meter) is 2 x 10-7 N/m when I=1 A; that
is an operational definition of the Ampere. Now, a Coulomb is the
amount of charge which passes through a wire carring 1 A of current in
one second (s), so 1 A=1 C/s. That defines 1 C. Now, as you know,
electric charges exert forces on each other. It may be determined that
the force F (in N) felt by a particle with charge q1
(in C) due to a charge q2
(in C) which is a distance r (in m) away is F=9x109(q1q2/r2);
this is called Coulomb's law. Now that you know the force law, you can
find the charge on an electron by measuring the force between two
electrons separated by a known distance. This charge turns out to be
1.6x10-19 C. If that is the number of coulombs per electron,
then the number of electrons per coulomb is simply the reciprocal,
1/1.6x10-19=6.24x1018.
QUESTION:
What would be the final
speed of an electron as it passes though a field generated by 1V
potential difference and expressed in km/s. We would assume the
same conditions as those in which an electron would gain an energy of 1
eV. Is it possible to determine this speed experimentally?
ANSWER:
Technically, one should use relativity to answer this
question but 1 V inparts, as we shall see, a velocity which is much
smaller than the speed of light, so I will use classical physics. You
are right, the kinetic energy of the electron will be 1 eV and since
1eV=1.6x10-19 J and the mass of an electron is 9.1x10-31
kg, we can write that 1.6x10-19=9.1x10-31v2/2.
Solving, v=5.93x105 m/s=593 km/s (which is much
smaller than the speed of light, 3x108 m/s). And, yes, of
course, it is possible to measure this experimentally.
QUESTION:
What is the difference
between Linear velocity and Angular velocity?
ANSWER:
Linear velocity measures the rate of change by virtue of
translation (moving in a line). For example, when we speak of a car
going with a velocity 60 miles/hour it is going that fast down the
road. Angular velocity measures the rate of change by virtue of
rotation. For example, the earth is rotating on its axis with an
angular velocity of 1 revolution/24 hours. The wheels of the car have
both linear and angular velicity.
QUESTION:
If I set up a laser that
sends a beam out to a mirror and then the beam is reflected back upon
itself, is it possible to adjust the the distance between the laser
source and the mirror so that one could see interference effects?
ANSWER:
Yes. You can set up a standing wave. But you could not "see"
it because the nodes would only be half the wavelength of the light
apart. This is a technique which is used to create a diffraction
grating out of light.
QUESTION:
Why is parking a car in a
measured space easier while reversing than when moving forwad?
ANSWER:
I don't know if this is really physics; more common sense. It
is because you steer with your front wheels. If you steer your front
end into a parking space, the rear end is left outside and there is no
way to get it in. If you steer so that your rear end goes in first (you
have to go in reverse to do this) then your front end is left outside
but now you can steer it in.
QUESTION:
Can a person get shocked
from the electrical charge that comes up from the ground during a
lightning strike or is it from the charge coming from the cloud?
ANSWER:
Usually the bottom of a cloud is negatively charged, so when
lightening occurs it will result in a large electric current of
electrons flowing to the ground. This is what will kill you (I find
"shock" too mild a word!) Also, you will get burned by the extremely
hot plasma which is the path through which the electrons flow. There is
a lot more detail at
http://science.howstuffworks.com/lightning.htm
QUESTION:
What causes light bulbs to
glow? Is it the gas inside? Do different kinds of bulbs
contain different gases? Like neon gas in neon light bulbs.
ANSWER:
Usually when we refer to "light bulbs" we are talking about
"incandescent" lights. Here electric current is passed through a very
thin filament (wire) and it becomes white hot; that is the source of
the light. The bulb is filled with an inert gas so that the wire will
not burn as it would in air. There is a disadvantage to this kind of
light, however--only a small fraction of the energy it uses is
converted to light, only about 10%; most of the rest of the energy
becomes heat. "Flourescent" lights are much more efficient. These
devices have a mercury vapor inside them which is caused by a very high
voltage across the ends of the tube to emit radiation; unfortunately,
most of this radiation is in the ultraviolet region which is not
visible. To remedy this, the tube is coated with a material called a
phosphor. When ultraviolet radiation strikes the phosphor, it is
absorbed and reemitted as visible light. (You may have seen a "black
light" which makes some clothing, posters, etc. glow; it is an
ultraviolet light and the the glowing things are phosphors.) If you
fill the tube with other gasses it will often glow with visible colors
without a phosphor, e.g. neon will glow orange.
QUESTION:
If I hold a bicycle wheel (with an axle) that the wheel is free to spin
about, and I hold the two ends of the axle in each hand, how would I
would I find the minimum rpm (rev/min) that would allow me to hold it
in one hand & it not fall to the ground? There has to be a minimum
value for this horizontial gyroscope's angular velocity.
ANSWER:
This is a very complicated question. In fact the top (I will
refer to your wheel as a top) begins to drop the instant you let go of
it regardless of how fast it is spinning and then "nutates" as it
precesses. However, there is a simple formula which tells you the
minimum angular velocity of the wheel (which is only valid for the
angle q with the vertical
unequal to 90o): Smin=[4mgIs
cos q ]1/2/I where
Is is the moment of inertia of the top about
its symmetry axis and I is the moment of inertia about an axis
passing through the pivot point and perpendicular to the symmetry axis.
For example, a top straight up spins in a vertical direction until the
the angular speed drops below Smin= [4mgIs]1/2/I
and then falls. You should get a book on intermediate
mechanics (e.g. Marion and Thornton or Fowles and Cassiday) to
study this very beautiful problem.
QUESTION:
Why is the difference between the deviation produced by a prism onto
red light and blue light called angular dispersion, and that of yellow
light called mean deviation?
ANSWER:
I will venture a guess. Red and blue light are approximately
the longest and shortest wavelengths of the visible spectrum, and the
angle between them is therefore a measure of the total dispersion of
the system over the visible spectrum. Yellow light is in the middle of
the spectrum and so its deviation is about halfway between red and
blue, so this is the approximate mean (average).
QUESTION:
I read on your website that electrons flow on the surface of a
wire/conductor for AC currents, known as the "skin effect". I was
wondering why this happens for AC currents and not for DC currents?
ANSWER:
The thing which pushes the electrons out to the surface is
the magnetic field. There is always a magnetic field for any current,
but its effect on DC currents is small. However, if the current changes
with time, then so does the magnetic field. It is beyond the scope of
this site to work out the details, but new phenomena appear with time
varying magnetic fields which result in much less negligible effects of
the magnetic fields on the electrons; this does not become important
until the frequencies are radio frequencies (MHz or more).
QUESTION:
Why are objects as seen using mirrors closer than they appear to be?
ANSWER:
The blunt answer is that they generally are not. If you stand
in front of a mirror, your image is precisely as far behind the mirror
as you are in front of it. You are probably referring to the sideview
mirrors in cars which have a warning imprinted on them about objects
being closed than they appear. The reason is that the mirror is convex
rather than being flat like your bathroom mirror; convex means that the
mirror has a curvature such that it is a portion of the outside of a
sphere. (A concave mirror has a curvature such that it is a portion of
the inside of a sphere.) The reason for this is that you will get a
wider view of what is behind and beside you. I cannot give you a
tutorial on optics here, but you can read about it in any elemtary
physics text or many web sites, for example
here.
QUESTION:
Is there any instance (hypothetical or not) that only one of the four
fundamental forces is at work or acting?
ANSWER:
Since most particles in nature have mass, gravity is always
at work there (even if negligible). Furthermore, if you are in a region
of space which contains mass, even a massless particle (nowadays only
photons are thought to be massless). But there is a hypothetical
situation. Suppose that you have an electromagnetic wave propogating
through totally empty space. Then there will be electric and magnetic
fields so only the electromagnetic force exists. You could split hairs,
of course, and say that it is a field, not a force, which is in the
space through which the wave travels.
QUESTION:
Ok, i know what E=MC2 is, but do you have a DETAILED
description explaining it ? Do you have any examples of it that i can
teach to a senior (College) class ?
ANSWER:
It basically says that mass is a type of energy. You need to
know, of course, what energy is. In a nutshell, energy is what changes
about something if you do work on it, that is if you push on it over
some distance. For example, if you push hard on a baseball at rest over
a couple of feet (i.e. pitch) you do work and impart to the
baseball kinetic energy which it did not have before. If you take a
baseball on the floor and lift it up to a table top, the kinetic energy
has not changed but work has been done lifting it; here you have
imparted gravitational potential energy to the baseball.
With that said, let us give an example of an experiment which could be done to prove that mass and energy are interchangeable. Suppose that we take an atomic nucleus, for example the nucleus of the most common isotope of carbon which consists of six protons and six neutrons, and rip it all apart into its 12 constituent pieces. Will this take work? Of course, because otherwise this nucleus would not exist since there would be nothing holding it together. Before we rip it apart we should measure its mass; I will call that MC. After we have finished, we have done an amount of work W and have six protons, each of mass Mp and six neutrons, each of mass Mn. Does MC=6Mp+6Mn? Someone who has studied chemistry is very likely to answer affirmatively to this question but the answer is no and it is not a hypothesis, it can easily be done. In fact the mass of the sum of the parts is larger than the mass of the nucleus and E=Mc2 gives the result: W=(6Mp+6Mn-MC)c2. The mass gained is not some trivially small amount--it is on the order of 1%. Nuclear energy, of course, is where the energy from nuclear power plants and nuclear bombs comes from.
QUESTION:
If spring scales (bathroom scales) measure weight (force), and Dr.'s
scales measure mass, why do I "weigh" the same on both the spring scale
and the balance in the Dr.'s office?
ANSWER:
Both scales measure weight, they just do it in different
ways. The spring scale measures the force necessary to compress (or
stretch) a spring by a certain amount; knowing the properties of the
spring, the scale can be calibrated. The doctor's scale is essentially
a balance where your weight is compared with a known weight; the idea
of torque is also used where perhaps 1/10 your weight is needed to
balance your weight. Weight is a force, and mass is, conceptually, a
very different thing: mass measures the resistance (inertia) which
something has to acceleration when you push on it. Because of one of
the most fortuitous "accidents" of nature, it just so happens that
weight is exactly proportional to mass, so measuring weight turns out
to be equivalent to measuring mass. The "accident" is that inertial
mass is precisely the same as gravitational mass (which is a property
of matter which measures how strongly its gravitational attraction to
other bodies is). We now understand that this is not an accident; the
theory of general relativity fully explains this equivalence.
QUESTION:
I teach AP physics in a high school in michigan, and can't seem to
reconcile these two facts: The electric field due to an infinite
conducting sheet with surface charge density sigma is
E=sigma/Epsilon_0. If I introduce an oppositely charged infinite
conducting sheet facing the original, by superposition, I get that the
field between them should be double in strength, i.e. E=
2*sigma/epsilon_0. However, gauss's law, using a cylinder with one flat
face between the sheets and one face within one of the conducting
sheets still gives me E=sigma/epsilon_0. Where is the flaw in my logic?
When I look at the field lines, I see that the oppositely charged
infinite sheet doesn't introduce more, since every positive charges
field line on the positive sheet must end on a negative charge, either
at infinity or on the negative sheet, but that doesn't explain to me
why superposition doesn't seem to work here.?
ANSWER:
The problem you are having is rather straightforward. You are
correct in saying that with two sheets the field is twice as large
between the plates; however, the field outside the plates, also by your
superposition argument, is zero. Thus, when Gauss's law is applied
there is no flux leaving the surface outside, which gives twice the
field inside: e0E1*(2*A)=s A with one plate
and e0E2*A=s A with two, so E2=2*E1
QUESTION:
Since the orbital period of a satellite in near-earth orbit is much
less than 24 hours, why does the earth itself rotate only at that rate?
If the earth had formed from a collection of infalling particles,
wouldn't they have been rotating at the average orbital period based on
their distance from the centre of mass?
ANSWER:
The orbital velocity has nothing to do with the earth's
rotation. Suppose that when the earth formed it did so from a large
number of rocks all at rest. Each would fall toward the center of mass
and the resultant earth would have no rotation; the near-earth orbit
would still be the same, though, because it depends only on the mass
and radius of the earth. The real key to understanding how the earth
rotates is to understand that how it ends up depends on how it starts
and the operative concept is angular momentum. Angular momentum of the
earth is the same as it was before the earth was formed; as the
distribution of mass changes the angular momentum stays the same but
the angular velocity changes. If the present day earth were suddenly to
shrink to half its current radius, the length of a day would shorten by
a factor of four, 1 day = 6 hours.
QUESTION:
Is the amount of matter in the universe constant? A related question is
can new matter be created?
ANSWER:
Matter and energy are interchangeable, so matter can be
created by adding energy to a system. The best known example is called
pair production: a photon (quantum of light) may spontaneously create
an electron/positron pair (a positron is the antiparticle of the
electron). Another example is that the mass of the nucleus of an atom
is less than the mass of all its neutrons and protons, so when that
nucleus was made (probably in some star) a little bit of matter
disappeared from the universe. Obviously, the amount of matter in the
universe is not constant, but the amount of energy, we believe, is.
QUESTION:
While helping my daughter in grade 5 with a wind power project I was
wondering how to measure in a simple way the wind speed of the fan. We
thought to try the approach where you suspend a ping pong ball from a
thread. A table exists which relates angle of swing to wind speed.
However I was wondering about the physics of it.
If you have a ping pong ball suspended from a 30 cm thread and the ping pong ball weights 0.0027 KG then if a wind blows the string at an angle of 30 degrees from the vertical, then what would be the wind speed. What formula would you use if you ignore the aerodynamic effects of the wind going around the ball etc. I am helping Raeann with her wind power project. So we have a room fan that we use to drive a wind turbine (propeller hooked to a motor). It would be nice to measure the wind speed of the fan. It is expensive to buy a real anemometer so people on the net have published a table that relates ping pong ball angle to wind speed. However I was interested if you could calculate this. So the force downwards is mg for the ping pong ball. The tension onthe thread would have a downward force and a sideward force component. The sideways force would have to be matched by the pressure of the wind. Wind pressure would include the density of air and the cross section area of the ping pong ball I would imagine. Also there is the potential energy of the ball lifting up so many meters would be matched by the kinetic energy of the ball. So any thoughts. [Questioner also included data which came from http://marsville.enoreo.on.ca/mission/challenges/anemometer.htm .
Angle
kph
90 0.00
85 9.30
80 13.20
75 16.30 
70 19.00
65 21.60
60 24.00
55 26.40
50 29.00
45 31.50
40 34.40
35 37.60
30 41.50
25 46.20
20 52.30 .]
ANSWER:
If you plot your data, angle as a function of wind speed, it
will not be particularly enlightening. Before plotting anything you
should think about the physics. This is the simple pendulum problem
except with a horizontal force which keeps the ball at a particular
angle. I will not do the details which you can get in any elementary
physics text; I will give the results. Let us call the (horizontal)
force of the wind on the ball F, the (vertical) weight of the
ball W, and the tension (along the string) in the string T,
and the angle the string makes with the horizontal q. Then, solving this
problem we find that T=W/sinq and F=Tcosq =Wcosq /sinq . To understand the
physics, therefore, you should plot F as a function of cosq /sinq . I have done
this in the plot on the right. The black crosses are the data, the red
line is a fit. In essence, what you find by fitting the data is that
this is almost a perfect parabola, that is the force is proportional to
the wind speed squared.
If you want to now calculate the force of the wind on the ball, it is approximately F=W 0.001 v2 where v is the speed of the wind in km/hr. Once you know the force, you can deduce the angle q =arctan[W/F]=arctan[1000/v2]. For example, if v=24, q = arctan[1.74]=60.1 degrees, in pretty good agreement with the data above. It is interesting that the length of the string is irrelevant; also, you do not need to know the weight of the ball as long as you have the quoted data. Probably more useful to you would be the inverse of this equation, v=[1000/tan q ]1/2 ; for example, if the angle is 30 degrees, v=41.6 km/hr.
For common wind speeds on things about the size of ping pong balls, wind resistance is roughly proportional to speed squared. This is not always the case and it can also be proportional to the wind speed or to some combination of linear and quadratic.
So now you understand things. Perhaps the simplest thing to do is just take the given data as the "calibration" of your instrument and then, having measured the angle, interpolate.
QUESTION:
Someone in an internet forum claims that Einstein's Theory of General
Relativity shows that a geocentric model of the Universe is entirely
equivalent as a heliocentric one. Is he right?
ANSWER:
First, it is the solar system which we should talk about, not
the universe. Geocentric has a specific meaning, namely that the earth
sits still and the sun goes around us. But, as we all know, this is not
a possible explanation using the laws of classical physics. What this
person was probably referring to is that the principle of general
relativity states that the laws of physics are the same in all frames
of reference. That is, you may equally well understand the motion of
the solar system from the perspective of a coordinate system tied to
the earth as tied to the sun. This does not mean that geocentric and
heliocentric are equivalent but rather that the question of who is at
the center is meaningless.
QUESTION:
Have scientists been able to accelerate a particle past the speed of
light? Or at least up to the speed of light?
ANSWER:
A "massless particle" necessarily must travel with the speed
of light (like a photon, the particle associated with light itself).
But, if a particle has any mass, it may become arbitrarily close to and
below the speed of light, but never equal to or greater. The easiest
way to understand this is to understand that the mass of a particle
increases as its speed increases in such a way that the mass approaches
infinity when speed approaches that of light. It therefore would
require an infinite amount of energy to accelerate a particle to the
speed of light and, of course, there is not an infinite amount of
energy in the entire universe.
QUESTION:
I am a sophmore in high school and i have a question that i thought of
while in chemistry. Is it possible to trap light(laser beam would
probable work best) using mirrors that would continually bounce the
light off each other without letting the light escape? If this is
possible will the light still be there even after the original light
source was shut off?
ANSWER:
Yes, that is possible but you need to devise a mirror which
is perfectly reflective and that is not such a trivial thing.
Think about a one-dimensional trap, two plane mirrors one meter apart.
Suppose that they are 99% efficient at reflecting the light (much
better than your bathroom mirror). And, you have trapped a beam
of light in there (easiest to think of it as a very short pulse moving
back and forth. Each time it reflects it will loose 1% of its
intensity. Now, the speed of light is 3x108 m/s, so
the time between when the pulse leaves one mirror until it hits the
other is 0.33x10-8 s, that is it has 300,000,000 collisions
per second. If it loses 1% each collision, there will not be much left
after a second.
QUESTION:
What causes say, wood or metal, to bend and break? If I were to put a
board on bricks and hit it hard/fast enough it would break because it
causes shear (I believe) but what would cause the board to break, say I
was in space and I hit it extremely hard? It would definetely still
break but nothing is pushing on the outsides of the board so why
wouldn't the board just go forward rather than bend and break?
ANSWER:
Suppose you have a board of length 2L and you
exert a force F in the center. Then there will be a
torque FL about one end. You should think of this torque
which breaks the board. If the ends of the board are held fixed,
there will be four forces on the board, a force N up on each
end of the board, the applied force F, and the weight W
as shown in the figure. So, you can see, the torque about one end
is (F+W)L-2NL=(F+W-2N)L. Now,
if you are in empth space, the forces N and W go away,
but there is still a torque about the end due to F. So
pushing on the object will do two things: accelerate it (because of the
unbalanced force) and deform it (due to the unbalanced torque).
QUESTION:
A friend and I were discussing ballistics at a 1000 yard target
shooting match and need some expertise on a question. If two bullets
leaving the same caliber rifle with the same ballistic coeffiecents are
fired with the only difference being the weight of the bullet (for
example, 300 gr versus 150 gr), which bullet will incur the most wind
deflection?
ANSWER:
I know little about ballistics. As best as I can tell (with a
cursory internet search), a ballistic coefficient tells what
the air drag on a bullet is for a particular velocity. So, imagine that
you have two bullets which have the same speed and therefore experience
the same force F due to air friction. Newton's second law tells
us that a=F/m, so the one with less mass has a greater
acceleration and so it will lose its velocity more quickly. For the
same reason (a=F/m), the lighter bullet is likely to have
a larger muzzle velocity (the speed it exits from the rifle) if the
force propelling the two bullets is the same. The air friction force F
depends on the speed v, probably approximately like F=cv2
where c is a constant (probably related to your ballistic
coefficient). Therefore, the lighter bullet probably experiences a
bigger F than the heavier one. Looks to me like the heavier
bullet wins on both fronts.
QUESTION:
My husband and I may get divorced over this question! We both
have our positions, so maybe you can help us out. It's extremely
cold here in Calgary, about -20 celcius, and this came up on the way
home after starting up a very cold car. When starting the car,
after the temperature gage starts to indicate that the engine is
warming up, my husband cranks the heat full blast. It's my
position that if he were to keep it at a low setting, the air coming
out of the vent would be warmer, just not as much of it. If we
select the high setting, the temperature coming out of the vent will be
cooler. I think this is because it is forced air, and cold air is
being added to warm air from the engine that is not so warm yet,
thereby diluting it. Once the engine has warmed up enough, the
effect of the forced air created by cranking the heat is irrelevant
because the engine is very hot (meaning that the temperature of the air
coming out of the vent is the same at any setting, once the engine is
hot). He says that the setting of the fan does not change the
temperature coming out of the vents at the same engine temperature.
Which one of us is right????
ANSWER:
Well, it depends on what you want. If you want the air to be
as warm as possible coming in, you are likely right. On the other hand,
what you probably really want is to maximize the rate at which your car
is heating up and, in that case, your husband is likely right. I say
"likely" for the following reasons. Heat will be transferred from the
heating coils to the air passing over them at some rate and that rate
may or may not depend on the rate at which air is flowing over
them. One possible scenario is that the rate is about the same
regardless of whether the fan is high or low, i.e. maybe the
same amount of heat per second is achieved with either fast or slow air
flow; in this scenario, the air from the slower fan will be warmer than
from the faster fan but each will warm the car up in the same amount of
time because each carries the same number of calories or BTU or
whatever per second. It is my guess that your husband is right if you
want to heat the car up as soon as possible since I would guess that
fast air blowing across the heater coils would take the heat away
faster from the coils.
But, as is often the case in science, there is nothing to take the place of a measurement and that might be what you have to do to get a definitive answer. Let us make up what an experiment might measure so you can see how you could definitively do a measurement. The first thing you need to know is that the energy contained in a gas is proportional to its absolute temperature, i.e. E=a(T+273) where a is some constant, E is the energy, and T is the celcius temperature; the 273 is to convert T to absolute temperature (-273 C is absolute zero). Suppose that the temperature of your air is 15 and your husband's is 5. Then, the same volume of air contains energy in the ratio Ewife/Ehusband=288/278=1.036 (you are winning so far!) But, the volumes of air are not the same--your husband's method moves, let's say, twice as much air, so Ewife/Ehusband=1.036/2=0.518, your method now losing out by nearly a factor of two. This is not definitive since it depends on the relative temperatures and air flow volumes. I expect, as I said above, that the way to warm up the car the fastest is your husband's.
QUESTION:
Hey, do different frequencies of light have different amounts of heat
energy attributed to them? In otherwords, is UV light hotter/cooler
than visible light?
ANSWER:
Any frequency of light may carry any amount of energy--that
is what the intensity of the light is. However, we know that light is
made up of many photons, each carrying the minimum energy that such a
frequency can carry. The energy E of a photon is
determined by by the frequency f by the relation E=hf where
h is Planck's constant (an extremely tiny number).
Therefore, one photon of UV carries more energy than one photon of
visible light because its frequency is higher. So, UV light of
the same intensity as light in the visible range has fewer photons.
QUESTION:
I was told to ask this question to a phyicist, so here goes. Where did
air come from?
ANSWER:
Well, how far back do we want to go? All heavy elements
(essentially heavier that hydrogen) were produced in stars and then,
when the star was "all burnt up", it exploded and sent all the heavy
elements flying into space and then they eventually come together again
to form planets, etc. (Scientists like to say that we are
all made of "star dust". Then, depending on the chemistry of the
planet, its temperature, and other factors, some of the planet will
become an atmosphere, i.e. gases will escape from the surface
somehow In some cases (like the moon) the gravity is not strong
enough to hold the atmosphere and it eventually "leaks" off into
space. In the case of the earth, there is virtually no hydrogen
or helium in the air because it has all leaked off. The detailed
composition of the atmosphere depends on chemistry and biology.
For example, it is thought that originally the earth had much more
carbon dioxide in its air but that evolution of green plants resulted
in there being much more oxygen now.
QUESTION:
Hi. I was just wondering if you could give me an explanation on
why cars cannont fly? I realize that the gravitional pull has an effect
on it, but I want to know more specifically all the reasons.
ANSWER:
Anything can fly. You simply need to exert an upward
force equal or bigger than the weight of the object. An airplane
has wings and the air is made to flow over the wings such that the air
pressure on the bottom is greater than the top so there is a force up
which, if big enough, can lift it off the ground. A car could fly
if you gave it some upward force; for example, lift it up with a
crane! Or fit it with wings and an engine to keep it moving
forward.
QUESTION:
I am trying to get an estimated maximum wind speed that it would take
to blow over a 500 lb security tower that stands 10 feet tall. Can you
help me find ways to determine this?
ANSWER:
You have not given enough information. The force which
the wind exerts depends on the geometry of the tower. Also, is
the tower anchored to the ground in any way? Look here where I will put a very
rough calculation. The answer, about 60 mi/hr, is about what
you would expect, and it is likely that any other calculation would not
be a much better predictor.
QUESTION:
being that energy is conserved, what becomes of a sound wave in a
vacuum (i.e. space)? where does the energy go that would
otherwise go to produce the sound?
ANSWER:
Imagine that you have, as an example, a vibrating reed.
It has, as you imply energy. As it vibrates, it loses energy in
several ways:
- As you note, the sound carries away energy.
- As it moves through the air, it experiences air friction which takes energy away; this ends up as heat in the air and in the reed itself.
- There is "internal" friction because the reed is not perfectly elastic; for example, if you have something like a piece of thin metal and repeatedly bend and unbend it, it will get hot because of internal friction.
Those three will be the main modes of energy loss. Now, if you take away the air, the first two modes of energy dissipation are no longer available and so the reed will simply lose its energy more slowly, i.e. the reed will vibrate much longer before it stops.
QUESTION:
Hi, my friend and I were discussing rolling objects down a ramp.
I said that if one object is a cylinder and the other is a sphere, with
both the same radius and mass, that they both would have the same speed
at the bottom of the ramp. But my friend said that no she thinks
the sphere would be going faster. We are very interested to find
out which is right?
ANSWER:
The rolling object with the smallest moment of inertia will
win the race (and hence be going faster at the bottom of the
ramp). A solid uniform cylinder has a moment of inertial I=mR2/2
and the solid uniform sphere has I=2mR2/5.
So the sphere is the winner since 2/5<1/2; but it is a pretty close
race. You can try this experimentally but since it is so close,
the results will often not be definitive because of other factors
(rolling friction, nonuniformities, air friction, bumpy ramp, etc.).
If you want to try it experimentally, try a race with a cylinder and a
hoop (hollow cylinder) which has a moment of inertial I=mR2.
Here the solid cylinder should be the clear winner since 1/2<1 by a
pretty good factor.
QUESTION:
how do waves "rob" energy from one another very rarely to form massive
"killer" waves that rise somewhere around 100 feet in the middle of the
ocean?
ANSWER:
I am not really sure here. Usually "killer waves" refer
to tsunamis (tidal waves) which are caused by earthquakes, volcanoes,
or other geological catastrophes. However, water waves, like any
others, are subject to the superposition principle which
states that if two or more waves come to the same place the net
disturbance will be the sum of all the individual disturbances.
Simplistically, think of two waves, each 40 ft high and one comes from
the southeast and one comes from the northeast. If they collide
someplace where they are in phase (both are up at the same time) then
an 80 ft high wave will appear there. However, if they collide
someplace where they are out of phase (one is up and one is down), that
point will be calm. It is also what happens with a lens which
focuses waves: it gets very bright where the light comes to a focus
because many waves are adding up. But it is not a case of
"robbing" energy; it is more a case of combining their forces.
QUESTION:
I just got my last test back in college physics and I got this question
wrong, and I want to find out what was the right way to do
it.
A pitcher accelerates a .14 kg. ball from rest to 42.5 m/s in .06
seconds.
a.) How much work does the pitcher do on the ball?
b.) What is the pitcher's power output during the pitch?
c.) Suppose the ball reaches 42.5 m/s in less than .06 seconds, Is the
power produced by the pitcher in this case more than, less than, or the
same as the power found in part b. Explain.
ANSWER:
a.) The work done is equal to the kinetic energy
change. Since the ball started at rest, W=mv2/2=126.4
J.
b.) The average power is the work divided by the time it takes to
deliver that energy. P=W/t=2,107 W.
c.) If the same amount of energy is delivered in less time, the power
will be greater.
QUESTION:
If a puck slides across ice, and slows from 45m/s to 44 m/s in 25 m. ,
why does after another 25 m. does it slow to less than 43 m/s?
ANSWER:
The force on the puck is approximately constant and so its
acceleration is constant. Suppose that it takes a time t1
to go that first 25 m. Then it will take it t1
until the speed decreases to 43 m/s. But, it is, on average,
going more slowly during the second t1 and so it
will go less far than 25 m.
QUESTION:
Reflected light wave will have a phase change of 180 degrees at denser
medium, say when it travels from air to glass. The speed of light in
glass is smaller than that in air and we define glass as a denser
medium. For sound wave, its speed in air is smaller than that in glass.
Should we define air is a 'denser' medium for sound?
ANSWER:
For any kind of wave, reflection at a boundary will have a
phase change if the speed in the medium from which the wave is
reflected is smaller than the speed in the medium in which waves are
traveling.
QUESTION:
At the most fundamental level, exactly where does the energy from
fusion and fission come from? I know e=mc^2 describes how much energy,
but not the process itself. I know about the curve of nuclear binding
energy. E.g, when four hydrogen nuclei fuse, the resultant helium atom
has less mass and the excess is released as energy. But where exactly
does the energy come from? Is it correct to say the strong nuclear
force ultimately provides this? Or is simply an intrinsic process we
accept ("it just happens")? At the lowest level, is there a describable
mechanism by which matter stores energy, or by which the
mass->energy conversion releases energy?
