QUESTION:
I'm just confused because according to newton's third law of motion, " any action has an opposite but equal reaction", hence forces in nature are balanced, and balanced forces means bodies are at rest, so how come in reality, we can observe motion?

ANSWER:
You do not understand the third law. If there are two objects, A and B,
and A exerts a force on B, Newton's third law says that B exerts a force of
equal magnitude and opposite direction on A. So, you see, the two forces are
not on the same object and are therefore not "balanced".

QUESTION:
If a man was to be proportionately increased in size, would it be more difficult for him to perform a pull-up? All his body percents are exactly the same including his muscle, fat, ability, ect. The pull-up bar is also high enough that his feet would not touch the ground in either situation.

ANSWER:
Scaling of biological organisms has been a topic of interest for a long
time. I think Leonardo de Vinci was the first to think about it. Anyhow, I
think the answer to your question is that it would be harder to do a pullup.
Here is why: The weight of something is proportional to the cube of its
dimensions. So, if you doubled all the dimensions of a man, his weight would
increase eightfold. But, the strength of bones or muscles is dependent on
their cross sectional areas, a bone or muscle doubled in size would have
about four times the strength.

QUESTION:
I am confused if I put a book from ground to a table at height 'h' ,note that the book is at rest before and after putting , total work done on book is 0, as change in k.e. Is 0 ,therefore total transfer of energy on book should also be 0 but after putting the book on the table at height 'h' the book gain a p.e. Of 'mgh'.how?

ANSWER:
Yes, you are confused! The idea of a potential energy function is that
it keeps track automatically of the work done by a force always present in a
particular problem. The gravitational potential energy represents the work
done by the weight mg of an object. So, if you use potential energy,
you do not count the work weight does, that is
ΔK +ΔU =W _{ext } where W _{ext
} is the work done by all forces other than gravity, in your example
that is you. So, to lift the book you must exert a force upward of magnitude
mg for a distance h and so the work you do is W _{ext} =mgh ;
since kinetic energy has not changed, ΔU =mgh.

QUESTION:
Potential Energy + Kinetic Energy = 0. However Potential Energy = negativeKinetic Energy. Could please explain?

ANSWER:
First of all, the equation you write really does not make any sense and,
if it does, it does only under special circumstances. You are clearly
talking about a system where total energy is conserved. This means that
there are no external forces doing work on the system, it is what we call an
isolated system. In that case, the total energy never changes. But, that
means that the change of the total energy is zero
ΔE =0. But, the total energy is the sum of the kinetic and
potential energies, ΔK +ΔU =0. So, it is the sum of the changes
which is zero, not the sum. Now, it should make sense to you. If you write ΔK =-ΔU,
it simply means that if the kinetic energy increases, the potential energy
must decrease by the same amount to keep their sum constant.

QUESTION:
if you where able to make a hot air balloon that
could contain a vacuum would it rise like one containing He or would the
air pressure pressing from all side cancel it out ?

ANSWER:
The reason a hot-air balloon (or a hydrogen- or helium-filled balloon)
rises is buoyancy. The buoyant force is equal, in magnitude, to the weight
of the air which is displaced by the volume of the balloon. So, if you
create a vacuum in a rigid balloon which has the same volume and weight as
one filled with hot air, hydrogen, or helium, it would have a greater
lifting force because you would not have the weight of the gas. The problem,
though, is that it would be hard to make it sufficiently rigid that it would
not collapse when you evacuated it; atmospheric pressure, acting on the
outside of the balloon, is about 2000 lb/ft^{2} .

QUESTION:
Of the four formally recognized fundamental interactions, which is most responsible for causing elements from the environment to be drawn into living cells (a reduction of entropy) and, when the cell dies, appears to cease, allowing a reversal of the organization and decay back into the environment (an increase in entropy)?

ANSWER:
Anything involving microbiology is basically chemistry. Anything
involving chemistry is basically atomic and molecular physics. The only
important interaction in atomic and molecular physics is electromagnetism.

QUESTION:
If two beams of light pass one other, moving in opposite directions, at what speed are they moving away from each other? I believe the answer is at light speed but I guess the real question is, and briefly if you can. Why? ...or why not twice light speed?

ANSWER:
The velocity addition formula for special relativity is v _{ab} =(v _{ag} +v _{gb} )/(1+(v _{ag} v _{gb} /c ^{2} ))
where v _{ag } is the velocity of a with respect to the ground,
etc . For your question, v _{ag} =c and v _{gb} =c
so v _{ab} =(c+c )/(1+(c ^{2} /c ^{2} ))=c.

QUESTION:
I understand the part that all matter stays constant in a confined space, none can be lost no matter what happens to it. But I have trouble with the energy part. It is stated that if matter is changed into energy then that energy must be able to change back into matter for the whole of the conservation of matter & energy theory to work. My question is: has a scientifically qualified experiment ever, at any time, been done in the lab to show this? I mean taking pure energy of any type & changing it into matter of any type? I can find nothing on this on the net. Do you have the answer?

ANSWER:
You understand wrong: matter can be created or destroyed. There is no
such law as conservation of matter. See an
earlier answer .

ADDED ANSWER:
I see that I did not answer the main part of your question! Energy to
mass happens all the time. The best known example is pair production where a
photon (no mass) creates an electron and positron. See an
earlier answer .

QUESTION:
What causes atoms to oscillate and do they oscillate infinitely or do they absorb energy from a limited source?

ANSWER:
Think of each atom in a solid to be attached to its nearest neighbors by
tiny springs. This is a reasonable model which approximates much of the
behavior of solids. Then each atom is a tiny harmonic oscillator. In
classical physics, all those tiny oscillators could simply have zero energy,
each would be at rest. However, quantum mechanically a harmonic oscillator
cannot have zero energy, it has some minimum energy called the ground state.
Therefore, it can oscillate forever even without any energy from outside. At
room temperature all those oscillators are in higher states than the ground
state and the energy to put them there came from the environment, that is
the solid was heated up to the temperature of its environment.

QUESTION:
I have a pressure washer that pumps 5 Gallons of water Per Minute and sprays water at 800 - 3500 Pounds Per Square Inch, at the nozzle. The Inside Diameter of my high pressure hose is 3/8". The diameter of the circle of the nozzle is 1/16". My question
is: Is there a formula to convert PSI at the nozzle to Miles Per Hour (MPH) at the nozzle?

ANSWER:
If you give me the gallons per minute, you are essentially giving me the
velocity in the 3/8" pipe—9.89 mph.
Now, if the diameter drops to 1/16", the velocity increases by the ratio of
the cross sectional areas, ((3/8)/(1/16))^{2} =36, so the speed at
the nozzle is 36x9.89=356 mph. Wow, there is a surprise! Surely it does not
have the same flow rate for different pressures?

QUESTION:
I was doing some water sports and was wearing a wet suit and a 50N (50 Newtowns) rated buoyancy aid. I weigh 93Kg. Cliff jumping into the sea, after resurfacing I bobbed gently on the surface with no swimming effort.
Later wearing same equipment on fresh water river, fell of a kayak and promptly sank! To get to and stay on surface I had to swim or actively tread water. The question is what value N rated buoyancy would I need to "bob gently" in fresh water?

ANSWER:
You do not weigh 93 kg, your mass is 93 kg and your weight is 93x9.8=911
N. Suppose that you just float with your head above water in salt water.
Then the buoyant force the water is exerting up is 911-50=861 N. Archimedes'
principle says that this is the weight of the displaced water, and let's say
salt water has a density of 1030 kg/m^{3} ; so, the volume of your
body less head must be 861/(1030x9.8)=0.085 m^{3} . So, the buoyancy
you can expect in fresh water if just your head is up is 0.085x1000x9.8=836
N because the density of fresh water is 1000 kg/m^{3} . So, to have
identical buoyancy to the salt water situation, you need an additional 25 N
(861-836) buoyancy.

QUESTION:
.. is there a theory/postulate/law that explains how the strong nuclear force keeps the like-charged particles in a nucleus from flying apart - maybe some sort of virtual exchange particle?

ANSWER:
There simply is the strong interaction which is responsible for holding
nuclei together. Is there a field theory which attempts to model how the
force "works"? Yes, the virtual particles exchanged are various mesons. The
force itself is much more complicated than the relatively simple
electromagnetic force. But, you should simply say that, empirically, there
is a fundamental force in nature which is far stronger than the coulomb
force at small distances. A theory is itself more of a model to explain what
you observe.

QUESTION:
When you face a wall, place your toes against it, and try to rise up on the balls of your feet, why can't you?

ANSWER:
When you stand up straight, your center of gravity is above a point
between your heel and toes and you do not tip over because there is no net
torque which would tip you over. When you go up on your toes and are unable
to lean forward, your center of gravity now exerts a torque about your toes
which rotates your body backward. Try standing on your toes when you are not
against the wall; in order to remain standing you will find that you must
lean forward so that your center of gravity will be directly over your toes.

QUESTION:
If p=mv/(sq. root 1- v^{2} /c^{2} ), does this mean massless photons have infinite energy?

ANSWER:
If v=c, p= 0/0 which is indeterminate. The important thing to know
is that E= √[p ^{2} c ^{2} +m ^{2} c ^{4} ],
so if m =0, E=pc .

QUESTION:
Why is light attracted to black holes, if photons have no mass?

ANSWER:
One of the first verifications of Einstein's theory of general
relativity was to observe that gravity bends light. This was done by looking
at a star whose light had to pass close to the sun during a solar eclipse
and seeing that it was not where it was supposed to be. Read more about this
by looking at links on my FAQ
page.

QUESTION:
I write an article on car safety for my company. My topic this month is objects hitting windshields and the probability of cracking, breaking or injury to the windshield/driver. So if two cars are driving toward each other at a low speed (25mph) or speed limit on a highway (70mph) and a bottle, water balloon, egg (Halloween), cup filled with ice is tossed out of car A and hits car B's windshield. What is the damage to the windshield? I'm looking for the lowest speed/heavy object, highest speed/lightest object.

ANSWER:
I doubt that I can be of much help to you. There are too many variables for me to be able to focus on any physics principles. Here are a few things that you should be sure you understand

Relative velocity, e.g. if cars are approaching each other at 70 mph, an object thrown from one would hit the other at a speed of about 140 mph,
i.e. as if a 140 mph cup of ice (or whatever) hit a car at rest.

Newton's second law should be understood to understand the collision itself
(see faq page). In particular, the average force on the windshield will be approximately mass x striking velocity/time for mass to stop.

If you have to work in English units (lb,ft,s),
and do anything quantitative, the mass to use in calculations should be
weight(lb)/32 and all velocities converted into ft/s from mph. That way,
forces will come out in lb.

I have no way of knowing the physical properties of
glass used and so cannot possibly know threshhold forces for damage to
windshields.

QUESTION:
what is weight ? explain why we feel lighter while travelling in a lift which is accelerated downwards by ''a'' m/s2

ANSWER:
What "weight" is is a semantic problem. (I say this because a textbook which I am familiar with defines weight to be what a scale would read which, to me, is wrong.) To me, weight means the force the earth exerts on me. When I stand in an elevator there are two forces on me, the force of the floor up and my weight down. I judge how heavy I feel by how hard I feel the floor pushing up on me. When I accelerate up, Newton's second law tells me there must be a net force up and so the floor pushes up with a force bigger than my weight, so I
feel heavier than usual. When I accelerate down, the net force must be down and so the floor pushes up on me with a force smaller than my weight and I
feel lighter than usual. Actually, our feeling heavier is a much more
complicated thing. For example, your internal organs are supported by
muscles and, when accelerating up the muscles must pull harder than usual
and this is transmitted to your brain which it interprets as your being
heavier all of a sudden.

QUESTION:
My young son asked me what would happen if there were no more temperature in the world. It made me wonder what the theories are about what would happen inside an atom, if all movement ceased.

ANSWER:
First of all, whatever goes on inside an atom is independent of
temperature; the properties of a hydrogen atom at room temperature are
exactly the same as those at 1 degree above absolute zero. We should define
temperature: temperature is a measure of the average energy per atom in the
object you are looking at. It is easiest to visualize a gas where the atoms
or molecules are zipping around with a variety of different speeds and the
temperature is a measure of the average of all the kinetic energies of these
moving particles, ½ mv ^{2} , and so the slower they are going on average, the
colder the gas is. As you correctly surmise, absolute zero (or "no more
temperature" as your son would say) is when they all come to a grinding
halt. But that is the motion of the atoms. If you like to think of atoms as
having little electrons orbiting the nuclei, the electrons do not change
their motions as the temperature goes down. Finally, here is the catch—it is
physically impossible to achieve absolute zero and the reason is the
Heisenberg uncertainty principle which says that you cannot simultaneously
know with absolute precision both the position and the speed of any
particle. So, if an atom is "at rest" you are totally ignorant of where it
is, it could be anywhere in the universe. Or, if you happened to know
exactly where the particle was, you would be totally ignorant of how fast it
was moving. So, the question might better be "what would happen if the
temperature of the universe were 1 degree absolute?" It would be a really
cold universe, no gases, just solids and a few liquids. There would be no
stars, no galaxies, no you, no me.

REVISED ANSWER:
I have been advised that my answer regarding the reason absolute zero
cannot be achieved is wrong, that the uncertainty principle is not the
reason. Here is the correction: "HUP has absolutely nothing to do with the 3d law of thermo. You use HUP in effect to say that there's some ground-state kinetic energy. Fine, but that has nothing at all to do with the reachability of T=0. So that part is completely wrong."
The third law of thermodynamics is essentially that absolute zero cannot be
achieved. Similarly, my definition of temperature is apparently not general
enough, particularly at low temperatures. Still, for the nonphysicist, I
think that average energy per particle is a good way to think about what
temperature measures.

QUESTION:
what is the relationship between pressure and density for a given amount of a gas at a constant temperature

ANSWER:
The ideal gas law may be stated as PV /NT =constant where N
is anything which specifies the amount of the gas present. Suppose that N is
the mass of the gas; then V /N =1/ ρ
and, if T is constant, Pρ= constant.

QUESTION:
I have a question about Einstein’s paper on The Special Theory of Relative (STR). In the introduction there are some references to phenomena that I have never heard of. Are there differences with what happens when moving the magnet verses moving the conductor? Is there a phenomenon where there is a difference in the electrical field around the magnet and a difference with an electromotive force without a corresponding energy? Or is Einstein saying that the customary view (I assume the Maxwell’s view) is wrong and that what is observable is the same for moving a conductor or moving the magnet?

ANSWER:
I believe that it was well known that it does not matter whether you
move a coil toward a magnet or a magnet toward a coil, in either case an EMF
around the coil will occur. But, it was puzzling because in the first case
the electrons in the wire are moving through a magnetic field and therefore
experiencing a force which propelled them around the coil and, if the coil
was open, caused a voltage across the ends; in the second case, there was no
apparent law of physics which said that if the field moves and the charges
stand still the charges will experience the same force as if it wer them
which were moving. Relativity takes care of this because there is no
preferred frame of reference and the two should be identical.

QUESTION:
Lets assume you have a charged particle that travels with velocity into a square region with a constant B field. By F=q(vxB) we know that the force is upward for negative charge or force is downward for positive charge. Also we know there will be a rotation due to the B field lines. When it leaves such a field, does it trajectory continue to "loop"? How does the velocity change? First i thought that maybe the force that acts on the particle outside comes from the lorentz force, but then i was not sure. F = qE + qvB, but outside B = 0 and there is no E field so E = 0. It doesnt seem reasonable that after the particle has experienced a B field it travels on a straight path with the same velocity as it was doing so initially (before entering the B field).

ANSWER:
First, you are using velocity wrong because velocity is a vector, it has
both a magnitude (called the speed) and a direction. So, if you have two
cars, one going north and one going south, both with a speed of 60 mph, they
do not have the same velocity. Acceleration is the rate at which velocity
changes and there are two ways that can happen: the magnitude can change
(speed up or slow down), or the direction can change (move in a curved path)
at a constant speed (or, of course, a combination of both). So, when a
particle enters a magnetic field which is perpendicular to its direction, it
experiences a force perpendicular to its direction and the direction of the
field. A force perpendicular to a velocity changes its direction but keeps
its speed constant. So, the particle will move in a circular path. So, the
charged particle will continue until it leaves the "box" at which time it
will move in a straight line with the speed it came in with but not in the
same direction.

QUESTION:
I noticed in a textbook that 'g' varies directly with distance inside the Earth's radius but is inversely proportional to it outside. I understand the reason for the latter but would like to know more about the former - thanks! Please keep it simple.

ANSWER:
Here is as simple as I can make it: First, the assumption must made that
the mass of the earth is uniformly distributed, that is, the density is
constant everywhere. This is roughly true on a large scale. Second, you must
know that Newton's universal law of gravitation says that the force of
gravity from a spherical object is proportional to the mass M of the object
you are being attracted to and inversely proportional to r ^{2}
where r is the distance to the center of the sphere. [You erred when
you said g is inversely proportional to r , it is r ^{2} .]
Third, there is one more thing you need to know, called Gauss's law which
states that only the mass inside where you are attracts you, the mass
outside of r
exerts no gravitational force on you. So, as long as you are outside, F~M _{total} /r ^{2} ;
but, if you are inside, only part of M _{total} , M _{partial} ,_{
} exerts the force on you, F~ M_{partial} /r ^{2} .
But, think about it for a minute: M _{partial } is proportional to
the volume of the smaller interior sphere and that is proportional to r ^{3}
(V =(4/3) πr ^{3} ),
so F~r ^{3} /r ^{2} ~r. [~ means
proportional to here.]

QUESTION:
I have seen videos where astronauts release a small handful of objects like M&Ms, and they immediately drift away from one another, why do they do this rather than staying clumped together?

ANSWER:
When they are released, it is inevitable that each will have some random
small initial velocity and, in the apparent absense of gravity, they will
all just keep moving exactly like they were when released. These speeds will
be rather slow but very apparent over several seconds. When on earth, the
same experiment is over in less than half a second as they fall to the
floor.

QUESTION:
If a block placed on an inclined plane having base length say,L,moves towards right by distance x,what will be the horizontal displacement of the inclined plane?Why would it be towards left?

ANSWER:
I assume the plane is free to move, there is no friction.
Before the block starts to move the system has zero momentum. If there is no friction, the only
way the system can have zero momentum after the block is released is for the two to move in opposite directions.
To know the distance, you need to know the relative masses. For example, if
the block mass is m and the incline mass is M , the plane will
move mx /M when the block has moved x .

QUESTION:
Since kinetic energy = 1/2 mv^2, it follows that, for a given mass, a doubling of velocity requires a quadrupling of energy.
It seems to be certain then, that 2 joules are required to accelerate a 1 kg mass from zero to one meter per second in one second (ie accel=1m/s2)

Change in kinetic energy is equal to work. W _{1} =½ x1x1^{2} -½ x1x0^{2} =½
J, not 2 J.

Now, according to the KE formula, to further accelerate the mass to 2 meters per second requires an additional 6 joules... because at that velocity, the energy now possessed by the mass is 8 joules - 4 times higher than when it was moving at 1m/s.

W _{2} =½ x1x2^{2} -½ x1x1^{2} =3/2
J, not 6 J.

Here is my question: (in the form of a thought experiment)
Scenario: I am in space
Let's suppose I use 2 joules of energy to accelerate the mass to 1m/sec, relative to me.
A spaceship, passing by with velocity 1m/sec and going in the same direction as the mass, has a second person onboard who intercepts that mass. (Because it's and the ship's speed are equal, the mass is effectively at rest with respect to the ship.
Now let that person - on the moving train - apply 2 joules of energy to again accelerate the mass to 1 m/sec relative to HIM. So the actual velocity of the mass is now 2m/sec, relative to the first observer (me) (Assume same direction...)
So, between us, only 4 joules of energy have been required to get the mass to 2 m/sec instead of the 8 implied by the KE equation.
I have no difficulty in grasping that energy goes up as the square of velocity, but for the life of me, I cannot understand why both observers cannot use 2 joules of energy to effect an increase of 1 m/sec each
Surely, what is true for one must be true for the other... ? (ie. that it only requires 2 joules to effect an increase in velocity of 1m/sec, relative to that frame)
I have posted this on a Physics Forum and the only responses I have gotten allude to the fact I am getting it all messed up by changing the frame of reference and hence, the numbers won't add up. They say KE is not frame invariant and I can understand that - but no-one can explain where my logic is wrong...
Can you possibly enlighten me ?

ANSWER:
You have made some errors in your preliminary calculations above which I
have corrected. Those who told you that kinetic energy is different in
different frames are certainly on the mark. But, I will show that the work
done in one frame is also not the same, that is, if you do W Joules
of work, somebody moving relative to you will see you do a different amount
of work, that is, the energy expended is not something everybody will agree
on. I am going to set up a scenario which I hope will show you where you
have gone wrong. The ship goes by with a speed of 1 m/s and the guy on the
ship exerts a 2 N force on a 1 kg mass, initially at rest in the ship, for 1 s. The acceleration is
therefore F /m =2 m/s^{2} . The distance (in the ship)
the mass travels is x =½ at ^{2} =½ x2x1^{2} =1
m and the speed it ends up with is v=at= 2x1=2 m/s. The work done is
therefore W=Fx =2x1=2 J. Now, view this whole thing from your
perspective. You still see a force of 2 N being applied, the mass at the
start has a velocity of v _{0} ' =1 m/s, the acceleration
is still a'=F /m =2 m/s, and the time the force acts is still 1
s. So now, the distance (in your frame) that the force acts over is
x' =v _{0} 't+ ½ a't ^{2} =1x1+½ x2x1^{2} =2
m and the speed at the end is v'=v _{0} '+a't= 1+ 2x1=3
m/s. The work done is therefore W'=Fx' =2x2=4 J. Note that for both
observers, the work done equals the change in kinetic energy: W =2
J=½ x1x2^{2} -½ x1x0^{2} =2 J and W' =4 J=½ x1x3^{2} -½ x1x1^{2} =4
J. You were willing to accept that kinetic energy was different in different
frames but failing to see that energy expended is different in different
frames.

QUESTION:
I need you to settle an argument between my wife and I.
We are traveling down the road at 60 mph. My wife throws a piece of balled up gum out the window. When it hits the ground, my wife thinks it will roll backwards. This is not possible and I know it will roll forward at a speed less than 60 mph due to friction.
Who is right?

ANSWER:
You are …unless, of course, it
bounces back from some obstruction like a big stone in the road or
something.

QUESTION:
If a 180 pound person "plops" onto an object (chair, couch, etc) forcefully by just letting their feet go out from under them, instead of sitting down properly, how much force is this creating on the object (equivalent to pounds)?

ANSWER:
As I have explained many times before, the
force is the change in momentum divided by the elapsed time. Momentum is
mass times velocity. Mass, in English units is weight/32, so 180/32=5.6 lb s^{2} /ft.
(Don't worry yourself about the units; using this for mass, seconds for
time, and ft for distance, things will come out in a sensible form.) The
easy one is lowering yourself gently into the chair. In this case you are
moving very slowly and there is very little change in momentum and the net
force on you must be zero. The forces on you are your own weight and the
force the chair exerts up on you, so the force the chair exerts up is equal
to your weight, 180 lb. Newton's third law says the force the chair exerts
up on you is the force that you exert down on it. Now, if you are moving
when you hit the chair, you will have momentum and that will be taken away.
You need to know how fast you are moving to start and how long it takes you
to stop. Just as an example, suppose that you hit the chair going 8 ft/s
(about how fast you would be going if you dropped from rest one foot) and
you stopped in
ј s. Then your change in momentum would be 8x5.6=45 lb s and the average
net force on you would have been 45/ј lb=180 lb. But there is your weight
down, so the chair exerted an upward force on him of 360 lb; so he exerts a
force down on the chair of twice his weight while stopping. This is just an
example and the stopping time is crucial; a soft chair will have a longer
stopping time and smaller average force and a hard chair will have shorter
stopping time and a bigger force.

QUESTION:
if black holes keep sucking in matter, wouldnt the entire universe eventually be sucked up in one big black hole?

ANSWER:
Not necessarily. If there were two black holes orbiting each other
outside their Schwartzfield radii, for example. Or, if there were two black
holes moving away from each other at speeds bigger than the escape velocity.

QUESTION:
what happens if the constant "c" in e=mc squared is faster than light

ANSWER:
The entire universe collapses instantaneously to a purple singularity.
JK. What happens is that you have a meaningless and incorrect equation.

QUESTION:
As a gas bubble rises in a column of liquid, it expands due to the decrease in pressure from the liquid above it. Intuition tells me that the diameter of the bubble will increase at a decreasing rate; but is that true? Does the bubble's diameter increase at a decreasing rate over time, does it increase linearly with time, or does it expand at an increasing rate?

ANSWER:
Assuming temperature stays constant, volume is inversely proportional to
the pressure. Since the pressure increases linearly with depth, the volume
of the bubble will decrease linearly with depth. But the volume is
proportional to the cube of the radius, so the radius does not decrease
linearly with the depth. I have derived the following equation to
approximate how the radius R varies with the depth d : R /R _{0} =^{3} √[P _{A} /(P _{A} +ρgd )]≈ ^{3} √[1/(1+0.1d )]
where
R _{0 } is the radius of the bubble at the surface,
ρ is the density of water, and P _{A } is
atmospheric pressure. The graph at the right shows the variation to a depth
of 25 m.

QUESTION:
The electromagnetic spectrum consists of so many waves which are made up of photons. We know that photons do not have mass. But they say that all the electromagnetic waves possess different energies. How does a massless photon possess energy?

ANSWER:
Well, even if you never thought of light as photons, you would not think
it had any mass. But certainly you would not think that light carried no
energy? It is well known in classical electromagnetism that electromagnetic
waves carry both energy and momentum. So, how does a photon carry energy?
The energy of a particle in special relativity is E= √(p ^{2} c ^{2} +m ^{2} c ^{4} )
where p is momentum and m is mass. So, if m= 0, E=pc . In quantum
mechanics, a photon has energy E=hf , so the momentum is p=hf /c ;
here, f is the frequency of the associated wave.

QUESTION:
If a marble and bowling ball free-fall at a constant rate, why does a car falling tilt toward the side of the engine (landing vertically)? In other words, if the marble and bowling ball were connected by a stick, would it still land flat (horizontally) or would it land with the bowling ball down and the marble up (vertically)?

ANSWER:
The marble and bowling ball fall the same only if you do the dropping in
a vacuum or do not drop them from too high an altitude. To simplify things a
bit (since you are interested in the effect of mass, not size), suppose we
drop two bowling balls but one with
ј
the weight W as the other. It turns out, to an excellent
approximation, each experiences the same amount of air drag at a given
speed. But, the bigger the velocity v , the bigger the drag; so,
eventually as the balls speed up they will reach a velocity where the air
drag is exactly equal to the weight and then the weight and drag cancel and
the ball falls with constant velocity after that. It is pretty easy to show
the velocity this happens at is v = √(W /c )
where c is just some constant determined by the shape and size of the
falling object. So, the heavier bowling ball ends up going twice as fast as
the lighter one. If they were tied together, the pair would have a net
torque on it trying to move it from horizontal to vertical and they would
end up going faster (and aligned vertically) than either would have gone
alone because c is determined by the area presented to the wind and
the pair falling vertically would be about the same as one ball with the sum
of their weights. I guess that tells you why the car falls motor first.

QUESTION:
Could you please explain Einstein's theory of time travel to me?
I recently watched a program that explained he postulated the theory based on the idea that if he moved away from a clock tower at the speed of light the hands of the clock would apear to slow and even reverse.
Though I am willing to accept that this may be true I would argue that he is only catching up to the light that was reflected when the hands were in that position and that he would not actualy be traveling backwards in time.
Can you help me to understand this idea better?

ANSWER:
There is no such thing as "Einstein's
theory of time travel". However, Einstein's theory of special relativity
shows us that the rate at which time moves forward is determined by the
motion of the clocks, it is not a universal rate for everybody. The result
of this is that it is possible to travel forward in time. For example, you
can go traveling out to space and back for 10 years, arrive back home, and
find that 100 years have elapsed here. To get the gist of this, look at the
earlier answer on the twin paradox .
As far as physics knows, though, there is no way to travel backward in time.

QUESTION:
I have just started a new job and have been asked to do a presentation on the ABS (Avalanche Bag system). Upon reading about the product they kept quoting that the system works on the "Law of Inverse Particles." I figure ok, it's a Law, it should be fairly simple to research, but when I started typing it into search engines I am getting no feedback on this Law at all. So my question is does this Law even exits? If so is there maybe another name that it goes by? If it does exist who created it? Any help that you can off me on this topic will be very much appreciated, in the mean time I'll keep on checking books, sites and asking old teachers!

ANSWER:
I have never heard of the law of inverse particles. However, just
looking over how this thing is supposed to work, it just looks like
Archimedes' Principle to me. Archimedes' principle says that when an object
is placed in a fluid it experiences an upward force, called the buoyant
force, which is equal to the magnitude of the weight of the displaced fluid.
Hence, if the object has a density less than the density of the fluid, it
will float, if it is greater it will sink. I am guessing that the snow in an
avalanche acts much like a fluid and, if you can significantly decrease your
overall density, you will "float" on the surface, much like a cork floats on
water. The way you decrease your density is to increase your volume with the
inflated airbags. Why they invent this screwy name for this escapes me, must
be an effective marketing technique.

QUESTION:
I am not getting the idea of length contraction.Does length of an object actually shrink when in motion or is it just apparent to the observer which is at rest.Is moving rods shrinking a measurement problem?

ANSWER:
First you have to define how you are going to measure the length of
something. In physics, length is the difference between the positions of two
points in space which are measured simultaneously. So, if a stick is moving
by you, you have to be careful to measure the positions of the two ends of
the stick at the same time. If you agree that this is a reasonable
definition of what length is, then the answer to your question is yes, the
object actually is shorter if it is moving. What is apparent to the observer
is an entirely different thing. How long a stick looks is not the
same thing as how long the stick is . For much more detail, see an
earlier answer .

QUESTION:
If two identical golf balls fall from the same height what differences (if any) would it make if one of the golf balls was spinning?

ANSWER:
Normally a projectile moves in a vertical plane because the only forces
are gravity down and air drag opposite the velocity direction. But, if the
ball is spinning, the air drag can be asymmetrical so there is a force
perpendicular to the velocity. This is how a curve ball in baseball works or
why some golf drives "slice" or "hook". So a spinning ball dropped might not
fall straight down.

QUESTION:
At the S.F. Airshow when a jet airplane reached a very fast speed over the bay an aura of a white circle, with black around the edge appeared momentarily around the speeding jet. What caused this display? Was it from the speed of sound? I have also noticed that when the planes fly by they seem to have a steam like trail coming off their wings momentarily. Is that the same as the white aura? What causes it?

ANSWER:
Did it look like the picture to the right? If so, it was a supersonic shock wave
which causes air vapor to rapidly condense. The trail off a subsonic jet is
not the same thing. Here the turbulence at extremeties causes the water
vapor to condense. This is what you usually see with passenger jets going
by; it is condensation, not exhaust fumes.

QUESTION:
I know that as something travels closer to the speed of light, time dialates for it, like at 90% time passes at a 7:1 ratio (7 days outside:1 day inside) and at 50% it's a 2:1 ratio (I think). My question is if the two ships, 90% and 50%, were travelling 1 lightyear to the same place, what would be the difference in their arrival time? Would one actually be faster than the other? Or would they just age differently on their respective ships but arrive at about the same time?

ANSWER:
The factor for time dilation is
√[1-(v /c)^{2} ].
At 90% (v/c=0.9), elapsed time on the ship, compared to one year on earth,
is 0.436 years. I do not know where you got 1/7. So, the time, as measured
on the ship, when one year elapses on earth is 0.436 years. If the speed is 50% the
speed of light , a year on earth is 0.866 years on the ship. The first ship
would arrive at its destination, as observed from earth, in 1.11 (1/0.9)
years, the second ship would arrive in 2 (1/0.5) years. So, the speedier
traveler would have aged 1.11x0.436=0.484 years, the slower traveler would
have aged 2x0.866=1.73 years.

QUESTION:
I'm trying to understand the permittivity and permeability of free space and I've simplified the concepts to mean the resistance of a vacuum to electrical and magnetic fields (respectively). Assuming that's more or less correct my question is, why are these two constants non-zero values?

ANSWER:
That is not "more or less correct". They are simply constants which
quantify the strength of electrostatic or magnetostatic forces in vacuum.
Once you have defined what a unit of electric charge is, the permittivity
tells you how strong the force between two charges 1 m apart is. Once you
have defined what a unit of electric current is, the permeability tells you
how strong the force per unit length between two currents 1 m apart is. In
practice, the ampere is defined and the coulomb is than defined as the
charge carried by 1 A in 1 s.

QUESTION:
In a theoretical situation where an object is free falling in a vacuum: without any wind resistance and consequently without a terminal velocity, what would stop the object from reaching the speed of light?

QUESTION:
Would an object in an infinite free-fall in a vacuum reach the speed of light or would its mass bring it to a maximum velocity?

ANSWER:
The question is what happens if a particle experiences a constant force
"forever". The reason it does not accelerate forever is that Newton's laws
(like F=ma ) are only approximately true and are not correct for velocities
comparable to the speed of light. Linear momentum has to be redefined. I
have answered this question before, so for much more detail see an
earlier answer .

QUESTION:
I shoot clay targets with my daughter.
I used to be pretty good at physics at high school but I've been puzzling over this for months without success so here goes.
I was trying to create a "lead" calculator so she could roughly calculate the amount to lead the target by to take a more accurate shot.
The more you look at it, the more complicated it gets.
Straightforward to start.
Take the middle station on a skeet field where the shooter is trying to break the target directly in front of them. Then we can assume that the target is traveling perpendicular to the direction of the shotgun. If the target speed is 20m/s & the shot velocity is 396 m/s (1300fps) & the target is 15m from the shooter how much lead do you need?
But then I started to think, the target is decelerating & ascending/descending, the shot is decelerating and the shot is not all in the same plane. It is spread out an the shot "string" is quite long.
It gets more complicated then as you move around the arc of the skeet field and look at the target, and shoot it from different angles.
The target speed & distance from the shooter is fairly constant in skeet from each station. ie. target 1 at station 1 is the same each time, 2 from station 1 is the same each time etc.
In simulated field target speed varies from station to station. Target size may change. Trajectory can be varied. Angle to the target also changes. I'm not even going to think about wind velocity.
My head hurts enough already.
Got any suggestions? Please?

ANSWER:
Maybe there is an app for it! Seriously, though, there is science, art,
and skill. Science is wonderful, and, given modern computing capabilities,
one could do the calculations you want but it would have to be done
numerically, not analytically, because it is a very complicated, real world
problem. But, what is the point of what you are doing? Isn't it to have fun
and to acquire a skill which comes from practice and lots of trial and
error? Why try to put that into an equation? You should just encourage her
to have fun and get better over time. Of course, explain to her what the
considerations are, but qualitatively, not quantitatively. You can't compute
a home run or a hole in one or a field goal or a three-point basket, you
gotta do it! Go blast those clay pigeons!

QUESTION:
I am considering an electrical motor, which can
also be used to generate electricity. From what I understand, an
electrical motor consists of coils of copper wire rotating inside a
magnetic field surrounding the rotating coils. My question is: take a
cylindrical object - a metal rod base with copper coils, like in the
center of a motor - in orbit around the earth, free from friction and
gravity, could we start this object rotating, with the center of
rotation being the metal rod going through the center, with the rod and
copper coils being placed within a hollow magnetic cylinder and both
objects orbiting at the same speed, so that the copper coils are
spinning within the magnetic field to produce a constant electric
current? This seems to be a source of near endless energy, but I am not
a physicist and surely I am missing something, I just don't know what it
is.

ANSWER:
Ah, the perennial question "why can't I get energy for free"! Forget all
the orbiting stuff, and just cut to the chase. Imagine a coil which we get
rotating in a magnetic field and there is no friction. Imagine that the coil
is not closed but has two ends which I will call the terminals. It is
happily rotating and will do so forever, provided there is no friction. And,
if I connect an AC voltmeter across the terminals, I will find that there is
a voltage there. But, voltage is not a measure of energy. However, I can use
a voltage to drive an electric current and I can use that electric current
to deliver energy to me, for example to make my breakfast toast. So, can I
make toast forever for free? No, because the instant I start taking energy
away from my motor physics is going to demand that the motor give it to my
toaster, the coil will therefore quickly stop after all the kinetic energy
it had has been given to my toaster. And, I haven't even gotten that little
amount of energy for free because I had to put that much in to start it
spinning in the first place.

QUESTION:
Does hot water freeze faster than cold water and if so are the ice cubes clearer?

ANSWER:
See an earlier answer . I have no idea about the clarity of the ice
cubes.

QUESTION:
In my textbook time dilation formula was calculated using a special clock with a light beam,mirror,detector arrangement.So it was easy to calulate mathematically.But can you calculate the same using a simple clock,biological clock or any other clock mathematicaly or should i accept it only based on experimental evidences?

ANSWER:
What you suggest is a pointless exercise. What is required is that you
accept the light clock to be a perfectly good clock which will run at the
same rate as any clock at rest with respect to it. Your watch would tick out
100 seconds exactly like the light clock would. An atomic clock would tick
out 100 seconds exactly like the light clock would. A bunch of bacteria
would double its population in a way that could be perfectly measured by the
light clock. A man would become 100 years old when the light clock ticked
out 100 years. The beauty of the light clock is that all clocks run with it
and it can be easily shown to depend on motion.

QUESTION:
While in a 0 gravity, vacuum environment you hold a 600kg weight.
You weigh approximately 80kg
If then you push against the weight, does the significant difference in mass between you and the weight affect the reaction?
A) you and the weight move away from one another equally
B) you move away from the relatively stationary weight.
Thank you in advance for ceasing this argument :)

ANSWER:
Nothing has weight in zero gravity. But you and the object have mass, 80
kg and 600 kg. You simply use momentum conservation to solve this problem.
If you push such that you leave with a speed v , then the speed u
the object has is computed as: 0=80v +600u , u =-(80/600)v=- 0.13v ,
that is, the object has a velocity about 13% of yours in the opposite
direction (that's what the minus sign signifies). Neither A) nor B) is
correct because I would not call 13% of your speed relatively stationary .

QUESTION:
If an object were ascend 10 feet, perpendicular to the surface it was on, and then return to earth through the force of gravity, would it land in the exact physical location from whence it left?

ANSWER:
Look, let's just say we drop it and ask if it falls straight down, straight
down being in a straight line toward the center of the earth. The answer is
only if you drop it at the north or south pole. The reasons are complicated,
but result from the fact that the earth is not an inertial frame of
reference, that is, it is rotating. I have ignored any effects from the moon
or the sun in this answer.

QUESTION:
My wife and I had an argument this morning over E=MC^2. She proposed the following thought experiment: If you push down on a spring, you have put energy into it. Correct? I said, yes. Correct. And, when you let go of that spring, it bounces, correct? I said: Correct. Then, she said this: Have you increased the mass of the spring? I said no. She said – well there, Einstein was wrong. You have increased the energy, but not the mass of the object, therefore, mass and energy cannot be equivalents.
SO… I said, well, you have increased the constant. And, she told me that I was a moron, because the constant is… well, the constant and does not change.

ANSWER:
Let's just talk about the pushing it, forget the bounce back part for
now. You have done work on it so, yes, you have added energy to it. Where is
that energy? It is, indeed, in mass. However, the increase in mass is so
small that you could never hope to measure it. Suppose you do 1.0 J of work
compressing the spring. Then you have added
Δm =1/c ^{2} =10^{-17} kg. So, it looks like you are
both wrong. It sounds like you were actually agreeing with each other but
she concluded that Einstein was wrong which seems to have offended you!
Incidentally, when you did the work to compress the spring, you actually got
lighter by the same amount the spring got heavier—total energy was
conserved. "The constant"—does that refer to the spring constant of the
spring? Compressing the spring does not change the spring constant which is
a measure of how stiff the spring is. If you let it bounce back it does work
on you and so loses the mass it gained.