ANSWER:
It does indeed come from E=mc2. And
yes, it comes from the strong interaction. The example you state
(4H going to 1He) is not a good one because it is incorrect because two
of the protons have to turn into neutrons + electrons which complicates
things (but happens ultimately). Better to fuse two deuterons
(nuclei of "heavy hydrogen" which consists of a bound neutron and
proton) into an alpha particle (a He nucleus). As you correctly
state, energy is released because the mass of an alpha particle is
smaller than the mass of two deuterons. It comes from the process
of their becoming bound together so, as you suggest, the strong force
is responsible. It is perhaps easier to understand to think of
the reverse process: in order to pull apart an alpha particle into two
deuterons, you must supply work, right? Where does the energy
that you put in go? It goes into mass.
QUESTION:
If two trains,one loaded with lead, the other empty, are travelling at
60 miles per hour on identical flat tracks and at the same time their
engines were put in neutral which one would travel further and why?
ANSWER:
It all depends on the friction which each train
experiences. Normally we think of friction between surfaces
sliding on each other as increasing as those surfaces are pressed to
gether harder. The wheels are not slipping and furthermore being
steel are not very deformable, so their contribution to the friction is
rather small and probably similar for the two trains. However,
there are bearings which have some friction and the friction will
surely get larger as you increase the force (coming from the train's
weight) on them. So, the heavier train will have more friction
and therefore go less far.
QUESTION:
I learned that the interior of canons when smooth provided less
accuracy. When we learned to machine a spiral on the interior it
increased the accuracy much like a football through is more
accurate if you put a spin on the ball when releasing it. Why
does setting a spin on a projectile increase the accuracy of that
projectiles aim?
ANSWER:
You are right--the rifling (which is what the machined
spirals are called) imparts spin to the projectile. Why should
that help accuracy? Well, if something is spinning it will
continue pointing in the same direction forever unless there is some
external torque on it (this is how a gyroscope works); this is called
conservation of momentum. It is well known that such a projectile
is more accurate. But the reason is not just that it is spinning
and not "tumbling". It has to do with the interaction with the
air; a tumbling projectile will tend to be deflected by the air it is
moving through more during its trajectory than one which is not
tumbling. If there were no air, any projectile, no matter how it
spun or tumbled, would be equally accurate since the center of mass
would move as if it were a simple point.
QUESTION:
why is Ke=1/2mv^2; especially if it takes onlytwice as much rocket fuel
to accelerate it to twice the speed?
ANSWER:
I do not know where you got the idea that twice the fuel
results in twice the speed. That is incorrect. But you are correct in
your implication that twice the fuel will not result in twice the
kinetic energy. The problem is that when you burn fuel, a certain
amount of energy is released; but, you also "throw out" the burnt fuel.
In order to conserve the momentum of the system, you cannot give all of
the released energy to the rocket--the spent fuel gets some of
it. So looking at the relation between energy released to the
energy gained by the rocket is not a useful thing to do. To
answer your question, though, the reason we define kinetic energy as mv2/2
is because if you do work W on a particle, its kinetic energy
increases by exactly W.
QUESTION:
In rotational motion of a rigid object why are torque, angular
momentum, angular velocity and angular acceleration as vector
quantities defined along the axis of rotation? Is it due to the
lack of a better way to express those values/quantities graphically or
is there real meaning? For example when calculating angular
momentum (L=rxp) the linear momentum can be given a value directed out
of the rotational plane into the third axis or parallel to the axis of
rotation. This would seem to mean that linear momentum would be
directed away from the plane of rotation and directed around the axis
of rotation. Like the linear momentum would make a spiral around
the axis of rotation.
ANSWER:
Well, first of all they are not all necessarily along the
axis of rotation, only angular velocity and angular momentum are.
Torque and angular acceleration must be in the same direction because
of Newton's second law. But torque about an axis is defined as rxF where r is the "moment arm" and
F is the applied force.
Your definition of angular momentum is valid only for a point mass
which has linear momentum p in some well-defined direction. For a
rigid object, you must look at each infinitesmal part of the object (of
mass dm and velocity v having momentum dp=vdm and therefore angular
momentum dL=rxdp ) and then add up all
the infinitesmal angular momenta by integrating to get the total.
There is indeed real meaning since a complete description of rotational
physics results.
QUESTION:
Assume a V-slot is cut across the diameter of a turntable, and a ball
placed near the center. As the turntable spins, the ball moves away
from the center under the influence of an unbalanced centrifugal force.
Since "w" is constant (the ball stays in the V-slot), and "r" is
increasing (the ball is moving away from the center), then "v" must
also also be increasing (the rotational speed of the ball). Since "v"
is a squared term in the formula f = v * v / r, then the centrifugal
force "f" pulling the ball away from the center is INCREASING over
time. So the acceleration is not being caused by a constant force upon
a mass over time - as we normally think of acceleration - it is being
caused by an increasing force upon a mass over time. Can this be
(should this be) called "hyper-acceleration"? Also, since this ball
described above now has gained significant "radial momentum" (what is
the proper term for this?), the "exit" path of ball would NOT be
tangential - correct?
ANSWER:
You are analyzing the problem all wrong. For starters, there
is no such thing as centrifugal force (it is what is referred to in
mechanics as a fictitious force). I guess by w you mean the angular
velocity of the turntable. I am going to assume that the ball slides
frictionlessly in the slot; you could solve the problem if the ball
rolled without slipping, but it would be more complicated and obscure
the important features of the physics for you. The problem is identical
to a bead on a straight, smooth stick which is rotating. To use a
centrifugal force implies that you will work in the (accelerated) frame
at rest with the stick and if you do this, you must also introduce
something called the Coriolis force (another fictitious force). I take
it that your level is such that we should stick to the inertial frame
in which the stick rotates. Then there is no force along the stick so
the acceleration along the stick must be zero. This will be puzzling to
you because the bead starts not sliding out the stick but certainly
ends up sliding out. This is really a subtle problem which requires you
to work in polar coordinates. There is a force in the tangential
direction, however, which causes an acceleration in the tangential
direction. Again, this may be surprising to you because the
particle does not have an angular acceleration! The final result
is that the particle will, if it starts at some position r0,
then when it is a distance r from the axis of rotation it will
have a radial speed vr=w[r2-r02]1/2
and a tangential speed of vt=wr. So, you are
right, it will come off the end with a velocity which is not
tangential. If you want to see the details (you have to be familiar
with calculus and polar coordinates, send me another question and I
will give you the details.
ANSWER:
I just realized that I have previously answered this question
and the details are available here. That
problem looks a little different but upon reflection you will see that
it is identical to your problem.
QUESTION:
Could a person stand the pressure it would take to make a relatively
heavy gas (e.g. Argon) dense enough to enable bouyancy?
ANSWER:
Well, the density of our bodies is on the order of that of
water, so the density of the gas would have to be greater than water. I
don't think we could endure such a pressure.
QUESTION:
Ok, i know Bernoulli's equation 'says' if you reduce the cross
sectional area, you will increase the velocity of the fluid and
decrease the pressure. While i was watering my garden, i sometimes
squeeze the end of the hose to increase the velocity to make the water
go further. However, when i went to turn off the water from the tap,
the velocity of the water reduced even though i was decreasing the
cross sectional area by closing the valve. So, why does the velocity
increase when i squeeze the end of the hose, but reduce when closing
the valve, when both are reducing the area. Bernoulli's equation 'says'
the velocity should increase.
ANSWER:
When you are closing the valve, the water velocity through
the valve is increasing but the rate at which water is entering
the hose is decreasing. Hence the velocity through the hose
decreases.
QUESTION:
How does an inverted image of any object apperes in hot areas causeing
to feel like water pond in that area. or what happens to the ray of
light coming from that object to us after total internal refraction?
ANSWER:
I believe what you are asking is how a mirage is formed.
Basically what happens is that the layer of air near the ground becomes
much hotter than the air higher up so you have a thin layer of very hot
air which has a lower density and therefore a smaller index of
refraction. You may treat the layer of air as a slab of material from
which, since it has lower index of refraction, can have light reflected
from it so it acts like a mirror. It is really noticeable for a very
glancing angle of reflection and may be thought of as total internal
reflection.
The above is one good way to think about what is going on qualitatively, but it is not what really happens. There is not really a surface where index of refraction changes from one value to another. Rather, the change from hot to cool air is continuous over some small distance so you have a continuously changing index of refraction. The light gets bent as is moves through this region (that is, refraction and not reflection is what happens) so that, if the angle of incidence is very glancing, the light will get bent up and never actually hit the ground, so the effect is the same as reflection.
QUESTION:
When you jump off a chair, you bend your knees on landing. i
understand that this reduces the shock to your body. But how so?
ANSWER:
When you hit the floor, the floor exerts an upward force on
you. Newton's second law says that the average force you
experience must be your mass times your average acceleration. The
average acceleration is defined to be your change in speed divided by
the time it takes your speed to change. What flexing your knees
does is to increase the time it takes to stop, thereby decreasing the
average force you experience. For example, if your mass is 90 kg
(about 200 lb) and you jump from a 1 m high chair, then you will hit
the floor with a speed of about 4.5 m/s. If you keep your knees
straight you might stop in about 1/10 second, so the force you feel
will be about 90x4.5/(1/10)=4050 Newtons which is about 910 pounds; but
if you flex you knees you might make the time be 1/4 second, decreasing
the average force you feel to 90x4.5/(1/4)=1620 Newtons=364 pounds.
QUESTION:
This is a basic question. f=ma. If I push on a concrete
wall, I am applying a force but there is no accleration of the
mass. What am I doing to the wall?
ANSWER:
For all intents and purposes, the mass of the wall is
infinite and so, however hard you press, the acceleration of the wall
will be zero. What you are doing to the wall is exerting a force on it
and the wall will experience this force. Of course, if the force
is big enough, the wall will be damaged by your force. For example, if
the wall is made of plaster and you hit it with a hammer you are likely
to knock a hole in the plaster with the force.
QUESTION:
How does a neutron change into a proton?
ANSWER:
The physical process is called beta decay. It is not
actually as simple as a neutron changing into a proton because that
would not conserve electric charge (a neutron has no charge, a proton
is positively charged). What actually happens is that the neutron
turns into a proton plus an electron plus an antineutrino. The
antineutrino is a nearly massless particle which interacts very weakly
with matter; therefore, although its presence was predicted
theoretically in the '30s, it was not experimentally observed until the
late '50s. For a classic description of beta decay, link here.
For a description which goes deeper and describes the decay in terms of
what is happening with the constituent quarks, link
here.
QUESTION:
how can i design and conduct an experiment to show that dark surfaces
are better absorbers of radiation than shiny or white or bright
surfaces?
ANSWER:
The easiest experiment is to take two identical objects and
paint one black and one white. Then put them in the sun for a little
while. Then feel them. The black will be warmer. If you want to get
more quantitative, get two bottles, one painted black and one painted
white. Fill the bottles with water, cork them, and put a thermometer
through a small hole drilled in each cork. Then place them both
in the sun and record the temperature of the water in each bottle as a
function of time (maybe every 5 minutes or so) and graph your results.
QUESTION:
I know that the speed of light can be measured in miles per second or
meters per second, etc., but is there any way of expressing the speed
of light according to one of the other physical constants? In other
words, since things like seconds and miles are arbitrary, how might
nature express the speed of light, if it could?
ANSWER:
All nature "knows" is that the speed of light is a
fundamental physical constant. As far as we know, it cannot be
expressed in terms of other fundamental constants. (Actually, that is
what "fundamental" means; if it could be expressed in terms of others
it would not be fundamental.) In order to quantify that, however,
one needs operational definitions of length and time. The history of
all this is interesting. Prior to 1960, the meter was defined to be the
length of a standard stick kept in a vault in Paris and the second was
defined defined as a fraction (1/86,400) of a "solar day"; with these
definitions, the speed of light could be measured. Then, in 1967, the
second was redefined by using what is called an atomic clock. With this
new definition of time, measurements of the speed of light could be
made again. Then, in 1983, the meter was redefined again, but now as
the distance which light travels in 1/299,792,458 seconds; you see that
what this definition does is to set the speed of light to be
299,792,458 meters/second in our system of units and tailor the length
of the meter to agree. It is all very nice because length is defined in
terms of time and a fundamental physical constant. You can read more
about these details here
if you are interested.
QUESTION:
Is time really a measurable property?
ANSWER:
Yes, of course. Anything for which there is an
operational definition may be measured. A second is defined
in terms of an atomic clock. What time is not, however, is a
universal quantity; that is, just because clocks run at one rate on the
earth does not mean that identical clocks which are not here run at the
same rate. For example, if a clock has a very large velocity
relative to the earth (large means not small compared to the speed of
light, 186,000 miles/second) it will run slower than our clocks.
This is the famous result of the theory of special relativity and is
very well verified experimentally.
QUESTION:
A horizontial U-tube of horizontial length (L) & vertical columns
(each column y and y in static equilibrum & open to atmosphere).
The U-tube is filled with water (or alcohol or whatever fluid) such
that the horizontial section is full & the columns of the tube are
filled to some height y. The U-tube & the fluid inside are now
given a purely horizontial acceleration to right which results in an
unbalanced force that causes the fluid in left column (end of the tube)
to rise vertically some amount delta y. Assume this is an ideal fluid
with no friction losses & ignore the U-tube's mass. My question is
this in F net = mass x acceleration...........2nd law, but what is the
fluid mass to be accelerated? Is it the entire mass or just the mass
contained in the horizontial length (L)? Since there's no vertical
acceleration, I think it's the mass in the horizontial section only and
not that contained in vertical columns. Correct?
ANSWER:
If you focus your attention in the fluid in the horizontal
section of the tube, it is being accelerated. What force is
accelerating it? The only force is that due to the pressure difference
across its two ends. That is why the fluid rises in one side and falls
in the other, to provide this pressure difference at their bottoms. The
mass of the fluid in the vertical parts of the tube is being
accelerated by the sides of the tube pushing on it.
QUESTION:
I understand that Rayleigh scattering is associated with particles
about the size of 1/10th the wavelength of the incident light and that
Mie scattering occurs from particles about the wavelength of light and
larger. I have two main, related questions:
- Where does the 1/10th wavelength value come from? Is this a real constraint, or is it simply empirical reflecting the type of experiments that people do to investigate Rayleigh scattering?
- What happens in the region between where Rayleigh scattering is dominant and where Mie scattering is dominant? Is it a mixture of the two, or is there another scattering mechanism occurring? Where is this described mathematically (i.e., what can I use to model my data)?
ANSWER:
Calculation of scattering of electromagnetic radiation is one
of the classic topics of electricity and magnetism. It is very
complex and not easy to understand intuitively. As often happens
in theoretical physics, the results are most easily expressed
mathematically in certain limits, i.e. when one or more
parameters of the theory are very small or very large. Rayleigh
scattering is the expression of the scattering theory in what is called
the long-wavelength limit, meaning long compared to the size of the
scatterer (usually taken to be a sphere). How long is long?
It is not unusual to use one order of magnitude as a rule-of-thumb to
test bigness or smallness. There is certainly nothing magic about
1/10, it is just a place where you can expect a pure Rayleigh
scattering calculation to do pretty well. You should not think of
Rayleigh scattering and Mie scattering as two different things but as
the result of scattering theory in two different domains. I am
not an expert at these things, but as you now make the particles get
bigger, the theory becomes much more complicated and a smooth transiton
is made from the characteristics of Rayleigh scattering (where
intensity depends strongly on wavelength) to Mie scattering where it
does not; finally at very short wavelengths (large particle size) the
scattering just becomes geometrical optics. You can get some
detail on Rayleigh scattering at a Lawrence
Livermore site. You can do Mie scattering calculations here. Doing a Google search on Mie scattering or
Rutherford scattering will bring up a wealth of useful site.
QUESTION:
Regards rotation dynamics about a fixed axis, the torque & angular
velocity vectors are directed "along the axis of rotation" using the
right hand screw rule. But what is the direction of the angular
acceleration vector? I know from F = ma, that the acceleration of a
body is in the direction of the net force. Hence, I think it
(acceleration) should be in the direction of the net torque, which for
a disk is tanget to the surface assuming no slippage. Correct?
ANSWER:
Angular acceleration is not the same as usual acceleration
and so F=ma is not what you want to be
thinking about. Instead, Newton's second law takes its rotational
form t=Ia where I is the moment of
inertia. So, since this is a vector equation, the torque and
angular acceleration vectors must be in the same direction. Your
question is what is that direction? Now, you already know the
direction of the angular velocity vector and, if angular velocity
changes, there is an angular acceleration which is defined as a=[w2-w1]/Dt where w2 is the angular
velocity at the end of the time interval Dt
and w1 is the angular
velocity at the beginning of the time interval Dt, that is the
angular acceleration points in the direction of the vector difference [w2-w1]. But, this difference
is either parallel or antiparallel to the angular velocity direction
depending on whether the disk is speeding up (parallel to w1) or slowing down
(antiparallel to w1).
QUESTION:
Can it be proven that a massive particle/body, once accelerated to
close to light speed, would still adhere to the laws of physics as we
know them? Aside from light, I have not been able to find evidence of
any objects that travel remotely close to 180,000 miles per
second. I ask because the laws of physics that we know of seem to
break down as an object approaches a large massive body (black hole),
and I wondered if it has been proven that objects approaching light
speed would not behave similarly.
ANSWER:
The theory of special relativity is one of the best verified
(experimentally) of all physical laws. There are many things which are
not massless and move with velocities (relative to you) that are not
small compared to the speed of light. Any modern particle accelerator
has electrons, protons, or atomic nuclei which move with very large
velocities; a typical speed at an electron accelerator would be greater
than 99% the speed of light. Cosmic rays, which rain down on us all the
time have speeds comparable to the speed of light. The most distant
galaxies move with speeds more than half the speed of light. In
the vicinity of a black hole, you must invoke the general theory of
relativity which is the accepted theory of gravity.
QUESTION:
#1. you have two rectangular magnets (or four or more) each close
enough (and held in position) that they are exerting an equal force
against each other (same poles toward one another obviously). Now; what
is it like in the middle of the opposing fields? Is there a
"dead" center? Would it create properties that are creating either a
"nothingness" or a "greatness" of force??
ANSWER: 
Almost always when you have two or more sources of any field,
there will be places where the field is zero. For example, if you
put two equal electric point charges some distance apart, the electric
field will be zero halfway between them. If you put two bar
magnets (which is what I presume you mean by "rectangular magnets") so
the bars are in a line with north poles facing each other as shown in
the figure above, then the field will be zero where the smiley face
is. But there is no particular profound significance to this; it
just means that if you put a north or south pole of another magnet
there that it will experience no force.
QUESTION:
A block of mass (M) is sitting on a turntable that's rotating with some
value of omega. A smooth cord runs from the block thru the center of
the turntable and connects to another block of mass (m)..M>m. The
upper block is at rest relative to the turntable(it doesn't move
sideways nor is it pulled to the turntable's center by the cord-smaller
block system). I want to understand the direction of the upper block's
frictional force, i.e. is it sideways due to the turntable's rotation
(where it has a torque) or is it along the radius (parallel to the
cord) and hence has no torque about the axis of rotation. I wanted to
apply the conservation of angular momentum, but force's direction is
not apparent. Can you help?
ANSWER:
First of all, if the angular velocity is constant, there is
no angular acceleration and therefore no net torque on M.
Thus there is no tangential (sideways as you say) component of the
frictional force. Since m is at rest, the tension in the
string is mg. But the net force on M must be
the centripetal force, Fnet=Mv2/R=MRw2, the direction being
toward the center and R being the distance from the
center. Hence, Fnet=mg+f, so f=MRw2-mg. So
the answer to your question depends on both R and w; if MRw2-mg>0, f
points radially in and if MRw2-mg<0, f
points radially out; also, if MRw2-mg=0, f
=0.
QUESTION:
1) how can positrons be artificially produced
2) how can anti-protons be artificially produced
3) how can both be stored for long periods of time
ANSWER:
Positrons are naturally occurring in nature and come from
radioactive nuclei which undergo b+ decay; such nuclei are
unstable because they have too many protons and a proton decays into a
positron, a neutron, and a neutrino. They are also found in
cosmic rays (which is where they were first observed). But, if
you need them in copious quantities, they are created in accelerators
by what is called pair production where a very fast charged particle
(often an electron) strikes a nucleus and spontaneously creates an
electron-positron pair. Antiprotons are made in a similar
fashion, by creating a proton-antiproton pair; this is usually done by
shooting high-energy protons at a metal target. Antimatter is
most often stored in what are called storage rings in which
magnetic fields cause the particles to move around donut-shaped tubes
which have been evacuated to very high vacuum. If antiparticles
come in contact with their particle counterparts they annihilate and
disappear in a tiny flash of energy.
QUESTION:
Is it possible that just like gravity is a function of mass, so too,
ANTI-gravity is a function of vacuum (or ABSENSE of mass)?
ANSWER:
I am not really sure what you are trying to ask here since
there is no such thing as antigravity that we know about. One
interpretation of you question would be would a massless object
experience no graviational force; it turns out that the best known
massless object, a photon (a bundle of electromagnetic energy, e.g.
light), does experience the gravitational force, so light is bent when
it passes a massive object (a star, for example). If there were
antigravity, it would more likely be due to the existance of a
different kind of mass (call it antimass) which, instead of being
attracted to usual mass would be repelled. This would be like the
electric force which can be either attractive or repulsive because
there are two kinds of electric charge (called positive and
negative).
QUESTION:
I have read that it would take an enormous amount of fuel for a rocket
to approach the speed of light because mass increases as the speed of
light is approached. My question is: why wouldn't the increased
in the mass of the fuel compensate for the increase in the mass of the
rocket?
ANSWER:
How would it compensate? The fuel is in the rest frame
of the rocket and, in that frame, has its rest mass. In a frame
in which the rocket is moving, the mass of the fuel is very large, but
this in not due to an increase in the amount of fuel; rather it
reflects that it took work (energy) to accelerate this fuel up to its
large speed. This work shows up as an apparent increase in mass.
One of the biggest problems in carrying a large amount of fuel with you
is that you need to accelerate much of it up to high speeds which is,
in some sense, wasted energy because the fuel is not your
payload.
FOLLOWUP
QUESTION:
What I had in mind was that it takes mass thrown out the back to
accelerate a rocket and as you accelerate you have more mass to throw
out the back. Another thing you said was "This work shows up as an
apparent increase in mass". I'm guessing that your word 'apparent' is
the key to why it would not compensate, am I right?
ANSWER:
The mass of fuel which the rocket sees is unchanged, so if he
throws out some of the mass the remainder of the rocket will acquire an
added momentum equal and opposite to that which thrown mass carried
away. The thrown out mass as seen from the "rest frame" is much bigger
so it carries much more momentum; however, the mass of the rocket is
bigger by the same fraction so the increase in speed will be the same
in both frames. When I say apparent it is because many physicists
prefer to think think of mass as the inertia of an object at rest (rest
mass) and, in order for conservation of momentum to be conserved
(thereby preserving Newton's laws), the momentum must be redefined as p=m0v/[1-(v2/c2)]1/2
rather than the usual p=m0v. Here, p
is the momentum, m is the rest mass, v is the speed,
and c is the speed of light. So, you see, you may interpret
this redefinition of p to mean that the mass increases as m=m0/[1-(v2/c2)]1/2
and momentum is still mv. But in mechanics, the important
thing is momentum, not mass.
QUESTION:
At the top of a ramp there are a solid sphere, a hollow sphere, a
hollow cylinder, and a solid cylinder are all released at the top of a
ramp. They all have the same radius and mass. Which one reaches the end
of the ramp last? It was given that the hollow cylinder comes in
last, but that seems backward to me since the moment of inertia for the
hollow cylinder is the greatest.
ANSWER:
The kinetic energy at the bottom of the ramp must be shared
by the translational and rotational motions of the object. And
since the potential energy of all four objects are the same, the
kinetic energies at the bottom must be equal. To find out which
arrives last we need to figure out whose velocity at the bottom is
least. The kinetic energy is given by K=1/2mv2+1/2Iw2=1/2[m+(I/R2)]v2.
So the speed v at the bottom is smallest when the moment of
inertia I is the biggest (and vice versa).
QUESTION:
I understand that the surface area of a car's tyre in contact with the
road surface is entirely dependant on the weight over the tyre and the
pressure within the tyre. A given weight with a given pressure will
provide a given SA. What I don't know is: If you add more weight
over the tyre does the tyre's surface area increase or does the tyre's
pressure increase? Or maybe a bit of both ? Does the tyre's elasticity
play a critical part?
ANSWER:
The key here is the deformation of the tire. For a
given tire, the harder you press it down the "flatter" it becomes so
the surface area in contact with the road increases. Similarly, a
tire with a high pressure will be less responsive to being flattened so
it makes sense that the area depends also on the pressure. If
adding more weight to the load actually changes the volume of the air
in the tire, then the pressure would increase accordingly.
QUESTION:
I have heard that there are several wavelengths of light that never
reach our planet. What if a material could be constructed with the
specifics to reflect that wavelength? What would we see? Everything, or
nothing at all?
ANSWER:
If you could make an object which reflected one wavelength l and absorbed all
others, then if the light illuminating it did not include the
wavelength l, it would appear black.
QUESTION:
Can you help me understand what this term is or is not? It is used in
regard to calculation of moment of inertia, but texts seem to give it a
passing few words & never develop the concept. A study of
physical pendulums requires the calculation of moments of inertia (I),
so would (or could) this constant be used to compute the I? For
example, if the pendulum had a moveable disk mounted to a thin rod
& you wanted to know how far the disk should be moved (either
way) to achieve a certain period (T), can this be used in the problem
solving?
ANSWER:
As you probably know, the moment of inertia of a point mass m
a distance of s from an axis is ms2.
The radius of gyration is a number which expresses the moment of
inertia for any object about some axis in this form, that is I=mG2
where G is called the radius of gyration and m is the
mass of the object. For example, the moment of inertia of a
uniform disk of radius R and mass m about an axis
perpendicular to its surface and through its center is I=mR2/2=mG2,
so G=R/[2]1/2. For the axis shifted by
an amount d, I=m(R2/2+d2)=mG2,
so G=[R2/2+d2]1/2.
It is not clear to me how introducing G would make problem
solving any easier.
QUESTION:
I am an armchair physicist with a Master's degree in Metallurgical
Engineering.
My question pertains to Einstein's famous E=mc2 equation: Why doesn't E = ONE-HALF mc2, which is the sum of the momenta of all of the particles moving at the speed of light? Any elementary physics student knows that kinetic energy (which, when 100% kinetic, there is zero potential energy) = the sum of the momenta of the masses moving at velocity v. In other words, the integral of mv of the masses is 1/2 mv2.
Wouldn't it follow that when all of a given mass's energy is completely converted to energy, that it is ALL kinetic energy -- whose sum of all of the elementary constituent particles moving at the velocity of light would equal 1/2 mc2. In light of string theory, I would think this would be an obvious result.
Please tell me why I am wrong -- why am I off by a factor of 2? I have asked physics professors at a number of universities, who give me typical BS answers as "well, it's just a constant of the universe" (it isn't), or "it's been proven experimentally" (it hasn't -- even man-made nuclear explosions have resulted in only a few percentage points of the entire mass being converted to energy; I am not sure that even the energy of matter-antimatter annihilations have been measured to an accuracy to distinguish mc2 from 1/2 mc2).
If you can explain the answer, I am intelligent enough that you can explain it in technical terms.
If you cannot explain the answer, it this a bit of original thinking that has not been previously addressed?
ANSWER:
Well, gosh, you gotta get out of that armchair and learn
something about the theory of special relativity. Your
classically-based ideas of momentum and kinetic energy are wrong.
In my discussion, I will call the rest mass of a particle m
and the speed of the particle is v. Linear
momentum is not mv but rather p=mv[1-v2/c2]-1/2.
Note that as the speed approaches the speed of light, p becomes
infinite, not mc which is a whole lot less than infinity even
if c is a big number! Now, to calculate the kinetic
energy of the particle, we must calculate the work necessary to
increase the speed from zero to v. You do this by
integrating Fdx; if you now write F=dp/dt
(Newton's second law which is still valid as long as it is understood
that p is not mv) and do some calculus and algebra, you
need to integrate mu[1-u2/c2]-3/2du
from u=0 to u=v, and you get K=mc2{[1-v2/c2]-1/2-1}
for kinetic energy. We now make the identification that
the kinetic energy is the total energy minus the energy the particle
has by virtue of its mass (called the rest mass energy). So rest
mass energy of a particle is mc2 and total energy of
a particle is mc2[1-v2/c2]-1/2.
To make this even more plausible, kinetic energy should reduce to its
classical value if v<<c; this is easy to show using a
binomial expansion for the square root, [1-v2/c2]-1/2=1+v2/(2c2)+...,
so K=mv2/2+...
QUESTION:
do AM and FM radio waves travel at the speed of light?
ANSWER:
Yes. Radio waves are electromagnetic radiation, just
like light but different wavelengths. All electromagnetic
radiation propogates with a speed of about 186,000 miles/second in
vacuum (the speed in air is very close, but slightly less).
QUESTION:
When a spring is compressed and suddenly reliesed it's going to
accelerate and will achieve maximum speed when it reaches its normal
lenght. From that point on it will decelerate. Is it possible to
calculate it's accelaration based on its physical caracteristics as
diameter, twist, wire diameter and material properties?
ANSWER:
I suppose it is possible if you are an engineer, but I
wouldn't know how to do it. What I do know, though, is the
physics. I would first do a measurement of the spring constant k,
the length of the unstretched spring L, and the mass m.
Then I would compress it against a wall by an amount d and let
it go. If the floor is smooth (frictionless), the kinetic energy
of the spring would have to equal the potential energy the spring had
at the beginning, 1/2 kd2.
However, the acceleration is not a single number because one end of the
spring is at rest. If you call the velocity of the free end of
the spring v0 (when the spring returns to
its unstretched length) and assume that the speed of each part of the
spring increases linearly from one end to the other, then v=v0(x/L)
where x is the distance from the wall and L is the
unstretched length of the spring. Assuming that the mass of the
spring is uniformly distributed along its length, the kinetic energy of
a piece of length dx at position x will be 1/2
[(m/L)dx][v0(x/L)]2
= 1/2 [(mv02/L3)x2dx];
integrating, the total kinetic energy (which must equal the potential
energy at the beginning) is 1/6 mv02=1/2
kd2 so v0=[3kd2/m]1/2.
This is not what you wanted, but it is what I can do!
The above gives you the velocity of the end of the spring (as well as every other point) when the spring has just reached its unstretched length L; at this instant, the acceleration of every point of the spring is zero. You could extend the above analysis to find the velocity at the end for any arbitrary length L' <L.
QUESTION:
What can I use to make a magnet repel? I can't find the magnets
with opposite ends so they attract and repel, is there anything else I
could use to show how this works?
ANSWER:
I would guess that you are probably trying to use
refrigerator magnets and they are manufactured in such a way that they
consist of many tiny bar magnets but there are both north and south
poles on the surface. You can find a detailed discussion at MadSci.net.
So what can you use? You need to find some simple bar
magnets. When I was a kid we had little toys which were dogs
(terriers) sitting on little bar magnets. These would attract or
repel each other and lots of kids had them so I guess they were readily
available in toy stores. I haven't seen them in years, but a
quick internet search revealed that they are still available and may be
bought at the Restless
Mouse Shop for $2.49 a pair (I don't usually do commercials
here!) Good luck and have fun!
QUESTION:
When calculating the moment of inertia(I) of a meter stick that is
pivoted about its end, the parallel axis theorem would be used. If a
weight is then clamped (tack welded or whatever) to the stick's
midpoint, would the new total (meter stick + new mass) mass be treated
as if the meter stick had gained mass, or would this additional mass be
treated separately when calculating I?
ANSWER:
The parallel axis should be used only if you already know the
moment of inertia about the center of mass of the stick. Of
course, the well-known result is ML2/3. Now,
attach a point mass m to the center. The moment of
inertia of this mass relative to the end is m(L/2)2.
The net moment of inertia is L2[(M/3)+(m/4)].
If you assume that the effect is just increasing the mass of the stick,
then you would find (M+m)L2/3, clearly not the
same. The reason is that m does not have the same spatial
distribution that M does and I of anything depends on
not just how much mass there is but how it is distributed around in
space.
QUESTION:
An electromagnet is turned on and left on for one hour. At the instant
the magnet is turned on, I assume the magnet becomes surrounded by a
static magnetic field which grows larger at light-speed? - with
electromagnetic radiation present somewhere? If so, what's the "shape"
of the electromagnetic radiation? I mean, does the EM radiation have
the "shape" of a "thin shell" which is located at the boundary of the
magnetic field? After one hour the magnetic field has a radius of one
light-hour? Now the electromagnet is turned off. Does the entire
magnetic field instantly disappear when the magnet is turned off? While
EM radiation, somewhere, keeps propagating?
ANSWER:
Let us assume that there is a very short time during which
the field is changing when you turn it on and then a very short time
when it is changing again when you turn it off. During the first
time, there will be a pulse of electromagnetic radiation which moves
out with speed c and the space between it and the magnet will
be filled by a static magnetic field; the static field will become
smaller farther away from the magnet. Then when you turn it off
there will be another pulse of radiation and the space between the
second pulse and the magnet will have zero field. As time goes
on, this shell which consists of two shells of radiation and the
intervening space with static field all propagating away from the
magnet. However, the strength of the static field falls off much
more quickly as you get farther away than the strength of the radiation
fields does, so after not too long you would only see two pulses of
electromagnetic radiation.
QUESTION:
An operating pendulum is dropped from a great height above the earth.
It would be in freefall until it hit the surface.
ANSWER:
The pendulum bob would move in uniform circular motion about
the support point with the speed it had at the instant you released it.
QUESTION:
What is a "matter wave"? For example, if it's said that a rock is both
a material thing and also a wave, how is the rock a wave?
ANSWER:
DeBroglie's hypothesis is that any particle also has the
properties of a wave. The wavelength of the particle is given by h/p
where h is Planck's constant and has a value of 6.6 x 10-34
J s and p=mv is the linear momentum of the particle.
Suppose that you had a rock of mass 1 kg which was moving with a speed
of 100 m/s. Its wavelength would be 6.6 x 10-36
m. In order to observe this you would have to diffract the wave
around some obstacle which was about this size. But, the smallest
thing we know is an elementary particle like a proton which has a size
of approximately 10-15 m, far too big. However, the
wavelike nature of particles is well established by looking at much
smaller particles than your rock, say an electron which has a mass of
about 9 x 10-31 kg and so an electron with a velocity of 104
m/s would have a wavelength of about 7 x 10-8 m. Since the
spacing of atoms in a crystal is about 10-10 m, on may
easily see electron diffraction. This is the famous Davisson-Germer
experiment.
QUESTION:
A permanent magnet is travelling East at a constant speed - it bounces
off something and is now travelling West - did the magnet accelerate
when it changed directions? If so, I know two other things that cause
acceleration are rotation and a change of speed - anything else?
ANSWER:
Acceleration occurs whenever there is a change in velocity;
don't forget that velocity is a vector so that it can change by either
its magnitude (speed) or its direction changing (or both). To
actually compute the acceleration, you would have to know something
about the details of the collision. The average acceleration in
your example is change in velocity divided by the time it took to to
reverse the velocity. For example, if the speed in is 20 m/s and
the speed out is 20 m/s, the change in velocity is 40 m/s and it points
in the direction of the outgoing magnet. If the collision took
.01 seconds, the average acceleration would be 4000 m/s2.
If the magnet had a mass of 1/2 kg, then Newton's second law tells you
that a force of 2000 N (about 450 lb) had to be exerted on the magnet
during the collision to cause the acceleration.
QUESTION:
I've read that even "non-magnetic" things, like the human body, a tree,
a piece of plastic, and anything else, all have extremely weak magnetic
fields surrounding them. If all things do have magnetic fields, what
causes that?
ANSWER:
Magnetic fields will exist only if there is an electric
current distribution of some sort. That is a true statement if
you think of magnetic moments of electrons, atoms, molecules, etc.
as resulting from local current distributions. In the human body
there are many electric currents—all
nerve impulses are primarily electrical, the heart beats in response to
electric currents, the blood moves through the body and doubtless
carries many ions and therefore constitutes an electric current, etc.
Similarly, any living thing is likely to be a source of magnetic
fields.
Virtually any atom which has an intrinsic angular momentum also has a magnetic moment. As examples, any atom which has an odd number of electrons will have a magnetic moment; any nucleus which has an odd number of protons or neutrons will have a magnetic moment. Therefore, many atoms have magnetic fields around them because they are, in effect, tiny bar magnets. The catch is, however, that a chunk of stuff is made up of a very large number of atoms and almost always the magnetic moments of the individual atoms are randomly aligned so that the net effect is that there is zero field. The few "ferromagnetic" materials in nature (like iron) have magnetic fields around them because atomic magnetic moments have a tendency to align with their neighbors. On the other hand, all materials are "magnetic" to some extent—this means that the atomic moments will have a tendency to align with an external magnetic field. Therefore, I would suggest that the piece of plastic you ask about would have a field of its own only if it were in an external field (which it is since the earth has a magnetic field), but it will be almost immeasurably small unless the external field is very strong.
QUESTION:
If a permanent magnet is lying on a table, is it accelerating because
of the orbital and rotational motions of the planet? If so, is it
constantly emitting electromagnetic radiation? If so, it seems like
there wouldn't be any frequency or wavelength involved? Would there
just be a single magnetic field, and orthogonal to that a single
electric field - both fields constantly expanding larger and larger at
the speed of light? If all that is true, then are all things constantly
emitting that type of electromagnetic radiation, since all things are
made of atoms that are composed of electrically charged
particles?
ANSWER:
Technically, yes. Any accelerating magnet or electric
charge will radiate electromagnetic radiation. However, the
intensity of this radiation would be immeasurably small. Since
the accelerations you consider are not linear but centripetal, there
would be frequencies of 1 cycle/day and 1 cycle/year associated with
the radiation. There is an exception to this rule, however.
For very small systems of electric charges such as atoms, where quantum
mechanics is applicable, the system in its normal (ground) state will
not radiate even though my might think of atoms in terms of the Bohr
model in which we envision electrons as orbiting the nucleus; in fact,
the Bohr atom is a naive model which is not really an accurate
description of nature.
QUESTION:
what is edi current?
ANSWER:
I presume that you mean "eddy current". If you expose a
conductor to time varying magnetic fields, electric currents will run
around inside the conductor; these are called "eddy currents" because
they will often take the form similar to little whirlpools which is
what the word eddy originally means. You can find a more detailed
answer at physlink.com
. A nice example is "magnetic braking" of which there is a video
at http://www.pa.msu.edu/educ/lectdemo/index_e/3/E3_01.htm
QUESTION:
is electron particle or wave? is there experiment in this
matter?
ANSWER:
This is the fundamental question which leads to quantum
mechanics. The answer, surprisingly, is that it is both and it
will display the properties of either, particle-like or wave-like,
depending on what you look for. There are many experiments which
show that the electron behaves like a wave; these are essentially the
same type of experiments you use to determine that light is a wave—diffraction.
I suggest you do a Google search
on electron diffraction to learn more. You might also search on
wave particle duality which is the term for the fact that all things
(not just electrons) are both particles and waves.
QUESTION:
A permanent magnet has been lying on a table for one hour. Now, it is
rotated by 45 degrees. How large of an area do the new field lines
occupy? Is there anything left of the original field lines, beyond the
area occupied by the new field lines?
ANSWER:
First of all, the field occupies the whole volume around the
magnet, not some area as you imply. If the magnet is static (not
moving), the field is called magnetostatic; so when you finish moving
the magnet the field around it will be exactly the same (relative to
the magnet itself) as before. The real issue then is what happens
when the magnet is actually moving and how long does it take for the
field to transform from the old to the new? During the time it is
moving, the magnetic field is changing and therefore, according to
Faraday's law, there is an induced electric field also; the moving
magnet, since it is accelerating, will actually radiate electromagnetic
radiation (i.e. it is an transmitting antenna). However,
the information about the magnetic field progogates at the speed of
light, so the field, as accurately as you would be able to measure
unless you make the change extraordinarily rapidly, is essentially
smoothly and continously changing.
QUESTION:
If two magnets are brought near one another, do their fields completely
blend together to form one field?
ANSWER:
Yes. Magnetic fields obey the superposition
principle which states that if several different sources of field
individually cause fields at some point in space, the net field there
is the vector sum of all the individual fields. Since it
is a vector sum, it is possible that the sum will be zero even thought
the individual fields are not. For example, if two identical
magnets are positioned so that their north poles are some distance
apart, the field will be zero halfway between them.
QUESTION:
Does the speed of terminal velocity change on a planet with the same
ratio of gravitational pull and atmospheric gases as earth, assuming
that the air is made of the same gases in the preportions as on earth?
ANSWER:
Terminal velocity is when the velocity is just right that the
downward force of the weight of an object is just cancelled out by the
upward force of the air resistance force. Normally we compute the
air resistance force by assuming that it is proportional to the
velocity F=bv. (Note, however, that
this is also an approximation and the dependence on velocity is much
more complicated.) Since we want the weight W to equal F
when v=vterm, then vterm=W/b;
therefore, vterm depends on what the constant b
is. This proportionality constant depends on lots of things, the
nature of the fluid through which the object is falling and the
geometry of the object being the most important. I do not know
what you mean by you mean by "...the same ratio of gravitational pull
and atmospheric gases as earth..."; it is clear that what must be the
same is the ration W/b, so if b increases
proportionally to W, then the terminal velocity will be the
same. Also note that, since W=mg where g
is the acceleration due to gravity, g/b needs to be
constant. If we keep the geometry the same and go to a different
planet, then b will be primarily determined by the nature of
the atmosphere. If g is bigger on the new planet, then
the atmosphere must be more dense if vterm is to be
the same.
QUESTION:
An object that appears white (like a piece of paper) appears white
because it reflects all the colors of light that are incident upon
it. If that is true, then what is the difference between a white
piece of paper and a mirror? Doesn't a mirror also reflect all
the light upon it?
ANSWER:
The answer has to do with how the light is
reflected. Reflection from the paper is called diffuse
reflection; a ray of light which comes in making some angle
relative to the normal to the surface is reflected in a random
direction. Reflection from the mirror is called specular
reflection; a ray of light which comes in making some angle
relative to the normal to the surface is reflected in a direction at
the same angle relative to the normal (angle of reflection = angle of
incidence, "the law of reflection"). The result is that your
brain, interpreting the light it sees, can tell, by observing the rays
reflected by the mirror, where the rays came from but you cannot do
that with the paper. The net result, as anyone who has looked
into a mirror can tell you, is that an image of the "real world"
appears behind the mirror.
QUESTION:
Do electrons flow on the surface of wire/conductor or through it?
ANSWER:
It depends on the nature of the current. For DC
currents, the current is uniformly distributed across the whole volume
of the conductor. For AC currents, however, there is a tendency
for the current to be more concentrated toward the surface of the wire
so that, for very high frequencies, the current is mainly at the
surface; this is called the "skin effect". This effect tends to
make the resistance of a wire a function of frequency of the current
and it is therefore something hi-fi enthusiasts worry about; it is not
a big effect at audio frequencies, though, and becomes important at
radio frequencies (MHz). Very short pulses of current have high
frequency components. For example, you are safe inside a (metal)
car during a thunderstorm, even in the event of a direct hit, because
the current will flow only on the outer surface of the metal.
QUESTION:
Are magnetic lines of force literaly "strings" or "filaments" of energy
or are the lines shown in illustrations only there to show the size
& shape of a uniform field? i.e. is the force stronger directly on
the lines & weaker between them or is it a uniform field?
ANSWER:
The magnetic field lines are not "filaments of energy".
The field is a representation of the force which would be felt by the
north pole of a magnet (if you could separate it from its south pole
partner). The field lines you see drawn represent a uniform field
if the lines are all straight, parallel, and evenly spaced. In a
uniform field, a north pole would experience the same force, magnitude
and direction, no matter where you put it. If the field is not
uniform, the field lines either curve or have varying "packedness" (i.e.
some lines are closer to their neighbors than others) or
both; the more closely the lines are packed, the stronger the
field. Electromagnetic fields do contain energy, but you should
not think of the field lines as somehow representing this property.
QUESTION:
When working with projectile motion problems, the range (R) =
[(2*Vo^2)*(sin theta)*(cos theta)] / g, but when the launch angle is
horizontial (theta = 0) the sin theta is equal to 0, hence the range
would = 0. Since this is not the case, how do you calculate the range
when theta = 0?
ANSWER:
Ah, the dangers of using formulae instead of using your
head! You need to know the applicability of the formula and
understand precisely what it means. The range equation tells you
how far the projectile travels horizontally before it returns to the
same vertical position from which it was launched. Clearly, since
the projectile starts falling the instant it is launched, it never
comes back to that same height again.
QUESTION:
Do gravitons exist?Or are they only a theory?
ANSWER:
There is no working theory of quantum gravity and so a
graviton is a purely hypothetical particle. A graviton has never
been detected experimentally. Trying to unify general relativity
(the working theory of gravity) with quantum mechanics (the working
theory of everything else) is one of the holy grails of theoretical
physics.
QUESTION:
electrons are emiited from a conductor when the conductor is ?I think
the answer is cooled rapidly, but I AM NOT SURE.
ANSWER:
When: the conductor is heated up, when it is placed in a
strong electric field, when it is bombarded by photons of sufficient
energy, but I have never heard of rapid cooling causing emission.
In essence, it takes energy to remove an electron from an object and
cooling is energy being taken away. You can get more detail here.
QUESTION:
What exactly is the work function in the photoelectric effect? I know
it is in einstien's equations because electrons use energy in order to
get to the surface of a metal and thus be emmitted, but what exactly is
it measuring and determining?
ANSWER:
Suppose you have a perfect conductor from which you wish to
remove an electron. When you pull it out a little way, there will
be an induced charge distribution on the surface of the conductor
because the electron will push nearby other electrons away. This
problem is a classic problem in electrostatics and may be solved by
imagining an "image charge" inside the conductor which is oppositely
charged (positive). So, you see, work will be required to pull
the electron to where it is far away. This is essentially what
the work function is, the work necessary to remove an electron from a
conductor. There is one little catch--if you press this image
charge thing too far and try to calculate the work done it will be
infinite because eventually the two charges are on top of each other
(when the electron is exactly at the surface); this occurs because the
conductor is not really perfect and homogeneous but made up of atoms,
so when you get very close to the surface you see the individual atoms
which are separated by approximately 10-10 m. The
method to calculate the work function should then require you to
calculate the amount of work necessary to move an electron from r1=10-10
m to infinity; you will find that this gives you a ballpark estimate of
the work function, on the order of a couple of volts.
QUESTION:
Friction and air resistance cause kinetic energy of an object to
convert into heat energy. Why does this happen? Also, what kind of
"heat energy" is this? Photons of radiation or movement of particles?
ANSWER:
The detailed nature of friction at a microscopic level is not
very well understood. But the mechanism for heating must surely
be collisions between atoms resulting in increased kinetic
energy. For example, if a block slides across a table it loses
its kinetic energy but many individual atoms pick this energy up and
the block and table heat up.
QUESTION:
With the photoelectric effect are you hitting the free electrons or the
electrons in the atimic shells with photons? What would the difference
if you struck the different ones?
ANSWER:
The photoelectric effect is a phenomenon which occurs in
metals and the photoelectrons are the result of the conduction
electrons interacting with photons. In a conductor, to an
excellent approximation, the conduction electrons move around in the
material much as a gas moves around in a box and these electrons may be
thought of as free within the material.
QUESTION:
Whe you heat up an object by radiation (ie - hit it with infrared
photons) it gets hotter. Therefore, its atoms/ molecules must be
vibrating faster. However, I have also been told that when an electron
absorbs a photon it moves to a new evergy level and then releases the
photon again as it returns to its ground state. So, when do these two
different phenominum occur? ie - what must the photon in the 1st senari
strike, etc
ANSWER:
The photons scatter many times, elastically, from the atoms
but, since linear momentum must be conserved, each collision reduces
the energy of the photon a little while that atoms absorb that kinetic
energy and eventually share it with all their neighbors.
QUESTION:
Atoms go up in discrete energy levels. But what happens if a photon
hits an electron but does not posess the correct energy for it to move
up any level?
ANSWER:
If the photon has too little energy to excite the atom to any
level, the photon will just bounce off elastically from the atom,
leaving the atom in its original state but giving the whole atom
kinetic energy and slightly changing the energy of the photon (energy
and linear momentum must be conserved). If it has enough energy
to completely knock the electron out of the atom (ionize the atom), you
will have essentially Compton scattering. Things get more
complicated in a solid where the atoms are bound in a lattice; do some
research on the Mössbauer effect.
QUESTION:
Voltage or potential difference is "the difference in the energy per
unit of charge between two points". eg - If the pd across a resistor is
6V, then each Coulomb of electrons has lost 6J of energy and converted
this into other forms, such as heat. However, why do we then say that
if static electricity builds up on an object that the potential
difference between it and the earth or an uncharged object increases?
Where is the energy, voltage, etc?
ANSWER:
Perhaps a more fundamental way to think about potential
difference between two points is to think about the work it takes to
move a charge from one point to the other (if it takes one Joule of
work to move one Coulomb of charge from one point to another, then the
potential difference is one volt). It should then be obvious why
an object being charged is having its potential relative to some
reference point constantly increasing: it clearly will take more work
to move a charge from the reference point to the charged object as the
charge increases.
QUESTION:
"energy conservation requires that any energy entering the system via
the primary coil must equal the energy leaving the system via the
secondary coil." - But as far as I'm aware no energy is entering the
system. As electrons flow through the coil, they do not give electrons
to the core of a transformer. Therefore, how is energy entering the
system and consequently why does the conservation of energy apply
between the two coils? (ie - why does power in = power out? )
ANSWER:
See answer to the following question.
QUESTION:
Refering to others question on power and transformers - but surely
energy is not actually entering the system via the primary coil. After
all, the coil is not giving its energy to the core, is it?
ANSWER:
Of course energy is entering via the primary
coil. Do you think the transformer would operate if you did not
plug it into some power source? There is a time-varying current
in the coil which means that there is a time-varying magnetic
field. If it were just a coil it would be a time varying magnetic
moment which is a kind of antenna; electromagnetic waves would be
emitted from it resulting in energy leaving via those
waves. If you make a transformer, there will be a time-varying
field through the secondary coil which will cause a current to flow
through any load which is connected across it. Energy will thus
leave the system via the secondary coil and it comes from the power
supply which is driving the primary coil. The two will be equal
if you neglect losses due to heating of the transformer or the energy
which will be radiated away.
QUESTION:
Why is a transormer said to obey the equation Pprimary=Psecondary?
Surely, the two coils are completely seperate apart from their magnetic
fields, so why should the power (ie - V*I ) of one equal the power of
the other? If you take the concept of conservation of energy it
makes sense to say that the power in the primary coil of a transformer
is equal to the power of the secondry since energy cannot be created or
destroyed. But where does the energy change occur?
ANSWER:
I have consolidated two questions from the same person
here. Essentially, he/she answers the first question with his/her
second question: energy conservation requires that any energy entering
the system via the primary coil must equal the energy leaving
the system via the secondary coil. Of course, in the real
world the two will not be equal because of energy losses in the
transformer; for example, note that a transformer is always warm or hot
as some of the energy in is converted to heat.
QUESTION:
With a transformer, we say that all of the magnetic field of the first
(primary ) coil is trapped within the core. But if we look at a
solonoid containing an iron core, there is clearly a magnetic field
around the core. ( hence the familiar diagrams) How come that with a
transformer core we only get a magentic field within the core, and not
one surrounding it or around the coil?
ANSWER:
Well, the "familiar diagrams" show the core and a coil around
it, not a field. But a solenoid does have a field outside the
core, in fact the field looks, on the outside, just like the field of a
permanent bar magnet. Imagine now taking a bar magnet and bending
it around so that the two poles meet to make a closed loop of iron; now
nearly all the field will be confined to the inside of the iron.
This is the way the core of a transformer looks, closing on itself.
QUESTION:
The air conditioner in my apartment is terrible and I live on the top
floor of an a building with a BLACK roof. It's really hot.
I need an alternative method of cooling the apartment. My idea:
Fill metal pipes with dry ice, seal both ends, attach them near the
ceiling. I figure the pressure should keep the dry ice solid, the hot
air near the ceiling should cool near the pipes and fall creating minor
circulation and cool the room. Might it work? Could the pipes
explode?
ANSWER:
This is a sure recipe for disaster. Do not do it. Get a better AC
unit. Also, if you think about it a little, your idea attempts to
get something for nothing. If your pipes absorb heat from the
room, where is it going to go?
QUESTION:
if the psi is known along with the diameter of the tube the gas
is flowing through can the linear velocity of the gas be
determined? In my case the psi is 40 psi and the diameter is 1/8 in.
ANSWER:
No. You do not have enough information. If you
think about it, it is obvious that one possible solution is zero
velocity. Just add gas until the pressure is 40 psi.
QUESTION:
Why when an object gains an amount of charge ( eg - when charged with
static electricity) does its voltage increase?
ANSWER:
That is just the definition of electric potential.
Voltage is work done per unit of charge moved. So if it takes you
2 Joules of energy to charge something to 10-6 Coulombs, the
voltage is 2,000,000 volts relative to the place where the charge
originally resided.
QUESTION:
With capacitence, why is it that different bodies need to gain
different amounts of charge to gain a different voltage? ie - C=Q/V.
Surely, it would make more sense to think that since a body is gaining
xC of charge (ie - Q) and that each electron forming xC has a certain
amount of energy, that by gaining this xC any object's electrons would
possess this same extra amount of energy compared to earth. (ie - all
the objects would have the same potential difference or voltage)
Also, what does it mean when we say that a capacitor has a certain
value? What does this say about the particular capacitor?
ANSWER:
As you charge a capacitor, the amount of work to move each
electron from one plate to the other is constantly changing as the
capacitor charges up. It takes much less work to move the first
few electrons from one plate to the other than the last few. The
equation C=Q/V tells exactly what capacitance means--it is the
amount of charge moved from one of the two elements of the capacitor to
the other per volt applied. A capacitor is always composed of two
conducting elements which are separated by some nonconducting medium.
QUESTION:
1) Why can a capacitor only gain a voltage equivolent to the supply
voltage across it - no more?
2) What do you actually change about a capacitor if you increase the
value of it? Why does this mean it takes longer to charge?
ANSWER:
1) The voltage across a capacitor depends on many things
about the circuit in which it is placed. If you have in mind that
you just connect a battery across an uncharged capacitor, then it is
just the definition of potential difference which is the reason why the
two are equal when all current ceases to flow--the work it takes to
move a charge all the way around the circuit (including across the
battery and the capacitor) is zero. However, if you began with a
capacitor which already had a charge such that the voltage across it
was, V1 and you connected it across a battery which
was V2, then the voltage across the capacitor would
be either V1+V2 or V1-V2
depending on how you hooked it up.
2) To change the capacitance you need to change the geometry or put a
dielectric material in the intervening medium. For example, if it
is a parallel plate capacitor, you can increase the size of the plates
or decrease the distance between them to increase capacitance. In
a simple circuit with resistance of R and a capacitance or C
in series, the time it takes to charge depends on the product RC.
So, if you keep the resistance and battery constant and vary the
capacitance, the time is proportional to C. This should
make sense to you because as C increases, the charge which it
can hold at a particular voltage increases and it would take longer to
acquire that larger charge. Actually it takes infinite time in
principle to charge a capacitor completely, RC is the time it
takes to acquire 1/e=1/2.718 of its total final charge, but in
practice, wait a few times RC and it will have, for all intents
and purposes be totally charged.
QUESTION:
As I've understood it, Quantum entanglement of particles seems to mean
that information can be transmitted instantaneously between an
entangled pair, across any distance... Would you be able to set up an
instant link with the mars rovers - or a distant spacecraft - in this
way?
ANSWER:
No, because you have no control over the information being
"transmitted". You make a measurement on one of two particles
which puts that particle in a particular state and then the other
particle is "instantaneously" required to be in a particular state
depending on how the pair was originally prepared. But you cannot
send any information that way because you cannot predict in which state
you will find the first particle.
QUESTION:
Does work being done always require a force moving an object a
distance? eg - Work = force*distance, but suppose the transfer in
energy did not involve something being made to move a distance due to a
force, such as the conversion of chemical energy to light energry - is
work still being done? If so, what would it be equal to?
ANSWER:
The word "work" in physics generally is intended to mean
mechanical work, that is force*distance as you say. But work need
not be done in order for energy to be transferred. A simple
example is to add heat to a container of gas which has a fixed volume;
then no work is done but the energy of the gas increases, that is the
average kinetic energy of the molecules increases, that is the
temperature increases.
QUESTION:
what is a formula on how fast a roller coaster goes.
ANSWER:
Well, there is no formula because it is a complicated
problem. However, if friction is negligible and the roller
coaster starts from rest at a distance h above the bottom, you
can estimate the speed at the bottom using energy conservation, mv2/2=mgh
where m is the mass and g is the acceleration
due to gravity. Thus, v=[2gh]1/2.
QUESTION:
What is the volume differance between Diatomic Hydrogen(H2) and
Monoatomic Hydrogen(H1)?
ANSWER:
Assume that each is an ideal gas, so PV=NRT. (I
assume you know the ideal gas law.) If the two gasses have equal
pressure (P) and temperature (T) then V1/V2=N1/N2
where N is equal to the number of moles of each gas. If
we have one (1 gram approximately) mole of H1 (N1=1)
and 1/2 mole (1 gram approximately) of H2 (N2=1/2),
then V1=2V2; so equal masses will
have the H1 occupy twice the volume as the H2.
What you should realize, however, is that H1 never exists at
normal pressures and temperatures because it wants so badly to be H2.
QUESTION:
If I could temporarily deviate the gravity lines (shield it against
gravitons) from a body on the surface of the earth and move it to a
higher position than I would obtain 'free' potential energy?
ANSWER:
Yes. However, there is nothing profound here.
Your question is essentially "if I could suddenly turn on a force field
would something which feels that force suddenly acquire potential
energy?" Yes, it would.
QUESTION:
Planck said that the energy of a photon is h / f. What are the smaller
and bigger frequencies a photon may have?
ANSWER:
First of all, energy is Planck's constant times
frequency, not divided by. There is, in principle, no limit on
the possible frequency a photon may have. Of course it could not
have zero or infinite frequency because zero energy would mean it does
not exist and infinite frequency would mean it had infinite energy and
that much is not available in the universe!
QUESTION:
Lets say I have a 60 watt light bulb and a 40 watt light bulb. If
the 40 watt light bulb is the one with the highest resistance, and the
composition of both the filaments are the same. Then what would
be the difference between the two filaments to make the 40 watt have
higher resistance?
ANSWER:
Most likely the length of the filament of the 40 watt bulb is
longer to make its resistance higher. Resistance of a wire is
proportional to its length and inversely proportional to its cross
sectional area. Therefore, you could also increase the resistance
by decreasing the cross sectional area, i.e. use a thinner wire.
QUESTION:
If a gun is fired from a speeding car in the direction of travel, the
bullet takes on the speed of the car in addition to its own muzzle
velocity, correct? If the bullet is fired backwards at a bystander, the
velocity of the bullet is reduced by the speed of the car, correct? So,
if the car could travel as fast as the bullet, then fired backwards,
the bullet would never hit a stationary observer (absent gravity, the
bullet would be stationary?). So, if light is emitted from a star as it
is propelled outwards within the expanding universe, is the light
travelling at 200,000 miles per second if the star is already
travelling at 14,000 miles per second (relative to a stationary
object)? ie. light would be travelling faster than the speed of light.
Conversely, is the speed of light reduced by the speed of the star's
retreat, from the point of view of a stationary observer? If this is
true, how does that affect E=Mc2?
ANSWER:
The principal postulate of the theory of special relativity
is that the speed of light in vacuum is a universal constant. A
beam of light, regardless of who views it, travels at the speed of c=186,000
mi/s=3 x 108 m/s. So if you are in a car which is
approaching me with half the speed of light and you are shining a
flashlight at me, I will measure the speed of light to be c and
you will also measure the speed of the light to be c and a
third observer moving in the opposite direction as you with speed c/2
will also measure the speed of the light to be c, and so
forth. As paradoxical as this may seem, it is true and has been
verified experimentally numerous times!
QUESTION: 
We know that in the case of earthquake waves, P longitudenal waves can
pass through the liquid core but S transverse waves cannot - why is
that?