QUESTION:
How might buckshot propagate in a vacuum? I've found scores of websites and posts where questions about firing a gun in a vacuum (the vacuum of space) have been addressed, but not a single one about shotguns. What I am trying to figure out is if the pellets will maintain a tight grouping in a vacuum or if they will spread in the same fashion as they would at one atmosphere. Will the shot cup expand without a substantial atmosphere? Will the shock of the gunpowder blast transfer enough transverse energy into the pellets thus causing the shot cup open. Will the shot cup travel at the same speed as the much heavier pellets? Will the gasses vented out of the muzzle continue to move forward or will they instantly dissipate into the vacuum? I've been trying to figure this out for almost two weeks now and every time I think I've got it, some other variable or contradiction comes up to complicate matters. Help me Obi Wan Kenobi.

ANSWER:
How anything moves is determined by two things: the position and
velocity it had at the beginning (initial conditions) and the forces it
experiences after the beginning. We will call the beginning when a pellet
emerges from the barrel. When it leaves the barrel, the scatter depends
mainly on the initial conditions; a longer barrel would tend to have the
pellots traveling more or less in the same direction (which is why sawed-off
guns have a bigger scatter) but, the farther you go from the gun the more
noticeable the scatter due to different directions becomes. But, there are
forces the pellots encounter —gravity
which makes them fall and air drag which slows them down. I think neither of
these tends to increase the spread very much because the air drag is
determined by the speed and they are all around the same initial speed. If
you took away all the air, the pellets would all go farther but, not spread
much differently (although going farther necessarily means spread more at
larger distances). If you took away gravity and air, each pellet would simply
go in a straight line with constant speed and likely spread about as much as
in air with our without gravity. So, my belief is that the spread is more
determined by the initial conditions (speed and direction when emerging from
the barrel) than anything else.

QUESTION:
Sound is what we here when a disturbance is created in air by a vibrating body. At microscopic level, we see that air molecules have very high speeds and they collide with each other. Our ear drum feels sound if a disturbance is created in air . But what about already moving molecules?? Even they are disturbed every second as they collide with each other that doesnt create sound but a small vibration does. Why so??

ANSWER:
When there is no sound, our ear experiences many tiny percussions very
closely spaced and the net effect is that there is a nice, smooth pressure
on the outside of your eardrum. But, inside your eardrum there is also air
at the same pressure. What your ear detects is variations in pressure. Sound
is a wave of pressure variation, so you ear is subjected to not a constant
pressure but varying pressure and, if the variations have frequencies within
our hearing range, our brain interprets the pressure variations as sound.

QUESTION:
If a person were thrown from a vehicle traveling a given speed, with no resistance (for arguments sake let's say they are riding on the luggage rack) what would be a good formula to determine how far they would travel before hitting the ground, assuming the ground is level?

ANSWER:
If they fall from height h , the car is going speed v , the
time from falling off to hitting is t , and g is the
acceleration due to gravity and, if air friction is truly negligible, the
person will go as far as the car, s , in the time t . So,
s=vt ; but h = Ѕgt ^{2}
so t =√(2h /g ); finally, s=v √(2h /g ).
For example, if h =2 m, v =60 mph=27 m/s, then s =27√(2x2/9.8)=17.25
m.

QUESTION:
Please can you explain me or tell me something energy-time uncertainty principle and what is del E and del t in it?

ANSWER:
Generally the uncertainty principle involving position and momentum is
pretty easy to visualize what it means —you
cannot simultaneously know the position and momentum of a particle to
arbitrary accuracy. However, there are other pairs of conjugate variables
for which the uncertainty principle occurs, two examples are angle/angular
momentum and energy/time. The energy/time uncertainty principle is a little
harder to get a grasp of, as you have found. There are two examples I can
think of which are examples of physical consequences of ΔE Δt ~h.

As you know, one
of the truths of classical physics is that the total energy of an
isolated system must be conserved. However, in quantum mechanics it is
perfectly ok for energy conservation to be violated, as long as you do
it for a very short time. So, suppose a photon with energy E is
suddenly emitted from a charged particle. This violates energy
conservation. But, if that photon disappears, gets reabsorbed or
absorbed by some other charged particle in a time on the order of Δt~h /E ,
that is ok. This gives rise to the so-called vacuum polarization where
particles pop into and out of existence even in completely empty space,
so we do not really think of a vacuum as containing nothing.

Another way to
look at ΔE Δt ~h is that to know an energy absolutely
perfectly, the measurement will take forever. So, the ground state of an
atom has a well defined energy because it never changes and is
"available" forever. So, apart from experimental uncertainties, you can
measure the energy of a stable atom perfectly accurately. However, the
excited state on an atom does not live forever, it decays and the decay
is characterized by a half life, τ _{Ѕ} . Therefore, if you
measure the energy of such a state by measuring the energy of the
emitted photon, if you make many measurements you will find a spread of
energies ΔE=Γ~h /τ _{Ѕ} . Because an excited atom
does not last forever, that energy state does not have a specific energy
and spectral lines are not really perfectly sharp. Very short-lived
states have very noticable widths. Measuring the widths of states is
often a more accurate way of measuring half lives.

QUESTION:
I was asked the question "How can anyone think Einstein could have been wrong about E=mc2 and the speed of light when we know that atomic bombs work based on the equation?" I did not think the atomic bomb was inherently based on the speed of light's importance in the equation, but I did not know for certain. Can you help? Does the fact that nuclear bombs exist and work serve as proof of the validity of E-mc2 and the speed of light? Or are the two in fact not related?

ANSWER:
The fact that the speed of light is the same regardless of the motion of
the source or of the observer is the starting point for developing the
theory of special relativity, its cornerstone. One of the results is that
E=mc ^{2} . What this says is that even if you are able to convert
a tiny amount of mass into energy, you get a hugely bigger yield of energy
than you can from, say, chemistry. The atomic bomb is based on nuclear
fission which is heavy nuclei breaking in two. When that happens, you end up
with two lighter nuclei and a few neutrons. If you now measure the mass of
all the fragments you find that their mass is less than the original
nucleus, mass has been converted to energy. The H bomb (and the sun) is
based on nuclear fusion which is light nuclei fusing together —again
the final fused nucleus has less mass than the two which combined to make
it. You can read an earlier answer if you want more detail on
fusion and fission .
There are a huge number of other experiments which have verified special
relativity and therefore the constancy of the speed of light.

QUESTION:
I'm trying to get a grip on quantum mechanics. What was the original question theoretical physicists such as Planck, Heisenberg, Bohr, Pauli & Schrodinger had that lead them to the uncertainty principle. (If Einstein never could accept it.... do I stand a chance?) Do the particles in question have to be traveling at high speeds for this question to meaningful? Why would their be uncertainty of measurements of a car hitting a solid wall at 60 mph.

ANSWER:
The uncertainty principle is the inevitable result of the deBroglie
hypothesis — that
particles sometimes behave like waves. It turns out that the frequency of a
wave is related to its linear momentum. In order for a wave to have a
well-defined frequency, it must be infinitely long. That is, if you take a
wave train made up of only 10 waves of wavelength
λ _{0} , it turns out that the "frequency" is really
a distribution of frequencies which is peaked at where you would expect it (f _{0} =v / λ _{0} )
but really contains all frequencies. Mathematically, this comes from the way
in which momentum and position are related, called a
Fourier transform .
Note that the position of the "particle" in my example is uncertain by an
amount 10λ _{0}
and the frequency, and hence momentum is uncertain by an amount reflected by
the width of the distribution of frequencies. As you make the length of the
wave train longer and longer, say 1,000,000 wavelengths, your uncertainty in
the position gets bigger, the distribution of frequencies gets narrower and
if you make the wave train infinitely long, the frequency becomes just the
single frequency f _{0} . So, for this situation, you know
momentum exactly but are totally ignorant of position. If you try to know
position better by shortening the wave train, the distribution of
frequencies gets broader such that if you know position exactly, you are
totally ignorant of position. The speed of the particle is not relevant for
understanding the uncertainty principle, that is in relativity where you
have to worry about that. None of this is relevant to the car hitting a wall
because the particle wavelength of a macroscopic object is so incredibly
small that it behaves like there is no uncertainty principle.

ADDED QUESTION:
why were these scientist so smart and at such a young age? It seems like the best of them were Jewish/Germans, e.g. Einstein and Oppenheimer .

ANSWER:
Well, one reason that young guys have an advantage if we are talking about revolutionary breakthroughs is that they are not so brainwashed with the status quo. Why Jews and Germans? Because, in the 19th century these cultures respected and rewarded intellectual activity, including science, unlike the antiscience and antiintellectual bozos today (read Tea Party).
Bohr was Danish, Fermi was Italian, Yukawa was Japanese, Rutherford was from
New Zealand, Feynman was American, so Germany did not have a monopoly.

QUESTION:
Imagine I have a 10000 ft long water filled vertical tube (1" ID though I dont beleive it matters) and I have a cap on both ends. If I now remove the cap on the bottom of the tube how much fluid will come out due to density chages in the water column (assuming no air comes into the tube thru the bottom)

ANSWER:
Your question is not about compressibility. Water is extremely
incompressible, only a tiny change in density occurs for a huge change in
pressure. The simple fact is that atmospheric pressure, about 15 psi, is
only able to hold up a column of water of about 34 ft. So, as soon as you
unstoppered the bottom of the column, water would flow out until the column
was 34 ft high (with the remaining 9,966 ft a vacuum). This has nothing to
do with compressibility of water. You have made a barometer.

QUESTION:
I have been conducting some simple experiments in my local park (i have been launching carbon-fiber rockets of between 3-6kg using compressed air out of a tube) and have come across a confusing contradiction relating to the projectile in motion:
with a near constant starting velocity (speed and angle of projection) and near constant shaped objects, the distance they are thrown varies with the mass of the objects in a positive manner (that is, make them heavier, they go further). I am guessing this is because Ke=Mass x Velocity squared, therefore the heavier ones, having more energy, travel further, even though the angle and speed are more or less constant. However, I have noticed that after increasing the speed above a certain point, the relationship between increased mass and distance travelled has a negative relationship ie. the distance starts decresing if I increase the mass too much. Why is this?

ANSWER:
I guess my first question is how do you get constant velocity using
compressed air? Surely you need either bigger pressure or longer tubes for
the more massive rockets. Anyhow, assuming they are all launched with the
same velocity, the reason the heavier ones go farther is explained in the
answer just two answers down from yours:
they all experience about the same air drag force but this force is more
effective in slowing the lighter ones because of inertia. Regarding your results
if you continue to increase mass, I would guess that you simply do
not give the big mass nearly as much initial velocity as the lighter mass.

QUESTION:
Suppose there is a completely white room with a source of light. The room is white, therefore the light from the source will, at least mostly, reflect off all it's surfaces. Now let's say the light was suddenly turned off and the room gets dark; well if all that light from when the light source was on was reflected and not absorbed, shouldn't the room still stay lit? But it doesn't, so where does the light "dissapear" to?

ANSWER:
There is no such thing as a "completely white room" which reflects all
the light; even really good mirrors reflect only on the order of maybe
90-95% of what hits them. But, let's say we have an incredibly white room
which reflects 99.9% of the light which hits it. Now, let's say the walls of
the room are 3 m apart. Then, since the speed of light is 3x10^{8}
m/s, the time light takes to cross once is 10^{-8} s. So, in 1 ms=10^{-3}
s, the light bounces back and forth 10^{-3} /10^{-8} =10^{5}
times, each time losing only 0.1% of its intensity. So, how much is left
after 1 ms? 0.999^{100,000} =3.5x10^{-42} % of the original
intensity. That is really gone!

QUESTION:
Why do I go faster than my lighter weight friends when on a bike? I have noticed this downhill, but also on flats, I'm pedaling less. Assuming we both have top of the line bikes (we do). I thought things fell at the same rate regardless of mass. I think momentum and friction have something to do with this, but can you remind me of the principles involved here?

ANSWER:
If it were not for air drag, everybody should be pretty much equivalent.
However, the air drag F depends mainly on the cross sectional area
A presented to the wind and the speed v ; a useful approximation I
often use in these answers is F ≈јAv ^{2} .
(This works only if you use SI units.) So, assuming that you are not a whole
lot fatter than your friends, you all experience the same amount of drag
force. But, wait a minute, the acceleration is given by Newton's second law,
a=F /m so, if you have a mass bigger than somebody else, your
acceleration will be smaller, that is you will slow down more slowly. What
it boils down to is that you have more inertia.

QUESTION:
I recently heard about the clever creation of Hydrogen 4.1 and its use to study reaction kinetics. However, I cannot find an explanation in any of the reporting, or original Science paper on why the muon orbits so close to the nucleus beyond "it's 200x heavier than electron." Surely they did not apply gravity to a Bohr model of a muonized atom, so why does the muon orbit closer?

ANSWER:
I actually never heard of hydrogen 4.1. Apparently what it is is a helium
atom with one of the two electrons replaced by a negative muon. The radius
of an orbit is inversely proportional to the mass in the Bohr model of the
atom. It has nothing to do with gravity, it essentially enters from the mass
in Newton's second law. So the muon, having about 200 times larger mass has
a ground state orbit 200 times smaller. So, you may think of the "nucleus"
as being of atomic mass of around 4 amu and charge of +1. Therefore this
will look like a hydrogen atom with a much bigger mass.

QUESTION:
In a simple series resistance circuit current remains constant through out the circuit.A resistor simply reduces the current drawn by the circuit.Just at the instant when the switch is pressed current is already determined by the value of the resistors,even before it encounters a resistor.I think the resistor should alter the value of velocity of charge & therefore current but current remains same even after encountering resistor ?

ANSWER:
If the current were different in the resistor than it is in the wires
connected to it, charge would be building up in the resistor. But, the value
of the current is not solely determined by the drift velocity. Also
important are the number of charge carriers and the geometry. So, almost
certainly, the velocity in the resistor will be different, but the current
will be the same.

QUESTION:
Why is it that if you jump from a airplane you will stop acceleration at some point of going down?

ANSWER:
When you move through the air, you experience a force due to air
friction. This is a force which depends on how fast you are moving, the
faster you move, the bigger the air drag. Usually it is a good approximation
to say that the drag force is proportional to the square of the speed —if
you double the speed you quadruple the force. So, when you first jump from
the plane the forces on you are your weight (down) and a small air drag
(up); since the net force is down, you continue accelerating down. As you
speed up, the air drag gets bigger so your acceleration gets smaller.
Eventually, you reach a speed where the air drag is equal to your weight and
you then move with constant speed called the terminal velocity. You can
derive an expression for the terminal velocity: weight is mg and drag
is cv ^{2} where c is a constant characterizing the object falling (c big
for a parachute, little for a marble). So, v _{terminal} =√(mg /c ).

QUESTION:
My wife bought a medical device that claims to heal joint pain by emitting photons into the affected area is it possible that such a device purchased at the holistic pharmacy could actually shoot photons into your body?

ANSWER:
And, I bet she paid a bundle for it! A flashlight will shoot photons
into your body. Lying in the sun will let you get a bunch of photons. For
joint pain, heat is probably good and the purchase of a simple infrared heat
lamp would probably be just as good as this device and a lot cheaper. On the
other hand, if this thing does not plug in and create heat or light you can
see, like the copper bracelets you can buy, it is likely a fraud. Also, see
a recent answer .

QUESTION:
If antimatter and matter collide, they would annihilate each other. The definition of annihilate is to destroy completely. If these two meet and annihilate each other, wouldnt that make the law of conservation of matter - which states that matter cannot be created or destroyed, only changed - obsolete or false?

ANSWER:
I periodically get questions regarding the "Law of Conservation of
Matter". In fact, there is no such law; it has been understood for more than
a century that matter can be turned into some other form of energy or
vice versa (E=mc ^{2} ). Where does this misconception come
from? I guess that it is taught in chemistry courses where the change in
mass of something, to or from which energy has been added or taken, is a
superbly good approximation. If matter were conserved, we would not be here
because that is where the energy from the sun comes from —mass
disappearing.

QUESTION:
Can I have a non - mathematical, and simple intuition on why the speed of light is constant? Thank you.
(I am 13 years old.)

ANSWER:
First, do not expect it to be intuitive. Our intuition tells us that if
we run with a flashlight that the speed of light is what it was when we
stood still plus however fast we are running. Nevertheless, the speed of
light is always the same, regardless of the motion of the source or the
observer. I have answered this question many times, nonmathematically, which
you could have found on the FAQ page.

QUESTION:
I have been reading up on general relativity and i have a question regarding it. If as matter approaches the speed of light it compresses along the axis it is travelling in (please correct me if i got this wrong). why wouldnt the same be true for time? let me clarify.. if travelling at near the speed of light through time, why doesnt your displacement in time compress, or conversely if you are travelling slowly through time but quickly through space, why wouldnt you have a larger displacement in the time dimension? Basically, i am asking, since matter displaces space, and travels through time, why doesnt photons, or electrons displace time and travel through space? from what i have read (so far) i dont see anything that touches on displacement, and i dont think this violates anything (yet).

ANSWER:
This is not general relativity, it is special relativity. Rather than go
through your rambling, let us just say that moving clocks run slow. To
understand why, see the FAQ page.

QUESTION:
When we say we consumed electricity at home ?
what actually we consume Voltage,current or energy of the charge flowing ?

ANSWER:
You pay for energy and that is what you should think about consuming.
The electric bill is billed at some rate per kilowatt-hour. A watt is a
joule per second, so 1 kW-hr=1000 J hr/s (3600 s/hr)=3,600,000 J.

QUESTION:
Here is the scenario: a pool ball is traveling at certain pace with an enormous amount of spin on it. Once it hits the rail at about a 45 degree angle is it possible for it to shoot off of the rail and travel faster than it was going before when it was on it's way to the rail?

ANSWER:
It does not even need to have "an enormous amount of spin", a modest
amount will probably do it. The ball, coming into the rail, has
translational kinetic energy and rotational kinetic energy. The friction
between the ball and the rail can transfer some of the rotational energy to
kinetic energy, so after the collision it will have a higher speed and a
smaller spin.

QUESTION:
If you take a rock and weigh it, then take the exact same rock and crush it into fine powder and weigh all the powder (assuming no powder escaped), will the powder weigh the same as the solid rock?

ANSWER:
Two possible answers:

Classically, you have the same number of atoms
after and before, so you have the same amount of mass.

Relativistically, you had to do work (W )
to crush the rock and so you added energy to it. Therefore you added
mass (Δ m ) when you crushed
it,

Δm =W /c ^{2} where c is the speed of
light. Suppose you did 30 J of work. The change in mass would be 10^{-15}
kg.

QUESTION:
Why is there angular momentum in empty space? I.e., if there are no frames of reference in empty space, how does a spinning top "know" it's spinning?

ANSWER:
A constant velocity in an inertial frame of reference can always be
viewed from another inertial frame in which the velocity is zero. So, we say
that velocity is relative, absolute rest has no meaning. Why can't we extend
this to rotating systems? Quite simply, because if we view the rotating
object from the frame where it is at rest (which would be rotating with it),
it is not an inertial frame, that is, Newton's laws are no longer valid.
There is such a thing as "absolutely at rest" in a rotational sense. If an
object has an angular velocity in one inertial frame, it has the same
angular velocity in all inertial frames.

QUESTION:
How do I calculate the impact force of an 800 gram object dropped 2m?

ANSWER:
Ah, the perennial question which demonstrates how folks do not
understand force! See below .

QUESTION:
Would a "fuzzy" tennis ball make any difference in the distance the ball travels?

ANSWER:
The fuzz on a tennis ball plays the same role as dimples on a golf ball,
to reduce air drag. So, indeed, a fuzzy ball will travel farther than a
smooth one with the same initial speeds.

QUESTION:
A thermos bottle consists of two glass vessels, one inside the other, with air removed from the space between them. The vessels are both coated with thin metal films. Why is this device so effective in keeping its contents at a constant temperture?

ANSWER:
Removing the air prevents heat loss or gain by conduction or convection,
although the vacuum is never perfect. There is still the possibility of heat
loss or gain through radiation, and the silvering reduces this.

QUESTION:
if earth's mass was , say 10 times lighter, would the gravity be 10 times weaker?
if so , why as it would still occupy the same amount of space.

ANSWER:
Yes.
The gravitational force on a mass m a distance R from the center of a spherically symmetric mass distribution
M and at or outside the surface is given by GmM /R ^{2}
where G is Newton's universal gravitational constant. It has nothing
to do with the size of the sphere. So, if G, m , and R stay the
same, making M 10 times smaller makes the force 10 times smaller.

QUESTION:
To what degree does buoyancy affect actual weight as opposed to apparent weight? The support for a 2nd story soaking tub/shower takes into account the weight of a person standing to take a shower and the weight of however many gallons of water it holds, but does the weight of the person plus the weight of the water need to be calculated for the amount of building support necessary or does buoyancy mitigate the person's actual weight?

ANSWER:
The weight of tub, water, and bather must all be supported by the floor.
True, the weight of the floating bather is supported by the water, that is
the water exerts an upward force on the bather equal to her weight. But,
Newton's third law says that if the water exerts an upward force on the
bather, the bather exerts an equal and opposite force down on the water. So,
your question would be like asking: if we have a 5 lb box on top of a 10 lb
box, does the floor only have to hold up 10 lb because the 10 lb box holds
up the 5 lb box?

QUESTION:
when electric field is applied in a circuit ,charges experiences a force in a direction opposite to it.So why doesnt current accelerate in a circuit?why do they move with constant drift velocity?

ANSWER:
Because the electric field from the potential difference is not the only
force on the electrons. The wire is full of atoms and the electrons keep
bumping into these, getting stopped or scattered, then accelerated again,
etc . So it is a constant stop-start process which averages out to some
average drift velocity.

QUESTION:
Please can you tell me how fast force is transmitted in a solid object? I am asking as I am wondering if it is possible to transfer information faster than the speed of light using a piece of string?! That is, for a (very) tight piece of string between two points A and B, when you pull the string at point A, does it not move instantaineously at point B?
In answers by others I have been told it relies upon the speed of sound in the material - if you agree with this then please could you elaborate upon this for me?

ANSWER:
When you pull or push on the end of a long rod or string you are
actually only pulling or pushing on those atoms you are holding on to. Then
(let's say we are pulling, as in your example), the bonds between those
atoms and the next one down the string get stretched and that stretch (think
of all the atoms as being attached to their neighbors by tiny springs) pulls
that group of atoms toward you, but that stretches the next layer of atomic
bonds and so a pulse travels down the string. That is what sound in a string
or a rod is, compression/stretch between bonds traveling through the solid.
So, the information traveling down the string travels much slower than the
speed of light, the speed of sound. You could have read an earlier
explanation if you had gone to the FAQ
page.

QUESTION:
Is there a measure of time that does not make use of, or reference any planetary body?
As an example, a day on earth is based upon its period of rotation as a year depends upon how long it takes to make a circuit about the sun. I am curious if there is a system of time measurement that does not use of any planetary reference.

ANSWER:
The unit of time, which is the basis for all time measurements, the
second, is not based on an astronomical measurement. Widepedia says "Since 1967, the second has been defined to be
the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom."
Atomic clocks are used for precision time measurements nowadays.

QUESTION:
My question relates to atomic theory in regards to how I was taught that a nucleus is surrounded by electrons depicted by concentric circles. However I was listening to a Pink Floyd song on Youtube and in the
video it depicted the Flower of Life symbol in a unique way. It struck me the Flower of Life could be another way of explaining atomic theory. Notice in the video how the electrons zip through the nucleus and reaching various atomic levels in a more believable manner. This depiction seems more real than what I was taught in grade 12. It shows both wave and particle in its depiction. Your thoughts would be appreciated.

ANSWER:
This is less real than what you were taught in school. Admittedly, well
defined electron orbits which go around the nucleus are a simplistic picture
of reality, but such a model (the Bohr-Sommerfeld model) was quite
successful in many respects and led to better descriptions of atoms,
particularly those using quantum mechanics. This picture, as you note, has
the electrons moving in well-defined orbits which pass throught the nucleus;
although there is a small probability that an electron could be found inside
the nucleus, your picture would have us believe that all the orbits passed
right through, simply not true. The best way, as determined from solving
Schrodinger's equation, to think about the electrons is as clouds
surrounding the nucleus. Where the cloud is dense, the likelihood of finding
an electron there is high and where it is thin the probability is low. See
an earlier answer .

QUESTION:
While the vehicle is sitting on the launch pad at zero velocity place a
detector that will light when the photons hit it, at the position the
photons hit the sidewall.
Launch the vehicle and accelerate to 0.5 c. After reaching a constant speed
of 0.5 c cause the laser pen to fire the photons.
Will the photons hit the detector and cause it to light or will the photons
travel in a straight line from the point emitted and miss the detector?

ANSWER:
Hit it.

FOLLOWUP QUESTION:
Now could you please explain what causes the photon to travel 2 feet in the direction of the vehicle while traveling 4 feet independent of the emitter across the vehicle?
If the laser pen is pointed in the direction of the travel of the vehicle the speed of the vehicle can not be added to the speed of the photon.
So why can the speed of the vehicle be added to the photon when the photon is traveling at a 90ï¿½ angle relative to the travel of the vehicle?

ANSWER:
Your example is a variation of the old
light clock problem. In the
spaceship, you have no way to know you are moving, so it is a no-brainer
that the light goes straight across the ship and hits the detector, just
like you aimed it. In the earth-bound frame, the ship moves with speed c /2
but the light must still hit the detector (it cannot both hit and not hit).
So the path it travels in the earth-bound frame is, as you note correctly, 4
ft across and 2 ft along, a distance of
√20 ft>4 ft. But, as relativity postulates, the speed as seen by the
earth-bound observer is still
c
and so it takes longer than the time the guy on the ship measures. This is
time dilation, moving clocks run slow.

For completeness we should briefly discuss velocity addition and how the
speed of light remains constant even if the light is not originally along
the direction of travel. In fact, velocity addition equations in special
relativity are not just for velocities parallel to the direction of the motion
but also for velocities perpendicular to the direction of motion but, unlike in
Galilean relativity, the two are not of the same form because there is
length contraction along the parallel direction but not along the
perpendicular direction but both directions have time dilation. Most
elementary textbooks do not address this. The usual (parallel) equation is
u _{x} ' =(u _{x} +v )/[ 1+(u _{x} v /c ^{2} )]
but the corresponding equation for transverse components is
u _{y} ' =γ u _{y} /√ [1+(u _{x} v /c ^{2} ) ]
where
γ= 1/ √ [1-(v ^{2} /c ^{2} ) ].
Using these, you can take a beam of light
which you have going in some direction (with speed
c ,
of course) and show that, in another coordinate system moving with speed
v
in the
x
direction in the first, the speed is still
c
but that it does not go in the same direction unless the light was
originally in the
x
direction.

QUESTION:
I have a question regarding special relativity. I
was pondering over a thought experiment that I came up with using a
modified version of a light clock. This modified light clock, with every
tick, sends a signal to a space ship which is moving towards it at a
uniform speed. This is where I get a little confused. It would seem that
since the spaceship is moving towards the light clock that the frequency
of the signals being received would increase. Intuition would tell me
that the rate of the ticks has therefore increased. And yet I find it
hard to reconcile with the fact that, since the light clock is in motion
relative to the spaceship, that the ticks ought to be slowing down. I
realize that my argument is flawed somewhere as special relativity has
already been proven through countless experiments. I ,however, need help
understanding where exactly that flaw is. Do you have any thoughts on
this?

ANSWER:
You have stumbled upon, as many of my loyal readers will tell you, one
of my favorite points of emphasis in special relativity. How things are
is not necessarily how things appear to be . To answer your question,
read my explanation of the twin
paradox where I emphasize that the rates of clocks as you see them is
not the same as the rate at which clocks are ticking. See also
another answer . Time
is defined as the time interval between two events at the same point in
space and the clock coming toward you is not at the same places when it
sends you sequential ticks, so the ticking you observe is not the time
interval between ticks on that moving ship. If the space ship is going away
from you the observed frequency of ticks would be less than the real
tick rate. The same kind of phenomena are observed for length contraction.
Length is defined as the distance between two points in space measured at
the same time. But if a stick is moving toward you, it appears longer
than its rest length even though it is actually shorter and the stick moving
away from you looks shorter than its actual contracted length. Again,
see an earlier answer and links
there.

QUESTION:
I was just thinking ... I'm about to fix up a very old bicycle that has one of those generators on the rear tire which supplies power to the headlight. Does anyone out there wonder if something similar could be applied towards a hybrid automobile? Perhaps it could be installed in a switching primary/secondary battery powered vehicle to give a charge to one while the other provides power to the vehicle.

ANSWER:
Well, this is exactly what cars have been doing for generations, that is
what the alternator or, on older cars, the generator does, generate energy
and store it in a battery. However, your idea of having two and have them
swap off isn't realistic because somewhere something has to put energy into
the system (to make your car move in the first place). On the bike it is you
putting energy in, in a conventional car, it is the engine putting the
energy in. A hybrid goes one step beyond and that is to get energy from the
braking of the car. In conventional cars, when you brake you use friction
and all the kinetic energy of the car (energy it has because it is moving)
gets turned into heat (brakes get very hot) and is just lost. In hybrids,
the same ideas as generators allow us to capture that kinetic energy and put
it into the car's batteries.

QUESTION:
What would you consider to be a good representation of the distance between the earth and the moon, and the relative scales involved, in terms of household objects, with regards to a total arm span of 6 feet, so as to be easily replicated without the need of a measuring tape?

ANSWER:
Let's work in units of the radius of the earth, R _{E} .
With this length unit, the radius of the moon's orbit is about R _{O} =60R _{E}
and the radius of the moon is about R _{M} = јR _{E} .
Now, you want
R _{O} to be scaled to 6'=72"; then R _{E} =72/60=1.2"
and R _{M} = јx1.2=0.3". So
hold say maybe a golf ball in one hand and a marble in the other.

QUESTION:
In a step up transformer,the number of windings in the secondary side is greater than in the primary side.And this increases the voltage by decreasing the current(satisfies law of conservation of energy).how is this even possible if we consider this in terms of work done?How can a huge potential create only a small current?Rate of flow of electrons should be proportional to the work done?Or where is this huge voltage hiding without producing current?

ANSWER:
Why do you associate huge potential with huge current? If you connect
nothing across the terminals, you have a huge potential and zero current. If
you connect a huge resistance across them, you will have a small current. If
you connect a small resistor across them, you will have a big current. But,
in every case you have to ask whether you have the energy which allows that
to happen. In the transformer, if you ask for more energy out (by forcing a
large current) than you are putting in, the machine will refuse to oblige
you. That is, the potential will simply drop. Here is a simple DC example.
Suppose you have a capacitor which has been charged up to some high voltage.
Can you simply attach a resistor across it and use it as a battery to give
you a big DC current? Of course not because when you put energy into the
capacitor by charging it, you only put in so much and you can only get that
much out.

QUESTION:
I have a question I can't seem to find. I was wandering what the orbit would look like if there were three objects of equal mass all orbiting each other. I can only find simulations of two objects orbiting each other. So my question would be first, is this possible? and if so, what would the orbit look like?

ANSWER:
Your question cannot be answered because the three-body problem cannot
be solved in closed form. There are special cases which can be solved, for
example if one of the three is held in place and the others move around. I
believe that the orbits are chaotic, that the objects do not have single
closed orbits. And the behavior which could be simulated depends sensitively
on the initial conditions. That is not to say that there cannot be such
systems in nature. A recently discovered planet has been found to orbit a
binary star system. But if you looked at the planets orbit this year, it
would not be the same shape as next year. For more information, google
"three body problem".

QUESTION:
How do moving electrons or charge particles produce their own magnetic field?
For instance when the current in the wire produces its own magnetic field. How is it happening?

ANSWER:
Well, in classical electromagnetism, where the electric and magnetic
fields are treated as different things, it is just an empirical fact that
magnetic fields are produced by moving charges. But, in relativistic
electromagnetism, electric and magnetic fields are just components of the
electromagnetic field tensor, all just a single field. Then, if you start
with a tensor which has only electric fields in it (that would be for a
charge at rest, for example) and then transform the tensor to a frame which
is moving, the transformation causes magnetic field components to appear.

QUESTION:
I'm designing an experiment for my physics teacher, involving gas density and how it effects a ball's acceleration. The problem is finding a ball that is light enough to be effected by the change in density, but heavy enough to fall no matter what. Would a tennis ball be appropriote?

ANSWER:
In order for the ball to not fall, the weight of an equal volume of your
gas would have to be more than the weight of the ball; there is always a
buoyant force when you put something in a fluid (Archimedes' principle). I
do not know how big pressures you propose using, but I would think something
less massive than a tennis ball might give you results easier to measure.
When you are analyzing your results, be sure to include the buoyant force in
the forces acting on the ball.

QUESTION:
If neutrinos are traveling through the earth what distance from each other would be typical? Can they be "seen" say at a light spectrum not typically seen by the human eye.

ANSWER:
The solar flux at the earth is about 10^{11} neutrinos/cm^{2} /s,
100 billion pass through a postage stamp in one second! If you really care,
you can compute the average distance between them but it is not that
interesting a number, flux is what physicists care about. They are not part
of the electromagnetic spectrum (like x-rays, gamma-rays, radio waves,
etc . are). That is, they are not photons, they are neutrinos; a neutrino
interacts extremely weakly with matter whereas photons react very strongly.
As a result, neutrinos are notoriously hard to detect. It was on the order
of 50 years between the time we understood that neutrinos exist until the
time when we actually directly observed them.

QUESTION:
is it possible to see an image of the past (ancient earth) reflected from a surface in space millions of light years away. if no then how far away can an image be reflected and still be seen with some degree of clarity?

ANSWER:
The issue would be intensity. Since the light from earth spreads out, by
the time you get "millions of light years
away" there will not be enough photons per square meter per second to form
an image; like, with a single photon you could not tell if it came from a
dinosaur or a bowl of jello. I have done some rough estimates in an
earlier answer for tv
signals being received at a much closer (4 light years) distance and found
the size of the detector to be impractically large.

QUESTION:
Is it possible for a projectile to travel in more than one direction, while in a single unobstructed flight ?
I.E. Vehicle traveling at 30mph, an antelope running parallel to vehicle at say 50yds. distant and also moving in same direction at 30mph. A firearm is aimed at the antelope from in or outside the car. A firearm is discharged toward the antelope with the projectile leaving the weapon at say 2,000fps. The firearm is exactly 90o from front of vehicle and nose of antelope. Will the projectile move at the speed of the car and equal speed of antelope once it leaves the muzzle of firearm, therefore having it move forward with car and animal at 30mph, while still moving at 2000fps in direction of the antelope?

ANSWER:
I will assume there is no appreciable air drag on the bullet; that is
probably not true but is not the issue I believe you are trying to address.
When you shoot the gun it gets the speed relative to the barrel of the gun
but it also gets the speed of the car, 30 mph=44 fps. So the velocity is a
tiny bit bigger 2000 fps and has a component which carries it forward with
the speed of 30 mph. Therefore, the bullet, if you have corrected for
gravity, should hit the prey. The path still will be a parabola in a
vertical plane.

QUESTION:
If 2 people the same size and weight are racing the exact same car on a flat service, but one car weighs 10 pounds lighter, which car would win? We think it's the lighter car, but my husband says the heavier car because it would get more traction.

ANSWER:
The thing which drives a car forward is static friction between the
drive wheels and the road. There is a maximum amount of force you can get
and if you try to get any more your wheels start to spin. The maximum force
you can get on a level road without slipping is f _{max} = μmg
where mg is the weight (m is the mass, g is the
acceleration due to gravity) and μ is a constant which depends on the
type of tire and type of road. The acceleration a of a car with a
forward force of
f _{max} is a = f _{max} /m =μmg /m= μg .
Now, the heavier car will be able to get a bigger forward force, but a
bigger force is needed to give it the same forward acceleration; so, you can
see that the acceleration of the car turns out to be independent of its
mass.
So, as you see, it will be a tie.

QUESTION:
I have heard that Newton's laws fail at some points like at the speed of light or at the quantum level. Please explain how they fail in the above cases.

ANSWER:
Newton's laws are true to an excellent approximation for everyday
situations — cars,
planets, moons, satellites, baseballs, etc . However, they turn out to
be pretty useless if you try to apply them to situations where sizes are as
small as atoms and smaller. However, the important conservation concepts
(conservation of energy, conservation of momentum, conservation of angular
momentum) originally derived from Newton's three laws turn out to hold to
all known extrapolations. Newton's second law, F =ma , turns out to
not be true at high speeds, speeds at appreciable fractions of the speed of
light. But, here is the interesting thing: Newton never wrote F =ma ,
rather he expressed it as F = dp /dt , which,
with a judiciously chosen definition of p which approaches
mv for everyday speeds, Newton's second law turns out to be
true at all speeds. Newton's third law as usually stated turns out to not be
true for electromagnetic forces; but, if you generalize the law by including
the energy and momentum densities of the fields, all is well. The concepts
of energy, momentum, and fields, all having their roots in Newtonian
mechanics, continue to be important in physics.

QUESTION:
suppose there is a toroidal device and u take cylindrical coocrdinate (r,theta,z) system in to solve it. now the magnetic field in this toroid is non uniform and its along theta direction. iwant to find the curl of magnetic field i.e. theta component.so hw could i find it?? will u b able to give a soltion to it?

ANSWER:
A torroidal coil has a field which is of constant magnitude everywhere but
has only an azimuthal component, as you state. When you state that it is
"nonuniform" I presume you mean that its direction is not everywhere the
same. Therefore, B = φ B_{φ
} where φ is a unit vector in the azimuthal direction and
B_{φ } is a constant. If you look up the curl in cylindrical
coordinates, you will find that the only nonzero term for this problem is
curl[B ]= z (1/ρ )[∂(ρB_{φ} )/∂ρ ]=z B_{φ} /ρ
where z is a unit vector in the z direction. Actually,
this is a rather technical question for the purposes of this site which is
more intended for laymen.

QUESTION:
A particular type of string of say 1 metre length (and mass 10 grams per metre) can handle 100N of force being applied before breaking. But if a string of the same substance was the length of the solar system, just floating in space and tethered to nothing, I'd assume that it couldn't be pulled because the total mass of the long string exceeds the mass that could be pulled by the string. Perhaps my assumption is wrong? Perhaps that string could be pulled extremely slowly with less than 100N of force? Please consider it if there was no gravity acting on the string from planets or the sun, so there's (hypothetically) no minimum force required to overcome gravity.

ANSWER:
As long as you do not pull with a force bigger than 100 N, the string
will not break and will move. Newton's second law tells you that the
acceleration a=F /m, so unless m is infinite, which it
will never be, there will be some acceleration.

QUESTION:
My friend is writing a childrens book and somethings have come up in his book and he would like to make it as accuratly as possible even though this make believe he would like the children to learn some physics.
First question is, in theory how long would someone have to travel faster than the speed of light to go back in time a month? And would this vary depending on how much faster than light you can travel?
Second question, in theory what would someone experiance while traveling faster than light?

ANSWER:
Well, if by "in theory" you mean according to the laws of physics as we
know them, then none of your questions are possible. Nothing can travel
faster (or even as fast as) the speed of light. It is possible to travel
forward in time but not backwards in time. So, if he wants children to learn
some physics, this is fantasy or science fiction, but certainly not physics!

QUESTION:
I am curious to know. A brick with a crack in it was left outdoors in a cold country. Water that was collected in the crack froze into ice. When the temperature rose a little, the ice melted into water, only to freeze again when the temperature fell. This continued for a long time until the brick finay broke into two. Why is this so?

ANSWER:
When water freezes it expands. So freezing water in the crack tries to
pry the brick apart. As it expands, the crack enlarges a little bit and when
a thaw comes, water goes deeper into the crack. Eventaually, as this process
recurs, the crack expands far enough that the brick breaks. It is sort of
like slowly driving a wedge into the crack.

QUESTION:
Could a simple aluminum spiral planar array be expected to equilibrate or harmonize our bodies electromagnetic field? I am a family physician in Columbia, SC and am looking at the C prime bracelet which touts said planar array to create a balanced EMF field around our bodies. I am most interested in the effects on the cellular level for release of stress hormones that lead to anxiety and illness and a heightened pain response. Studies in EMF seem to suggest increased release of stress hormones and somthing simple like a basic planar array worn in the clothes or on the wrist, if it worked, would be a very interesting idea to me. Thanks for any help.