ANSWER:
For a longitudinal wave to traverse a medium, the medium must
be compressible. This is how sound travels through air or water
or steel. The medium is displaced in a direction parallel to the
direction the wave is moving and does that by compressing closer to
gether. For a transverse wave to traverse a medium, the medium
must have a shear elasticity;
I am not sure that
this is the right wording but if the medium is displaced in a direction
perpendicular to the direction the wave moves, it must "spring
back". A solid does this, a liquid or a gas does
not. Representations of the two types of waves are shown at
the right. The S wave could not exist if there were not a
restoring force in a vertical direction in the figure. The P
wave, however, has a natural restoring force because the pressure
increase in the regions of compression push the material back.
QUESTION:
i've noticed that sometimes if a rock falls off the bed of a truck, it
might, at some point, reach a height which is higher than the height of
the previous bounce. What energy is being transferred to
potential to allow this to happen? (ie. rotational, kinetic due to the
speed of the truck, etc.) and how can it be described
mathematically?
ANSWER:
When it leaves the truck, it has a velocity about equal to
the truck's velocity so it has a significant amount of kinetic
energy. Now, if the road were perfectly smooth and elastic, it
would keep bouncing always returning to the same height. However,
suppose that the road had an imperfection in it such that the collision
with the rock sent the rock straight up (or at least more up than its
downward velocity vector before it hit). It is easiest to think
of it going straight up: suppose the stone hit the ground with a
speed of 88 ft/s (60 mi/hr) and bounced straight up; also, to
make it more realistic, suppose it lost 1/2 of its speed in the
collision so it bounced up with a speed of 44 ft/s; I calculate that it
would bounce about 30 feet up which is pretty high. So, it is
easy to believe that a stone will frequently bounce higher than the bed
of the truck.
QUESTION:
Let's say I have a thermonuclear rocket, and I want to take my friends
on an extended tour of the solar system. My plan is to
continually accelerate at 1 G to keep us all healthy and
comfortable. Will I have to keep adding to my thrust to maintain
this if we approach relativistic velocities? For the return trip,
the plan is to cut the motor, turn around, and decelerate our way back
like they did in Tintin: Explorers on the Moon. But lets say
there is a malfuntion of the motor cut-off, so the throttle is
stuck open. Would I still be able to turn around and decelerate
toward home?
ANSWER:
Acceleration is not a useful quantity to talk about in
relativity. However, it is not possible to have a constant
acceleration of g for the following reason. Suppose that you
had accelerated to within 8 m/s of the speed of light; if your
accereration is and remains 9.8 m/s2 then you will exceed
the speed of light before the next second passes which is a
no-no. Anyhow, until you get very close to the speed of
light, you may have an acceleration of g but it will cost you
increasingly more energy as you go faster (finally requiring infinite
energy to achieve c). The details are a little technical,
but if you know a little special relativity and calculus, they are not
hard. I have done a brief explanation which you can link to here. I
really don't see what your last question has to do with
physics--whether you can turn around with the motor running is a
function of how the craft is designed.
QUESTION:
Can you explain to me how to find the mass in grams, and volume in
grams of a rock? I have a project in school to do involving many
steps and the teacher is only giving us a week to do it. Please give me
hints!!!
ANSWER:
If you weigh the rock, it will probably be in pounds, so you
need to know how many grams in a pound: 1 lb = 453.6 gm. There is
a nifty little program called Convert which you can download from http://www.joshmadison.com/software/convert/
which will let you look up conversion factors like this one. Your
second question is nonsensical because you cannot measure volume in
grams. Volume must be measured in either a length cubed (e.g.
cubic centimenters, cm3) or in a unit defined as a volume,
like a gallon or a liter. The volume of a rock will not be real
easy to measure if it is not some regular shape like a cube or a sphere
whose volume you can calculate if you can measure the length of a side
or the diameter, respectively. What you will probably want to do
is to see what volume of water is displaced by your rock. Then
that is the volume of rock. For example, suppose that you find
the rock displaces 2/3 cups of water and you want the volume in cm3.
Then since 1 c = 236.6 cm3, your rock has a volume of
157.7 cm3.
QUESTION:
Why in inelastic collisions is momentum conserved but kinetic energy
not? Energy is lost as heat, sound, etc, but why is momentum not lost
too?
ANSWER:
In a collision, if there are any forces which do work then
these forces change the total energy of the system. For example,
internal frictional forces will do negative work during the collision
taking kinetic energy away and changing it to heat energy. On the
other hand, linear momentum changes only if there is a net force on a
system. The easiest way to see why is to express Newton's second
law (usually force equals mass times acceleration) as net force equals
the time rate of change of linear momentum. For colliding
particles we generally regard the system as isolated, that is the only
forces which the constituents feel are forces internal to the system
itself. But Newton's third law states that all the internal
forces on a system must add up to zero (the force which particle 1
exerts on particle 2 is equal and opposite to the force which particle
2 exerts on particle 1). Thus the time rate of change of linear
momentum is zero, i.e. linear momentum is conserved.
QUESTION:
My question regards friction and determining reaction forces. If we do
the simple ladder problem, to determine the angle of slip, the typical
assumption is frictionless wall. What if I do have friction between the
ladder and the wall? If I use the static equilibrium equations, (Fx=0,
Fy=0, Mz=0), I get three equations for only two unknowns.
F1x - mu F2y = 0 mu F1x + F2y - mg = 0 mg cos(theta) x s = F1x
cos(theta) x L + F1y sin(theta) x L
where s is the length to CG and L is the length of the ladder, theta is
the angle the ladder makes with the ground, F1 reaction force on the
wall, F2 reaction force on the ground. I could solve the moment
equation for tan(theta) as a function of mu, s, L, and mg but when I
put a ladder up, it will stay up for a variety of thetas. Should this
be an equality relationship? I seem to be having a mental block
grasping it.
ANSWER:
The reason you are getting into difficulty is that for static
friction, the relation f=mN is not true. The
static friction force is whatever it has to be in order for the system
to be in equilibrium but may not exceed mN
(that
is when slipping will occur). The problem you describe is
thus an underdetermined rather than overdetermined one; there are four
unknowns and and three equations. There is no way to determine
them without additional information. Incidentally, the way you
describe the problem is valid if the ladder is just about to start
slipping. Then the angle q is the third unknown
and you can predict the minimum angle the ladder can make with the
floor without slipping.
QUESTION:
If one had 2 droplets of water (1 with twice the radius of the other or
8x the volume) and one were then to put them on a flat surface, what
happens to the perceived radius from above? Does the larger
become more than double that of the smaller or not? Why?
Where can I read about this? Does it depend on the surface?
ANSWER:
It does depend on the surface. If the surface is
"wettable" then each droplet will spread out making a very thin film
and the thickness of each will be about the same. Therefore, each
flattened drop will have an area proportional to its volume, so the
larger will have an area 8 times that of the smaller, so the ratio of
the radii will be square root of 8 or about 2.83. If the surface
is not wettable, the shape which each drop takes on will depend on the
surface tension of water which keeps it from flattening out
completely. The surface tension depends on the temperature so,
using hot water with smaller surface tension will result in flatter
droplets. The flatter the droplet, the closer the radius ratio
(from above) will be to 2.83. The rounder the droplet, the closer
the ratio will be to 2. To learn more about this, you obviously
need to research surface tension.
QUESTION:
When perfect materials are specified in regards to a collisions, why is
there a "loss" of energy during inelastic collisions and not elastic
ones?
ANSWER:
Elastic and inelastic are just words. In usual
classical mechanics, an elastic collision is defined to be one in which
the kinetic energy of the colliding particles is the same after the
collision as it was before. But, the total energy of any isolated
system never changes, this is the principle of the conservation of
energy. So an inelastic collision, in which kinetic energy is
either lost or gained, must have kinetic energy changing into some
other kind of energy or vice versa. Examples are:
- A bomb at rest has no kinetic energy. Now the bomb explodes and pieces of the bomb fly in all directions. Where did the kinetic energy come from? It came from the chemical energy which was stored in the explosive until it was released by a chemical reaction.
- Two putty balls collide and stick together (a so-called perfectly inelastic collision). It is easy to show (using momentum conservation) that kinetic energy is lost. Where did it go? It takes work to squash a putty ball; this work is like friction and the kinetic energy lost ends up as heat, i.e. if you carefully measured the putty balls after the collision, they would be a tiny amount warmer.
For macroscopic particles, one never sees a perfectly elastic collision because there will always be friction-like forces to take energy away. However, for microscopic particles elastic collisions are the dominant process. For example, if you shoot protons at protons, since protons are very elementary particles (with no internal degrees of freedom which would let them "heat up" as it were) they bounce off each other elastically. (If you give them a huge amount of energy, it is possible to cause the proton to "heat up".)
QUESTION:
Ok. Well my question is, if the earth was suddenly to lose it's
atmosphere, where would the water go? Would it evaporate into space?
And then what? Will it seek a place with less density? What can
be less filled with atoms than space?
ANSWER:
No, it would not evaporate into space. It would
evaporate and form an atmosphere of water vapor. This process
would continue until the pressure of the atmosphere was just right so
that the water and the water vapor were in equilibrium at the
prevailing temperature. Atmospheres evaporate into space if the
atoms in the atmosphere have velocities larger than the escape velocity
from the surface of the earth. (The escape velocity is how fast
something must move in order to completely leave the earth, the
velocity you would have to give a rocket to send it to Mars, for
example.) At the temperatures present in the earth's atmosphere,
heavier molecules (like O2, N2, CO2, etc.)
do not have velocities large enough to escape but lighter molecules and
atoms, most importantly hydrogen and helium, do and that is why there
is no hydrogen or helium in the earth's atmosphere; the only way to get
helium is to get it from inside the earth where it is trapped, for
example in natural gas wells. If the earth were to get a lot
hotter, the atmosphere would have larger velocities and would escape
into space.
FOLLOWUP
QUESTION:
But if there isn't an atmosphere, then does the temperature
get hotter or how then will the planet's temperature be? And what
I meant in the other question is, doesn't molecules travel from crowded
places to more open places?
ANSWER:
What determines temperature of a gas is the average kinetic
energy of the molecules (essentially the hotter the gas, the faster the
molecules are moving). What determines what the temperature will
be is a large variety of things like what energy is coming in (e.g. from
the sun), what energy is going out, etc. But any given
system will tend toward some equilibrium where the energy coming in
will equal the energy going out and some equilibrium temperature will
be reached. Regarding your other question, yes, there is a
tendency for the gas to spread out as far as it can moving to fill the
surrounding vacuum. But in the case of an atmosphere, this is not
the only thing to take into consideration. The atmosphere is
being pulled on by the earth's gravity which has the opposite effect of
the tendency of the atmosphere to spread out in space. If the
earth wins this tug of battle, it has an atmosphere; if not, it does
not (like the moon has no atmosphere).
QUESTION:
Let's say there is a stationary observer A and a stationary flashlight
B some distance away. Beyond B there is another flashlight C moving
toward B at a constant velocity. At the point when flashlights B and C
are equadistant to A, they are switched on. Which flashlight will A see
first?
A
B <------- C
ANSWER:
I presume that you mean that B and C are at the same location
when they turn on. Then A will see both flashlights
simultaneously. The speed of light is independent of either the
motion of the source or the motion of the observer; this is a basic
postulate of the theory of special relativity. The light from C,
however, will be blue shifted because of the motion of the source.
QUESTION: 
Why is the formula for a moment "Moment = force * perpendicular
distance"? WHy does the perp.dist. have an effect?
ANSWER:
The easiest way to think about this is to define Moment=distance
* perpendicular component of force
where distance is the distance from the axis about which moment is to
be calculated to where the force is actually applied. This makes
sense because if the force has no component perpendicular to the
distance it will not tend to rotate the object as moments do.
(For example, think of trying to close a door by pushing on the edge
straight in toward the hinges.) Now, if you know a little
trigonometry, drawing a few sketches should convince you that the exact
same moment is achieved if you define moment as Moment=force
* perpendicular component of distance.
In the figure I have shown the vectors d and F and their components
perpendicular (red) and parallel (blue) to the other. In one
case, moment is d(Fsinq) and in the other it
is F(dsinq).
QUESTION:
Why does increasing the number of turns a coil has increase its
magnetic field? Also, why does increasing the current and adding an
iron core do the same as this?
ANSWER:
One coil of wire carring a current I will cause a
field B, so two will cause a field 2B, three will cause
a field 3B, etc.; and similarly, the same coil carrying
a current 2I will cause a field 2B, so two will cause a
field 4B, three will cause a field 6B, etc. The
question about iron has been previously answered.
QUESTION:
In an ac generator (ie - an alternator) you apparently get the ac wave
pattern because when the coil is in the horizontal position it is
cutting the magnetic field at the fastset rate, etc. Why is it cutting
it at a faster rate? Surely it is being turned at a constant speed,
meaning it shoud be a constant rate?
ANSWER:
The induced EMF in a loop rotating in a magnetic field is
proportional to the rate of change of magnetic flux through the
loop. The flux is defined as the area of the loop times the
component of the field perpendicular to the loop. If the
field is B and the area is A, then the flux is BA if
the field is perpendicular to the loop, 0 if the field is parallel to
the loop, and BAcosq
if
the normal to the loop makes an angle q with the field. So as the loop rotates with
constant angular velocity the flux through it will look like a cosine
function.
QUESTION:
In radioactivity, what causes the difference in the absorption of
alpha, beta and gamma? (ie - alpha is stopped by paper, beta by
aluminium, etc) Is it the size of the particle/ wave, its ionizing
power, or what?
ANSWER:
Radiation is stopped by losing energy and energy is lost by
interacting with matter. If we have one of each type of particle,
an a, b, g of the same
energy. The a has a charge of +2e
(it is the nucleus of a helium atom) and a mass about 8000 times bigger
than the b which has a charge of +e
or -e (it is either an electron or a positron); the g has no mass and no
charge (it is a photon). They lose energy by interacting with
atoms and the slow, heavy, doubly charged a
loses it fastest while the massless, chargeless g loses it most
slowly. Learn more at HyperPhysics.
QUESTION:
How does a capacitor actually "store charge" and make a current flow?
Also, why does it take longer for the capacitor to charge and discharge
if the resistance is high or the capacitors value is greater?
ANSWER:
Basically, a capacitor is just two pieces of metal (I will
call them 'plates' below, but they need not be plates), not touching
each other. If you cause there to be an electric potential
difference between the plates, for example by connecting them together
with a battery and wires, then the electrons will flow off the plate at
the lower potential and go to the plate with the higher potential
(positive charges go from higher to lower potential). So, that is
how to "store charge"; actually you see that it is not storing some net
charge but has charge moved from one to the other. If you now
remove the battery and wires, a charge of +Q will be on one
plate and a charge of -Q on the other; a potential difference
will exist between the two plates, the one with +Q being at the
higher potential. So now, if you connect them back together, the
electrons will again move to higher potential, that is they will go
back to where they came from. If there is a resistance in the
wire, then the current will be inversely proportional to the resistance
and proportional to the potential difference. So a big resistance
means a small current which means a smaller rate of transfer of charge
(which is what current is) which means a longer time. Learn more
about electricity and magnetism and about capacitors at HyperPhysics.
QUESTION:
If you charge a piece of polythene through rubbing it, can you
discharge it by connecting it to earth?
ANSWER:
This question has been previously
answered.
QUESTION:
Does the physical hand positioning on a hockey stick affect the speed
of the hockey puck?
ANSWER:
The speed of the puck will be determined by the speed of the
stick where it touches the puck. This speed will depend on how
much acceleration you can give to that end of the stick. There
are two ways you can accelerate the end of the stick: either accelerate
the whole stick forward by pushing on it or give the stick a rotational
acceleration by rotating it. I believe that second way is much
more important and that it plays the dominant role in getting the puck
going very fast. Now, in order to impart a large angular
acceleration to the stick (get it rotating fast) you have to exert a
large torque on it. Imagine that your left hand holds still and
that your right hand exerts the force which causes the torque which
causes the stick to rotate; since torque is force times distance, and
since the maximum force you can exert is limited by your strength, you
can exert a bigger torque by having your hands as far apart as
possible. Of course, it isn't really quite that simple since if
your hands are at each end of the stick (as far apart as possible),
they are awkwardly placed and you would not be able to exert as much
force as if they were somewhat closer together. As I recall, a
powerful slap shot has the player's hands about half the stick length
apart.
QUESTION:
what does it mean to "trickle charge" a battery?
ANSWER:
Many types of rechargeable batteries cannot handle being
charged up too fast; trickle charge is charging them up at the right
(slower) rate.
QUESTION:
Logic Gates - what exactly are they? Ive been told the truth tables and
circuit symbols for AND, OR and NOT by my teacher but he didnt actually
say what they are, how you make one, what they look like in real life,
etc? Why for logic gates is it called a "Truth Table?
ANSWER:
This is really way too extensive a question to answer on this
site. You need to do research. I would suggest you start at
Hyperphysics.
QUESTION:
Is there a type of light which makes you see in black and white? If so,
how?
ANSWER:
No, not really. However, under very low intensity
situations, like at night, we cannot see colors very well because the
color sensors in our eyes are not sufficiently sensitive so we see
mainly in b&w in very low light.
QUESTION:
In terms of electronics - why would you ever need to turn an ac supply
into dc? (ie - why do you sometimes need to rectify it?) Also, when you
half wave rectify it using a diode and place a lamp in the circuit does
it go on then off, on then off, etc? is that the same if you full wave
rectify it using 4 diodes?
ANSWER:
Well, almost all electronics, your computer, tv, radio, etc.
use mainly dc. There is something in every one of these things
called a power supply whose job it is to supply dc to the
electronics. You should know that most electronics operate on dc
because many of them are supplied by batteries which are dc
devices. The rectifiers you refer to are seldom the last word in
making a dc source but are only the first step in conversion to
dc. The power from a dc power source also comes in ups and downs
and the light does not flicker because the change is too rapid and the
filament does not have time to cool down.
QUESTION:
What is the "intensity" of a wave? Is it the same for light waves as
other waves?
ANSWER:
Intensity is normally defined as energy per second per square
meter, the rate of energy flow per unit area. Often, for periodic
waves, this is defined as the average over one period of the energy
flow. For most waves the intensity is proporional to the square
of the amplitude of the wave.
QUESTION:
Firefighting test questions...PLEASE help
Which wagon is easier to pull a heavy load
ANSWER:
I think these questions might require more than physics, some
practical common sense.
Problem 2 is pretty straightforward: pull parallel to the ground and all of your force will be applied to the wagon in the direction it has to go. Pull at other angles and part of (the vertical component of) the force will be wasted.
The first problem is a little trickier. If you are on a hard, flat surface, it should not make any difference at all. However, if the surface is somewhat soft, the smaller wheels would sink in more deeply because the force would spread out over a smaller area (like a needle goes into your finger a lot more easily than, say, an unsharpened pencil). Also, if there were bumps on the surface like rocks or ruts or the like, big wheels would go over them more easily. I can think of no advantage to having different wheel sizes front and back.
QUESTION:
What is the effect of increasing the amplitude of a light wave, since
this usually shows the amount of energy of the wave but for light this
is instead affected by the frequency?
ANSWER:
This question has been previously
answered.
QUESTION:
In reading Stephen Hawking's "The Universe in a Nutshell," I came
across a section about the spin of particles. It states that a particle
with a spin of 1 will look the same only after a full rotation of 360
degrees. A particle of spin 2 will look the same after a rotation of
180 degrees. The example given for this was the face of a playing card,
as it looks the same after turning it upside down. The section
continues to say that some particles have a spin of 1/2. This means
that the particles must complete 2 rotations of 360 degrees before it
looks the same. This is where I am lost. How could it be that an object
could rotate 360 degrees and look different but rotate another 360
degrees and look the same.
ANSWER:
The easiest way I can think of to see this is to think of a
mathematical function which has the properties you seek. Consider
the function sin(x/2); when you get to x=3600
you have only gone through one half the cycle for the function, you
must go to x=7200 before the function came back to
where it started from. Incidentally, spin 1/2 particles are the
most important ones, and include protons, neutrons, and electrons from
which matter is composed. I cannot think of a good geometrical
example from everyday life but I can invent one: imagine that you had a
playing card which, when you turned it, changed suit from hearts to
spades and then back to hearts and it took 3600 for each
transformation to occur.
QUESTION:
I was wondering, is there a linear relationship between the steepness
of a slope and the rate at which an object falls down that slope?
For example, suppose I were timing the descent of a marble down a slope
with an angle of 10 degrees. If I doubled the number of degrees
of slope, how would the time change? I know it will take less
time, but is it an exact amount of time?
ANSWER:
If we denote the angle of the incline as A, then the
acceleration (a) of an object sliding or rolling down the
incline (and which loses no energy to friction) is proportional to sin(A).
Since the distance traveled if the object starts from rest is given by x=at2/2,
it follows that if the distance x traveled by the object is the
same, the time to go x must be proportional to the square root
of 1/sin(A). For the example you give, sin(100)=0.174
and sin(200)=0.342, so t(200)/t(100)=[sin(100)/sin(200)]1/2=0.713
QUESTION:
Can a sound wave in air bend light? Since sound waves are
compressions of air, and compressed air has a different index of
refraction than less dense air, I think the answer is yes. But is
it something noticeable without sensitive equipment or earth shattering
loudness?
ANSWER:
You are right, the answer is yes. The most important
examples of sound refraction result from sound speed differences in air
due to temperature changes. A nice example is given at HyperPhysics
FOLLOWUP
QUESTION:
thanks for the quick reply, but I was wondering if the sound wave can
*refract light* passing through it, not sound itself. My
assumption is that the sound will change the index of refraction of air
momentarily, and the light will bend just as it would in the case of a
mirage on a hot summer day?
ANSWER:
Again, the answer is yes but now I must say that I believe it
would be an extraordinarily small and difficult-to-observe
effect. Here is why:
- If the density variation in air is due to temperature differences, let's estimate the effect on the density: treat the air like an ideal gas (PV=NRT), imagine the temperature (T) to increase by 300, the pressure (P) to remain constant, the amount of gas (N) to remain constant, and the volume (V) is allowed to increase. (R is just a constant.) So if the temperature were to increase from 3000 to 3300, the volume would increase by 10% so there would be a 10% decrease in density. As you note, this is how mirages happen and mirages are easily observable.
- If the density variation in air is due to sound waves, let's estimate the effect on the density: If the intensity of sound is at the level of the "threshold of pain" (120 dB), the maximum change in pressure is about 30 N/m2. But atmospheric pressure is about 105 N/m2. Hence the pressure variation is only about 0.03%; hence the gas will compress by about 0.03% so the density will increase by 0.03%.
So the bottom line is: sound of normal intensities has very little effect on the air.
QUESTION:
In terms of static electricity 1) Why do tv screens become very dusty?
2) Why are floors tiles in operating theaters made from conductiong
meterials?
ANSWER:
There is a lengthy explanation of dust on a tv screen at MadSci
Network.
The floor tiles conduct electricity because if someone were to become electrostatically charged, a spark could ignite flammable gasses which might be present because of the anesthetics used. The floor provides a path to ground to keep the personnel from acquiring an electrostatic charge.
QUESTION:
Why is this formula true for a transformer: Vs/Vp = Ns/Np ?
ANSWER:
This requires too much space to explain. Look in any
elementary physics text for details. It is basically because of
Lenz's law and Faraday's law. There is a nice discussion at HyperPhysics
.
QUESTION:
If you have a solenoid coil (ie - which has a current passing through
it and thus a magnetic field around it similar to that of a bar
magnet), why if you place some iron in the middle does that increase
the magnetic field strngth?
ANSWER:
Because the iron itself becomes magnetized when placed in a
magnetic field. See the answer to another
recent question.
QUESTION:
Why when a wire is moved in a magnetic field does this induce a
current/ voltage? Also, why does a current in a magnetic field cause
motion?
ANSWER:
A moving charge in a magnetic field experiences a force
(unless that charge moves along the field lines). Atoms are made
up of positive and negative charges. In a nonconducting material
all the charges are tied down to their location in the material, but in
a conducting material, there are electrons free to move so they do and
those moving electrons are the current. Conversely, if you put a
current (moving charges) in a magnetic field, they will experience a
force which will cause there to be a force on the current carrying wire
which will then move if free to do so.
QUESTION:
Is there a four dimensional space-time coordinate system? If yes,
please explain this.Does not time remain a constant state of reference?
ANSWER:
The most elegant way to do the theory of relativity is to
introduce a fourth dimension which is, essentially, time. I say
"essentially" because the fourth dimension should be measured in the
same units as the other three, that is the fourth dimension should be
in units of length. The way this is done is to define the fourth
dimension to be x4=ct where c is the
speed of light. It is not absolutely necessary to do this, it
just is a very sensible way to do it. Regarding your second
question, the one of the main results of the theory of special
relativity is that time is relative, there is no absolute time in the
universe and clocks of different observers (those in motion with
respect to the others) run at different rates.
QUESTION:
How can you measure the diameter of the sun using a pin hole?
ANSWER:
The ratio of the diameter of the sun over the distance to the
sun is equal to the ratio of the diameter of the pinhole image of the
sun over the distance to the image.
QUESTION:
I am in the 6th grade in Denver, Colorado. I made a machine out
of 2 by 4s that will kick a ball. I want to see if air pressure
in a ball has an effect on the distance the ball will travel. I
did some experimenting and found that the lower the amount of air
pressure in the ball, the farther the ball went when kicked with my
machine. I thought that the opposite would happen. Why did
the ball with the lower psi go farther??
ANSWER:
I would have expected it to go farther too. This is
kinda tough since I do not know the details of your machine. What
I would do first is determine the coefficient
of restitution of the ball at different pressures; do this by
dropping the ball from some height and measuring the height of
rebound. This tells you the fraction of energy lost when the ball
collides with something which is the most important thing.
Another thing I would do is carefully weigh the ball at different
pressures and see if there is a significant change in mass which would
certainly be an important factor; of course, the mass increases when
you increase pressure (add more air) but I just haven't the time to
compute it. Also, I would use a soccer ball, not a football
because of the odd shape of the football which could mess up your
experiment. Also, worry about your machine--does it impart impact
to the ball because of its own inertia or is it designed to always
deliver a constant force (a spring loaded device, for example). I
have not answered your question, but I have given you some things to
think about.
QUESTION:
If time dilation is measurable to an observer by its effects on the
actual physical properties of a very fast spaceship (eg people would
actually age more slowly), does increase in relativistic mass also
change the actual physical properties of an object? I mean, if a bullet
were travelling down a gun barrel at .99c, would the bullet swell up
and get stuck in the barrel?
ANSWER:
In a sophisticated formulation of special relativity, the
only mass that one talks about is the rest mass of an object.
Dynamics is done in terms of the four-momentum which, in any particular
frame, consists of the three components of the linear momentum plus the
energy of the particle. The catch is, however, that the linear
momentum is not p=m0v like in
Newtonian physics, but p=m0v/[1-(v2/c2)]1/2.
One possible way to interpret this is to say that the mass has
increased to m=m0/[1-(v2/c2)]1/2
so that we can write p=mv but that is only a way
to think about it and the important quantity is momentum, not
mass. Regarding your second question, increasing mass does not
imply that all the spatial dimensions increase somehow. The
bullet would shrink along the direction of motion, the usual length
contraction, only; this is regardless of what happened to the
mass. Essentially, the mass would increase by the factor 1/[1-(v2/c2)]1/2
and the length (and therefore the volume) would decrease by the factor
[1-(v2/c2)]1/2 so the
density would increase by a factor 1/[1-(v2/c2)].
QUESTION:
Why the color of sky is blue?
ANSWER:
The reason is that (white) light from the sun strikes the
molecules in the earth's atmosphere and is scattered via a process
known as Rayleigh scattering. Rayleigh scattering is most likely
to occur for light of short wavelengths and blue is at the short
wavelength end of the spectrum of light we can see. You can read
a little more detail in the answer to an earlier
question I answered. You can read a much more detailed
discussion at http://www.sciencemadesimple.net/sky_blue.html
QUESTION:
When people talk about magnetism they often talk about domains - what
exactly are the domains? ie - are they refering to particles, etc?
ANSWER:
First, read the answer to the question just after
yours. Electrons in iron tend to line up their magnetic moments
with their nearest neighbors. However, when solid iron is
forming, cooling down after being very hot, electrons in different
locations tend to line up in different directions. So there will
be many little permanent magnets in a piece of iron all pointed in
random directions so that the iron, as a whole, does not look like a
magnet; these little locally aligned regions are called domains.
When you take this piece of unmagnetized iron and place it in a
magnetic field caused by something external, the electrons have an
additional bias to also want to line up with that field; therefore,
domains aligned with the field tend to grow at the expense of other
domains and, when you take the external magnetic field away, the iron
tends to stay with a net magnetization. That is why you can take
an iron nail, for example, and make it into a little magnet using
another magnet.
QUESTION:
Magnetism occurs when an electric current flows. ( ie - there is an
electrostatic field around a still charged particle and a magnetic
field when it moves.) So, when is iron often magnetic?
ANSWER:
First, a moving electric charge also has an electric
field. You are usually not aware of this in a current carrying
wire because the net charge of the wire is zero even though the
electrons are moving. To answer your question, you must first
understand that an electron itself, at rest, has a magnetic field; this
is essentially because it looks like a little spinning ball of negative
charge and therefore looks like a tiny bar magnet (called a magnetic
moment). In most materials, the jillians of electrons line up
their magnetic moments in random directions so the magnetic properties
of the material are not easy to see. In a very few materials,
some miracle of nature causes all of the outer electrons in atoms to
align their magnetic moments with those of their neighbors so that the
whole macroscopic thing becomes a magnet. Such materials are
called ferromagnetic and iron is the most dramatic and best
known.
QUESTION:
If you place a book on a table, what actually is (ie- causes) the
reaction force of the table?
ANSWER:
You are still saying this wrong. I cannot tell for sure
what you want to know. Is it the force which the table exerts up
on the book or the force which the book exerts down on the table?
If it is the former, then the origin of the force is essentially
electromagnetic: when the atoms in the table get very close to the
atoms in the book, then the outer electrons in the atoms on the surface
of the book feel a repulsive force due to the presence of the outer
electrons in the atoms on the surface of the table. If it is the
latter, just replace the word "book" by the word "table" and vice
versa in the preceding sentence.
QUESTION:
Will an identical sound travel FURTHER in air or water?
ANSWER:
Two things determine the distance sound will travel.
One is that the sound spreads as it moves away from the source (unless
the source has been designed to send sound in only one direction);
hence it gets quieter and quieter as the energy available is spread out
over a larger and larger area, until the intensity gets below our
threshold of hearing. The other is that there is absorption in
the medium through which the sound is moving; here the energy is
actually removed from the sound by "frictional" effects in the medium
which results in heating of the medium. For normal sound, this
heating is imperceptibly small but it is possible for the heating to be
significant--ultrasound may be used to destroy tumors by heating.