ANSWER:
I have done some research into this kind of thing (my wife has some
serious medical conditions and somebody was trying to sell her a magnetic
matress pad) and I have to tell you that what you are talking about is
pseudoscience. "Harmonize our body's electromagnetic field", "spiral planar
array", "create a balanced EMF field around our bodies", what on earth do
these things mean? Nothing in physics I can assure you and nothing in
biology that I have ever heard of. The companies who market these things
are, in my opinion, composed of nothing but charlatans and the trash they
peddle is worthless. I have heard of cases where such devices may help a
patient, but you may be assured that nothing more than a placebo effect is
in play.

QUESTION:
Ok me and my brother had a very long argument over this hopefully you can set this right.
A helicopter that can lift 10,000 lbs is attempting to lift off. A man weighing 200 lbs but can lift 20,000 pounds grabs a hold of the helicopter. The man begins pulling down, what happens?
I said that the helicopter would lift the man off the ground and the man would basically be pulling himself up.
My brother believes that the man would be able to hold the helicopter on the ground.
And this is all without an anchor on the ground for the man to grab a hold of or somehow bolt himself to the ground.

QUESTION:
If a super-hero was standing on the ground and a helicopter was taking off next to him, he reaches out to grabs the helicopter to the ground. Would it take him up in the air or would he be able to hold it down?
He weighs about 200lbs and can exert more force in his muscles than the 10,000lbs helicopter can lift.

ANSWER:
Looks like both you and your brother decided to "Ask the Physicist". You
are right, the strong guy could not hold down the helicopter unless he was
tied to the ground, and even then, even if he can lift 20,000 lb he would
not necessarily be able to pull down with a force of 20,000 lb.

QUESTION: ^{}
There are numerous problems connected with air drag acting on the falling object. What if it acts on the object thrown up with a given initial speed v(0) and has a given terminal velocity of v(term)? What height would an object rise to?

ANSWER:
This is a standard problem, and if the object is such that we can
approximate a drag force of the form F=-cv ^{2} , the solution
is h =(v _{t} ^{2} /(2g ))ln(1+(v _{0} /v _{t} )^{2} ).
The graph to the right shows heights achieved for various initial velocities
if the terminal velocity is 10 m/s compared to the case of no air drag.

QUESTION:
I hold a tennis racquet by the handle out in front of me, with the flat head parallel to the ground.
I flip it upwards, the head rotating toward me, to do a complete flip. I catch it by the handle again.
The odd thing is that it rotates, or twists, parallel to its long axis halfway through.
My friend says I'm spinning the racquet. I think it's some quirk of physics. Maybe the Coriolis effect, though I have never understood it, with all its "conservation or angular momentum" stuff.

ANSWER:
This is a complicated problem only dealt with in intermediate or advanced classical mechanics, not really feasible for me to try to reproduce it here. The reason it works is that the object you are tossing has three (x,y,z) different moments of inertia along its principal axes. If it is rotating freely, it rotates about its center of mass and if you start it with it rotating about the axis with neither the biggest nor smallest moment of inertia,
it will do what you observed.
The picture at the right shows this problem illustrated in an
intermediate-level classical mechanics textbook. If you rotate it about the
axis with the smallest moment of inertia, the axis running through the length
of the handle, it will be stable, i.e. not do the flip you observe.
If you rotate it about the axis with the largest moment of inertia, the
axis running through the center of mass and perpendicular to the paddle, it
will be stable, i.e. not do the flip you observe. But, if you rotate
it about the axis with the intermediate moment of inertia running through
the center of mass and in the plane of the paddle and perpendicular to the
handle, it will be unstable and the axis of rotation will precess so that it
does exactly a 180^{0} flip in a 360^{0} rotation as you
observe. You are not spinning it, just get your friend to try it. It works
nicely with a hammer also which is a little more compact and easier to toss.

QUESTION:
why do you keep your legs far apart when you have to stand in a fast moving bus ?

ANSWER:
If you are standing in a bus, when the bus accelerates the friction
between your feet and the floor is the force which accelerates you forward.
But this force causes there to be a torque which tries to tip you over. To
tip over, your center of mass must go beyond either of your feet. If your
feet are close together, you do not have to tip very far to fall over, but
if they are far apart you have to tip much farther before you fall over.

QUESTION:
Some people been debating about the impact force on a bike of a bike rider riding off a 6 foot drop to a flat landing. Mainly the debate is about if speed would lessen the impact force on the bike because of the impact angle.
So, if a bike rider was to ride off a 6 foot drop, would it lessen the impact force on the bike jumping off the 6 foot drop with faster speed rather than slower speed and just drop off? In both examples the bike rider is landing on both tires exactly at the same time.

ANSWER:
Both bikers hit the ground with the same vertical component of their
velocities. The one dropping straight will experience only a force up when
he hits. The other will experience a force up and a force backward due to
the friction which would eventually stop him if he just sat there; the
horizontal force is very small compared to the vertical force and I will
ignore it. The vertical forces are determined by two things, how much the
velocity changes (the same for both, from the speed just before they hit to
zero) and how quickly the velocity changes. It seems to me that the only
thing which will determine the time the collision with the ground lasts is
the pressure in the tires which I will take to be identical. Therefore, I
believe there will be no significant difference.

QUESTION:
I want to move 20000 lt of water up hill every hour. the drop is 30 mt. the didtance uphill is 60 mt an the ange is about 30 degrees. can this be done using solar or wind. How much power would it take in watts or kW to move this much water .

ANSWER:
The energy needed to raise 20,000 kg of water (the mass of a liter is
1kg) a distance 30 m is mgh =2x10^{4} x9.8x30 ≈6x10^{6}
J. To do this in an hour requires a power P=mgh /t =6x10^{6} /3600≈1.7
kW. I am not an engineer, so I cannot tell you whether you can practically
do this with solar or wind, but I sort of doubt that it would be cost
effective.

QUESTION:
If I spin an object, and a piece of that object comes off - will it continue to spin in the same direction as the original object was spinning? Do you have any ideas of how I could demonstrate this to a class?

ANSWER:
As long as the piece does not come off in a direction opposite the
direction it was going before coming off (which it would only do if
something impelled it rather than its just coming off), the spin of the
object has to be in the same direction as the object+piece were rotating
beforehand. This is simple conservation of angular momentum. To make a
demonstration, you just have to have a spinning object where you can release
a piece of it remotely so that it just leaves. How about a big disk rotating
and a mass on the surface of the disk rotating with it; but the mass is also
attached to a string which passes through a hole in the center and which you
are holding on to. Let go of the string and the mass will slide off the
disk? You can probably come up with something more clever than that! Be
careful that the piece coming off does not fly into your students!

QUESTION:
Hi, Hope you can help! I am driving a 6000 truck at 60 Mph. I am going to T-bone another vehicle (3500 pounds). Would I be better off putting my foot to the floor and hitting him at 80mph OR hitting the brakes and impacting at 40 mph??? My thought is that I should speed up and try to drive right through the impact.

ANSWER:
Sorry, I cannot say for sure! However, I can say this: the force you
feel will depend on two things, how much your speed changes and how long it
takes to change (i.e . how long the collision lasts). The force is
proportional to the velocity change (bigger change means bigger force) and
inversely proportional to the time (smaller time means bigger force). With
certain assumptions you could maybe get some idea. Suppose the collisions
are perfectly inelastic, that is, the two stick together. Then using
momentum conservation you can show that the final speeds are 50 and 25 mph
for the initial speeds of 80 and 40 mph, respectively; so your speed will
change by 30 mph if you are going 80 and 15 mph if you are going 40. The
bigger speed change would result in the larger force if the times of
collision were equal. The times of the collisions will be determined by how
long is taken to crush the colliding surfaces (the front of your truck and
the side of the other vehicle). But that probably depends on the average
speed during the collision, 65 mph if you are initially going 80, 32.5 mph
if you were going 40; again, 80 in the bigger force since the time of
collision would probably be shorter. So, if the collision is inelastic, I
estimate that the average force during the 80 mph collision will be about 4
times bigger (30x65/(15x32.5)).

QUESTION:
Why is it that I can sit on my front porch and get no wifi signal on my iPhone, but if I step inside and pick up the signal, then walk back out to my front porch I will continue to have the signal when I then sit inthe same place on my front porch?

ANSWER:
I guess because these devices scan for signals with a particular threshhold.
Your wifi is there when on the porch, but the strength is below threshhold.
When you move to where there is a stronger signal, it locks in and connects. Once you are connected, there is probably no threshhold level to stay connected, so you can reduce the signal but still stay connected when you go back to the porch. If this is the explanation
(I'm really just guessing), a good analogy would be scanning your radio for stations. A weak station will not stop the scan, but if you manually tune to that station, you can listen to it.

QUESTION:
Hello, my science class is currently studying air, and we have gotten to a section on air pressure. An interesting question was brought up, and the debate became so wild that my teacher asked me to google the question and come back with the answer. But, it seems that google has failed me, so I turn to you for advice. The question is: What would happen if all the air was sucked out of a room in the middle of no where? Would the room collapse? Has this been tried by scientists? If so, what happened? If not, what do you think would happen?

ANSWER:
The issue is more what would happen to the room before you pumped out
the air. If there is air inside and none outside, the pressure in the room
is trying to blow the room apart; so what happens all depends on how well
the room is designed, how strong the walls are. When you pump the air out,
it is no longer in danger of exploding. Actually, you do not need to "suck"
it out, just punch a hole in one of the walls! Has it ever been "tried by
scientists"? Every time we put up a spacecraft without a pressurized
compartment (essentially all but those carrying people) we are trying it.

QUESTION:
If Newton's first law of motion is true, why do you have to keep pedaling your bicycle to maintain motion?

ANSWER:
Because there are other forces other than the ones you bring about by
pedaling. Always there is air drag and friction in the wheels. If you are
going up a hill there is gravity trying to pull you down the hill. These
forces are opposite the direction of motion and, if you want to proceed with
constant speed, you are obliged to provide an equal forward force (as
Newton's first law stipulates).

QUESTION:
Yesterday, i studied about the spin quantum number and i came to know that electrons spin around themselves under magnetic field in clockwise or anti-clock wise. But my question is that why will the electrons not loose any energy when they rotate around themselves? what actually happens?

ANSWER:
First of all there is no energy associated with the electron's intrinsic
angular momentum, nobody had to do any work to get it spinning, its presence
is simply a property of electrons, that is what intrinsic means.
Second, you should not really think of them to "rotate
around themselves" since this is not your classical angular momentum and if
you try to imagine a little sphere rotating you will get into all sorts of
problems. Finally, there is no reason to expect an angular momentum to
change unless there is a torque acting on it; a magnetic field will exert a
torque on the angular momentum (because the electron has a magnetic moment),
but the result is to change the direction of the angular momentum, not its
magnitude.

QUESTION:
My question starts with fundamental laws: gravity, electromagnetism, etc. As I understand it, these are the forces with the capacity to exert work on the physical world. I'm curious whether the work that biological systems produce are the product of fundamental forces, and if not, whether physics defines the source of the force that produces work in biology.

ANSWER:
Biology is essentially applied chemistry and chemistry is essentially
applied atomic and molecular physics. The four fundamental forces (gravity,
electromagnetic, nuclear, and weak) are the only ones we know in nature.
Biology, at root, is mainly electromagnetism because that is the force which
governs interactions among atoms and molecules. The other three forces play
little role in biology.

QUESTION:
A metal weighing 10kg and wood also weighing 10kg are both dropped from a building which would hit the ground first? If they are of the same shape,which would hit the ground first?

ANSWER:
If air drag is negligible, they would hit the ground simultaneously. If
air drag is a factor, then the one with the larger density will hit first. A
good approximation to air drag force is F = јAv ^{2}
where A is the cross sectional area presented to the wind and v
is the speed; this equation is only true if SI units are used. The one
having the larger density will have the smaller volume and therefore the
smaller A ; hence, at any given v the denser will have the
smaller air drag force. The question, then, is how high does the building
have to be in order for air drag to be important. I assume the two are
spheres, the density of the metal is 8000 kg/m^{3} (near iron), and
the density of the wood is 800 kg/m^{3} (near oak). You can find the
terminal velocity of each by finding the speed at which the drag force
equals the weight. Without any details, I find the terminal velocities of
the metal and wood balls are 167 and 78 m/s respectively. The plot at the
right compares the velocities of the oak and iron balls after they have
fallen a given distance and also shows how each would move if there were no
air drag. As you can see, even dropping from 100 m neither ball is anywhere
near terminal velocity and neither velocity is much reduced from free fall.
If you drop the balls from less than about 30 m, they will, for all intents
and purposes, hit simultaneously.

QUESTION:
i want to ask that how are neutrons obtained to bombard on U235 in fission reaction

ANSWER:
It depends.
Usually in reactors there is no neutron source. There is a small probability that a fission will occur spontaneously. When that happens neutrons are released and these can be used to cause more fissions.
See an earlier answer .
However most modern reactors have a starter source consisting of an alpha
particle source inside a coating of Be or Li; the alpha particles react with
the nuclei and neutrons are released.

QUESTION:
How do atoms, which are 99% empty space, create the solid world around us? Is it simply that they get packed very close together?

ANSWER:
Most of the mass is in the nucleus, a tiny fraction of the size of the
whole atom, but most of the size is outside where the electrons zoom around.
The electrons form a cloud of negative electric charge around the nucleus.
For more detail, see the FAQ page.

QUESTION:
I have been following football since 88, and I have often heard that its easier to kick a long Field Goal at Denver, because of the thin air...
I believe the regulatives says that a football has 13 psi, so if they check the ball-pressure at a higher altitude, and corrects it, it will still have 13 psi.. or at least the same relative difference to its outer surface ..
Then there is the fact about lesser resistance in thin air, but since the pressure has been evened out compared to sea level/high altitude, the object should have the same relative weight..

ANSWER:
In an elementary physics class, we almost always do examples where one
stipulation is "ignore air friction", the drag force on the object passing
through the air. But, in the real world, air drag is seldom neglegible. This
is particularly true in sports when balls often have relatively high speeds
because, approximately, the drag force is proportional to v ^{2} .
When you say that the air is "thin" what you are saying is that it is less
dense, that is there are fewer air molecules per cubic meter than there are
at sea level. The air drag force also depends on the density of the fluid
(how far do you think you would be able to kick the ball if the air were as
dense as molasses?). Less retarding force on the football guarantees that it
will go farther than a ball with the same initial velocity at sea level.
This same effect is seen in baseball where more home runs are hit and curve
balls do not break as much at higher altitudes.

QUESTION:
If there is a magnetic bar with a strong magnetic field pointing from the floor to the ceiling. am I correct that if protons, neutrons and electrons are brought near it the protons will be least effected?

ANSWER:
You have to be careful to specify how the particles are moving. If they
are at rest or moving in the direction of (or opposite) the field, then the
neutron will experience zero Lorentz force because it has zero charge. Let's
simply ask about putting each at rest in a uniform magnetic field. None will
experience a force but all will experience a torque because each has a
magnetic moment, that is, each looks like a tiny bar magnet. The electron
has, by far, the biggest magnetic moment and will therefore experience the
biggest torque; the proton has a magnetic moment larger than the neutron's
and so it will experience a larger torque than the neutron. If you put them
in a nonuniform field (which would be the case for the bar magnet you
specify), each will feel a force in proportion to its magnetic moment, so
again, the electron would feel the strongest force and the neutron the
weakest.

QUESTION:
what the direction of motion of electron around nucleus clock or anticlockwise?

ANSWER:
I guess that depends on whether you look from above or below! Once you
choose a coordinate system relative to which you can make measurements, you
might find it going in many possible directions. The direction of the
angular momentum vector is a topic which is too involved to go into here.

QUESTION:
I fly alots of transatlantic flights and a friend of mine in discussion have a disagreement.
Here's the problem- If I am flying in a plane at 60,000 feet and he is flying in a plane at 20,000 feet and our airspeeds are 400 mph.
Who covers more ground faster the higher aircraft or the lower aircraft?
I contend the higher aircraft will have a higher ground speed than the lower aircraft even though they are both travelling at the same speed of 400 mph. I believe the further you are away from the earth the distance you travel in a straight line is lengthened by the curvature of a sphere. Whose right?

ANSWER:
I think it is a good idea to define "ground speed". Let's say it is the
speed of the plane's shadow as measured by somebody on the ground. Then if
you have a speed v and are a distance h above the earth's surface, the ground speed v_{g
} can be shown to be v_{g} =v (R /(R+h )) where R
is the radius of the earth. Alas, you are wrong! (By the way, I do not know
what you mean "travel in a straight" line because if you fly at some
altitude you are flying in a circular path.) To make this plausible,
consider the following: two planes, each with speed 400 mph, one at altitude
1000 ft, the other at altitude 100,000 miles and both are keeping their
altitudes constant; which will go all the way around the world first?
Because the altitudes of your example are so small compared to the radius of
the earth, there is practically no difference. I calculate the ground speeds
to be 398.9 for the higher plane and 399.6 for the lower. [If you care, here
is the proof. The ground speed and the plane both have the same angular
velocity which is essentially the rate you are going around the circle, for
example 1 revolution per month is the angular velocity of the moon around
the earth, 1 revolution per day is the angular velocity of the earth on its
axis, 1 revolution per year is the angular velocity of the earth around the
sun. If angular velocity is measured in radians per second, then the angular
velocity is v /r where r is the radius
of the circle v is going. Then, v _{g} /R =v /(R+h ).]

QUESTION:
Please settle a bar room argument. 1. An ice cube is floating in a bowl of water: how does the water level change as the ice melts? 2. An ice cube containing the same amount of water as before but this time also containing an air bubble is floated: what happens to the water level as the ice melts and how does this level differ from the change before? 3. The same ice cube as in 1 but this time with a nail (or some heavy object) embedded in it is sunk; what happens to the level of water as the ice melts and how does this level differ from the previous 2 cases?

ANSWER:
Case #1 is a classic problem. Archimedes' principle states that the
buoyant force equals the weight of the displaced water. Since the ice just
turns into water, it exactly fills the void it originally created and the
water level stays the same. Note that it should be clear that the density of
ice is less than water since it floats. Case #3 is essentially to hold the
ice totally submerged until it melts. But since the melted ice has a smaller
volume than the unmelted ice, the water level will go down. Case #2 is
essentially the same as case #1 because the buoyant force must still hold up
the weight of the ice so the weight of the displaced water is still equal to
the weight of the ice; so, when the ice melts it is water of that same
volume. I have assumed that the weight of the air in the bubble is
negligible.

QUESTION:
although you can't see an atoms through a microscope, at some point a clump of atoms is large enough to see as a dot through the microscope. What determines when the clump of atoms is big enough to be seen?

ANSWER:
The best optical microscope can resolve objects as small as about 0.2
μm=2x10^{-7} m. The diameter of an atom is about 1 Е=10^{-10}
m. So the diameter of your clump would be about 2000 atoms. That would
correspond to a total number of atoms of about 4 billion.

QUESTION:
How fast do you have to travel before you experience significant relativistic effects. For the purposes of this question, let's say 1 hour of travel at "X" velocity equals 1 hour plus 10 seconds.

ANSWER:
The quantity which tells you how important relativistic effects are is
called gamma and is
√[1-(v /c )^{2} ] where v is the speed and
c is the speed of light. I am glad you gave me a concrete example (10
seconds for an hour) since "significant" depends on how accurate you need to
be. For example, although satellites used for GPS navigation are nowhere
near the speed of light, not correcting for relativity leads to
significant errors in the computation of where you are. Your situation is
about 0.3% time difference, so √[1-(v /c )^{2} ]≈0.997≈1-Ѕ(v /c )^{2} +…
where I have used a binomial expansion of the square root because I assume
that v /c is small compared to 1. Solving, v /c ≈0.055,
so the speed would about 1/20 the speed of light.

QUESTION:
Suppose that two water droplets detach from a liquid, is it possible that bigger droplet reaches to its spherical shape sooner than smaller droplet. Because as you know, after detaching, the droplets have elliptical shape because of drag force.

ANSWER:
Well, actually, water drops do not reach a spherical shape as they fall. Nor
are they ever shaped like the traditional raindrop/teardrop shape. If the
drop is very small, it will stay approximately spherical but if you now look
at bigger drops, first it becomes oblate (doorknob shape) then oblate with a
dent in the bottom and finally like a parachute. See an
earlier answer for more
detail.

QUESTION:
I was doing some physics revision and completely off the topic I wondered if theoretically a pendulum could travel faster than light. although our given formula: T = √(l/g) does not mention anything about velocity, there seems to be no reason that it couldn't (theoretically), so i searched around and round a formula to find the velocity of the pendulum at the bottom of the swing:
v = √(2gl(1-cos(a)))
v is the velocity of the weight at the bottom of the swing
g is the acceleration due to gravity
L is the length of the wire
a is the angle from the vertical were the pendulum is released
cos(a) is the cosine of angle a
assuming that there is no change in gravity as the pendulum comes down (its theoretical! im keeping it simple), that the length is not able to bend, friction is ignored... etc... a few simple calculations found that:
√(2 Ч 9.8 Ч 4.6Ч10^15 (1 - cos(90))) = 300266548.3 m/s
AHHHH!!! (even though this assumes we could have a pendulum with a length of 4.6 quadrillion meters... its theoretical alright!)

ANSWER:
Well, the formula for the period of a pendulum or for its velocity at
the bottom of the swing are classical and simply incorrect for very large
velocities. In addition, the equation for the period itself is not even
exact, it is based on a small angle approximation. It is possible to get the
speed at the bottom exactly (classically, that is) using energy
conservation; but, again, this is not applicable to high speeds because as
the speed of the pendulum approaches the speed of light, the mass increases.
Finally, where do you propose getting a uniform gravitational field whose
extent is quadrillions of meters?

QUESTION:
I am trying to prove that the energy of a ball in its trajectory is conserved. I believe that a correct approach is to prove that when the derivative of E = K + V (potential energy) is taken, it will equal zero because E = constant. I know that I need to find h(t) and v(t) but I am having difficulty doing this. I seem to be coming for circle. Am I missing something? I feel that I can't use typical mechanics equations here because then I have to deal with components and angles. It's been a while since I've done physics so I feel a bit stumped.

ANSWER:
What you are doing is right. You need to keep in mind that the ball has
zero acceleration in the horizontal direction (a _{x} =0) and
a _{y} =-g in the vertical direction (y ) and
remember that a _{y} =dv _{y} /dt=-g, a _{x} =dv _{x} /dt= 0,
and v _{y} =dy /dt. So, 0=dK/dt +dV/dt.
Now, dK /dt =(d/dt )( Ѕmv ^{2} )= (d/dt )( Ѕm [v _{x} ^{2} +v _{y} ^{2} ])=Ѕm (2v _{x} a _{x} +2v _{y} a _{y} )=-mgv _{y} ;
and dV /dt = (d/dt )(mgy )=mgv _{y} =-dK /dt.
Guess that about proves it!

QUESTION:
A friend of mine and I were discussing crowds and applause. We could not quite figure out why a crowd of people clapping is louder than one and if indeed it even is louder. Assuming that everybody clapped at the exact same volume, would one thousand people clapping be louder than one person, even though they are producing the exact same sound at the exact same volume just from a thousand different sources?

ANSWER:
Sound, like most waves with not too large an amplitude, obey the
principle of superposition; this principle states that if there are two
different sources both radiating waves, then the sum of those waves at any
point of space is what the total wave is. Clapping is a sort of special case
because it is a series of short pulses, so if there are three people
clapping randomly, you just hear more claps, usually not louder sound. But
if all three clapped in unison and were equidistant from you, you would hear
sound with three times the amplitude of a single clap. So, if your thousand
clappers are not happening simultaneously, you will hear a more or less
uniform sound at about the same volume as each clap is.

QUESTION:
This is my understanding: They (Einstein I think) say that we cannot travel at the speed of light due to the fact that we will be traveling faster than time. Couldn't they have said the same thing about traveling faster than sound? I understand that the light we see from space is old. Like when watching someone hit a baseball from 1/4 mile away. You will see the bat hit the ball then you will hear it. BEcause sound travels slower than light. If I were to travel at twice the speed of sound going towards the ball being hit and then back to my current position would that alter time in any way? Then why do scientists think If I were to travel towards old light, faster than the speed of that light and then back again anything would happen to time? Is there a simple answer for my pea brain?

ANSWER:
Light is a perfectly unique wave. Unlike all other waves, it can
propogate without any medium. A sound wave needs air, for example, a water
wave needs water, earthquake waves need the ground, a guitar needs the
string. I cannot derive all of special relativity for you here, but the
result of the fact that light can move through a vacuum ends up completely
shaking up our concepts of space and time. I do not think that saying you
are "traveling faster than time" is fruitful, in fact it does not make much
sense because time does not travel. Why you cannot travel the speed of light
is that it takes an infinite amount of energy to accelerate something there
and, of course, there is not an infinite amount of energy in the entire
universe. See earlier answers in my FAQ page.

QUESTION:
does the earths speed around the sun stay the same? Or stay constant during the entire earth year orbit? It seems that it speeds up during fall and spring.

ANSWER:
Kepler's second law says that a line drawn from the sun to a planet
sweeps out equal areas in equal times. That means that the closer the earth
is to the sun, the faster it goes. The figure to the right shows that the
earth's orbit is slightly elliptical and it is closest to the sun in the
winter (January 3) and farthest from the sun in the summer (July 4). It is
not much of a difference, though, just about 3%. What puzzles me is how "[i]t
seems that it speeds up during fall and spring". How would you feel that?

QUESTION:
Would adding a greater amount of force acting against the force of a spring slow the movement of the spring? So say i have a compressed compression spring with weights in front of it (the weights get pushed as the spring expands), would adding more weights slow the rate at which the spring expands?
I know this is kind of a boring question, it's for a sculpture i'm working on, and the only other way i can find out is to build the whole sculpture and spend hours and hours fooling around with weights and springs.

ANSWER:
The force which a spring exerts is F=kx where k is the
spring constant (how stiff the spring is) and x is the amount of
stretch or compression. If an object with mass m is attached to a spring,
Newton's second law says F=ma=kx and so the acceleration of the mass
is a=kx/m . So, the bigger m is the smaller a is which means
that it will slow down how fast the compressed spring pushes out the masses.

QUESTION:
I have a question of which I cannot find the answer on the internet. Hope you can help me. I'm writing a book and I need to know if it's yes or no possible for an 11 year old boy to push a car over (back) that is on it's side. i have no idea and it's not like I can try this at home. My instinct says no, but I want to make sure.

ANSWER:
It is certainly possible. It depends on the geometry of the car and how
it is sitting on its side. All he has to do is push it far enough so that
the center of gravity is beyond the furthest point of support; once that was
achieved, gravity would do the rest. Imagine a 10 ton object sitting,
carefully balanced, on a base of the diameter of an apple. It would be
incredibly easy to push it over.

QUESTION:
My question is about space and objects. If you were to lay a sheet of steel on top of another sheet of steel, assuming that they are both perfectly flat, is there a measurable space betwen the sheets. My intuition tells me that there is space there, which however small keeps the objecst seperate, but I am not sure.

ANSWER:
To measure this space measure the thickness of the two sheets and show
that it is larger than twice the thickness of one. However, this is likely
to be a difficult measurement and could maybe be achieved by stacking many
sheets. What is between the sheets is air, a thin film of it. I have been
told that if you press two smooth sheets of metal together in a vacuum, they
will weld together.

QUESTION:
let us say we have a string passing through a circular hole on a frictionless table with on end of the string(that lying on the tabe) attached to a body which is rotating with certain constant angular velocity initially and the other end (that hanging through the hole) is being pulled with a constant velocity.now since torque on the body is zero its angular velocity will start increasing as it slids towards the hole on account of conservation of angular momentum.my question is what is the internal torque (analogous to internal forces in conservation of linear momentum) responsible for the increase in the angular velocity of the body?

ANSWER:
If there is zero torque (as is the case here), angular momentum must be
conserved but angular velocity is not necessarily conserved. The angular
momentum is given by L=I ω.
I is the moment of inertial and is approximately mr ^{2}
for this problem, m the mass of the body, and r the distance
from the hole. So, the angular velocity is ω =L /(mr ^{2} )
which, if L is constant, clearly gets bigger as r gets
smaller. What is not conserved in this problem is energy, Ѕmv ^{2} =ЅIω ^{2
} because work is being done by whoever is pulling down on the string.

QUESTION:
What spring constant would be needed for a tension spring to give us a 18000 J of energy?

ANSWER:
A spring does not give energy, it stores the energy you give it. If you
compress (or stretch) a spring with spring constant k by an amount
x , then the amount of work you will do (and therefore the energy stored
in the spring) is
Ѕkx ^{2} . Therefore any spring, as long as it can be
compressed or stretched by the amount x without destroying the spring, can
store 18,000 J of energy. Suppose that you wanted to stretch the spring by 5
cm=0.05 m, the required spring constant would be k =2x18,000/0.05^{2} =1.44x10^{7
} N/m; the force necessary to hold it there would be F=kx =1.44x10^{7} x0.05=7.2x10^{5}
N. I think you can see that it will be quite an undertaking to store that
much energy in a spring.

QUESTION:
I was playing with my son's wooden train set and it got me to thinking. Each train car has a magnet on each end used to hook up the train cars to each other. Turn the cars one way and they hook up, turn them the other way and they repel. Now my question is, does it take the same amount of energy to push the train around the track whether the cars are repelling or attracting or does it take less with the cars in the repel position with the magnets doing some of the work, like winding up a spring and letting it go?

ANSWER:
You may be sure that if you (or the little spring motor) does a certain
amount of work, the same amount of energy is used either way (pulling
coupled or pushing decoupled train). The magnets do not do any work because
if the magnet of car #1 exerts a force on car #2, the magnet of car #2
exerts an equal and opposite force on car #1; hence the net work done by
that pair of forces is zero. It is also true, incidentally, that a magnetic
field never does work , but maybe that is a bit too
technical here.

QUESTION:
why do people say to shut down electrical appliances during lightning?l?

ANSWER:
Due to lightning strikes, the voltage can sometimes surge, become much
bigger than your appliance is designed to operate with. Turning it off is
one level of protection for your stuff because the power switch opens the
line between the outside and the appliance. However, it may not be enough
and to be really safe you should unplug it. Another option is to use "surge
protectors" which are designed to not let the surge pass. Also, remember
that a surge could also come in your telephone line and whatever the line
goes to (fax, telephones, computers) may be damaged.

QUESTION:
You and a child half your height lean out over the edge of a pool at the same angle. If you both let go simultaneously, who will tip over faster and hit the water first? Or will they hit at the same time?

ANSWER:
The child will hit the water first. To understand why, model a person as
a stick of height L and mass M . Newton's second law says I α =(MgL sinθ )/2
where I is the moment of inertia about the end of the stick, α
is the angular acceleration, θ is the angle the person makes with the
vertical, and g is the acceleration due to gravity. The right side is
the torque exerted on the falling object and angular acceleration measures
how quickly the rate of falling is increasing. So, whoever has the biggest
α hits first. The moment of inertia needed is I=ML ^{2} /3,
so solving for α, α =[(MgL sinθ )/2]/[ML ^{2} /3]=3g sinθ /(2L ).
So, mass does not matter and the larger L is the smaller α is.
To convince yourself that this is reasonable, imagine a race to fall over
between a matchstick and a 100 ft high chimney.

QUESTION:
I recently performed the "Determining the Densities of Pennies" lab. In this lab, I know the answers to my percent error calculations should NOT be 0%, but what I don't know is WHY. Can you help? Your time is greatly appreciated.

ANSWER:
You really haven't given me enough information. If you were asked to
compute a percent error, you were probably being asked to compare your
result to the known accurate result. If a penny were pure copper then you
could compare your result with the well-known density of copper. In fact, it
depends on the year of your penny; pennies before 1982 were pure copper and
after 1982 were pure zinc plated with copper. The mass of pennies decreased
from about 3.1 g to 2.5 g in 1982. If your density is not exactly equal to
the known density of pennies, you compute the percent error as ( ρ_{measured} -ρ_{known} )x100/ρ_{known} )
where ρ_{ } is density. This provides a quantitative measure of the
accuracy of your result. But maybe you are asking why your measurement has
any error when compared to the known value. The reason is that no
measurement is absolutely perfect. You have to measure the mass, the
diameter, and the thickness of your penny and each measurement has some
uncertainty. For example, your scale might only be able to determine a mass
to the nearest 0.1 g and so if you measure 2.5 mg it might just as well have
been 2.4 or 2.6 or anything between. We would then say that m=2.5ï¿½0.1 g.
Similarly measurements of size can only be done to some accuracy depending
on how you do it. These are usually referred to as uncertainties, rather
than errors. And you have to compute density from mass and volume (which
itself is computed from more than one measurement), but there are ways of
doing this which I will not go into here. Finally, you would hope that the
experimental uncertainty is within the percent error; for example, if ρ_{known} =5
and ρ_{measured} =4.7ï¿½0.4, it would be a pleasing result whereas if ρ_{known} =5
and ρ_{measured} =3.7ï¿½0.4, it would not! There are also uncertainties
which are related to statistical considerations, but we won't talk about
that here.

QUESTION:
A mass amount of an object is traveling down on to a conveyor belt and then in to a basket. Maintaining the same amount of mass on to the belt, and then you slowed the belt down, would the basket fill up slower or stay the same?Also vise versa if belt speeds up would the basket fill faster?

ANSWER:
It is not clear what "maintaining the same
amount of mass on to the belt" means. If you mean you do not change the rate
at which mass is added to the belt when you change the speed of the belt,
then the answers to your questions are same and no . If you mean that
you adjust the rate at which mass is added to keep the total mass on the
belt constant, then the answers are slower and yes .

QUESTION:
If the distance between the south poles of two long bar magnets is reduced to half its original value, will the force between these poles be doubled?

ANSWER:
If there were such a thing as magnetic monopoles, they would behave like
electric charges, the force would be inversely proportional to the square of
the distance between them. When you say "long bar magnets" I presume you are
saying that the other poles are so far away to have negligible effect. In
that case, halving the distance would increase the force by four times.

QUESTION:
suppose an astraunaut landed on a planet where the acceleration due to gravity is 19.6 meters per second square,compared to earth,would it be easier,harder,or just as easy for her to walk around?
would it be easier,harder,or just as easy for her to catch a ball that is moving horizontally at 12 meters per second square?
assume that the astraunaut stay suit is a lightweight model that does'nt impair her movements in any way.

ANSWER:
Essentially, everything would weigh twice as much as you are used to. So it would be difficult to stand, let alone walk. If the ball were truly going horizontally, it would be just as easy to catch as on earth because it still has the same mass even though it has greater weight. If it were falling, though (high fly ball) it would be like catching something twice as heavy.

QUESTION:
Regarding gravity and light outside of a vacuum:
Assertion 1) Object A pulls on object B, as object B pulls on object A.
Assertion 2) Gravity from object A can alter the direction of travel of a photon (object B).
So the photon (object B) should then in turn alter the travel of object A. Given enough light, could you actually move object A?
If we adjust the scales of A and B to be ideal/non-realistic, the question can be reduced to "Would the light from a (very bright) flashlight pull (or perhaps it would also be pulled back from the opposite side as the light passed it) a (very light/non-massive) balloon?"

ANSWER:
I am not an expert in the fine points of general relativity. However, it is
my impression that, although it is usually mass which everyone thinks of as
"warping space-time" and hence creating gravity, that it is more correct to
say the presence of an energy density also does it. Since mass contains so
much energy (E=mc ^{2} ), that is what is, by far, the most
important source of gravity. However, it is not incorrect, in this context,
to say that a very intense beam of photons would also be a source of
gravity, but it would probably be totally impossible to measure it because
it would be so weak.

QUESTION:
How many rpm do i need to spin a rope 30 feet long with a 6inch piece of chain on end weighing 1lb. to keep rope straight out?

ANSWER:
Infinite! It cannot be done. You could get it nearly "straight out" but
never all the way.

QUESTION:
When driving a vehicle that has a camera equipped inside it and you are going around a curve in the road will the camera follow you in the curve or go straight making the line in the road appear as if you crossed it when in fact you didn't?

ANSWER:
A camera and your eye work just the same way: the image of what is aimed
at is focused onto the film (retina) for the camera (eye). The camera is
just a mechanical eye. Whatever your eye sees will be the same as what the
camera sees.

QUESTION:
We have ten fingers and so we made a base 10 number system. We have based everything from maths to physics to computers on this assumption. It seems a bit convenient. If I ask a calculator to calculate 1 divide zero it cannot cope. How therefore can we be confident that results from theories such as 'big bang' or theoretical physics are anything other than arbitrary?

ANSWER:
First of all, computers usually use other base number systems, binary,
octal, or hexadecimal. But the bottom line is that it makes no difference
what number system you use, any given theory would be the same. The
equations and such would look different but mean the same thing. What a
theory means is independent of the base of the numbers. The fact that you
cannot divide 1 by 0 has nothing to do with the fact that you are working in
a 10 base system, 1/0 is the same in any number system, undefined (which is
why, of course, your calculator cannot "cope"). You can put 2 into 6 just 3
times, but if you start putting nothing into 1, you will continue putting it
in forever and never get there. Finally, lots of things like the big bang
are not the result of a theory but the result of experimental measurements.
There is lots of evidence that the universe originated about 14 billion
years ago in a huge bang and the evidence has nothing to do with theories.

QUESTION:
If 225 pounds falls 24 inches straight down to the ground, propelled only by the force of gravity, what is the force of impact?

ALMOST
SIMULTANEOUS QUESTION:
what is the striking force of an object that weighs 300lb falling from a height of 3 feet

ANSWER:
My God, will I never stop getting this question?
It depends entirely on the details of the collision. In essence, what
you need to know is how long did it take for the weight to stop? If you had
checked the FAQ page you would have
found an answer. There is no such thing as "striking force".

QUESTION:
If you take coherent light of a fixed frequency and add more coherent light of the same frequency, but out of phase by 180 degrees then can you reduce the intensity due to canceling out the waves?

ANSWER:
Yes, you can change intensity at points in space by causing destructive
interference. But you cannot simply reduce the intensity everywhere, only at
places where the two waves interfere destructively. There will inevitably be
other places where they interfere constructively depending on path
differences. There is a very lengthy discussion of a similar question in an
earlier answer .

QUESTION:
I am creating an educational physics poster for a class assignment. I am representing terminal velocity of two ~10lb objects (let's call them spheres), one dropped from 100 feet, one from 50 feet. The sphere dropped from 100 feet hits terminal velocity and the one from 50 feet doesn't. I need to find out how long it takes the sphere to hit terminal velocity (i.e. the average human takes 12 secs), and where in the 100 feet it hits it (half way down, 3/4 way down, etc). There is one slightly confusing matter in that during this example the words "feet" and "stories" was interchanged, so if it turns out to be impossible for a 10lb sphere to hit terminal velocity in only 100ft, change it to 100 stories (1000 ft) instead!