It turns out that the energy loss from absorption in air (1.2 dB/100 m)
is much larger than in water (0.008 db/100 m). Therefore sound
will travel much farther in water. It is interesting that this
absorption also depends on the frequency of the source with low
frequency sound being much less absorbed than high frequency
sound. For this reason whales transmit low frequency sound and
are thought to communicate over tens if not hundreds of miles.
QUESTION:
Newton's 3rd law of motion says that there is alway "action/re-action",
but that these do not cancel out because the 2 forces are acting on
different objects. So, how come we say that eg a book on a table does
not move or accelerate up or down because the weight is the same and
cancels out the reaction force?
ANSWER:
Your first sentence is essentially the answer to the question
you are trying to ask in the second! And your second sentence is
all jumbled. Here is the ticket:
- focus your attention on the book and forget all about Newton's third law;
- there are two things which can exert a force on the book, the earth and the table;
- the earth exerts a force straight down called the weight (which we know) and the table exerts a force, call it N;
- this book is at rest, so Newton's first law stipulates that the sum of all the forces on it must add to zero;
- therefore N must point straight up and have a magnitude equal to the weight.
The solution of this problem has absolutely nothing to do with Newton's third law. Do you want to talk about Newton's third law? If the table exerts a force N on the book, then the book must exert a force -N on the table. If the earth exerts a force W on the book, then the book must exert a force -W on the earth. You might want to read an earlier answer to a similar question.
QUESTION:
How do we get the formula F= m(v*v)/r for centripetal force? I can
understand that it must come from F=ma, but why is a = (v*v)/r ?
ANSWER:
The centripetal acceleration of an object with speed v
moving in a circular path of radius r is indeed v2/r.
I am not going to derive it here because it is a little
involved and may be found in any introductory physics textbook.
There is a concise derivation here.
QUESTION:
How can we measure the half life of radioactive meterial when the decay
is completely random?
ANSWER:
The decay of any individual nucleus is indeed a random event,
but it will occur with a probablilty. For example, one nucleus
may have a 50% probability of decaying in the next hour whereas another
may have a 50% probability of decaying in the next century. You
learn this by watching a bunch of them and counting the number of
decays you see in some time. Now if you have a very large number
of radioactive nuclei (a typical macroscopic sample, say the head of a
pin, would have something like 1023 atoms in it), you can
compare the rate of decays you see with the number of nuclei and
thereby deduce the probability that any nucleus will decay in some
particular time. The half life is the time it takes N
nuclei to decay away to N/2 and, if you think about it, you can
convince yourself that this is essentially a probability.
QUESTION:
How come when we talk about the differing wavelengths of EM waves we
say that "the higher the energy of the wave, the smaller the
wavelength/higher the frequency", yet we also say with Sound waves that
the "more energy the Sound wave has, the higher the Amplitude"? Which
one is affected by more energy - Wavelength/frequency or Amplitude?
ANSWER:
For any wave, the intensity (which is essentially energy
delivered per second averaged over a complete cycle of the wave) is
determined by the amplitude of the wave--the bigger the amplitude, the
greater the intensity. Electromagnetic radiation, as you may or
may not have learned, is also a quantized wave. This means that
the smallest intensity possible is not zero but some tiny bundle of
energy called a photon. The energy of a photon is proportional to
its frequency, E=hf, where h is a (very tiny) constant
of nature called Planck's constant. Thus, the energy per photon
in an electromagnetic wave is what is being talked about when it is
said that "the higher the energy of the wave, the smaller the
wavelength/higher the frequency". (Incidentally, wavelength l and frequency f
are related by lf=c where c is
the speed of light.)
QUESTION:
I wanted to ask about the famous Galileo experiment to do with gravity
and the leaning tower of piza. It has come to my notice that in this
"legendary" experiment, both the lighter and more massive ball fell at
exactly the same acceleration and hit the ground at exactly the same
time. Now, what I find curious is surely would this not be impossible?
Air resistance would occur between the ground and top of the tower and
mean that the acceleration of the lower mass would be reduce quicker,
since the air resistance would be greater in comparison to its weight
than the larger mass. Therefore, they could not have hit the ground at
the same time. If so, then what was the actual outcome of the
experiment? Or am I wrong?
ANSWER:
We cannot report the results actual experiment because the
story may be apocryphal--maybe it never happened. When describing
this experiment, the proviso "if air friction is neglected" should
always be stated. You are absolutely correct, the experiment is
very likely to fail unless the air friction is negligible (or
comparable); if you drop a marble and a feather, there is certainly no
argument about who would win the race. You do err, however, in
assuming that the lower mass will always lose the race since the air
friction depends also on the geometry. For example, a man with a
parachute races a marble to the ground--the marble (less mass) will
certainly win. The experiment reputedly was between cannonballs
of differing weights, and for these the air friction would be fairly
negligible for drops of this height.
QUESTION:
I'm experimenting with a small solar furnace (made with mirrors) for a
science fair project. I want to do the experiment inside so I can
control the environment. What type of light bulb would best
simulate the solar radiation that would reach the earth's
surface? I found one site that mentioned a 100 watt bulb at a
distance of 10cm but that won't work very well with my design. My
dad has a 200 watt spotlight, and I'm wondering if I could use it from
a farther distance.
ANSWER:
A rough number for the flux of solar energy hitting the earth
is 200 watts/m2. Of course, this depends on your
latitude, time of day, time of year, cloudiness, etc. but,
let's just take 200 to get the idea. Suppose you have an
energy source which is radiating W watts (joules/s) equally in
all directions (like a light bulb). Then, if you go a distance r
away from it the flux will be W/(4pr2).
The catch is that an incandescent light bulb of wattage W is
not radiating W watts of light energy because much of the
energy is being wasted as heat. You can probably find the exact
numbers, but I would guess that a 100 W light bulb only radiates
something like 30 W of light. (Incidentally, that is why it is
energy efficient to use fluorescent instead of incandescent
lights.) So, to get 200 W/m2 what distance from a 100
W bulb would you have to go? Plugging in, 200=30/(4pr2);
solving, r=0.109 m which agrees (surprisingly!) well with what
you told me in your question. So, with the 200 W spotlight, let's
estimate that it radiates 60 W of light. But, unlike the light
bulb, the spotlight does not radiate in all directions equally but puts
its light in a forward direction (that is what the reflector
does). You need to estimate the flux at a distance r from
the light. Here is what I would do:
- place the light a distance s from the wall;
- there should be a rough circle of light on the wall of radius R, so the area is A=pR2;
- assuming that the whole 60 W falls on A, the flux at distance s is F=60/A W/m2;
- now you should repeat the whole thing several times with different values of s;
- now graph F vs. s. From your graph you should be able to approximate the distance s at which F is approximately 200 watts/m2.
This is what is called an empirical approach.
QUESTION:
In a college level classical physics text, there's a section on
"relativistic mechanics" that I had a general question about as
follows. If an electron is moving along a path with v = 0.67c (c=speed
of light) and is acted on by a forceand the electron is given is given
an acceleration
(a) = 1 X 10^10 cm per sec^2 , what is the force IF: a) force is @
right angles to path (ie it would then be moving in circle)
(b) force is in same direction ( ie is parallel)
Since the F = m(dv/dt) + v(dm/dt)......in general, the part a) is easy because v(dm/dt) = 0 .........moving in circular path & equation reduces to F = m(dv/dt) = ma & since the electron's rest mass ( 9.1 X 10^-28) is known the force can be calculated using the relativistic mass. The force is then directed toward the center of the circle & is the centripetal force.
My question is part b) & how I should think about the v(dm/dt) term? The force & acceleration are in the same direction ( i.e. parallel ). Appartently, I've not recognized something that would allow me deal with the (dm/dt) term. Can you give me some coaching on how to think about this? The answer is not important, only the thought process.
ANSWER:
You are right, you have to worry about the changing mass, not
just the changing speed. In order to calculate dm/dt you
have to know how m depends on v and, thus how m
depends on t (since you know how v depends on t because
you know a). What you need is m=m0(1-b2)-1/2 where
m0 is the rest mass and b=v/c.
Hence dm/dt=m0b(1-b2)-3/2(dv/dt)/c.
Perhaps part (a) is not as easy as you think: the answer is not 9.1 x 10-18
gm cm/s2.
QUESTION:
Assuming we compare samples of the four phases (excluding the plasma
phase) of any one given chemical element - i.e. Bose-Einstein
condensate, solid, liquid, vapour - will the size and energy of the
electron shells be identical in each phase or radically different?
ANSWER:
Not absolutely identical because the atoms interact with each
other, but not radically different either. The gas phase will
have atoms which are essentially isolated and may be considered on
their own. But, if the average spacing is on the order of the
size of the isolated atoms, the orbitals of one atom will be influenced
by the presence of neighboring atoms. Normally this is not more
than a few percent effect on what an atom might "look like".
QUESTION:
I was ask "How many cubic inches of water I would
displace?" I weigh 291 lbs and am 5' 9".
I would think it mattered what size of body of water you where in, if
it was salt water, how much you floated etc. or am I on the wrong
track?
ANSWER:
It does not depend on the size of the body of water unless
the depth is such that you rest on the bottom. If you are
floating, Archimedes'
principle says that there will be an upward force on you which is
equal to the weight of the displaced fluid (fresh or salt). Since
you are in equilibrium when just floating, this force (called the
buoyant force) must equal your weight; therefore, you will displace
exactly 291 lbs of whatever you are floating in. If you now look
up the density of whatever you are floating in, the volume is weight
divided by density (because density is weight divided by volume).
For example, the density of fresh water is about 0.036 lb/in3,
so the volume of 291 lb is about 8,083 in3; since salt water
has a different density, a different volume of it would weigh 291
lb. As you can see, your height is irrelevant.
QUESTION: 
I working for an attorney on the cause of a car accident and have a
question on light. I know that our vision depends on the amount
of light returned to our eyes, so my question is if you double the
light on the object, two sets of headlights instead of one, does the
object reflect double the light back because double the light is
falling on it? Or would it be more inverse square law? And
as a side note, why don't a set of headlights on a car interfere with
each other?
ANSWER:
If you illuminate something with twice the light, it will
reflect twice the light, i.e. it will be itself twice as
bright. Inverse square law has to do with the distance from a
point source of light; for example, if the object in question is small
compared with the distance you are from it (so that you may approximate
it as a point), then if you get twice as far away it will be four times
less bright. But you must be careful about the inverse square law
because the object must be able to be approximated as a point source
radiating in all directions uniformly; the headlight of a car is
designed to defeat the inverse square law by sending all the light from
the source in a single direction (see figure).
QUESTION:
You've said in your previous answers that with a conductor the surface
its charged whilst the insulators charge is throughout - shouldn't it
be the other way round since the electrons in a insultator cannot move
so must stay on the surface? Also, why does all charge go to earth? As
well as this, would you be able to discharge a +ve insulator by
connecting it to earth, since the electrons from the earth can move?
Finally, how can you discharge an insulator? - and if tyhe charges
can't move how does the insulator discharge over time in air?
ANSWER:
The first thing we need to do is establish which electrons we
are talking about. There are the electrons which "belong" to the
conductor or insulator and there are those which have been added (or
subtracted if there is a net positive charge). For simplicity, I
will only talk about electrons since they are really the ones moving
around even if the net charge is positive. In a conductor there
are some electrons (on the order of one per atom), even if there is no
net charge on it, which are essentially free to move around; it is
these which carry current when a voltage is applied across the ends of
a conductor. An isolated, uncharged conductor will have all these
conduction electrons uniformly distributed to keep the total electric
field inside the conductor zero. (If the field inside were not
zero, the conduction electrons would move in response to that
field.) But, the conductor can also have a net charge, for
example we could to add a bunch of electrons to it. Both these
and the conduction electrons are free to move; now, the extra electrons
must arrange themselves in such a way that there is no electric field
inside the conductor because if there were a field inside the
conductor, any electron inside the volume would experience a force and,
being free to move, would move! It turns out that the extra
electrons must end up on the surface if this is to be true. An
easy way to see this is to imagine two electrons added to a conducting
sphere; since they are going to get as far as they can from each other,
they will move out to the surface. (This simple example is too
simplistic, actually, because the presence of the two electrons will
cause other conducting electrons on the surface to rearrange themselves
to make the field inside zero, but it gives you the flavor.) In
an insulator, there is nothing equivalent to a conduction
electron--each atom holds on tightly to all its
electrons. And, it turns out, any electrons which you add to the
insulator also are quite immobile. Hence, they can be put
anywhere because, even if they feel the electric fields of other
electrons, they are held in place by the surrounding atoms.
(Again, this is simplistic since it is certainly possible, with strong
enough fields, to move the electrons around.) Hence, connecting
the insulator to "earth" with a wire does nothing because the electrons
cannot move to it to exit. The reason that electric charges ever
move is to reach a point of lower electric potential energy (which is
just a fancy way of saying that they feel a force). That is what they
are doing when they move to "earth". If you have a charged object
which has an excess of electrons, the electrons would "like to" move
away from the object because of the field they cause; they do that if
you give them a path they can follow. Electrons can leak off
where they are if the air they contact can accept electrons; for
example, many molecules are such that they can add an electron easily
and become a negative ion.
QUESTION:
You can discharge a metal conductor which has been charged by static
electricity by "connecting it to the ground with a metal strip" - can
this also be done a work for charged insulators? If so or not, why?
ANSWER:
No. On a perfect insulator the charges are not free to
move, so even if they have a path to a place with lower electric
potential, they are not free to move. Of course, there is no such
thing as a perfect insulator and charge will slowly leak off. For
a conductor, excess electric charge is perfectly free to move; that is
why all excess charge on a conductor is always located at the
surface.
QUESTION:
What exactly is a spark? ie - is it a flow of electrons jumping the gap
or is it due to an electron avalanche? Could you please go through the
exact mechanism of a spark for me? Is it the same mechanism for
lightning?
ANSWER:
Lightening is just a big spark as you speculate. The
following question and answer will point you toward a complete
explanation.
QUESTION:
Someone said to me that a +ve "streamer" meets the -ve leader during
lightning and said that this "streamer" came from the ground - how can
this be since the +ve protons are fixed in the nuclei of the ground
atoms? What actually is the streamer made up of and what causes
it? It would really help if you could give a step by step
explanation of lightning if you would please - there are SO many
different versions that people give.
ANSWER:
There is not a concise answer to your question. The
most complete explanation I have found is at the Howstuffworks
site.
QUESTION:
We always talk about charging up "insulators" via static electricty.
(ie - through friction) Does it work for conductors? Does it still
remain a surface phenominum? Does the metal remain charged afterwards?
Does it dissapate naturally?
ANSWER:
Yes, conductors can become charged via static
electricity. For a conductor, all excess charge must reside on
the surface whereas, for an insulator, you may have charge
inside. The metal, in the real world, will discharge by leaking
its charge either to the air around it or through whatever (imperfect)
insulators may be supporting it. Even if it is perfectly isolated
from its environment, a metal will leak its charge away from any place
on its surface where the electric field (due to the excess charge) is
strong enough (this is called corona discharge); for example, a very
sharp needle will have a very strong electric field at the point and
electrons will stream away (or toward) this point.
QUESTION:
I'm having some trouble understanding densities and liquids, and I was
thinking about this problem. I think it could help me understand
how liquids work a little better, but I'm struggling to understand
it. If I suspend a rock from a spring scale and it has a weight
of W, then I take the same rock and, keeping it attached to the scale,
I completely submerge it in a liquid of density p and it now has a
weight of w, how do I find the ensity of the rock? I know
density=m/V, but this doesn't seem to be enough.
ANSWER:
You need to know Archimedes'
Principle. An object submerged in a fluid experiences an
upward force called the buoyant force which is equal to the weight of
the fluid which was displaced when the object came in, that is the
weight of an equal volume of water. If the buoyant force is less
than the weight of the object, it will sink and if it is greater than
the weight of the object, it will float. For your problem, the
forces on the object are the weight W=rVg
down (r is the density
of the object and g is the acceleration due to gravity), the
force that the scale exerts up, w, and the buoyant force up, B=rwVg (rw is the density of
water). All these forces must add up to zero because of Newton's
first law: B+w-W=0=rwVg+w-rVg. But, rVg=W=Mg where M
is the mass of the object, and since you measured W
you know M. So rwVg+w-W=0
and solving, V=(W-w)/(rwg).
Finally, r=M/V=W/(Vg)=rwW/(W-w).
QUESTION:
In physics class we performed an experiment called projectiles launshed
at an angle (to the horizontal). Now we made a setup where we let a
marble shoot away with the help of an elastic band. We then measured
the distance the marble travelled when letting it shoot from different
angles, 20, 30, 40,45, 50, 60, 70 degrees. I would appretiate some
theory concerning this experiment because I thaught that the marble
would reach the longest distance from the angle 45 but the distance of
the angles 40, 45 and 50 were very similar....would you please explain
to me a little about what a "fair test" is and what factors do
actually affect the range of this elastic band.
ANSWER: 
Well, an elastic band is not a terribly good energy source
because it tends to get "fatigued" with repeated use so it will not
produce reproducible results. But, even if you were using a
better launcher, say a good steel spring, your results would not be
surprising. The expression for the range of a projectile is R=v2sin(2q)/g where v
is the speed at launch, g is the acceleration due to gravity,
and q is the angle of
launch. To the right I have plotted the range as a function of
the angle for an experiment where the maximum range is 5 m. The
maximum is at 450. But the thing to note is how slowly
the range is changing when you are at angles close to 450;
the right-most graph is the same as the other graph except that it is
plotted only over the range 40-500. As you can see,
any angle in this range will give about the same value for the range--5
m.
QUESTION:
what are some ways that atomic energy can affect the environment?
ANSWER:
The actual production of atomic energy is much less damaging
to the environment than traditional energy generation sources (coal,
oil). However, after the fuel is "burnt" in a nuclear power
plant, it is radioactive and will not become nonradioactive for
thousands of years. Still, this is not a danger to the
environment provided it can be reliably stored without leaking into the
environment. That, however, is a big "if". Imagine trying
to contain something with absolute certainty for thousands of
years. The containers would have to be impervious to corrosion,
earthquakes, fires, etc. People have thought about
sending this radioactive waste into outer space, but the small chance
of an accident at launch is too big to take. Some thought has
also been given to "transmuting" this waste into either radioactive
stuff which would decay away faster or not be radioactive at all; this
does not presently appear to be practical, though. The other
danger, obviously, is the possibility of a catatrophic accident at the
power plant causing radioactivity to leak into the environment.
There was one bad accident in the US in 1979 at the Three Mile Island power
plant in Pennsylvania; fortunately, most of the radioactive waste there
was contained. In 1986 there was a bad accident at the Chernobyl power plant in
Russia; this was a much worse accident since about 30 people were
killed immediately and bad radioactivity spread over a 20 mile radius
exposing many people to dangerous levels of radioactivity and many
cases of cancer were caused by this. Lower levels of
radioactivity were recorded all over Europe.
QUESTION:
what's the worst kind of radiation?
ANSWER:
This is an impossible question to answer. It depends on
the intensity of the energy and the damage that it will do to a
biological object (like you). Let me give you a couple of
examples:
- Neutrinos are particles which are released in certain radioactive reactions. There is a huge number which come from the sun; about a million billion per second strike you! But they do not interact strongly with matter and just zip right through you.
- Electromagnetic radiation can interact strongly with your cells. For example, if I were to put you at the doorway of a blast furnace, the radiation would literally cook you within seconds. This is just plain old fashioned infrared radiation which we use to broil foods and it can be very bad for you.
- The various types of nuclear radiation, alpha, beta, and gamma, can do damage if intense enough. Alpha radiation (helium nuclei) interact very strongly but that is good in one sense since they are easy to shield against: most alpha radiation can be shielded from with a sheet of paper. On the other hand, alpha emitters can be dangerous if in the environment because if ingested or inhaled; then the source is right inside you so you can't shield against it. Beta radiation is just high energy electrons and can be dangerous if intense enough. Gamma radiation is just very high energy electromagnetic energy (like light, radio, infrared, x-rays, etc.) and it can do serious damage if intense enough. Sometimes that is good because this is the kind of radiation usually used to irradiate cancer tumors.
- Another kind of electromagnetic radiation which can be very bad is ultraviolet radiation. This is the radiation responsible for sunburn and skin cancer. There is evidence that the layer of ozone which is high in the earth's atmosphere is being depleted by pollution. This will lead to greatly increased rates of skin cancers.
QUESTION:
How exactly can one reconcile or relate the two different descriptions
of Heat as infrared EMR on the one hand and Heat as high frequency
phonons (quantised molecular vibrations) on the other?
ANSWER:
Heat is just another word for energy and is normally used
only for energy related to thermodynamics. Infrared radiation certainly
carries energy and, when absorbed by something, usually results in its
temperature increasing. The kinetic energy of atoms in solids,
liquids, or gases is also a form of energy we generally associate with
temperature; in an ideal gas, for example, temperature is a measure of
the average kinetic energy per molecule.
QUESTION:
what chemicals are in coke-a-cola
ANSWER:
Don't you know that the recipe for coke is a deep, dark
secret?! All you need to know is that it contains enough sugar to
make it bad for your health and teeth if consumed more than a couple of
times a week. One of the flavorings comes from the cola nut and
carbonated water has carbolic acid in it. Also, although the
original recipe 100 years ago included cocaine (hence the name), it
certainly does not today.
QUESTION:
What is electrochemistry?
ANSWER:
Passing electric current through something may cause a
chemical reaction to occur; an example is the electrolysis of water
where electric current causes water to split into oxygen and
hydrogen. Sometimes when a chemical reaction occurs, an electric
current is caused; and example is a simple battery where chemistry is
the source of the electricity. The scientific study of such
phenomena is called electrochemistry.
QUESTION:
As is commonly known in science, things float according to differences
in boiling points. Helium rises upwards because it is less dense than
air. But why upward. Why not downward since the natural course for
anything to take is with the flow of gravity. why then does helium
resist gravity and is propelled upwards. What propells it in this
manner. even if it is less dense than air i do not see why it would
ascend up.
ANSWER:
Well, I do not understand your "commonly known" assumption
that boiling points have anything to do with floating! But then
you go on to say the real thing which determines
floating--density. The reason that a helium balloon will go up in
air is exactly the same reason that a block of wood goes up in water;
it is because the density of the object is lower than the density of
the fluid in which it is immersed. The "why" of how this happens
has been known for thousands of years and is called Archimedes
principle. A freely floating/falling object in a fluid has two
forces on it--its own weight and whatever force the fluid might exert
on it. The force which the fluid exerts on it, called the buoyant
force, may be shown to be (see any elementary physics textbook)
upward in direction and equal in magnitude to the weight of the fluid
which has been displaced by the object; thus, if the object is less
dense than the fluid, the buoyant force up will be greater than the
weight down and the object will accelerate upwards.
QUESTION:
Say the orbits of two celestial bodies intersect...we'll say at two
different points I guess. How do you calculate the point and time at
which these two bodies will collide if we know the starting position of
both objects and assume that they must collide at some point. If the
role of gravity makes the calculation too complex we can ignore it.
[Because of our groundrule requiring "single, concise, well-focused questions", I have considerably edited your question.]
ANSWER:
Well, for starters, the role of gravity certainly cannot be
neglected since it is gravity which determines the orbits in the first
place! What you probably mean is that maybe the gravitational
force between the two objects might be neglected. Of course, your
problem is way too complex to answer in detail on a site like
this. But, it is easy to do in a general qualitative way.
The objective of classical mechanics is, for a particular particle, to
specify its position as a function of time. For example, the
vector r1(t) tells us where a particle
is in some coordinate system for any time t assuming that we
have defined some particular time to be t=0 and that we know
the particle's velocity and position at that time (initial
conditions). This position vector is determined by solving the
fundamental law of classical mechanics, Newton's second law, for the
particle. This requires, of course, knowing what the forces
experienced by this particle are. Any intermediate-level
classical mechanics course includes the study in detail of the Kepler
problem in which we assume that the sun is infinitely massive and that
a planet only experiences the sun's gravitational force; this leads to
Kepler's three laws, in particular, of interest to you, elliptical
orbits for objects orbiting the sun. So, let's let r1(t)
be the position of the earth as an example; and, as the rest of the
example, let's let r2(t) be the
position of some comet which we fear might strike the earth some
day. Assuming we have done the physics to get these two position
vectors, we simply set r1(t)=r2(t)
which gives us one equation with one unknown, the time(s) when the two
particles are at the same position. Problem solved!
However, much easier said than done! First of all, assuming that
the earth and the comet never interact with anything but the sun is
simply wrong; it will only take a little tweak of the comet's orbit
when passing close to, say, Jupiter, to cause it to be at a very
different place later in its orbit. And, when we start making the
problem a many-body problem rather than a two-body problem, chaos
(unpredictability) becomes an issue. Also, predictions using
computers, the only practical way to calculate, will rely on accuracies
of initial conditions which are far beyond our ability to measure; a
tiny change in initial conditions can make a huge change in the orbits
a long time in the future. Thus, our ability to do such
computations are very restricted--we might be able to make some
predictions reasonably accurately, even centuries into the future (e.g.,
I can be pretty sure that Pluto and earth will not collide in
the next 100,000 years even without doing a calculation) but others
predictions (e.g. for an asteroid which is known to cross our
orbit) can only be reliably done for maybe a year or two.
QUESTION:
I saw a static generator (millions of volts) charging a sphere. As the
great ball charged, the belt slowed as the motor labored to push the
tiny electrons on the belt into the sphere. This contradicted the
explanations I was hearing about how the voltage was created at the
contact between the different materials of the belt and drive rotor. I
think now only the electrons are obtained this way and the voltage is
inconsequential beyond what is needed for the comb at the sphere to
collect them in. As the electrons were forces toward the charged ball,
some kind of work was done on the character of the electrons stuck to
the belt, raising their potential and slowing the drive motor as the
charge rose.
[Because of our groundrule requiring "single, concise, well-focused questions", I have considerably edited your question.]
ANSWER:
You have a basic misconception about how the device
works. The sphere actually becomes positively charged by giving
electrons to the positively-charged arriving belt. There is a
very detailed and easy-to-read explanation at howstuffworks.com.
Regarding your question of the belt slowing down, the half of the belt
which is moving toward the positively-charged sphere is positively
charged and therefore feels a repulsive force; this force will increase
as the charge on the sphere increases.
QUESTION:
I know that if you pump up a soccer ball to 9psi at sea level and bring
it to Denver, the pressure on the ball will change. Conversely, if you
pump it up to 9psi in Denver, and take it to sea level, it will also
change. (I think that it will pop going from sea level to high
altitude, and go flat when going to sea level, but that doesn't really
affect my question). My question is: Isn't 9psi at sea level the
same as 9psi in Denver? It might require a different amount of air
molecules placed inside the soccer ball to achieve 9psi at the 2
different locations, but aren't they both 9psi? If I use a pressure
gauge (like a tire gauge with a needle) to measure the psi, and it says
9, regardless of where I am, isn't it 9? Is the firmness the same
at both altitudes at 9? The flight of the ball and the bounce of the
ball might be different because of higher altitude, but the firmness of
both balls would be the same at 9psi, regardless of the altitude?
ANSWER:
Most pressure guages measure what is called "guage pressure"
which is the pressure above atmospheric pressure. Atmospheric pressure
is about 14.7 psi but varies fairly dramatically with altitude and also
somewhat with the weather. So a pressure of 9 psi is an absolute
pressure of 24 psi if the atmospheric pressure is 15 psi but only 20 if
the atmospheric pressure is 11 psi. But the absolute pressure is
not what matters; the guage pressure is what will determine the
"hardness" or "bounciness" of the ball because the properties of the
ball are determined by all forces on it, not just the forces on the
inside. A 9 psi ball in Denver should behave essentially
identically to a 9 psi ball in New York (with regard to "hardness" or
"bounciness").
QUESTION:
say i have a satellite that has a ridiculously low orbit of 6.4x10^6 m
and it passes at a high speed, i know the mass of the earth 6x10^24 kg.
and i need to know what the velocity of the satellite is?
ANSWER:
This is called a near-earth orbit and applies for orbits near
the surface of the earth. You do not even need to know the mass
of the earth, just that the acceleration due to gravity is g=9.8
m/s2 and that the weight of a mass m is mg.
Then the force (mg) must equal the mass times the centripetal
acceleration (mv2/R). Solving, v=[gR]1/2=7920
m/s which is about 17,700 mi/hr. This is only true near to the
earth since the weight, the force of gravitational attraction, gets
smaller as you get farther away.
QUESTION:
Why can't the flight of an airplane be fully explained using
Bernoulli's equation alone?
ANSWER:
I have previously answered this
question.
QUESTION:
I hope you have not already answered this question already, but I was
wondering what the basic concepts of string theory are.
ANSWER:
No I haven't. In fact, I do not know enough about
string theory to give a cogent answer and nobody in the department here
works in the area. I can tell you that a lot of physicists are
somewhat hostile toward a theory which has been fashionable for over a
decade yet has really few results or successes to date. Still, it
represents a fascinating glimpse of cutting edge yet controversial
research. There has recently been a series of three Nova shows devoted to string
theory which, I confess, I have not seen but which I have read is
really a marvelous production. The entire three hours is viewable on
the web.
QUESTION:
If the positive charge on the proton and the negative charge on the
electron cloud of an atom have an attractive force on each other, why
does the electron cloud not collapse onto the proton and stop "moving"?
What force prevents the electron cloud from being annihilated or
absorbed by the proton? If there is an attractive force between them,
why is the electron's momentum conserved? Why don't the two "particles"
merge or fuse together?
ANSWER:
I have answered this question before and you can find that
answer here. And your question about
momentum being conserved--it is not conserved which should be clear
since the electron's direction is continually changing and momentum is
a vector quantity; angular momentum, however, is conserved because the
proton does not exert a torque on the electron.
QUESTION:
If action & reaction are always equal in magitude & opposite in
direction, why don't they always cancel one another & leave no net
force to accelerating a body?