ANSWER:
This is a very well-known problem, the solution to which is only an
approximation. If an object presents a cross sectional area of A to
the air as it falls vertically, the force which it feels due to the air drag
is approximately proportional to A and the square of the velocity
v , F _{drag} ≈јAv ^{2} .
It is important that this equation is only true if you work in SI units (A
in m^{2} , v in m/s, F in N=kg-m/s^{2} ); get
rid of the pounds and feet! So, a falling object has two forces on it, its
own weight down, W=-mg , and F _{drag} . So, Newton's
second law is ma=-mg +јAv ^{2} . Terminal velocity is
when a becomes zero, and so v _{t} =2√(mg /A ).
Note that the bigger the weight (mg ), the bigger the terminal
velocity; this might suggest to you that it would be more practical to
choose an object lighter than 10 lb for your example. Also, note that the
terminal velocity is, technically, never reached but approached asymtotically. You
need to set some limit, like when the speed reaches 90% of the terminal
velocity. The solutions for the speed at times other than when it is v _{t}
are quite mathematical and you give no indication of what your level is. So
I will just give you the general results and let you apply them to your
situation. Maybe what you really want is how far the sphere falls rather
than how long it falls before achieving, say, 0.9v _{t} ; that
is given by v ^{2} =v _{t} ^{2} (1-exp(-2gy /v _{t} ^{2} ))
where y is the distance it has fallen. (The function exp(z ) is
the number e =2.718 raised to the z power.) The expression for
v as a function of time t is a little more complicated, v=v _{t} {[exp(2t /T )-1]/[exp(2t /T )+1]}
where T=v _{t} /g is called the characteristic time,
similar to half life for radioactivity. This may all be too much for you,
but gives you everything you need to work out any scenario. Here is a quick
example you can check yourself with. A spherical mass of 1 kg (about 2.2 lb)
has a radius of 0.1 m (about 4 in) and area A =0.031 m^{2} has
a terminal velocity of about v _{t} =2√(mg /A )=2√(1x9.8/.031)=35.6
m/s. The distance it has to fall to achieve 90% of this speed can be
calculated as follows. v ^{2} /v _{t} ^{2} =0.9^{2} =0.81=(1-exp(-2gy /v _{t} ^{2} ))=1-exp(-2x9.8xy /35.6^{2} )=1-exp(-0.0155y ),
so exp(-0.0155y )=0.19. To solve this, take the natural logarithm (ln
which is the logarithm to the base e ) of each side: y =-ln(0.19)/0.0155=107
m≈350 ft. Your 10 lb will almost certainly not get close to terminal
velocity in 100 ft unless it is huge in size, big A .

QUESTION:
I'm confused. If there's a ball at rest on the ground (floor), does that mean it has zero gravitational potential energy? When we say "zero", do we mean that it there is none or it just refers to the "position"? In relation to the center of the earth, does the ball have gravitational potential energy?

ANSWER:
Your last sentence indicates that you know what the answer. Potential
energy (PE) has no meaning, it is the change in PE which matters; you have
to specify where you choose PE to be zero. If you have chosen the floor as
your zero of potential energy, the ball has zero potential energy. If you
have chosen the basement floor for zero PE, and it is 10 m below the ball,
then the ball has PE=mgy= 10mg . If you have chosen the
upstairs floor for zero PE, and it is 10 m above the ball, then the ball has
PE=mgy=- 10mg .
(PE=mgy implies that up is the +y direction.)

QUESTION:
If the faster you go the more mass you achieve, the why dosnt light achieve mass while traveling at..... light speed.

ANSWER:
See a recent answer .

QUESTION:
an electron moving perpendicular to a magnetic field experiences lorentz force , here there is action but where's the "equal and opposite" reaction force . Is the newton's third law of motion being violated here.

ANSWER:
The field is being produced by a current somewhere and the "reaction
force" will be on that wire. However, that force is not necessarily equal
and opposite to the force on the electron. Newton's third law, as taught in
elementary physics, is not really absolutely true and magnetic forces, since
they are velocity dependent, are the best example to show this. The figure
shows two positive charges moving with velocities shown by black vectors.
The electrostatic forces are red and are equal and opposite. The magnetic
force felt by the particle moving horizontally is zero because the field
there (due to the other charge) is zero. The field due to the charge moving
horizonatlly points up out of the screen, so the magnetic force (blue) on
the other charge is to the right. The net force on each particle is shown in
green and they are neither equal nor opposite. Newtonian mechanics takes on
new forms in electromagnetism and everything still all hangs together,
but you need to include the energy and momentum densities of the fields.
This is way beyond the scope of this site to address.

QUESTION:
Is it possible to hover above the ground and then "travel" to other places by letting the world revolve beneath you?

ANSWER:
See earlier answer .

QUESTION:
If work is defined as force x distance then if you are pushing against an immovable object, no work is being done. Nevertheless, you are expending energy as you are pushing. It seems to me that there is some kind of paradox here--you are losing energy but you are not doing any work. Can you explain?

ANSWER:
Physics would say no work is being done because the force is not acting over a distance; you and I both know, however, that sugar is being burned to provide the energy necessary to
push on this stationary object. What is going on here, as I understand it, is that the muscle fibers in your arm are continually slipping and retensioning thereby doing lots of little parcels of work to hold your arm steady.

QUESTION:
I have several questions involving the law that energy cannot be lost or created, only converted.
Say a person was cranking an electric generator. it made electricity and lit a lightbulb.
Now, what if the generator wasn't a generator, but simply a container with an equally-hard-to-push crank. The person would crank it, outputting just as much energy, and creating movement but no electricity. This makes the assumption that the man is not making as much energy. ???

ANSWER:
Site groundrules stipulate "single, concise, well-focused questions" so
I will only answer your first. If a person does a certain amount of work
cranking the generator, that energy will show up somewhere. If your "container
with an equally-hard-to-push crank" is the recipient of the work, that
container will have some built-in friction to make it equally hard to crank.
When work is done against friction, the result is mostly heat. Your
container will get hot.

QUESTION:
How can we say that the Earth rotates the Sun?
Which one rotates in which the two or more other planets?
Which point is defined as the static point in the space to say that what rotates which?
Can there be the static point in the space without first thinking the Sun as a static point?
To say that an object is rotating or moving, there must be the static point first beside that as a criteria.
So,
What is the axis or the criteria in the space to describe an object is moving without regarding the Sun as a static point first?

ANSWER:
The sun is not static. The earth does not revolve around the sun. Rather
the sun and the earth each revolve around the center of mass of the
earth-mass system. However, the sun is so much more massive than the earth
that the center of mass is close to the center of the sun and therefore you
get excellent numerical results if you just assume the sun does not move.

QUESTION:
If we transfer compressed air from a one-hour bottle rated at 4500 psi into a 30-min empty bottle also rated at 4500 psi will the pressure of both bottles equalize at 2250 psi each? And will there theoretically be 30-min of breathing time left in both tanks?

ANSWER:
I presume that the volume of the 30 min bottle is half that of the 60
min bottle. Also, 4500 psi is, most likely guage pressure, the pressure
above atmospheric, the "empty" bottle is not really empty but has
atmospheric pressure in it. But, atmospheric pressure is only about 15 psi,
so let's just ignore that. What is operative here is Boyle's law, PV =constant,
so P _{1} V _{1} =P _{2} V _{2} where P _{1} =4500,
V _{2} =1.5V _{1} , P _{2} is the
final pressure. That is, you are increasing the volume of the original gas
by a factor of 1.5. So, P _{2} =3000 psi. I do not understand
what you mean that both will have the same breathing time left. Seems to me
that, for each, 4500 psi gives the full time, so each would have 2/3 of
their maximum times, 40 min and 20 min.

QUESTION:
This question was mentioned to me a long time ago by a physicist who I think was just trying to mess with my head - a task at which he succeeded.
One of the simplest ways of measuring the circumference of a circle that many children are taught is rolling out the circle; and this is basically the starting point of my question.
To make my question clear, I feel I have to describe the process :
Take a wheel or circle; draw a radius on it from the center to the edge; position the circle on a flat surface or number line so that the radius points straight down at a known point; roll the circle along the line until the radius is again pointing straigh t down at a second point and mark it; the distance between the two points is the circumference of the circle.
Now draw another circle on the first that is half the size of the original.
If you repeat the original procedure of measuring the circumference of the larger circle, the smaller circle also starts and ends at the same position despite travelling a larger distance than it's own circumference?
Why can't I get my head around this?
I hope this doesn't count as a maths question - it was originally put to me by a physicist...

ANSWER:
The small circle is not rolling on the dashed line, it is slipping on it. At any given time the whole circle is rotating around the point of contact
of the big circle with the ground, that is, the point of contact is at rest (that is what rolling without slipping
means). Therefore, a point a distance s above the point of contact is moving forward with a speed
vs /R where v is the forward speed of the center, and
R is the radius of the big circle. The bottom of the small circle is going forward with speed
v /2 and the dashed line is at rest.

QUESTION:
Does light, a photon, have mass, any mass? If so, and a photon by definition travels at the speed of light, why doesn't it have infinite mass? If not, how is it subject to gravitation?

ANSWER:
Answer to first question: a photon has no mass. Answer to second
question: m=m _{0} / √[1-(v /c )^{2} ]
which, if m _{0} =0, is m =0 for every velocity except
possibly c in which case m =0/0 which can be anything; the
limit as v approaches c is therefore 0. Answer to third
question: see FAQ .

QUESTION:
I'm puzzled by an Einstein thought experiment. The idea seems to be that gravity is the same as acceleration. If you are in a closed box you can't tell if you are resting on Earth or on a rocket ship being accelerated at 1G. Obvious nonsense; gravity is curved and acceleration is not. I can think of at least three ways to tell the difference - in gravity:
1 - weights hung from different sides of the box would not hang parallel.
2- gravity would be slightly (ok, very slightly) stronger at the bottom of the box.
3- time would proceed slower at the bottom of the box.
None of these effects are true for accleration. Is this 'thought experiment' just meant to be a crude approximation, and am I just picking nits?
It seems... unsatisfactory.
Oh, and an interesting side effect - if you apply a certain acceleration to a spaceship in interplanetary space and then applied the same accleration deep in a gravity field, say close to the sun, you'd get a greater increase in velocity from option two - ehh?

ANSWER:
You totally misunderstand the equivalence principle. The normal way of
expressing it is to compare experiments in a uniformly accelerating
frame with those in a frame at rest in a uniform gravitational
field. The earth's gravitational field, as you seem to appreciate, is
not uniform, but it is pretty darned close if you stay within a few
thousand meters of the surface. Hence, some authors will talk about the
equivalence principle using the earth's gravity as an example. However,
the equivalence principle is perfectly valid in a nonuniform field
provided that you adjust your acceleration (a nonuniform acceleration).
So, if you adjusted your acceleration to vary like 1/r ^{2}
instead of being constant, all the effects which you think you could
debunk the principle with would appear. By "gravity is curved" I presume
you mean it is not necessarily uniform. Your "interesting
side effect" makes no sense to me; it simply seems to say, in Newtonian
language, that if you apply two forces (sun's gravity and rocket engine)
you get a different result than if you just apply one force (rocket
engine) —duh!

QUESTION:
Why was work done defined as Force x Displacement? Is there a sound reason or was it simply derived from the work energy theorem?

ANSWER:
Work was not defined as Fd , it arises naturally from Newton's second
law (the source of essentially all classical mechanics). You start with
Newton's second law, F=ma , rewrite it as F =m (dv /dt ),
use the chain rule F =m (dv /dx )(dx /dt )=mv (dv /dx ),
rearrange mv dv=F dx. On the right of this equation is
work done if m moves the small distance dx ; on the left is what turns
out to be how much the kinetic energy changes when that work is done. If you
integrate this you get
Δ(Ѕmv ^{2} )=F Δx=W.
Maybe this was too much detail for you.

QUESTION:
Could you help me understand the concept that you can't measure a particle's speed and position at the same instant. It seems to me that this is a technical problem. We know that they have a speed and position at any instant so they are knowable, thought perhaps never to us, and therefore their future is determinable.

ANSWER:
You know why "it seems to" you that it is wrong? It is because your
intuition (that's where "seems to" comes from) is based on your experience
and you have never tried to measure these things to anywhere close to the
precision you would have to to see the effects. The "concept" is not that
"you
can't measure a particle's speed and position at the same instant", it
is that you cannot know both to arbitrary precision . The uncertainty
principle says
m Δv Δ x ≈h /2π
where m is mass,
Δ v is uncertainty in
velocity, and
Δ x is uncertainty in
position, and h /2π ≈10^{-34} m^{2} -kg/s is the
rationalized Planck's constant. Let's take an example you could have
measured, say a BB with a mass 1 g=10^{-3} kg sitting on your desk
(supposedly at rest). You measure its postion to an accuracy of the
wavelength of visible light, 5x10^{-7} m (that would be a tour de
force measurement!) Then, how accurately do you know that it is truly at
rest?
Δv ≈h /2π /( m Δx )=10^{-34} /5x10^{-7} /10^{-3} =2x10^{-26}
m/s, meaning it could have a velocity as big as this. So, we now test this
by waiting until we see it move one micron (10^{-6} m); it should
only take 10^{-6} /2x10^{-26} =5x10^{19} s=1.6
trillion years. That is about 10 thousand times the age of the universe! So,
you see, the uncertainty principle does not have any significant effect on
everyday life in our macroscopic world. That does not mean it is not true,
that means you cannot see it in everyday life. If you now look at electrons
(m ≈10^{-30} kg) inside atoms (Δ x ≈10^{-10}
m), we have a very different story.

FOLLOWUP QUESTION:
I would like to have another go at my previous question about the uncertainty principle. I'm just an old farmer, long in the tooth and have just gotten interested in this physics stuff in the last ten or so years so let me present a thought experiment.
Say I shot a cue ball at a racked set of pool balls, once that ball is rolling there is a definite route each ball will take and the outcome will not be altered whether I know the forces involved or not. I can't predict the outcome since I can't possibly measure the exact forces
but that doesn't interfere with the outcome. Now I understand that by trying to measure particles, no matter how close I measure them, over the course of the life of a universe there will be drastic changes in any outcome I would predict. However, at the time of measuring,
isn't there already a prescribed path set in concrete based on Newtonian physics and if it were possible to know the forces involved and we had the technical ability could we not define that path. I realize there are probably other factors that would make that impossible such as particles popping into and out of existence but I
don't understand why not knowing the data makes the outcome any different. Isn't that why Einstein commented that god doesn't play with dice?

ANSWER:
Your whole question continually assumes that position and velocity have
definite values simultaneously, and that is simply wrong if you accept the uncertainty principle. The range of outcomes for your billiard example is so small that you would not have any hope of measuring it. The issue is not that you cannot measure it, the issue is that at
any given time the particles actually have a distribution of momenta and positions. I know it is disturbing to say that a particle can have many different velocities
and positions simultaneously. The root of all this is in wave mechanics; at some level if you want to understand how a "particle" behaves, you have to think of it as a wave which is not all in one place
and is not moving in one direction with one speed. In spite of your insistence, the universe is not governed by Newtonian physics. Yes, Einstein was famously suspicious (to say the least) of quantum mechanics. General relativity
(Einstein's "baby") is (as I understand it) a purely deterministic field
theory — everything
which will ever happen is already written. You can think of yourself as in
good company with Einstein since your world view, that Newtonian physics
rules and every particle right now has a specific position and speed, also
says that the universe is determinisitic. Forget free will!

QUESTION:
if i drop a paper clip and a 500 pound weight out of a plane at the same time which one will hit the ground first?

ANSWER:
Unless the plane is flying very close to the ground and very slowly, air friction will be important. Therefore, the shape of the 500 lb weight will also matter. For example if the 500 lb weight is a 400 lb box and a 100 lb parachute, it is very different from a 500 lb ball of lead. Let's just drop a 500 lb box and a paper clip from a jetliner at 30,000 ft — the box hits first.

QUESTION:
I am a flight attendant on a private jet. Flying at 43,000 ft we passed over a lage cloud formation toward another. All of a sudden the plane dropped about 1000 ft before contacting 'pressure' again. I was standing in the cabin - there were no passengers on this flight. Pilots were in seatbelts and said there was absolutely no indication of change of pressure and no up or down drafts. I did not hit the ceiling but i hit the floor like a bag of rocks.
What speed was i falling and what were the g's on my body (5'5" 170 lbs)?
Several people have asked and i would really like to know.

ANSWER:
As is often the case, I do not have enough information to calculate what you
want. If the plane all of a sudden dropped in freefall, it would take about
8 s to go 1000 ft down; sound about right? If so, when you were 1000 ft down
you would be going about 250 ft/s=170 mph in the downward direction (as well
as about 500 mph forward). Now, if you hit the ground going
170 mph, you would certainly die. However, the plane did not suddenly stop
falling, it happened over some period of time. So, if it happened over 2 s,
for example, and you, having hit the floor, will also stop in 2 s. The force
you would experience is your mass 170/32=5.3 times your acceleration
250/2=125 ft/s/s; so the average force you would have felt over that 2 s
would be about 664 lb which would be about 4g. But all these are numbers I
basically made up, but I think they are the best you can do given the
information available.

QUESTION:
My son has a question that I can't seem to find an answer to researching on the web. WHAT FORCE IS BEHIND A GOLF BALL WHEN HIT ON THE MOON? I appreciate your time on answering this for my son. I think it has something to do with Velocity, but not sure!

ANSWER:
I think there is some confusion about what force is here. Here are all the
forces a golf ball experiences here on earth:

The club, traveling with some speed, hits the
ball and exerts a contact force on it for a very short time but it is a
very big force and it results in the ball acquiring a very large
velocity. As soon as the ball leaves the club, there is no force
"keeping it moving". If there were no other forces, it would keep going
forever with the speed with which it left the club.

Once it is started, gravity pulls down on it
which is the force which eventually does bring the ball back to the
ground.

As it flies through the air, it experiences air
drag which can be very complicated. Essentially, it is a force trying to
slow it down and the bigger the speed is the bigger this force is.

If it happens to be spinning, the air drag can
act asymmetrically so that the ball curves. This is what is called a
hook or a slice in golf (depending whether it curves left or right,
respectively, for a right-handed golfer).

Of course, when it hits the ground, it
experiences forces from the ground which ultimately bring it to rest.

What is different on the moon?

If the club is the same club with the same
speed, there is no difference for this force. Therefore, the ball
launches just the same as on earth.

The moon is much smaller than the earth and the
result is that the gravity on the moon is much weaker. Therefore, this
force (trying to pull the ball back down) is much smaller and the ball
will go a lot farther.

Since there is no air on the moon, there is no
drag and this also results in the ball going much farther.

The ball will not curve on the moon, regardless
of how much spin it has.

When it hits the ground, things are about the
same as on earth except that all the forces are smaller, again because
of gravity being smaller, so the ball rolls farther before it stops
(also because it is going much faster when it hits the ground than it
would have been on earth).

QUESTION:
In a projectile motion, which has a farther range (neglecting air resistance, or in a vacuum) fired from a catapult with constant force and same angle of flight, a paper(crumpled) or a rock? Knowing that the paper's mass is less than the rock, I thought that the paper would have a farther range since mass is inversely proportional to acceleration; but my teacher says otherwise, that they would have the same range. Who is correct?

ANSWER:
The problem is not specified completely, so either of you could be right.
Think about it: if the (constant) force acted on the rock for 5 minutes but
on the paper for 1 second, the rock would go way farther. You can go one of
two ways —specify the time the force
acts on each or the distance over which the force acts on each.

If the forces
exerted on both are the same and the time which the force acts for each
is the same, then you are right and your reasoning is right. Your
reasoning is right because the acceleration of the less massive paper is
larger and if the times of acceleration are the same, it ends up going
faster; the ratio of the velocities will be given by v _{1} /v _{2} =m _{2} /m _{1} .
If the rock is 4 times more massive than the paper, it will have
ј
the initial velocity of the paper.

However, if the
catapult exerts equal forces over the same distance (the distance of the
arc of the arm, presumably), the paper will still have a larger initial
velocity than the rock. But it will not be because of acceleration being
different since the two will spend different times experiencing the
force. Here each acquires the same kinetic energy because equal work was
done on them. In this case, the paper still goes faster but the ratio of
velocities is v _{1} /v _{2} = √( m _{2} /m _{1} ).
If the rock is 4 times more massive than the paper, it will have
Ѕ
the initial velocity of the paper.

The second scenario is
likeliest to be a description of a catapult. A very light stone will get
launched in a much shorter time than a very heavy stone and both will
experience the force over the same distance. In all fairness to your
teacher, he was trying to illustrate that, without air drag, two projectiles
with identical initial conditions will go over identical paths. This
requires stipulating equal initial velocities, though, not equal forces.