ANSWER:
Newton's third law states that if one object exerts a force
on a second, the second exerts an equal and opposite force on the
first. Therefore, the "action/reaction" forces are never exerted
on one body. If you select a body to study, its motion is
determined only by the forces exerted on it, not by forces exerted by
it. Students often make mistakes with this "action/reaction"
thing because they tend to identify any pair of equal and opposite
forces as being an "action/reaction" pair. For example, a 1 lb
book sitting on a horizontal table has two forces on it, its 1 lb.
weight pointing down and a force of 1 lb which the table exerts up on
it (usually called the normal force); these have nothing to do with
Newton's third law but are equal and opposite because the book is in
equilibrium and the force the table exerts is therefore required to be
1 lb up. If we now look at the table, the book exerts a 1 lb
force down on it because of Newton's third law; the "action/reaction"
pair is the force the table exerts on the book and the force which the
book exerts on the table. Lots of novice physics students want to
say that the weight of the book is the 1 lb force down on the
table--this is totally false since this is a force on the book, not the
table.
QUESTION:
Pardon me if this sounds a little far out, but I was just wondering if
time series analysis is ever used in trying to determine the origins of
the universe--that is, by studying the movement or motion of the
planets and other celestial bodies and trying to predict backward
toward the big bang or the creation or whatever got the ball
rolling. And if so, could you direct me to any books on the
subject?
ANSWER:
This is more than a little far out! Even if the laws of
nature which govern the motions of everything were really time
reversible, there are simply too many variables to be able to do the
calculations you suggest. And too many unknowns from the
past. Think of the complications--the solar system came into
existence long after the big bang and before that it was a cloud of
stuff which was the remnant of older stars which had lived out their
lives and exploded. Also, much information is lost as time goes
on. For example, suppose we had a big puddle of water on the
ground which, yesterday, was an ice sculpture of a swan. Can you
imagine trying to compute backwards from the puddle to the swan?
QUESTION:
I am having trouble understanding torque. What is it? A
force? A tendency? Something else? Also, how does it differ from
inertia? Does it not resist circular motion? I am very
mathematical, so maybe an explanation and/or derivation of the formula
t=Fd could help me.
ANSWER:
It is difficult to come up with an answer since you ask a
question, "...how does it differ from inertia?" which indicates that
you know very little about rotational physics! I will try, but
you must first understand translational physics, where everything may
be treated as a point mass. In this arena, you might very well
ask "...what is force and how does it differ from inertia?"
- Inertia, which is called mass (m), is the property which measures how resistant something is to being accelerated (a) if it is pushed or pulled; the amount of push or pull, called force (F), measures how effective the push or pull is at causing acceleration. All this is stated, of course, by Newton's second law, F=ma.
In rotational physics the whole scenario sounds very much the same:
- Inertia, which is called moment of inertia (I), is the property which measures how resistant something is to having an angular acceleration (a) about some axis; that "thing" which causes the angular acceleration, called torque (t), measures how effective the "thing" is at causing acceleration. All this is stated, of course, by Newton's second law in rotational form, t=Ia.
For translational physics, the only thing which determines inertia is how much stuff (mass) there is; for rotational physics, inertia is determined by how much stuff there is but also how it is distributed: an object with most of its mass near its axel is much easier to get spinning than is an object with equal mass but much of it far from its axel. For translational physics, the force is determined only by hard we push; for rotational physics, torque is determined by how hard we push but also where we push and the direction in which we push. For example, suppose you want to open a door and you push as hard as you can but you push at the hinged end of the door; the door doesn't open, does it? This is because the torque increases as you increase the "moment arm", the distance from the rotation axis to where the force is applied (your t=Fd equation). Actually, your equation is a bit too simplified to completely understand torque. If you have an open door and push at the edge of the door which is opposite the hinges but you push straight toward the hinges, you do not close the door even though F and d are both big. It is really only the component of the force Fp which is perpendicular to d which contributes to the torque, so t=Fpd.
I hope this has been helpful.
QUESTION:
Okay, I have a stick of mass M and length L held so that it makes and
angle theta with respect to the floor. The stick is not hinged on
the floor, and the contact between the end of the stick and the floor
is frictionless. I release the stick and it falls to the
floor. How do I find the horizontal distance the left end (the
end on the floor) travels during the fall? And with what speed
does the right end (the end initially in the air) hit the floor?
ANSWER:
This is easier than you think. Since there are no
forces which have horizontal components, the center of mass of the
stick must fall straight down. The distance the end will move is (L/2)(1-cosq). The second
question (not quite so easy!) should be addressed by energy
conservation. At the time the stick is released, it has only
potential energy E1=Mg(L/2)sinq . Just before it
hits, the speed of the end will be v and the speed of the
center of mass will be v/2. The energy, purely kinetic,
will be the kinetic energy of the center of mass plus the kinetic
energy about the center of mass: E2=(1/2)M(v/2)2+(1/2)Iw2=(1/8)Mv2+(1/2)(ML2/12)[(v/2)/(L/2)]2=Mv2/6.
Equating E1 and E2 now gives v=[3gLsinq ]1/2.
QUESTION:
I have been curious about this problem for some time. If a solid
ball of mass m and radius r is resting on a block of mass M and a
horizontal force is applied to the block, how would I find the maximum
value of the force F that moves the block without causing the ball to
slide. This assumes there is friction between the block and the
ball, but not between the ground and the block. I think F acting
on the ball has to equal the force due to friction between the two
objects, but I'm not sure if, when finding this frictional force, you
consider the mass to be m+M or just m. And I feel like I'm
leaving something out. Thanks!
ANSWER:
This is a tricky problem. We know the ball has a
frictional force forward on it given by f=mmg if the ball is about to
slip. Therefore, the center of mass of the ball has a forward
acceleration of am=mg.
However,
this is not the forward acceleration of the block. So now
consider the sum of torques about the center of mass of the
ball. This must equal the angular acceleration of the ball about
its center of mass times the moment of inertia of the ball about its
center of mass: t=fr=mmgr=(2mr(2mr2/5)(at/r)
where at is the acceleration of the rim of the ball with
respect to the center of mass of the ball. Therefore at=5mg/2. Now, since
the center of mass itself has an acceleration, then the acceleration of
the point of contact between the ball and the block as measured in the
laboratory is at-am=aM=3mg/2.; I have
called this aM because it is also the acceleration
of M since the ball is not slipping. So, finally, F-f=F-mmg=M(3mg/2), so F=mg(m+3M/2).
Note that the ball exerts a force on the block of magnitude f
backwards because of Newton's third law.
Note that I have used a theorem of classical mechanics: The sum of torques about the center of mass of a rigid body is equal to Ia even if the center of mass is accelerating.
(PS: In future I cannot answer a question without an email address.)
QUESTION:
Does the force of gravity influence the movements in ice skating ?
ANSWER:
Ice skating would not be possible without the force of
gravity because there would be no force to push your skates down
against the ice. Apart from that minor detail, any movement which
involves motion in a vertical direction would be entirely
different. For example, a simple jump would become a launch from
which you would never come down.
QUESTION:
Does a sunset/sunrise over a large body of water appear differently
than a sunset/sunrise over a large body of land? If so, why?
ANSWER:
There are lots of possible answers. I asked the faculty
here if anybody had a good answer; I got several replies:
- Yes, it's much more romantic over water!
- Maybe, if the land is desert. Then there is less aerosol, a less hazy sunset. If this is a 'green flash' question and was rephrased I might have a more complete answer.
- Yes, all our 1112 and 1212 instructors know that there will be a reflected beam into your eye that reflected from the water surface, and inteferes with light that traveled directly from sun to eye. The angle of incidence was so close to 90 degrees, that the path length distance for the reflected ray is much less than one wave length more than the path length for the direct ray. Since there is a phase change of pi for the reflected ray, it interferes destructively with the other ray. Consequently there are horizontal dark lines in the preceived image of the sun. Of course variations in the refractive index of air can also play a part. For sunsets/rises over land, there is no appreciable reflected ray, because the land surface is not smooth enough. If it were, such as flat ice, the same effect would show up. (This is known as "Lloyd's mirror" and is analogous to the Young's double slit experiment except for the phase change since there is an image of the edge of the almost-set sun under the water surface.)
QUESTION:
i have read somewhere that according to general relativity, only mass
undergoing acceleration will produce graviton. so if an elementary
particle does not undergo acceleration, does it still have a
gravitational field? according to general relativity, only a mass
undergoing acceleration will produce gravitons. then why does a
stationary ball have a gravitational field around it?
ANSWER:
First of all, there is no working theory of quantum gravity
and so a graviton is a purely hypothetical particle. It is best if we
do not involve gravitons in the answer to your questions. Where
your confusion comes from is that general relativity predicts that an
accelerating mass will produce gravitational waves. All
masses produce gravitational fields. A mass which is not
accelerating will produce a field but not a wave. A gravitational
wave propogates through space, most likely at the speed of light much
as electromagnetic waves do. There is a lot of analogy between
gravity and electromagnetism here: any electric charge produces an
electromagnetic field but it will only radiate electromagnetic waves if
it accelerates.
QUESTION:
what is the spin of an elementary particle?
ANSWER:
Elementary particles have an intrinsic angular momentum; this
is called spin. A classical example would be the earth which has
an orbital angular momentum because of its revolution about the sun but
also an intrinsic angular momentum because of the rotation about its
own axis.
QUESTION:
does a stationary elementary particle have a gravitational field?
ANSWER:
Any object which has mass has a gravitational field.
QUESTION:
If a celestial object has Surface Gravity = 43054.5 cm/s^2 and Central
Pressure = 2.116347946 * 10^16 dynes/cm^2 i.e. 21.16347946 Gbars, how
do we go about converting these parameters given in CGS units into
their MKS equivalents?
ANSWER:
I always recommend to my students to multiply by one so that
the units come out right:
1.0 cm/s2
x (1 m/100 cm)=0.01 m/s2
1.0 dyne/cm2=1.0 (gm cm/s2)/cm2
x (1 kg/1000 gm) x (1 m/100 cm) x (100 cm/1 m)2=0.1 (kg m/s2)/m2=0.1
N/m2.
Incidentally, there is a free software utility which is helpful for conversions of units called, appropriately, CONVERT. You can download it at http://www.joshmadison.com/software/convert/ . However, for reasons not clear to me, pressure units of N/m2 are not included in the pressure options. However, you can use it to convert dynes to newtons and cm2 to m2. Still, as seen in my little computation, unit conversions are generally pretty easy from CGS to MKS. The CONVERT software is useful for more arcane conversions, e.g. how many m2 in an acre?
QUESTION:
how do scientists decide whether an element particle has mass or not?
how do they measure the mass and with what kind of equipment do they
achieve that?
ANSWER:
Elementary particles are created by observing byproducts of
collisions of known particles. For example, one can smash a very
energetic proton into another proton and many things will come
out. In special relativity there is a simple relationship which
connects the mass of a particle, its energy, and its linear
momentum. Furthermore, energy and linear momentum are always
conserved in a collision. Therefore one can measure or deduce the
energy and momentum of each particle that comes out of the collision
and if you know the energy (E) and momentum (p) of a
particle you can deduce its mass. For a massless particle, E=pc
where c is the speed of light. If the particle has mass, E2=m2c4+p2c2.
QUESTION:
If you are in a sealed train car (no way to see outside, soundproof,
vibration-proof etc.), can you devise an experiment to determine if the
train is moving? I'm only a high school physics student, but this
sounds pretty near impossible to me. Since speed is relative and there
is no way of comparison in the sealed car, I doubt there is an answer.
Any input would be greatly appreciated. Thanks!
ANSWER:
One of our basic postulates about nature is that the laws of
physics must be the same in all inertial frames of reference. In
classical physics an inertial frame is defined to be one in which
Newton's first law is true (e.g. a particle at rest will remain
at rest if there are no net forces on it). Once you have
found one inertial frame, any other frame with constant velocity
relative to the first will also be an inertial frame--find one and you
have found them all! If we approximate the earth as an inertial
frame (it really isn't because of its rotation on its axis and its
revolution around the sun), then if you are in a train going with
constant velocity, there is no experiment you can do which will
determine that the train is moving. On the other hand, you merely
stipulate that the train "is moving"; if the train were accelerating,
speeding up for example, you would certainly be able to detect that
(because you would feel like you were being forced into the back of
your seat). Frames of reference which are accelerating relative
to inertial frames are not inertial frames. The theory of special
relativity has the same postulate, that the laws of physics are the
same in all inertial frames, but an inertial frame is defined somewhat
differently and Newtonian mechanics must be reformulated; essentially,
the test for an inertial frame is that Maxwell's equations be the
correct laws of electromagnetism.
QUESTION:
Since the Earth is already travelling through space at something like
30 miles per second, isn't it already a bit shortened along its length,
and isn't time already a bit slower, and isn't earth's mass already a
bit greater than it "should" be? In short, isn't our perception of time
and space already mired in some degree of relativistic effects?
ANSWER:
The mistake which you make is that you are assuming that the
earth has some absolute velocity relative to some "special" frame of
reference. But the whole idea of relativity is that it is not
possible to distinguish between inertial frames of reference, so saying
that we move with speed of 30 miles/second is meaningless. All
that matters is that we be in an inertial frame; we are at rest
relative to our own frame so everything is as it "should" be.
Incidentally, an inertial frame is one in which all the usual laws of
physics are correct. Once you find one inertial frame, any other
frame moving with a constant velocity with respect to the first is also
an inertial frame; any frame accelerating with respect to the first is
not an inertial frame. Obviously, the earth is only approximately
an inertial frame.
QUESTION:
Ok this is a relativity problem.. Lets suppose that there is a
cylindrical hole in the ground (3m deep) with a bug trapped in the
whole that is (.5 metters high). There is a cylindrical plunger
traveling at .99c that is (2m in length). The traveler on the
plunger will see the hole shrink and he will squash the bug, but the
bug see's the plunger shrink so he has no worries of getting
squashed?? Who is right? They have to see the same thing
right?
ANSWER:
I think the issue here is the position of the end of the
plunger. The bug sees it approaching with speed 0.99c and
the moving observer sees the bug approaching the end of the plunger
with speed 0.99c. So both have to agree that the bug is
doomed.
QUESTION:
Just a simple one that i can't seem to find the answer to anywhere. Is
Atomic oxygen naturally present in the atmosphere at ground level and
if so, in what quantity.
ANSWER:
I have talked to several people and all agree that atomic
oxygen at sea level will be rare and transient since it is so
reactive. However, because oxygen is so reactive, O2
molecules can dissociate on the surface of transition metals (which is
the principle behind a catalytic converter) and there might be
transient but detectable levels near such surfaces. It is
interesting, however, that there is a significant amount of atomic
oxygen at the very highest limits of the atmosphere where incoming
radiation from the sun dissociates O2 and the O has
difficulty finding other molecules with which to react. It turns
out that the reactivity of O is one of the most important issues in
corrosion of satellites and the shuttle in orbit.
QUESTION:
it is postulated that every object with non-zero mass produces a
gravitational field. now i know that photon has no stationary mass.
however, we can never observe a stationary photon since light travels
at the same speed to different frames. can we say that moving photon
has mass (since it has momentum) so there is a gravitational field
around every photon?
ANSWER:
One of the important findings of the theory of special
relativity is that a massless particle carries both momentum and
kinetic energy. The notion that momentum or kinetic energy
necessarily implies the presence of mass is a classical idea and not
correct.
QUESTION:
We are studying forces and their agents. The author of the text
states, "If you can't name an agent, the force doesn't exist!" I
have been unable to come up with possible examples where there appears
to be a force but no agent can be identifed. Any examples would
be helpful.
ANSWER:
Since Newton's laws are valid only in inertial frames of
reference, one often finds in accelerated frames of reference
(accelerated relative to an inertial frame) the appearance of
forces. For example, when you accelerate in your car you would
think that there is a force pushing you back in your seat, but there is
not. Such forces are called "fictitious forces" and the best
known is the centrifugal force; if you are in a spinning drum, you
think that there is a force pushing you into the wall of the drum
(centrifugal) but there is not. Always rememeber--there is no
such thing as a centrifugal force!
QUESTION:
i read from some book that the natural corollary from special
relativity is that every motion is relative, which implies that if you
are in a bus undergoing a uniform straightline motion, you cannot tell
your velocity or whether you are moving unless you look outside.
however, i think this phenomenon is also true under newton's law. if
special relativity does not exist, are there any ways, or by any
experiments that we can find whether we are moving or not in the bus
without looking outside?
ANSWER:
What you are referring to is what is called an inertial
frame of reference. In classical physics (Newtonian, as you
say), an inertial frame of reference is one in which Newton's first law
is observed to be true. Then it is an experimental fact that,
once you find one inertial frame, any other frame which moves with
constant velocity with respect to the first is also an inertial frame;
frames which accelerate with respect to inertial frames are not
inertial frames. Einstein extended this idea to all laws of
physics. He postulated, in formulating the theory of special
relativity, that the laws of physics must be the same in all inertial
frames of reference. Classical physics predicted that the laws of
electricity and magnetism would not be the same in all inertial frames,
and fixing that led to special relativity. The theory of general
relativity, the relativistic theory of gravity, postulates that the
laws of physics be the same in all frames of reference,
inertial or not; this results in your not being able to distinguish
between an accelerating frame and a nonaccelerating frame in a
gravitational field.
QUESTION:
As I drove to work this morning I noticed frost on the rooftops of
homes (it was 32def F). I remembered a friend of mine mentioning
that the frost forms due to black body radiation if the sky was clear
that night and the temperature is close to freezing. I'm still
wondering what that means. I understand that if it is cloudy,
then the clouds will reflect the escaping heat. But, isn't the
heat escaping just thermal heat. Why would the clear black
sky at night cool a rooftop or grass due to losing radiation?
ANSWER:
I have previously answered this question
in another context and in considerably more detail than I will answer
yours, so you might want to look there. In a nutshell, all things
radiate. Often everything in a particular environment is in
equilibrium with everything else which means that the energy it
radiates into its environment is exactly the same as the energy it
absorbs from the other things which are radiating. This is not
always the case though: E.g. something hot will cool down
because it radiates more energy than it absorbs. Similarly, the
roof will be radiating energy and if there is nothing to give it energy
back it will cool. In my previous example I
estimated that energy will be radiated from a blackbody at a rate of
about 30 W/cm2. That is pretty rapid and can result in
cooling below the freezing point provided that the air (which will give
energy to the roof by convection) is not too warm. For this
reason (and also because of evapoarative cooling), frost can form when
the temperature is above freezing.
QUESTION:
My physics teacher assigned us a problem to solve and i NEED
HELP. The question is : Why does it take longer for a plane
to fly east to west traveling on the same route and going from the same
two places? What should I do? How should I start it?
ANSWER:
The
groundrules of this site clearly state: "do not use 'the Physicist'
to do your homework"! I can only give you a hint: the
"speedometer" in an airplane measures airspeed, not groundspeed.
QUESTION:
Is 2.116347946 * 10^16 dynes/cm^2 = 21163.47946 Mbars =
211.6347946 Gbars?
ANSWER:
Since 1.0 bar is 106 dynes/cm2, and 1
Mbar=106 bar, the conversion from dynes/cm2 is
correct. However, since M means 106 and G means 109,
21163.47946 Mbar=21163.47946 x 10-3 Gbar=21.16347946
Gbar. Incidentally, there is a free software utility which is
helpful for conversions of units called, appropriately, CONVERT.
You can download it at http://www.joshmadison.com/software/convert/
. It would not have helped you with your problem, however, since
Mbar and Gbar are not listed (but it did help me check how a bar is
defined). You need to know the standard
prefixes for powers of 10
QUESTION:
I have spent a number of hours on the Internet over the last week
trying to find some information. I have been unable to find a
satisfactory answer to my question. My question relates to the
Bernoulli principal and conservation of energy. I would like to
understand conservation of energy at a molecular level, and how this
relates to a reduction in static pressure as air accelerates through a
venturi tube. (how it accelerates the flow). Perhaps, along with your
explanation you could give me some keywords that I can put into the
computer to help with my understanding in this area.
ANSWER:
You are right, Bernoulli's equation is a statement of
conservation of energy. However, it is not derived from
velocities of individual molecules but rather the velocity of the bulk
fluid. The velocity distribution of the molecules is determined
by the temperature of the fluid and has no relation to the velocity of
the fluid (which is added as a small drift velocity to each molecular
velocity). The things you are trying to understand
microscopically are best left as macroscopic.
QUESTION:
Is there an
advantage to having polarized lenses on sunglasses? I've heard
they are better; they are definitely more expensive. I have
polarized lenses now. The only difference I see is with these lenses I
see many different color patterns on car windows. Sometimes shiny
objects on the ground seem to "glow". What's going on?
ANSWER: 
The advantage is that polarized glasses can reduce glare
reflected from surfaces. Reflected light is rather highly
polarized as shown by the figure at the right. At just the right
angle, the light will be totally polarized. The picture shows
unpolarized light before reflection becoming polarized after
reflection. Think of the surface as the hood of your car; then if
you are wearing polarized glasses which filter out horizontally
polarized light you will not see much of the directly reflected
sunlight (glare). Light from a clear sky (blue) can also be
highly polarized depending on the location of the sun and so
photographers use polaroid filters to get striking effects in black and
white photos.
QUESTION:
I'm
interested in a Career at NASA as a research scientist. I was
informed that you should have a good physics background during your
undergraduate years. What major would you recommend for this
career and what major in Graduate School would you recommend?
ANSWER:
This is an impossible question to answer. A "research
scientist" is simply too broad a thing to be aiming at. I am sure
NASA has scores of different types of research scientists--physicists,
astronomers, biologists, computer scientists, chemists, etc.
When you decide what kind of science you want to do, then you can plan
your college career more sensibly. Also, why restrict yourself to
NASA? There are many kinds of employers who carry on fascinating
science research.
QUESTION:
Could you
please explain to me the thermodynamics/heat transfer of: a) toasting a
marshmallow on an open fire.....b) and heating a can containing pea
cream soup
ANSWER:
a) Properly toasting a marshmallow requires that it be
exposed to the radiant energy from a bed of hot coals. (I
am a marshmallow-toasting gourmet!) The radiant energy then heats
and melts the sugar inside (change of state) and toasts the sugar
outside (a chemical change).
b) Heat from the fire is conducted through the metal of the can making the surface in contact with the soup become hot. Then the heat on the inside surface of the can heats the soup in contact with it and this is mainly transmitted to the rest of the soup by convection (the soup moving around because of its heating). It is useful to stir the soup occasionally (particularly cream soups which do not convect so well) so that if the convection doesn't move it away from the can a chemical change won't occur (scortching to the can).
QUESTION:
how can I
simulate invasion percolation procees?
ANSWER: (provided
by M. R.
Geller)
There is a nice site at http://www.ucls.uchicago.edu/People/02/Beckett_Sterner/
. There is even some shareware software you can download (I think
it is just for Mac.)
QUESTION:
What would
the mode be for a group of numbers that only appear once? such as 3,4,5
ANSWER:
This is really math, not physics. But check out http://mathworld.wolfram.com/Mode.html
. I think the answer to your question is that 3 4 5 would be trimodal
with each as a mode. This site http://scienceworld.wolfram.com/
is really helpful in getting facts about math, physics, astronomy,
chemistry, or biographical information about scientists.
QUESTION:
Have
isolated quarks ever been observed? I am getting conflicting
information from texts about whether they have actually been discovered.
ANSWER:
An isolated quark has never been observed; we believe that
the reason is that the force which binds quarks together is such that
the further the quarks get from each other, the stronger the force
gets. But the theory based on quarks accounts for such a rich
variety of experimental results that most particle physicists believe
in them wholeheartedly. You might ask if something unobservable
in the sense of seeing it all by itself is "real"; I think this is a
question for philosophers, not scientists.
QUESTION:
This a not a
brain teaser in any way at all. Me and a friend got in to an
argument the other day. We were eating pizza and it was too hot,
so he said that we should put it in the freezer for a little
while. well i don't remember how the argument got started but he
said that hot pizza would freeze in the freezer faster than pizza that
was room temp. I think he's nuts. What do you think?
ANSWER:
No, the hot pizza will not freeze faster. The classic
question along these lines is will hot water or cold water freeze
faster. For a lengthy discussion of this problem, see an earlier answer.
States of matter are concepts used in both physics and chemistry.
QUESTION:
I have read
that the speed of light cannot be broken. I have also read that all
motion is relative. Suppose point A is a stationary reference point.
Point's B and C are start near point A and travel away from
each other at 51% the speed of light relative to point A. Does this not
mean that relative to each other, points B and C are traveling apart
from each other at 102%? Do you not add their speeds together as if two
automobiles are each traveling away from each other at 60 mph are
traveling at 120 mph relative to each other?
ANSWER:
No, your intuitive way of adding velocities does not work in
the theory of special relativity. Because of the fundamental
postulate of special relativity, that the speed of light is the same in
all frames of reference, the correct formula for velocity addition is v=(v1+v2)/(1+v1v2/c2)
where c is the speed of light. So, for your example, v1=v2=0.51c
so v=0.81c, 81% the speed of light. Note that the
formula reduces, to an excellent approximation, to your v=(v1+v2)
if the speeds involved are very small compared to the speed of light
because the term v1v2/c2
is extremely small compared to 1.
QUESTION:
I'd like to
know what charge is. Not the descriptive stuff like coulomb's law
and opposite charges attracting, but what is it? And how does it
work over a distance?
ANSWER:
This is a profound question. It is like asking what
gravitational mass is and how gravity works. Electric charge is
simply a property that some objects in nature have which allow them to
exert and feel electric and magnetic forces. Theories which try
to predict charges and masses of elementary particles are very complex
and only partly successful. It is known that quarks, from which
protons are made, have electric charge which is in multiples of one
third the elementary charge e. Regarding how the forces
act over a distance, classical electricity and magnetism postulates
what are called fields; and electric charge "distorts" the space around
it by placing a field there and then other charges feel this
"distortion". In quantum electrodynamics, a more complete theory
of electromagnetism, the picture is that charges exchange virtual
photons which are the mediators of the field.
QUESTION:
what is the
present device used in measuring acceleration due to gravity?
ANSWER:
Actually, the acceleration due to gravity is not considered a
fundamental constant because it is so variable. It depends on
altitude and where you are on the earth. The more fundamental
quantity which physicists are interested in measuring is the universal
gravitational constant G which appears in Newton's law of
gravitation, F=Gm1m2/r2
where m1 and m2 are two masses
and r is the distance between them. There is a very
readable article in the journal Physics Today in the November
2002 issue; the title is "Beam Balance Helps Settle Down Measurement of
the Gravitational Constant".
QUESTION:
What do you
think of the classical EM theory and why do we still use it when a lot
of it has been proven wrong? Tom Bearden - PhD, MS (nuclear
engineering), BS (mathematics - minor electronic engineering) claims
there are 34 flaws in the clasical EM theory which most people are
taught that it is still real ( http://www.cheniere.org/misc/flaws.htm
)
ANSWER:
Looking over the website you cite, many of these are bogus,
many are situations which simply point to the fact that classical
E&M is not properly quantized, and many I simply have no idea what
he is talking about. It is true that classical E&M is not
valid if quantum effects are important and that there are lots of
questions (self energy, virtual photon exchange, vacuum polarization, etc.)
which are not addressed. But quantum electrodynamics (of Feynman,
Dyson, Schwinger, Tomonaga, et al.) is a fully developed and
understood theory. To suggest that classical electromagnetism, a
beautifully complete, compact, rigorous, and precise theory is somehow
not "real" because it does not address quantum effects is
ludicrous. The first major breakthrough of modern physics,
special relativity, was discovered because classical E&M is a
completely correct relativistic theory. Surely you would not
throw out classical mechanics, absolutely imperative for computation of
orbits etc. in celestial mechanics, because it does not contain
relativity or quantum mechanics.
QUESTION:
I would like
to know if someone has conducted or tried to the following experiment:
with an isolated hydrogen atom, to confine the electron gradually
towards the nucleus restricting its orbital so that finally the
electron colapses onto the nucleus? What would be the main experimetal
difficulties with this and where is described such experiment?
ANSWER:
There is no way to do this and have the H atom retain its
identity. The ground state is the lowest one possible in such a
quantum system. Under extreme conditions (as in a neutron star)
it is possible to compress electrons and protons together to create a
neutron and a neutrino. You can also scatter very energetic
electrons from protons (this is done routinely at electron accelerator
laboratories) and the electron can probe the interior of the proton,
but this would certainly not be called a hydrogen atom.
QUESTION:
For a
science experiment, I decided to build three Newton's Cradles, each
with different ball materials, and observe the amount of times that
each would bounce. One would be rubber, one wooden, and one
steel. I concluded that rubber bounced the least amount of
times. I figured that more energy is absorbed into the balls with
each cycle, and so there would be less energy each time. I also
figured that the steel balls, being heavier, would have more potential
energy when lifted up. However, I'm specifically curious as to
why more energy is absorbed by rubber, and how that works
entirely.
ANSWER:
When you compress a ball, as happens when it collides with
another ball, you do work. Part of this work goes into the
compression like a spring where the work you do is stored as potential
energy; and part of the work is work against internal friction and that
work is lost, not stored, and shows up as heating the ball up. An
extreme example would be if you made your balls out of putty and all of
the work of compression would be lost. In your cases, the steel
balls are most elastic and energy comes closest to being conserved; the
rubber balls lose a considerable amount of the energy because it does
not come back out when the ball decompresses.
QUESTION:
I’ve read
that there is a maximum density limit for matter. If so, this would
mean an object made from the densest possible matter could not be
compressed any further. Isn't this a rigid body?
ANSWER:
As you probably know, nearly all the mass of an atom is
concentrated in the nucleus, an extremely small volume in the center of
the atom. Under extreme conditions it is possible to compress the
atom down to the size of a nucleus and compress all nuclei in the
object together. The mass density you come up with is called the
density of nuclear matter; this density is approximately 2x1017
kg/m3. Compare this with water which has a density of
1000 kg/m3 and you will appreciate that the density of
nuclear matter is almost unimaginably huge! But, does this ever
happen? Yes, there are special stars which, late in their
lifetimes, collapse under gravitational attraction to this density;
they are called neutron stars. Some stars collapse even further
such that their density becomes essentially infinite; they are called
black holes. So there is no maximum density limit.
QUESTION:
Can lines of
magnetic force be reflected or focused like rays of light?
ANSWER:
A beam of light is something which moves through space.
A magnetic field (or electric field for that matter) should not be
thought of as equivalent, so in some sense, reflection really has
little meaning. You can of course, distort things so that you
bend the field lines. If you want to call that reflection or
focusing, that is fine, but usually physicists don't use such
terminology.
QUESTION:
What are the
advantages/applications of a stem and leaf graph?
ANSWER:
Since you don't ask what it is, I assume that you know.
I didn't when I received your question! Actually, it has nothing
to do with physics, it is a reputed to be a convenient way to display
modest data sets (<100 points) which makes it easy to see some of
the qualitative statistical properties of the set like modality,
median, etc. It is more or less just like a histogram,
the only advantage that I can see being that you can actually pick out
each element of the set, not just see the lump sum of all the numbers
between 70 and 79, for example. Since I have not heard of it
before, it is safe to assume that phyiscists rarely if ever employ
it. Given modern computerized spreadsheets with automated
statistical analysis of virtually any kind you can think of, this kind
of plot looks real 19th century to me!
QUESTION:
How much
stronger is the force of Earth's gravity over the magnetic force of the
Earths poles?
ANSWER:
This question cannot be answered. It is the old
"comparing apples to oranges" thing. The force of gravity depends
on the mass of the object which is feeling it whereas the magnetic
force depends on the magnitude and geometry of an electric current
which will feel it. Since the mass and current distributions have
no relation to each other, your question has no meaning. One may
make the general statement, however, that the gravitational force is
nature's weakest force. The only reason you think that it is a
strong force is that you happen to be in the vicinity of a very huge
mass (mass being the source of gravitational forces).
QUESTION:
I would like
to know your explaination of HOW magentic fields in Space are being
created. EARTHS MAGENTIC FIELD, SUNS MAGENTIC FIELD, MY FRIDGES
MAGNETS MAGENTIC FIELD, ELECTROMAGNETIC FIELD, ACTUALLY EVERY "BODY"
THAT IS IN SPACE. HE WHO KNOWS THE ANSWER, UNDERSTANDS THE
"WORKINGS" OF THE UNIVERSE. HE WHO KNOWS THE ANSWER WILL EXPLAIN
T.O.E.
ANSWER:
There are only two ways to make a magnetic field: have moving
electrical charges (electric currents) or have time varying electric
fields. Understanding the origin of some magnetic field may help
you to understand some specific thing, but believe me that one will not
understand everything by understanding magnetic fields.
QUESTION:
If I hold a
device that measures the speed of a passing light beam as C, isn't it
more reasonable to assume that the light is not moving, and that I am
the one moving at "light speed". Since such a device only measures the
difference in speed between me and the light beam, and we know that I
am wizzing through space on earth, isn't it more accurate to say that
light is standing still, and everythng else moves at light speed other
than the other way around?
ANSWER:
What you are suggesting is fallacious because c is a
universal constant, so somebody moving by you with a speed of half c
will also measure the speed of the light to be c. Now,
applying your logic, each of you is moving with speed c while
the light stands still, so the two of you are at rest with respect to
each other. You can't be both at rest and not at rest with
respect to the other fellow.
QUESTION:
A box
sliding across the floor is soon stopped by friction and its motion
(velocity) soon ceases. The frictional retarding force is an external
force acting on the "box-floor system" and momentum is not
conserved. However, I can redefine the system to include the
frictional force by redrawing the boundry.The force is now an internal
force in my new system, but the sliding box still stops. Is momentum
now conserved?
ANSWER:
You are all muddled. I have no idea what you mean by
boundary. You need to choose an object or objects to focus your
attention upon. Say it is the box. What are the forces
acting on it? Its weight, a force the floor exerts up on it to
keep it from accelerating through the floor, and a frictional force in
a direction opposite the velocity. As I told you last time you
wrote me, if the system you have chosen to focus upon has a net force
acting on it, then momentum will not be conserved. There is
therefore no mystery as to why the momentum of this box
disappears. To find a system in which momentum is conserved it is
easier to understand if we imagine the box sliding on a very big
(massive) block like the floor and this block sliding frictionlessly on
a horizontal surface. The big block (surrogate floor) is at rest
at the beginning of the problem. What are the forces on this
system? Just the weights of the big block and the box and the
normal force up on the big block which add up to zero; therefore
momentum will be conserved. The momentum starts as mv and
ends as (M+m)V. Conserving momentum, V=[m/(m+M)]v,
so you see that if M>>m, then the final velocity
of the your floor and box will be zero. Of course, the floor is
not sliding frictionlessly on something, it is attached to the house
which is attached to the whole earth. Therefore M is the
mass of the whole earth plus the house plus the floor. I think
you would agree that that is a whole lot bigger than the mass of the
box!
QUESTION:
What is the
basic illusion with “movement” of any object? If I move one end of a
metal stick, the other end can not possibly move instantaneously
together, for that would be a faster-than-light phenomenon. With a
sufficiently long, hypothetically undeformable stick, a movement at one
end would reach the other end just after some time, therefore needing
some kind of “movement wave” or “wave of deformation” running along the
object. If I suppose a very long object, and a very large initial
movement, shouldn’t this “wave of deformation” be so strong as to
disrupt any kind of material? Where lies the problem here? Are objects
in some degree always deformed with any movement? How does the
relativistic question of simultaneity of events apply here?
ANSWER:
You are right, the information which tells the end of the
stick to have an angular acceleration cannot get there
instantaneously. The way the end of the stick gets accelerated is
by the force exerted on it by atoms attached to the part of the stick
you are looking at. This would be like a "shear" force.
This shear force must propogate from atom to atom along the length of
the stick at some speed which cannot exceed the speed of light since it
is carrying information. If the required shear force gets too big
it will be bigger than the material can bear and the stick will break
or undergo some other nonelastic deformation. There is no such
thing as an undeformable object, so you need not worry yourself with
that. Your question is not unrelated to the rotating disk problem
which has been answered earlier and you might want to look at and to
check out the links given.
QUESTION:
The extremes
of "perfectly elastic" and "perfectly inelastic" collisions are
somewhat confusing as most books show something between the
extremes. For instance, a billard ball that hits the billard
table's side bumper and moves away with the same constant velocity that
it approached with would have 0 change in the ball's momentum, because:
mass of ball * (+velocity) = (mass of ball) *(-velocity), hence net =
0, assuming only x component direction. Kinetic energy is conserved,
but momentum is not? s the way to think about this collision, or
think of a small object (the ball) colliding with a much larger object
(the table). Just seems "fuzzy"?
ANSWER:
Well, perfectly elastic simply means that kinetic energy is
conserved. Perfectly inelastic is a little more subtle since it
is defined as all the kinetic energy in the center of mass system as
disappearing in the collision; it is easy to recognize, though, since
it corresponds to the interacting particles being stuck together
afterwards.
The billiard ball collision you cite you have incorrectly analyzed. There is not zero change in linear momentum of the ball because linear momentum is a vector. The change is a vector away from the bumper with magnitude 2mv. The momentum of the ball is not conserved because there is an external agent which delivers an impulse (force x time) to it; that agent, of course, is the bumper. If you want to see momentum and energy both conserved, you must look at some isolated system: imagine the pool table to rest on a frictionless floor so that the system is the ball plus (the much more massive) table; then the table would recoil a tiny amount and the ball would have slightly less speed after the collision such that energy and momentum would both be conserved. In the limit that the mass of the table approaches infinity, the ball would have the same speed after the collision as before but the momentum of table would be nonzero after the collision even though it would be at rest! This is because zero x infinity need not be either zero or infinity. This assures momentum is conserved. Of course, there is no such thing as infinite mass and the real world has the table essentially at rest after the collision.
QUESTION:
Can Landua
damping occur for frequencies that are lower than 1000Hz?
ANSWER: (Provided by
M. R.
Geller)
Most likely not. Landau damping does not occur until the
plasmon dispersion relation crosses the single-particle excitation
continuum. The crossing point depends on Fermi momentum of the system,
but is usually at higher frequencies (above, say, a THz). See "Theory
of the Normal Fermi Liquid" by Pines and Nozieres.
QUESTION:
There is a question I have that's twisting my brain. Although it's in a
chemistry text, the question is really more about physics: Work is
defined : w=fd, or force applied over a distance. One set of simple
units for this would be kg.M, right? ((I am using capital M's for
meters so as to distinguish from the variable mass as in f=ma.)) Ok
now, The Joule, which is often used to measure work has units of
(kg.M^2)/(s^2)..((sorry I can't use superscripts in this e mail
format)) Now I understand that this is really mass x acceleration x
distance that (f.d=m.a.d = kg x M/s^2 x M), but my question is: how can
it be both? How can it be kg.M^2/(s^2) and kg.M, or I guess the real
question is this: My understanding that a mass unit IS a force unit.
The kg is a unit of force, right? so, how can force have units of kg
(only) in one case and kg.M/(s^2) in another? Acceleration is not
unitless, so how can you have both sets of units for the same quantity?
My understanding of the above assumption is based on the fact that I
have seen the units for work expressed both as kg.M and
(kg.M/(s.s). I hope you can help me, and I really appreciate it!!
ANSWER:
You make the classic beginning physics student mistake of
thinking that mass and weight (a force) are the same thing. In
fact, they are totally different things (although intimately
related). A force is a push or a pull and the result of exerting
an unbalanced force on something is an acceleration of that
thing. Mass, on the other hand, measures the resistance which an
object has to being accelerated (inertia). Newton's second law
related the two. Newton's second law is two things. First, it is
a statement of an experimental fact: acceleration is proportional to
the force and inversely proportional to the mass. Second, it is a
definition of the unit of force which we agree to use, viz. one
unit of force is that force which causes a 1.0 kg mass to have an
acceleration of 1.0 m/s; the unit is called the Newton (N) and has the
units of kg m/s2. Force can never have the units of kg
alone. So now, force time distance is N m = kg m2/s2
which is called a joule (J). I think that takes care of your question.
QUESTION:
If a bullet were fired into a wooden block and the bullet went thru and
exited the other side, would this be thought of as "elastic" collision
in linear momentum terms?
The block would then side along the horizontial surface until frictional force consumed all of the available kinetic energy and the block would come to a stop? If so, then the work of friction could be equated to the block KE and the coeficient of sliding friction calculated. Correct?
ANSWER:
Elastic has a very specific meaning--energy is
conserved. Most collisions, like the one you describe, are not
elastic. But momentum is conserved; there is no analogous word
for this conservation, but it isn't really necessary because in an
isolated system the total linear momentum is a constant of the motion.
What you are saying regarding a specific problem with a sliding block will work but it is not quite the correct way to frame it. What you should say is that the work done by all external forces (friction in this case) is equal to the change in kinetic energy. That is not exactly what you said because you implied that the friction does some amount of positive work when in fact the friction, taking energy away from the system, does negative work. Of course, the change in kinetic energy (final minus intitial) is also negative, the final being zero.
QUESTION:
Am I correct in assuming the gravity of an object is dependant on its
mass and density? A singularity in a black hole is small yet dense and
has infinetely powerful gravity. My question is, is it possible to
increase the density of an object and change its gravity? The reason I
ask this question is because of a situation that I was trying to
imagine. Suppose an experiment could be done where a relatively small
sphere who's density could be increased could produce more
gravity than the earth itself and thus pull nearby objects towards
itself. I'm assuming that theres no way to increase the density of an
object to perform the experiment but I wondered wether the actual
concept was valid.
ANSWER:
Well, first of all the gravitational force due to a black
hole is not infinite. If it were a true singularity, a point
mass, then the force would approach infinity as the distance from it
approached zero. In order to answer your question, I am going to
make a simplifying assumption: I will talk only about mass
distributions which have spherical symmetry; the shape of the
object is always spherical and the density depends only on the distance
from the center of the sphere. So, for example, a basketball has
spherical symmetry since all the mass is on the outer surface but the
density does not depend on where you are on the surface.
Spherically symmetric objects are, at least approximately,
representative of many objects of interest in the universe. A galaxy is
an example of something which normally does not have approximate
spherical symmetry since they tend to be disk shaped spirals. If
you have a mass, say a star, the gravity outside all the mass is
exactly the same as it would be if all the mass were a point mass in
the center. Therefore, if this star were compressed down to the
size of a pea (corresponding to a huge increase in density), the
gravity outside the original surface of the star would be
unchanged. However, the gravity halfway to the center of the star
would be smaller for the star than the same place for the pea-sized
star with the same mass. Generally speaking, the gravity will be
due only to the mass inside where you are measuring it. The most
dramatic example is a spherical shell, like a basketball. Imagine
a star with all its mass on its surface; if you were inside it there
would be no gravity whatever.
QUESTION:
I have a question that i really want to know the answer to. I'm
hoping you can help. Sometime when you're watching tv or driving
on the road and you look at cars it appears that the rims of the cars
are actually rotating in a backwards motion. Why does this happen?
ANSWER
:
If it happens on the road, I don't really know why; I have
never seen that. I can tell you why it happens in movies and
TV. The classic example is a wagon wheel in a cowboy movie.
The key to understanding is to recognize that a movie is a series of
still photographs in a sequence of time. Imagine a spinning wheel
with two of the spokes labeled A and B such that if you watch one place
you will see A there first and then, some time t0
later you see B at the same place; so t0 is the time
it takes the wheel to rotate the angle equal to the angle between two
spokes. Now suppose that you are taking a movie of this spinning
wheel and the time between frames is some time t; then, if t=t0
each frame would have all the spokes in the same position so the wheel
in the movie would appear to not be rotating! [An example is
shown to the right: here the wheel rotates as you can see from the red
dot on the rim of the wheel, but it does not appear to be if you don't
watch the red dot.] If t were a little more than t0,
the wheel in the movie would appear to be going in the correct
direction, but slowly; if t were a little less than t0,
the wheel in the movie would appear to be going backwards. There
you have it.
You can get a similar effect in "real life" if you view something with a strobe light; some light sources have a normally imperceptible flicker (often 60 cycles per second like the line voltage powering them) which can cause the source to be a strobe.
QUESTION:
I am researching the Hall effect for a possible innovention. I
understand the concepts well, but I have not been able to find any
information on the time sensitivity of a Hall effect measuring
device. I assume that once a B-field and voltage are applied,
that electric potential at the sides increases and then begins to level
off, approaching the final voltage. And that after some amount of
time, the measurement made would be close enough to the final voltage
to be accurate. But what kind of time delay are we talking
about? Milliseconds? Seconds? Minutes?
Hours? Could I alter ther B-field and make a new measurement
right away? Or do I have to sit on my thumbs and wait for my
little electrons to crawl to one side of the plate?
ANSWER:
Intrinsically the device should be very prompt,
microseconds. What will be much more important is the circuitry
in which the device is installed, often it will be an RC or RL
circuit which will have its own time constant. The probe, for
example, will have some capacitance which depends on its size.
So, if you are interested in superfast response, you have to worry
about all these things. Also, the response time of the device
should be listed in the specs of the probe when you buy it. I
have used Hall probes to map magnetic fields in magnetic spectrometers
and have never had to worry about how long to leave the probe before
making a reading.
QUESTION:
Is it possible to make a wind-up pendulum movement that can swing in an
arc of at least three feet and has a bob that is exposed to air as well
as water friction? I have an invention in mind and do not know if the
type of large pendulum movement I am envisioning is even possible.
ANSWER:
This is an engineering question, not physics really. In
any case, it is likely that it would be possible to do what you want
depending mainly on what you mean by water friction. It would
probably be hardest if the water the pendulum had to go through was not
a regular periodic event. But, if each swing of the pendulum
encountered identical friction, it would be straightforward to devise a
mechanism to put lost energy back into the system.
QUESTION:
I am doing self-review in electromagnetism. I was wondering how I
would set up a problem. A micron of oil with specific gravity =
0.9 and a net charge of 100 electrons is placed between two parallel
plates that are 5mm apart. I need to find the voltage required to
keep the oil suspended. I was thinking of relating the force to
the electromagnetic field. But I am confused about units.
How do I find the net downward force of the oil drop using specific
gravity? Would the charge simply be 100 times the charge of an
electron?
ANSWER:
The specific gravity of something is its density relative to
water, so this particular oil has a density of 0.9 that of water (which
is 1000 kg/m3), or 900 kg/m3. So the mass
is the volume times the density and I presume that 1 micron drops of
oil mean spheres of radius half a micron. The the weight of a
drop is its mass times the acceleration of gravity; I compute the
weight to be W=4.62 x 10-15 N. The magnitude
of the charge on an electron is 1.6 x 10-19 C, so the
(magnitude of) net charge on a drop is q=1.6 x 10-17
C. The force of a charge q in an electric field E
is qE, and this force must be equal in magnitude to the weight,
qE=W. I compute E=289 N/C. Since the
charge on the electron is negative, the electric field must point down
for the electrostatic force on the drop to be up.
QUESTION:
I am 45, a wee bit bright but scientifically and numerically challenged
in the traditional sense. Other than independent study, my last science
course was biology, 1973 and math, Geometry or Algebra 2, 1974. I'd
like to return to school for a degree in physics as it fascinates me.
Is there hope? As I am a believer in the school of thought "we are only
limited in our potential by our desires," I think it's possible. Yet,
it seems a bit far-reaching. What do you think? By the way, I am so
right-brained, I tilt slightly in my gait. Hmm. - Sunny
ANSWER:
Take it one step at a time. First of all, you cannot
really do much with physics without quite a bit of higher-level
math. Take a course at a local college in analytic geometry
(essentially the prelude to calculus); this course might even be
labelled "precalculus". Enlist some good advisor to make sure you
get into the right course. Then, if you like this and do well,
take two terms of calculus where you learn both differential and
integral calculus. If this goes well, take an introductory
physics course (it should be referred to as "calculus-based
physics"). You are probably two years into your program now, and
if you are doing well and loving it more than just about anything you
have done before, you are off and running! Good luck!
QUESTION:
Several questions regards SHM:
1) In using the SHM equations to find the time required for a body to move from one position to another in y = A*sin(w*t+theta), is it always permissible to set theta = 0? This allows the easy solution to "t" via the inverse sin assuming the amplitude & w (omega0 are known?
2) The net force acting at either extreme of motion can be found from F = m*a = m(w^2*A) where A = amplitude (or Xmax). This where the acceleration is max & velocity is 0; however, when the body passes thru the equilibrium position (say a vertical spring & on the way up) the velocity is max, but the accel = 0, hence the net force = 0. There is still a force acting (as it passes the midpoint) just to support the weight of the body at the end of the spring..is this correct?
3) Is the sign convention (+ or - ), how you would know whether a body in SHM is moving up or down? Example, when y = - 5 the acceleration becomes = -w^2*y = (-w^2)*(-5) = positive number & the body is moving up, while a negative would indicate that its moving down. Is this correct?
ANSWER:
1) The choice of when t=0 is always up to the person
solving the problem, just like the choice of coordinate system.
However, it may not always be the best choice for any particular
problem.
2) The net force is (by definition) zero at the equilibrium position of the spring. The equilibrium position of a vertical spring is determined by the weight of the mass such that when the spring is stretched by an amount s, ks=mg.
3) The signs take care of themselves if you do the problem carefully. For the example you give, it should always be such that the acceleration is opposite the displacement; this is the basic characteristic of a restoring force like SHM. For a given position you cannot determine the direction of the velocity because it is inherently ambigous. At any particular time, however, the velocity is well determined where v<0 means moving in the negative coordinate direction and vice versa for v>0.
QUESTION:
I'd like to ask for some help regards how to evaluate time
requirements when working with shm equations. Here's a problem that has
a time component asked for: "A body of mass = 100 grams hangs from a
long spiral spring. When pulled down 10 cm. below its equilibrum
position, it vibrates with period(T) = 2 sec. The problem then
asks for velocity as it passes thru equilbrum position & what is
the acceleration when 5 cm above the equilbrum position. These are easy
to solve with the basic equations. However, the next step is to find "
when the body is moving upwards, how long a time is required for it to
move from a point 5 cm below its equilibrium position to a point 5 cm
above its equilibrium position. How do I get the time out of the
wave equation y=A*sin(WT+theta)? I know thru differentiation the
velocity & acceleration will fall out, but getting the time isn't
as straightforward, but I think it's asking for the body to evaluated @
90 degrees & @ 270 degrees. Is this correct or how would I proceed?
Thxs.
ANSWER:
It is all a matter of what you are solving for. You
just have to be able to invert the equation x=A sin[wt+f] to get t=[sin-1(x/A)-f]/w. One
important tip is to always work in radians, not degrees. For your
particular problem, I would choose f=0 which means
that the mass is at x=0 at t=0 and moving in the +x
direction. In your problem, w=p and so
the time when the mass will first reach 5 cm is t=sin-1(5/10)/p = (p /6)/p = 1/6 s. So the
time it will take to go from -5 cm to +5 cm is clearly twice this, 1/3
s.
QUESTION:
This question is on the brink between biology and physics. When
we feel the sensations of "hot" or "cold", are we feeling heat tranfer,
or temperature? I ask this because temperature doesn't seem right
(bathroom tile feels cooler than the rug at the same temperature do to
their different heat capacities). On the other hand, it would
make sense if there were certain neural receptors that expanded as
their temperature increased. I cannot think of a mechanism by
which heat transfer could be transformed into a neural sensation.
I have had trouble getting a good explanation of this.
ANSWER:
I have no idea! However, I did a Google search on heat receptors and
many sites which looked very helpful regarding details of
heat/cold/pain receptors were returned.
QUESTION:
I have been pondering a question that I have no idea how to put into
mathmatical terms. I have postulate a very large disk, in a vacuum and
no gravity, with a slot running the entire diameter of the disk. In the
center of this disk you place a ball bearing so that if the disk is
spun, the two ball bearings would move apart. Now if the disk is spun
fast enough, the ball bearing should come close to the speed of light
provided the disk is big enough and the rotation (rpm) is great enough.
Would the velocity at which these 2 ball bearing are moving away from
each other not be greater than the speed of light? If not, is there
some relationship between the two ball bearings other than the spinning
disk that prevents them from accelerating past the speed of light? What
I mean is there some relationship that exceeds the normal relationship
of two ball bearings simply because they are close to light speed? Does
this state exist if the two bearings are not at light speed?
ANSWER:
Your question is essentially the spinning disk paradox.
You do not need to introduce your ball bearings, just ask if it is
possible for a point far enough out on a spinning disk to exceed the
speed of light. The answer, of course, is no. See the
answer to a previous question.
QUESTION:
Since light vibrates when it moves it makes a zig-zag line when it
travels through spacetime right?
EXample/\/\/\/\/\/\/\/\/\/\/\/\\/\/\/\/\/\/\/\\/\/\//\/\\/\/\//\/\/\/\/\//\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/
and the distance from the begining of that zigzag to the end is the
distance that it has traveled in a certain time right? But
shouldnt the actual distance be the distance of a stretched zigzag?
since it is the actual path that an atom makes through spacetime?
The zigizag I drew above if stretched could be twice as long.
Example---------------------------------------------------------------------------------------------------------------------
Therefore the actual distance light travels in a certain time is longer
so the speed of light should also be longer.
ANSWER:
Although the electric and magnetic fields are oscillating
sinesoidally, the wave itself is traveling in a straight line.
This is really no different from a water wave, for example, where the
water is oscillating up and down but the wave travels the straight line
distance between two points on the surface.
QUESTION:
I am 14 years old and my best subject is science. The question is
would I have to major in physics to get into UGA?
ANSWER:
I don't know what you mean. Getting into UGA is
determined by numerous things including SAT scores, high school GPA,
letters of recommendation, extracurricular activity etc. and
what you want to major in is relatively unimportant. Some
programs (but not physics) have strict limitations on how many majors
they will accept, but that is something which is more likely to be
enforced when you are already here than to affect your admission.
(By the way, you should include an email address with your question so
that I can communicate with you more effectively; you did not fill in
that box in the form.)
QUESTION:
If light (electromagnetic radiation) is known to be heavy,
why are the constituent photons they are composed of considered to be
massless (zero rest energy)?
ANSWER:
I do not know what you mean "known to be heavy". Viewed
either as electromagnetic waves or photons, light has no mass.
Perhaps your confusion is that electromagnetic radiation is known to
carry energy and momentum even though it has no mass. In the
(well verified) theory of special relativity, massless particles may
have kinetic energy and mass, and photons are the prototypical
relativistic particles. Furthermore, in the classical theory of
electromagnetism the electromagnetic field has both an energy density
and a linear momentum density.
QUESTION:
You mount a laser on a rigid tripod and shine the beam out a
few meters to a mirror (again, rigidly fixed) so that the beam reflects
back to a spot, X, near the laser. Will the beam continue to shine
exactly on X throughout the day (as the earth turns)?
ANSWER:
It seems to me that the only things which will affect the
path of the beam of the light are the gravitational field and the
acceleration of the reference frame (which are in some sense, the same
thing in the view of general relativity). The most important
gravitational field is that of the earth which does not change with
time and the most important acceleration is that due to the earth's
rotation which does not change. However, there are other
gravitational fields, in particular due to the sun and the moon which
do change over the course of a day because of the earth's
rotation. For example, suppose the laser were at the equator and
pointing straight down (toward the center of the earth); at dawn the
sun's gravitational field would point east and at dusk it would point
west, so the deflection of the beam of light would differ. So,
technically, the answer to your question is no. However, these
effects are so incredibly small that you could never hope to observe
them.
QUESTION:
I am a proffesional engineer who designs ASICS for the
telecommunication market. I hold a Ph.D. in Computer Engineer. I
have always enjoyed physics and am interested in finding like minded
people. Do you know of any group/club that meets on a regular basis to
discuss physics?
ANSWER:
I must say that I have never heard of a club which meets to
discuss physics; there are, of course, many clubs of amateur
astronomers which certainly will talk about many physics-related
things, so you might look in your vicinity for astronomy clubs.
Another possible avenue for what you want would be at local colleges
and universities. Two things you could do are:
- Check out the web sites of physics departments and check their program of colloquia; these are generally intended for nonspecialists (unlike seminars which tend to be more for specialists in subdisciplines).
- Look, again in physics department web sites, for Society of Physics Students (SPS) chapters at local colleges and universities; these often have regular meetings and programs at which I am sure you would be welcome.
QUESTION:
If an object hanging on a spring has its equation of motion
as y = A*sin W*t where A & W(omega) are constants & t = time,
how can I express the velocity as a function of the coordinate?
The velocity & acceleration are easy to express as functions of
time by taking the first and second derivatives of the equation of
motion. This results in v = dy/dt = WAcosW*t and accel = - W^2 A sinW*t.
It seems easy to find the accel as as a function of coordinate by
substiuting as follows: since accel = - W^2 (A sinW*t) & with Y =
A*sin W*t (original given equation of motion), then accel = -W^2*(y).
Applying the chain rule to get the velocity : a(accel) = v(dv/dy) hence
a (-W^2*y*dy) = v*dv. If this is intergrated over the limits (y= 0 to y
and t= 0 to v), I get v^2 = -(W^2)*(y^2). I now have velocity as a
function of its coordinate? Am I correct?
How could I determine the maximum velocity? Usually, this when
something is set = zero?
I've plotted the curve on an Excel spreadsheet and it is sinusodial
& by assuming different values for A & W (omega), I can easily
make the curve take on different shapes.
ANSWER:
If you are interested in velocity as a function of position
for the harmonic oscillator, it is far simpler to never get time
involved in the calculation. Just use energy conservation: kA2/2=mv2/2+kx2/2;
solving, v=[k/m]1/2[A2-x2]1/2=w[A2-x2]1/2
(this will be either positive or negative). The
maximum speed will be when x=0 and will be wA. You could
also get it from your answer, it being the coefficient of sin[w t] above, wA again; this is
because the maximum magnitude of the sine function is unity.
QUESTION:
Does general relativity explain an apple falling from a tree?
Even if space-time is curved, does that somehow result in an apple
falling? It seems like, for the apple to fall, a force grabs the apple
and pulls it to the ground - does general relativity explain that?
Thanks!
ANSWER:
If you are happy with the empirical explanation that the
earth has a mass and therefore causes a gravitational force which can
be felt by other masses, then there is no need for general
relativity. This is essentially what Newton did--he stated his
universal law of gravitation as an experimental fact and it then
becomes one of nature's laws if it describes nature accurately.
However, if you want to ask "why does the earth cause a
gravitational field?", that is what general relativity addresses.
And the answer to your question is that yes, general relativity does
explain why there is a gravitational field around an object with
mass--mass warps the space around it.
QUESTION:
Consider two charged particles traveling in equal but
opposite directions as shown on the website at www.hypercomplex.us/Questions/bforce.htm.
In a reference frame in which both charges are moving, the second
charge experiences a magnetic force with non-zero magnitude resulting
from the magnetic field produced by the first. The magnitude of this
force, however, becomes zero when the velocities are transformed to a
reference frame in which the velocity of either charge is zero. Why
does the force disappear given this transformation of coordinates?
ANSWER:
There are a couple of problems with this web page.
First of all, the magnetic field given is incorrect. This is not
a magnetostatic situation which means that the magnetic field is not
constant with time at every point in space. The actual field must
be computed at the 'retarded time', which takes into account the time
it takes the field information to propogate to the second charge (at
the speed of light); however, the given field is approximately true if
the speeds are small compared to the speed of light (which is called a
'quasistatic' situation). The second problem is that the Coulomb
force between the two particles is totally ignored. I would guess
that part of your problem conceptually here is that you are finding a
particle with no force in one frame and with force in another. If
you remember that there is also an electrical force, then you will not
have this problem. Because the magnetic force depends on the
velocity of the charge, then it should be obvious and not disturbing
that there will be some frame where the magnetic force is zero.
The correct transformation of magnetic and electric fields from one
frame to another is a topic in electromagnetism which is quite advanced
and the details are beyond the scope of what 'Ask the Physicist' is
supposed to be about.
Here is a similar example of magnetic forces going away. In this case it is the field itself which I can make disappear. Imagine a long line of electric charge at rest. Obviously, there is no magnetic field but there is an electric field. Now imagine making the whole line move with some speed v; you now have a long straight current and therefore a magnetic field has appeared. The electric field will increase because the charge density on the line will increase because of length contraction.
QUESTION:
Suppose there's a lamp next to a railroad track, and a train
begins accelerating away from the lamp, and someone on the train
measures the speed of the light from the lamp. My understanding is that
if the train were in uniform motion then C would be unchanged, but if
the train is accelerating away from the lamp, will C be found to be
less than usual? If the train is accelerating towards the lamp will C
be greater than usual?
If an observer is standing next to the track, will the accelerating train appear shorter to the observer, and will the people on the train appear to be moving slower than usual? If so, will those effects be more pronounced than if the train were in uniform motion?
Did Einstein have any theories about those things, and have experiments been done for them? Maybe it's extremely difficult to prove length contraction and time dilation? But maybe there's been some pretty convincing experiments to show if C is affected when measured from an accelerating frame of reference?
Are experimental findings in agreement with Einstein's theories? - if he had any theories about those things. Thanks
ANSWER:
The answer to your first question is that the speed of light is always c
as measured by any observer. The fact that the train is
accelerating is of no importance. If the train were accelerating
perpendicular to the direction of the light ray, the light would travel
in a curved path (not constant velocity) but it would move with
constant speed.
Lengths along the direction of motion are shorter than if at rest; therefore an accelerating train is shorter than its "normal" length and getting shorter as it accelerates to faster speeds. I do not know what you mean about the people on the train appearing to be moving slower than usual; what is usual?
Length contraction and time dilation are completely verified experimentally, but obviously not with trains and such. Rather, elementary particles moving with speeds comparable to c are used. Any good book about special relativity will discuss this. Regarding the accelerated frame of reference, the theory of general relativity says that an accelerating frame is indistinguishable from a gravitational field. Therefore, a beam of light bending when it passes a massive object like the sun or another star, is identical to what would happen if a beam of light entered the side of an accelerating elevator and was observed to bend down as it crossed the elevator.
FOLLOW UP QUESTION:
If a train is moving past an observer who's standing on the
ground next to the track, from the point of view of the observer, the
train would be subject to time dilation? - and from the point of view
of the observer time would be running slower on the train, and clocks
on the train would be running slower, and the people on the train would
appear to be moving in slow motion? Maybe I should be asking about at
rocket, because maybe then the speed of the rocket would be closer to
what is required for a noticeable effect?
ANSWER:
You should avoid using the word 'appear' because relativity
is not about how things look, it is about how things are.
Measurement of velocity involves measurement of both length and time
both of which depend on the frame of reference, so you have to be
careful. In your one-dimensional example (the velocity of someone
on the train is in the same or opposite direction as the velocity of
the train itself, i.e. walking up and down the aisle), there
is a simple equation which tells the velocity of the person as measured
by your 'stationary' observer. It is called the velocity addition
formula: u'=(u-v)(1-(uv)/c2)
where v is the velocity of the train, u is the
velocity of a person as measured by somebody on the train, and u'
is the velocity of the person as measured by somebody on the ground.
QUESTION:
I assume that the Surface Tension of Water is caused by
Electro-Magnetic Forces between Water Molecules? So it seems like Water
should be Attracted pretty strongly by a Magnet? - but it doesn't seem
to be Attracted very strongly, if at all?
ANSWER:
The word 'electromagnetic' encompasses all forces which
result from either electric or magnetic fields. The
electromagnetic force which is responsible for surface tension is
purely electrostatic, that is it has nothing whatever to do with
magnetism. The water molecule is a polar molecule, meaning it has
an electric dipole moment, and it is the linking of these dipoles which
results in surface tension. An electric dipole (or an electric
charge for that matter), when at rest, is unaffected by magnetic
fields. To ask what you would expect to happen if you put a
magnet nearby, you must ask what the magnetic properties of water
molecules are. I recently answered a
question about how water is affected by a magnet.
QUESTION:
If we put a 2 masses in the space (vacuum where there is not
any kind of gravity field) Is this 2 masses attract to each other? If
yes, Why?
ANSWER:
This is basic Newtonian physics. Newton's universal law
of gravitation states that any two masses exert gravitational forces on
each other which are proportional to the product of the masses and
inversely proportional to the square of the distance between them, F=Gm1m2/r2
. However, unless at least one of the masses is very large, the
force will be very small because gravity is nature's weakest
force. This is reflected in the constant G being very
small, G=6.67 x 10-11 N m2/kg2.
So if you put two 1 kg masses a distance of 1 m apart, each would
experience a force of 6.67 x 10-11 N and so each would have
an acceleration near the other of 6.67 x 10-11 m/s2.
To put this in perspective, it takes something with an acceleration of
6.67 x 10-11 m/s2 (and starting from rest) a
time of about an hour and a half to travel 1 mm!
QUESTION:
Here's a question about the Length Contraction Aspect of the
Special Theory of Relativity - say there's a Railway Car sitting
Motionless on a Track, with a Lamp on the East End of the Car - the
Lamp is Switched On and the Time is Measured for the Light to Reach the
West End of the Car - then the Car begins Rolling Eastwards, and again
the Light is Switched On and the Time Measured again - this time it
Takes Less Time for the Light to Reach the West End of the Car, so the
Special Theory would say that, since C is Constant, the Length of the
Car has Changed - the Car's Motion caused the Car to become Shorter?
Is all that right? - if so, then what about if the Car begins Rolling West, and again the Light on the East End of the Car is Switched On - this time it would take Longer for the Light to reach the West End, so by the same Reasoning, it seems like the Special Theory would say that now the Car's Length has Increased!
Is that a Flaw in the Special Theory? - maybe the Wording should be something like "Sometimes an Object in Motion becomes Shorter, but Sometimes it becomes Longer"
Do I
deserve a Nobel Prize? - or am I Overlooking Something?
ANSWER
I will answer your last questions first--no, you do not
deserve a Nobel prize! And it is not so much that you have
overlooked something, but that you have computed the length of the car
incorrectly. In order to measure the length of something, you
need to measure the positions of each end at the same time, not
different times. So first you must use special relativity to get
length contraction so that you know what the length of the moving car
is and then you can analyze your experiment.
You
cannot deduce the length of something by simply measuring the time it
takes light to travel between its ends (which is what you have
discovered here!). Let's do your problem but do it
correctly. Suppose we call the length of the car, in its own rest
frame L0. Then, if it has a 
speed v relative
to some observer, its length as measured by that observer is L=L0[1-v2/c2]1/2.
For your first experiment (car at rest) the time is t=L0/c.
For your second experiment, (car moving east) let us call the time t1.
Looking at the first picture (dashed lines show where the car has moved
to when the light, red, strikes the other end), it is clear that L=L0[1-v2/c2]1/2=vt1+ct1,
so t1=t[1-v/c]1/2 (you will
have to do a little algebra to get here). t1
is, as you state, indeed less than t but not because L
is shorter than L0, although it is. In your
third experiment, refer to the second picture and call the time t2.
Now we have that L=L0[1-v2/c2]1/2=ct2-vt2,
so t2=t[1+v/c]1/2. So
now, t2>t but not because the car is any
different length from experiment #2 (it has exactly the same length);
rather because the light had to travel farther than the length of the
car.
QUESTION:
How will light behave when entering a cone shaped
prizm? Assuming the light enters through the bottom of the
prizm. Will the light exit the prizm through the sides or just
bounce around and refract until it exits through the bottom of the
cone? If it does exit through the sides of the cone will it exit
with an even distribtion from the cones sides?
ANSWER
I will assume that you mean that the light enters the base of
the cone with the rays perpendicular to the base. The answer,
essentially, depends on two things--the angle of the cone and the index
of refraction of the material from which the cone is fabricated.
There are additional details which would depend on the polarization of
the light, so let us assume that the light is unpolarized; I won't go
into the details which depend on the polarization. A ray strikes
the inner surface of the cone with an angle of incidence qi which you can see is
determined by the angle of the cone a, qi=900-a. Part of the light will
be reflected (blue in the figure) at an angle equal to the angle of
incidence; this ray, it may be shown, will be (partially) reflected,
possibly bounce once more on the inner surface, but eventually reemerge
back out the bottom exactly opposite the direction in which the
incident rays are entering. But also, part of the incident ray
will emerge (red in the figure) from the cone at an angle qr which is larger than qi. This is called
refraction and the value of the angle depends on the index of
refraction of the cone, n: sinqr=n sinqi. The relative
amounts reflected and refracted depends on qi and n and on
the polarization. Also, if the incident ray is unpolarized as we
have stipulated, the refracted and reflected rays will not be, so the
fractions on the next bounce will not be the same. So, you see,
the details will be very complicated. There is one very important
special case. Suppose that n sinqi>1; then, since sinqr cannot be bigger than
1, there will be no refracted ray. This is called total internal
reflection and all the light will be reflected back in the direction it
came from.
QUESTION:
How does absorption occur in paper towels?
ANSWER
The short answer is capillary action, the tendency of water
to creep along very narrow passageways. This is how sap rises in
very tall trees and how blood moves through very narrow veins and
arteries in your body (called capillaries). You can find out lots
more about your question by doing a google
search on "paper towels" AND capillary.
QUESTION:
My question is rather simple, but has created much
controversey among my friends and I. When I am at an amusement park, I
often wonder... Which part of the rollercoaster goes faster? The front,
or the back? I immediately come to the conclusion that the
rollercoaster MUST be going the same speed, in the front and the back.
My friends disagree with me, and say that the back is going faster. I'm
pretty sure that this is untrue. Could you explain the reason our
confusion, and give a brief answer.
ANSWER
For a roller coaster, every car moves with the same
speed. The reason is that all the cars are constrained to move on
the track. If they were not constrained to move on the track,
however, one part of the train could easily move with a different speed
from another. Imagine, for example, that the train rotated around
the front car which was standing still (except for its rotational
motion); then each car, as you moved toward the other end, would be
moving faster than the one before. When I was a kid we used to
play "crack the whip" when ice skating: a long line of kids would skate
across a pond and the kid at one end would essentially stop and the
kids near the other end would get a very exciting and fast ride!
QUESTION:
If you bring a compass to the right side of a bar magnet (the
north is on the right, and the south is on the left of the bar magnet),
which way should the compass needle point? To the left (away from the
north pole of the magnet), or to the right (so it touches the north
pole of the magnet)
ANSWER
A compass is itself a bar magnet and the end which points
north is, by convention, called the north pole of the compass
(magnet). Now, as you probably know, like magnetic poles repel
and opposite magnetic poles attract. Therefore the north end of
the compass will be repelled by the north pole of the magnet so the
compass will point away from the magnet. Incidentally, note that
the convention means that the magnetic north pole of the earth is
actually a magnetic south pole if the north pole of the compass points
to (is attracted to) it.
QUESTION:
When an atom releases energy as a photon, does the photon
travel along only one direction? I am thinking that since light
can also be interpreted as wave, then the emitted light may be a
spherical wave front in 3D space. This is my dilemma.
ANSWER
The description of the properties of the photon depend upon
the quantum state in which it is put by some measurement. For
example, if you know nothing about the direction in which the photon is
traveling, it is equally probable that you will find it going in any
direction so it is like a perfect spherical wave. On the other
hand, if you know the direction perfectly then you will, according to
the Heisenberg uncertainty principle (HUP), be totally ignorant of the
position so it is like a perfect plane wave. The classic example
is to put a slit in the way of a beam of photons to try to pin down
where they are and in fact you destroy that information because of the
HUP. This is because a slit of width Dx means that the x
position of a photon which passes through is uncertain by an amount Dx so the momentum in the x
direction will be uncertain by an amount approximately equal to
Planck's constant divided by Dx. Therefore, the
smaller Dx is the more uncertain
you are of the direction in which the photon is going after passing
through, it is more spherical (actually cylindrical for a slit) like in
wave language.
QUESTION:
What are the physics behind the fun of an air hockey table?
ANSWER:
Well, I do not know much about air hockey, but the most
important feature of the game, physics-wise, is the fact that the pucks
slide with very little friction. This is achieved by having many
small holes on the board through which air is squirting. The puck
then slides on a very thin cushion of air which causes there to be
negligible friction. Thus, as the puck slides across the board it
moves with approximately constant velocity until something exerts a
force on it--a wall or a paddle.
QUESTION:
My name is Sam and I am 11 years old. Where can I find
someone to help me with my science questions? My mom told me to find
some web-sites and e-mail to see if I could get any suggestions. I'm
researching photons. My science book and most of my research states
that photons do not have mass. This interested me since photons have
energy, are affected by gravity, and (I'm still researching this) can
be split. If you have any suggestions for my research please
reply.
Thanks,
Sam and Sam's mom (who won't let Sam into cyberspace unchaperoned)
ANSWER:
Hi Sam! Answering questions of hungry young minds like
you is the primary reason for this web site! Here is a brief
overview of photons:
In the beginning there was light! You know what light is, electromagnetic radiation which includes visible light, radio waves, x-rays, etc. The study of the physics of light is one of the oldest areas of physics and is called optics. Optics existed long before anybody actually knew what light is (electromagnetic radiation). But the study of electricity and magnetism (E&M) yielded the wonderful four Maxwell equations (in the last half of the 19th century) which summarized everything there is to know about E&M. Now, here is the remarkable thing: when one manipulates Maxwell's equations there emerges equations which describe waves of oscillating electric and magnetic fields (in sound waves, air oscillates; in water waves, water oscillates, etc.). And guess how fast these waves travel--exactly with the speed of light. It was, before now, known that light is a wave, but nobody knew what was "waving". Now it turns out that the speed of light is a universal constant--no matter who measures the speed of a beam of light, everybody gets the same answer. This is amazing, if you think about it; for example, suppose that you were going 50 mi/hr down a road and somebody else going the other direction was going 100 mi/hr (the speed somebody standing by the road sees) toward you. Then, you see that other car approaching you with a speed of 150 mi/hr, right? That would not be right if 100 mi/hr were the speed of light: both you and the guy by the roadside would see the same speed. I know it is hard to swallow, but it is true. This gave rise to the theory of special relativity (by Einstein) which revolutionized the way we think about physics (around 1905). Around this same time another revolution in physics was happening, the discovery of quantum physics (by Planck) which found that nature is not a smooth continuous thing but rather discretized or kind of grainy. For example, a system of particles which is bound together (like an atom or a solid or the solar system or a galaxy) cannot exist in any old energy state it wants but only is certain states. You never see anything like this when you look, e.g. at the solar system because the states are very close together and we can't see anything but smoothness; but when you start to look at tiny things like atoms, you start to see these "quantizations". Anyhow, it turns out that one of the consequences of relativity and quantum physics is that light is both waves and particles (called duality by physicists). If you look for a wave you will find one; but if you look for a particle you will find a particle! In your research you should find out about the photoelectric effect which was the first instance in which the particle nature of light was actually apparent. (Again, it was Einstein who explained the photoelectric effect. It was this achievement, not relativity, for which he won the Nobel prize.) Particles of light are called photons. Now, by definition, photons must travel with the speed of light (they are light!). But relativity says that the only things which can travel with the speed of light are things with zero mass. Therefore, photons must have zero mass. There is no doubt whatever of this fact! Yes, photons do have energy and momentum even though they have no mass; this is allowed in relativity. And photons do feel gravity even without mass; this is a consequence of the theory of general relativity which Einstein formulated about ten years after the 1905 theory of special relativity. When you are doing your research, also look up the Compton effect (the change in wavelength of photons when they scatter from electrons) which is one of the seminal confirmations of the notion of photons. The energy of a photon is hf where h is a constant (called Planck's constant) and f is the frequency of the corresponding electromagnetic waves; you may think of this as the smallest amount of light of a particular frequency which is possible. Yes, a photon can be split in two but since energy must be conserved, the "colors" of the two new photons will be "redder" than the original. A recent reference for this is http://focus.aps.org/story/v10/st3.
I realize that this is sort of a rambling "stream of consciousness" answer to your question. Hope it is of some help!
QUESTION:
I have worked out enough "projectile motion" problems such
that I think I understand the principles, but there is one type that I
can't nail. It's where a ball rolls off the upper floor and then you're
asked to tell which step the ball will land on? For instance, a
ball rolls off upper landing at 5 ft/sec in the horizontial direction,
and the steps are 8" high by 8" wide. It asks which step the ball will
land on?
The problem tells us that initial velocity is horizontal & hence the launch angle (theta) = 0 degrees so sin (theta) = 0 & cos (theta) = 1. We also know that the initial velocity is constant along the X (horizontial) axis. We know that the Y axis velocity component = 0 initially, but begins to fall at rate(v) = -g*t (gravity x time) and the distance (y) it falls = -1/2gt^2.
We don't know time required & I can't make it fall out, and without the time I can't find the X distance... which is what I want to know, as this will tell which step the ball will land on.
What am I missing? I've not connected with something, but what?
ANSWER:
Imagine a ramp running down the stairs (just imagine a board
laid on the stairs). Its equation is y=-hx/w
where h is the height of a step and w its width.
The equations of motion of the ball are
x=v0t
y=-gt2/2.
Now the ball "hits" the ramp when -gt2/2=-hx/w. But, since x=v0t, we can get t=2hv0/(gw). Now, knowing t you can get x: x=2hv02/(gw). Once the ball crosses the ramp it must hit the next available step; x/w will be the number of steps when it crosses but it will (except for very special cases) not be an integer since it will have passed a fractional step. You just need to go up to the next integer from x/w=2hv02/(gw2) to get the number of the step which is hit. For your particular problem, v0=5 ft/s, h=w=0.75 ft, g=32 ft/s2, so x/w=2x0.75x52/(32x0.752)=2.08, so the ball lands on the third step.
QUESTION:
How does a magnet interact with water?
ANSWER:
Water is diamagnetic which means that it is repelled by a
magnet. However, the force is extraordinarily small and therefore
difficult to observe.
QUESTION:
Police agents flying at a constant speed of 200 km/h
horizontally in a low-flying airplane wish to drop an explosive onto
master criminal’s automobile traveling 130 km/h on a level highway 78.0
m below. At what angle (with the horizontal) should the car be in their
sights when the bomb is released?
ANSWER:
This sounds like a homework problem to me! Are you
telling me you just happened to wonder about this situation?
Anyhow, dropping means that the initial conditions for the velocity of
the explosive are an x-component of its velocity of 55.6 m/s
(200 km/hr) and a y-component of zero. I choose a
coordinate system with the origin at ground level and directly below
the airplane when the explosive is dropped, so the initial conditions
for the position are zero for x and 78 m for y.
We can now write the kinematic equations of motion (I presume you know
how to do that) for the explosive:
x=55.6t
y=78-4.9t .
Now, we are interested in where the explosive hits the ground because that is where the car is. So set y=0 to get the time. Solving, t=4 s. Knowing t, we can say where the car must be: x=55.6 x 4=222 m. Now, what is the equation of motion for the car? Since we do not know where it is (that is the objective of the problem) when the explosive is dropped, let's just say that it is x0. Then the equation of motion is
x=x0+36.1t
where 36.1 m/s=130 km/hr. Now, we want x=222 m when t=4s; solving, x0=76.4 m. There is probably some roundoff error, but this is close. Finally, to express the answer in terms of your question, the tangent of the angle of the line of sight to the car is 78/76.4 below horizontal; pretty close to 45o.
QUESTION:
I'm curious (and you may have been asked this recently,
but...) In martial arts, what are the components which influence
the damage done by a strike, what are thier relationships, and most
importantly, a laymans explanation as to why (sounds like a thesis
level question, don't it? :-) ?
ANSWER:
I haven't been asked this question in this context, but if
you carefully read the q&a just below you will get all the
information you need to answer your question. Just as in that
question, you want to deliver the maximum possible force over the
minimum area to do the maximum damage. [Below I only briefly
touched on the 'minimum area' aspect, but it is pressure, force per
unit area, that counts. For more on this see my discussion
on the 'penny falling from the skyscraper' question.] So, using
the edge of hand, as in karate, is more effective than using the flat
of your hand or than using your fist (other things being equal).
Also, to deliver the maximum force you need to maximize the momentum
transfer as described below (for example, it is better if your hand
'bounces off' than if it doesn't). And, the time of the blow
should be minimized to maximize the force. A good example of this
is using a hammer to drive in a nail: if you watch a master carpenter,
you will notice that the hammer bounces back with each blow and (of
course) each blow is of extremely short duration; someone less skilled
tends to have the hammer stop. Have you ever tried to push a nail
into a board with your hand?
My discussion has focused on the force which the hand or foot delivers to whatever it is attacking. However, a very important principle of physics is Newton's third law which states that if one thing exerts a force on another, the other exerts an equal and opposite force on the first. Therefore, the hand or foot experiences exactly the same force as it delivers. Which is why, of course, sometimes boxers break their fingers. So I guess that one of the skills in martial arts is knowing which areas of the body are most vulnerable so that a given force there will do a lot of damage but an equal force on the hand would not.
QUESTION:
A physics teacher asked a class this question, but I wasn't
around for the answer (not my class). Start with two objects
of mass m, travelling at velocity v, neither of which can penetrate a
particular target. Now, double the mass of one object, and the velocity
of the other, while leaving all other characteristics unchanged.
Which one is more likely to penetrate the target, or are they equally
likely? The class seemed to be divided into two categories, those
who thought that kinetic energy was the key, and that therefore
doubling mass doubled the impact, but doubling speed quadrupled the
impact force, and those who thought that the two were equal.
ANSWER:
What is important in breaking through the target? The
most important thing is the force which is applied to the target.
Of course, also of importance is the area over which the force is
applied, so a sharply pointed object which exerts the same force as a
blunt object is more likely to break through; but I assume that the two
objects have identical geometries and compositions. So we need to
focus on the force which a projectile exerts on something during the
time of impact. Newton's second law (N2) is usually thought of,
particularly by students just learning physics, as force equals mass
times acceleration, F=ma ; however, when collisions are being
studied, it is much more useful to think of N2 as force equals time
rate of change of linear momentum (which is mass times velocity, p=mv),
or the force (Favg ) averaged over the time of the
collision (Dt) is change in
momentum of the projectile (Dp) divided by Dt, Favg=Dp/Dt.
[Incidentally, it should be clear to you that the two forms of N2 are
the same since, if the mass does not change, the rate of change of mass
times velocity is mass times the rate of change of velocity and rate of
change of velocity is acceleration.] So, if we are interested in
force we have to think about both the change in momentum and the time
it takes for that momentum to change; big Dp
or small Dt means big
force.
My guess is that your physics teacher wants to teach the lesson here that the important thing is momentum, not kinetic energy, so let me talk about momentum and then I will add a little bit about time. What is the range of possible changes of linear momentum if the projectile does not break through (the first experiment)? The least it can change is if the projectile comes to rest after the collision, in which case Dp=mv ; the most it can change is if the projectile bounces back elastically, in which case Dp=2mv (don't forget that momentum is a vector quantity). For simplicity's sake, let's just assume that both projectiles stop in the first experiment (because if they bounce back they will never, by definition, go through!). Then, in the second experiment we will double the momentum of both projectiles so they are equally likely to break through if both collisions take the same amount of time.
But, are they likely to take the same amount of time? No, because the projectile with the doubled speed is likely to take longer to transfer the same amount of momentum as the projectile with the doubled mass and thus will exert less force. Therefore, I believe that the likeliest one to break through is the projectile with the doubled mass.
QUESTION:
Two capacitors in series have a neutral middle. The right
plate of the left capacitor is connected to the left plate of the right
capacitor. This section is isolated from the battery. Charges in this
section move because the plates connected to the battery are charged by
it. Free elctrons in the neutral middle are effected by the generated
field. The neutral middle becomes polarised and the charge on each
capacitor becomes equal.
How long does this polarization take given that drift velocities are
very low for electons in wires? Would the time be significant? The RC
constant in a series circuit seems to be reduced because C equivalent
is less)not increase becasue of drift through the middle???
ANSWER:
First of all, it is really inaccurate to say that the
"middle" is isolated from the battery. As soon as the plates
connected to the battery begin to charge up, they give rise to an
electric field which the middle plates feel. This field is what
causes charge to flow from one of these middle plates to the
other. I would say that the battery is causing the fields so the
middle plates are "seeing" the battery. The time constant has
nothing at all to do with drift velocity of the electrons. You
could just as easily have asked why the charges get to the plates from
the battery so quickly. The fields in the wires set up by the EMF
of the battery cause all the electrons in the wires to begin moving
instantaneously (actually, these fields propogate with the speed of
light, so it isn't quite instantaneous); you don't have to wait for the
electrons to travel the distance from the battery to a plate or, for
the "middle" plates from one plate to the other.
QUESTION:
what do we mean by friction? i.e-friction is defined as a
force that apposes the force applied by a person on a body, Many books
agree that when ever 2 bodies are in contact and if they rub together,
the junctures or the uneven surfaces, get smoother and even, so we can
say that friction is proportional to force applied if and only if 2
bodies are in contact, then how are we able to make precise
calculations about it.
ANSWER:
When two objects exert contact forces on each other there may
always be two components of the forces, one perpendicular to the
surface of contact and one parallel to the surface of contact.
The parallel component is usually called friction and the perpendicular
component is usually called the normal force. In many cases of
interest, the magnitude of the frictional force is proportional to the
magnitude of the normal force; the proportionality constant is
determined by the nature of the surfaces and is called the coefficient
of friction. This proportionality is only approximate, certainly
not a law of physics or anything like that. To understand
friction microscopically, at the atomic level or even at the level of
the roughness of the surfaces, is a very difficult problem and it is,
in fact, not well understood at all. I previously
answered a question about friction which you might want to look at.
QUESTION:
In physics we have classified all the forces into 4
categories as said in feynman's lectures-nuclear forces, gravitational
forces, electromagnetic and weak interaction..
But where do we categories forces like friction and push or pull. They
don’t satisfy any category.
ANSWER:
These forces are what are normally referred to as contact
forces, forces which depend on the contact between two objects which
are exerting forces on each other. For example, when your hand
pushes on a ball the atoms in your hand are interacting with the atoms
on the ball and the fundamental force by which atoms interact with each
other is the electromagnetic force; essentially the electrons in your
hand repel those in the ball.
QUESTION:
I wish to discuss a technicality in a sound wave problem. The
problem states: "Suppose that a sound wave is passing through a gas.
Does a change in the temperature of the gas affect the wavelength of
the sound wave if the frequency remains constant?" I understand that
wave speed=wavelength*frequency but changes in the temperature in a gas
must affect the speed at which sound travels through it and the
frequency and wavelength must change too right???? Is the problem
stated falsely?? Thank you for your time and consideration. I greatly
appreciate your help.
ANSWER:
The frequency of sound is determined by the source and is
completely independent of the medium through which the sound
passes. Even if there were no medium and therefore no sound, the
frequency of a 440 Hz tuning fork would still be 440 Hz. If the
frequency if f, the source causes the medium to compress every
1/f seconds. Now, it is well-known that the speed of sound
in a gas depends on the temperature of the gas. Since, as you
state, v=lf, if v changes
and f does not, then l must
change.
QUESTION:
I know that a rocket has to have enough thrust to accelerate
itself to about 25,000 mph to escape the Earth, but why that speed?
Let's say that in some futuristic vessel, it had 100 times more power
than the fuel-propelled rocket and had an almost unlimited source of
fuel. Couldn't this vessel just push it's way up gradually out of the
Earth's atmosphere at , let's say, 1000 mph?
ANSWER:
What escape velocity means is the speed a projectile would
have to have at the surface of the earth such that when it gets
infinitely far away it will just come to rest. It is understood
that "projectile" means something with no propulsion. However, if
you have propulsion, velocity is irrelevant. If you had adequate
fuel you could drive off to infinity going 1 mile per hour; it would
just take a lot longer than if you were going faster. If you got
going faster than escape velocity at the start, you would still have
some velocity (i.e. some kinetic energy) when you were
infinitely far away.
Incidentally, infinitely far is a "mathematical" construct here. What it really means is very far away, far enough that you could not detect the gravitational force of the earth with any reasonable experiment.
QUESTION:
At science centers, you can usually find the activity with
the bicycle wheel and a spinning chair? Can you explain the gyroscopic
precession principles involved with that activity and what makes you
spin when you change the angle of the bicycle wheel? Also, can you
please explain that using simple language...most websites I have
visited about this question are difficult for a layman to understand.
ANSWER:
Precession is not an issue here; rather, conservation of
angular momentum
must
be appreciated. Imagine that you are holding a spinning wheel
with its axel pointing vertically and that you are standing on a
turntable which can rotate without significant friction. Although
there are external forces on you (your weight, the weight of the wheel,
and the force the table exerts up), none of these forces exerts any
torque on you and the wheel. The total angular momentum of a system
which has no external torques on it must never change which is what it
means to say that the angular momentum is conserved. Angular
momentum is a vector quantity. If you look down on the spinning
wheel and it is spinning counterclockwise, it has an angular momentum
vector which points up and has a magnitude which I will call L.
You are not spinning and have an angular momentum, therefore,
of zero. Therefore the total angular momentum of you + the wheel
is L upward and this is what it must always be. Now you
flip the wheel over so that the angular momentum of the wheel is L
downward, so the angular momentum of the wheel has changed by an
amount of 2L downward. But, the total angular momentum
cannot change so an angular momentum of 2L upward must pop up
somewhere; the only part of the system which can get angular momentum
is you, so you must start spinning such that your angular momentum is 2L
upward which means, since up means counterclockwise as seen from
above, you will spin counterclockwise as seen from above. Here is
a movie.
Incidentally, if you care, conservation of angular momentum is simply a consequence of Newton's first law in its rotational form. In translational form, Newton's first law says that, if there are no external forces on a system, the linear momentum will be conserved; in rotational form it is that, if there are no external torques on a system, the angular momentum will be conserved.
QUESTION:
Xrays should be called VEXrays!
An electron is accelerated under the influence of an electrostatic
potential, generally in the neighborhood of 60-80 kilovolts under a
hard vacuum. The electron collides into a target atom, causing an
orbital electron to momentarily absorb this kinetic energy.This
energised electron relaxes back to its ground state producing a quantum
x ray photon.X ray photons have great energies compared to electons
accelerated to speeds of 60 kilovolts. Adding to this quandry,
the target electrode heats up, seemingly stealing energy, from this
process, generating Xrays. I try to envision this phenomenon using
billiard balls. Not a chance! Never hit an object ball to see it roll
away faster or spin faster than the ball I hit it with.
Sincerly! VEXED!
ANSWER:
Actually, you are wrong. If you send in 60 keV
electrons you will see all photons coming out as having less than or
equal to 60 keV. Energy must be and is conserved when electrons
interact with atoms. Also, classical physics has no problem with
the particles coming out having a greater velocities than particles
coming in: for example, imagine bowling balls with a speed of 5 m/s
colliding with bb's at rest. After the collision the bb's will
have speeds near 10 m/s.
QUESTION:
What is energy? A usual definition (the ability to do work)
only tells me what it can do.
ANSWER:
This is a nice succinct question. It certainly does not
have a succinct answer, though! Books have been written to answer
this question. I will hit a few high points:
- Perhaps you would like your "usual definition" better if it were turned around: energy is what you get if you do work. For example, if you do work by pushing on som