QUESTION:
Heat recovery ventilators are becoming very popular as homeowners weatherize their homes. They are simple units that intake air from outside, exhaust stale air from inside. Heat is transferred from the exhaust air to the intake air through a plenum of aluminum. All very simple so far.
Except, they work too well! They can take 0 deg F outside air, 70 degree exhaust air and instead of warming the intake air to 35 deg, like most of us would expect, they warm it to 50 degrees!
How do they do this?! Does the answer lie in the fact that 70 deg F contains more than twice the energy of 35 deg air?
(NB: No heat pumps, freon, electrical resistance, etc. are involved)

ANSWER: Suppose that
you mixed equal amounts of 0^{0} air and 70^{0} air. Then,
you are right, you would expect to get 35^{0} air. But, suppose that
you could take the heat out of 70^{0} air such that you cooled it
down to 0^{0} and then took that heat and put it into an equal
amount of 0^{0} air. You would end up with 70^{0} fresh air
coming in and 0^{0} stale air going out and have violated no laws of
physics —the energy was transferred from
the hot to the cold with no other expenditure of energy. This would be
called 100% efficient. I had never heard of these things, but my cursory
research reveals that they can be as much 85% efficient, in which case you
would end up with 60^{0} air coming in. (50^{0} would be
about 71% efficient; 35^{0} would be 50% efficient.) This is
achieved by forcing the hot exhaust air through tubes and baffles which get
warm and then blowing the colder intake air over them. So, you see, you are
not getting something for nothing, since with less than 100% efficiency you
are still losing energy (from the house) in the process.

QUESTION:
Does UV-C light delivered through fiber optics diminish the UV-C sanitising properties?

ANSWER: I do not know
about the "sanitising properties" of UV-C. What I know is that a
photon (of a particular frequency) is a photon. It does everything popping
out the end of a glass fiber that it could before it entered.

QUESTION:
Can A small cluster of conducting metal atoms on the nano size be given a positive electric charge using a laser (Preferably of modest power).

ANSWER: It depends on
how much the nano particle mimics the bulk metal. You are basically are
dealing with the photoelectric effect, so the laser photons have to have
enough energy to knock out an electron. Usually, ultraviolet would be the
best bet.

QUESTION:
How much energy is required to transmue a metal by removing a proton from the nucleus of the atom. For example remove one proton from mercury do I end up with gold?

ANSWER: The exact
amount depends on the particular nucleus you want to remove a proton from.
An average over many nuclei is about 8 MeV (million electron volts) which is
about 10^{-12} J. If this is a get-rich-quick scheme, forget it!

QUESTION:
I remember learning that momentum equals mass times velocity but this must not be strictly true since light is massless and has momentum. Why does light have momentum?

ANSWER: The reason is
that momentum is not really mv but rather mv / √[1-(v /c )^{2} ].
This leads to the relation between energy and momentum for a particle of
mass m ,
E =√[m ^{2} c ^{4} +p ^{2} c ^{2} ].
So, you see that for
m =0
this results in
p=E / c.
Also note that, for a particle of mass m at rest (p=0), E=mc ^{2} .
If you want to understand why linear momentum must be redefined in special
relativity, see an
earlier answer .

QUESTION:
I am a member of a high school debate team in Vancouver Washington, and I am currently writing a policy debate case on Nuclear Fusion. I was hoping you could give me some advice on the likelihood of scientist's ability to produce a nuclear fusion reactor in the near future. Any other opinions on the overall convenience in nuclear fusion and it's energy production potential would be greatly appreciated. Thank you!

ANSWER: I would not
want to be arguing in a debate that controlled nuclear fusion is imminent.
When I was a college student nearly 50 years ago, word had it that fusion
reactors would be operating within 10-15 years. It is a very hard problem
because you need extremely high temperatures for fusion to occur because the
electric repulsion between the hydrogen ions keeps them too far apart for
them to interact if they are moving around at normal room temperatures. It
is extremely difficult to design magnets which are capable of containing a
hot plasmas (essentially a 'soup' of electrons and positive ions) for longer
than milleseconds; nevertheless, efforts are still underway to build such
machines (tokomaks) at places like the Princeton Plasma Lab. An alternative
approach is to make tiny pellets containing hydrogen and implode them with
powerful lasers coming from all directions which is sort of like a tiny
H-bomb. I may be wrong, but I do not believe that anybody has yet achieved
"break even" which means you get as much energy out of the fusion as you put
in to cause or maintain it.

QUESTION:
If I drop a steel ball in a gaint never ending vacuum will the speed increase forever or level off at some point?

ANSWER: I guess you want this "never ending vacuum" to also be a never ending uniform gravitational field? In spite of the fact there are no such things, it would continue accelerating for a while and then,
when its speed became comparable to the speed of light, it would start
accellerating less and less as it approached the speed of light which is the
"speed limit of the universe." For details, see my
earlier answer
on this problem.

QUESTION:
If the matter from a collapsed star can create a black hole (tear in space or however you define it) then how is it that after the big bang, say when ALL THE MATTER in the universe had expanded to no bigger than a baseball and gravity took effect, an ultra massive black hole wasn't created stopping the expansion of the universe dead in it's tracks.

ANSWER: First, read
my recent answer discussing the ultimate fate of the universe. Then you will
see that, depending on how mass there is in the universe, it will either
expand forever or turn around and collapse. If it is the latter, then
everything, as you suggest, will coalesce into one giant black hole,
sometimes referred to as the big crunch.

QUESTION:
Why do magnets loose their magnetic properties on heating ?

ANSWER: Ferromagnetism results from electron spins (each electron is, itself, a tiny
magnet) all aligning themselves in the same direction; this results from
quantum mechanics and happens in relatively few materials. However, when the
magnet is heated the atoms all start vibrating more violently until a
temperature is reached where the alignment is shaken apart. This temperature
is called the Curie temperature.

QUESTION:
Given the right technology, could the vacuum of space, be used to lift material into space?

ANSWER: A vacuum does
not lift, the air pressure outside it pushes. Atmospheric pressure can only
support a column of water 33 ft high if the top of the column is at zero
pressure (vacuum). So, imagine a tube from the surface of a lake all the way
into space —water would only be "lifted"
to a height of 33 ft. It is being pushed up by the pressure of the
atmosphere on the surface of the lake. Pushed is more appropriate than
lifted. See an earlier answer .

QUESTION:
Is it the case that the isotopes of an element are found together in the same ratio (in natural deposits) because some process prevents their separation?It seems to me that some process must keep them together.

ANSWER: You ask why
are they together? I might ask why not? All the isotopes of an element have
exactly the same chemical properties and it is usually chemistry which
causes things to move around. Or, the masses could play a role in the
diffusion in a gas but the masses are so close that there would be no
mechanism to cause any significant separation. There is simply no natural
process which will cause isotopes to separate. I should point out that one
of the major problems which had to be overcome in making the atomic bomb
during WWII was separating the fissile isotope of uranium, ^{235} U,
from the more abundant ^{238} U. I can think of one process by which
a local abundance ratio might be different than other places. When
spontaneous fission happens we are left with fission fragments which have a
distribution of masses but tend to be around half the mass of the fissioning
nucleus, and that would favor particular isotopes which the more common
creation (in stars and supernovae) would not. Oh, I can think of another
example: cosmic rays can interact with the atmosphere or surface of the
earth and create isotopes there whereas they would not create them deep in a
mine, for example. Carbon dating relies on such a process; if you look at
the abundance of ^{14} C in a leaf of a tree you will find a certain
percentage. If you look at carbon in wood several thousand years old, you
will find significantly less ^{14} C because it is radioactive and
decays away (half life about 5700 years); it is created by cosmic rays. So,
if you look at carbon which is old or from deep in the earth, its isotopic
abundances will differ from "new" carbon.

QUESTION:
Does friction have fundamental roots or is it purely empirical? What I mean is like how gravity and electrical forces can be calculated at the most basic level as forces between point masses or charges. Do you know if there have been experiments that say tested a single hydrogen atom or some other particle moving along surfaces. Also is friction then really a fundamental force or is it merely a different version of the electrical forces of atoms?

ANSWER: Friction like
you are thinking about, sliding of one surface on the other, is mechanical,
little hills and valleys on the surfaces impeding the motion. But,
mechanical forces like this ultimately may be traced to forces between
atoms, that is to electromagnetism. So, you can say that friction is,
essentially, due to the electromagnetic force. As you know, it is often
taught in introductory courses that the force of sliding friction is
proportional to the normal force pushing the surfaces together, f= μN.
This is not a law of physics, it is empirical and only
approximately true for relatively smooth surfaces and modest normal forces.
See one of my earlier answer s
for more detail.

QUESTION:
If Helium-II has a viscosity of 0, does that mean you could accelerate it without friction? If so could you use an accelerator like the LHC (but smaller) to accelerate Helium-II instead of a particle beam? If so could it reach near light speed? If so what would happen?

ANSWER: Superfluidity
is a collective phenomenon, it depends on there being many atoms interacting
with their neighbors. In an accelerator it is not friction which limits
maximum speed, it is simply the design of the machine; friction plays no
role for individual particles. Anyway, to accelerate it as you suggest, you
would need individual helium atoms which would be singly or doubly ionized
and it would therefore not be superfluid helium.

QUESTION:
I understand that matter gains in mass as you accelerate it towards the speed of light, and that as you approach the sped of light, mass tends toward infinity, meaning you would need an infinite force to "push" it beyond the speed of light, and that's one of the reasons why nothing can exceed the speed of light. However, I also understand that with increased mass comes increased gravity, so... When they accelerate those sub-atomic particles to velocities approaching the speed of light at CERN, why don't they suddenly start exerting lots of gravity and sucking the entire CERN campus toward them?

ANSWER: First, a
little perspective. If I were to bring a baseball into the CERN campus, I
think you would agree that it would not "suck in" everything
gravitationally. So, an accelerated proton better be a whole lot more
massive than a baseball! Suppose the speed of the protons were 99.99999999%
(that's ten 9s) of the speed of light. Then you can calculate the mass of a
proton going this fast as m =1.7x10^{-27} / √(1-0.9999999999^{2} )=1.2x10^{-22}
kg. The proton has increased mass by a factor of about 70,000 and that is
still far from a baseball mass.

QUESTION:
I know water boils in a vacuum, but the question I can't seem to find an answer to is, does the temperature of the water remain constant? If I'm in a vacuum with a bowl of water, will the boiling water scald my hand or will it just be room temperature water in a boiling state?

ANSWER: If you do not
let any heat flow in or out and keep the volume constant, as the pressure is
decreased the temperature will remain constant. You can see from the
water phase diagram that the pressure will eventually reach the boundary
between liquid and vapor at that temperature and boiling will commence.

QUESTION:
What pattern does monochromatic light form when the double-slit experiment is performed in a vacuum?
If it forms the same interference pattern that is seen in double-slit experiments that are performed in the presence of air, does that mean that a vacuum can propagate a wave?

ANSWER: If you
performed the experiment under water the pattern would be noticeably
different because the speed of light in water is quite a bit different than
in air and therefore the wavelength would be different and, of course, n λ=d sinθ _{n}
so the pattern changes if λ changes. However, you would not see much
of a change if you did the experiment in a vacuum because the speed of light
in air is only a tiny bit slower than light in a vacuum. Which brings us to
your most interesting question, can waves propogate in a vacuum? Only one
wave, light, can propogate in a vacuum. Electromagnetic radiation needs no
medium in which to propogate.

QUESTION:
I'm a 16 year old and i would appreciate if you help me answer a question. I was taught that the colour black absorbs all the colours of light. However, i find it puzzling to see smooth, black surfaces reflecting light. Did i have a wrong concept or not understand something?

ANSWER: Well, this
should tell you that all the world is not just black and white unlike some
physics texts would have you believe. There is no such thing as an object
which actually absorbs all radiation which falls on it. So, even if an
object is not smooth, it will not reflect all the light on it. When you look
at a rough black object, you can still see features on its surface, right?
So some light must be reflected. If it is real smooth, it can work like a
mirror, even though it looks black "behind" the mirror, it still reflects
from the surface.

QUESTION:
Two bikes are traveling at the same speed of 10 mph towards each other. Starting at a distance of 20 miles away. A bee is flying from front tire to front tire of the two bikes at a speed of 25 mph. After several trips back and forth the bikes collide and the crush the bee. What was the total mileage of the bee in the many trips back and forth before the crush

ANSWER: I have
answered similar question s before. This is
probably your homework problem, not really allowed. Look at my earlier
answer and figure it out for yourself.

QUESTION:
I was playing paintball one day, and was shooting in a "range" for target practice while i as waiting for the next game. At the time i was reading a book called "The Big Bang" (I don't remember who it's by...). I had just read a part about space-time. Anyways, i was aiming for a particularly small target, and i just couldnt hit it. I would shoot at the target and it looked like it was going to hit it, but at the last second it would veer off to the left or right. I would adjust accordingly and the same thing would happen, but it would go to the toher side. I don't know why it was doing this. So my question is: Does space-time have any affect on small projectiles? I apologize if this is a stupid question, but im not particularly exceptional at physics...

ANSWER: Hey, no
apologies on my site! There are no stupid questions …OK,
there are, but yours does not qualify for the "off
the wall hall of fame "! You may be sure that space-time is not causing
your paintballs to behave erratically. The possible explanations include
your not being as good a shot as you think you are or, more likely, since a
paintball is not rigid, its interaction with the air at a high speed is
unpredictable like a knuckleball in baseball. Even a pitcher has no idea
what a knuckleball will do.

QUESTION:
Could the force of gravity be the direct, measurable result of the
universe expanding? I really hope you get back to me about this, i have
this idea in my head I just cant shake.

ANSWER: I do not see
how!

FOLLOWUP QUESTION:
everything has an equal and opposite reaction. What is the opposite of the universe expanding? because matter and atoms etc are the only objects in the universe, they collectively counter the expansion. Maybe not equally (as some of the energy is expended in other forms). Am I just stupid? i have not done physics since high school lol.

REPLY: "everything
has an equal and opposite reaction"? This has to be one of the most misused,
misunderstood, incorrect extrapolations in all of physics! Where does it
come from? Newton's third law states that if one object exerts a force on
another, the other exerts an equal and opposite force on the first. Some
jerk (sorry, I get a little "het up" about this!) thought that a good
paraphrase of this would be "for every action there is an equal and opposite
reaction". It just sounds so cool and profound, it must apply to everything
you might want to apply it to. And my years of of teaching have taught me
that it is a neverending source of confusion for students. I often get
questions like "If I push on something, why does it move? Why does the
reaction force not cancel out my force?" Well, duh, because the reaction
force is not on the same thing I am pushing on, it is a force on me. "What
is the opposite of the universe expanding?" Well, I guess it must be the
universe contracting? But, if it is expanding, it certainly can't be
contracting, can it? Then where the heck is that "opposite reaction"?

I usually don't answer astrophysics/astronomy questions, but let's talk
about the expanding universe a bit. So, the big bang happens. This imparts
energy to everything which has come into existance and it is all flying
apart right from the start. So here is the question which has been a
longstanding holy grail for astronomers, is what is the ultimate fate of our
universe? Since we observe the universe expanding, is it must be doing one
of two things: since there is only gravity acting on everything in the
universe, the gravity is trying to slow things down, either

if there is enough mass in the universe,
eventually it will turn around a collapse back into a "big crunch" or

if there is not enough mass in the universe, it
will keep expanding forever and, since everything is getting farther
apart, the gravitational force will just get smaller and smaller until
everything is just about drifting with a constant speed.

But, wait!

Several years ago an amazing discovery was
made. Measurements on the most distant objects in the universe have
shown that these objects are actually speeding up. So, there is some
force other than gravity, it seems, which must be repulsive instead of
attractive like gravity. This is what is often refered to as dark
energy .

Anyhow, this is a longwinded answer but here is the
bottom line: keep action/reaction to Newton's third law, not just anything
which comes to mind!

QUESTION:
If a train is moving with the speed of 180 km/h and a mosquito is flying inside the train with speed 10km/h. Will it collide with the train if yes! why yes ,if no! why no?

ANSWER: I always am
getting questions like this where velocities are given but the questioner
does not specify velocity with respect to what ? I am going to assume
that the train has a speed 180 km/h relative to the ground and the mosquito
has a speed of 10 km/h relative to the train forward. The bug will
eventually collide with the front of the train. If the bug's velocity is
backward, it will eventually collide with the back of the train. You ask why —you
do not need to know the speed of the train, all that matters it the bug's
velocity relative to the train. If you walk forward in the train, won't you
eventually get to the front of the train?

QUESTION:
If one has two vectors equal in magnitude and direction but with different units, then can the two vectors be considered equal to each other? For example, a displacement vector of 1m towards east and a velocity vector of 1m/s towards east.

ANSWER: No. You
cannot write x =v because it would not be
dimensionally correct. Does one watermelon equal one potato? However, it is
not really "different units" you want to ask about because, if A =1
foot north and B =12 inches north, you may write A =B ; these
vectors have the same dimensions (length) but different units.

QUESTION:
is
the force applied by a vertically faalling body equal to its weight?

ANSWER: If it is falling, it is not exerting a force on anything (except a little force on the air it is falling through). If you mean how much force does it exert on something it hits which stops it, there is no way to calculate that without knowing its speed and how long the collision lasted.
First, see a recent discussion of Newton's second
law. What determines the force is the mass of the object, its speed when it
hits, and how long it takes to stop. If it is 1 kg moving with a particular
speed, if it is a brick it stops quickly and hurts a lot (big force), if it
is a piece of foam rubber it takes a longer time to stop and does not hurt
(small force). But both were 1 kg moving with the same speed. If you drop
them from higher up, the force is bigger because the speed at impact is
bigger.

QUESTION:
If water was preserved in a very solid metal container where air was completely extracted from inside the container and u put it to freeze, how wud the water expand when there is absolutely no way??

ANSWER: See an
earlier answer .

QUESTION:
Is it ethical for cellphone companies to sell cell phones without a written health warning to the user?

ANSWER: I personally
feel that there is zero good evidence that cell phones are hazardous to your
health. See my recent answer on this topic; if you are serious about forming
an opinion, be sure to read the linked-to New York Times article.
Given the current status of the evidence, if cell phones must have a
warning, then so must coffee and pickles, both of which have been listed as
"possible carcinogens" by the World Health Organization.

QUESTION:
what is the difference in properties between dark matter, dark energy, and antimatter?
How does this relate to the universal law of "self preservation"?

ANSWER: Dark matter
and dark energy are related to puzzling observations in modern astrophysics
and have been discussed in an
earlier answer . Antimatter is a very different thing (and unlike dark
matter and energy, has been actually observed directly). Symmetries in
nature make necessary in nature particles which are the antiparticles of
corresponding elementary particles. The one commonly observed is the
antiparticle of the electron, called the positron, which occurs naturally
when a nucleus with too many protons to be stable decays by changing a
proton into a positron, a neutron, and a neutrino. However, positrons are
soon gone since if they encounter their nemisis an electron, the two totally
annihilate each other and result in electromagnetic energy corresponding to
all the mass energy the two had before annihilation. I have no idea "what
the
universal law of 'self preservation'" is, but it certainly is not physics!

QUESTION:
why do paratroopers roll on landing

ANSWER: When you hit
the ground, the ground exerts a force up on you to stop you. The force which
it exerts depends on how long the collision lasts. This is a result of
Newton's second law which says that F=m Δv /Δt
where m is your mass, Δv is your change in speed, and Δt is
the time to stop. You have no control over your mass or the speed you hit
the ground, but you can make Δt longer and rolling is one trick for
doing that. The longer you can make your collision with the ground last, the
less it will hurt. Also, just jumping off a table, for example, it is more
painful if you hold your legs straight than if you bend them while stopping.

QUESTION:
I hear all the wizards say that our H20 arrived on earth as a result of Comets and Meteors. I have trouble accepting that as the answer, because of the abundance of H20 on our planet? How many Comets and meteors would be required?
As a common-sense thought:
Something else seems to me, more a more practical and possible solution.
We are told that the early composition of Earth's atmosphere was mostly Hydrogen. Plus, it was violently hot, with an abundance of volcanic and lightning activity.
Fire + Hydrogen = BOOM!
The by-product of burning Hydrogen = H20.
Why wouldn't this be a better explanation for our abundance of H2O?
I listen and read constantly. I've never heard a theoretical physicist suggest this as a plausible explanation. So, I automatically assume I can't possibly be correct.
Then; I wonder . . . why not?
So . . . that's my question.
Why not?

ANSWER: As I say on
the home page, I usually do not answer astronomy or astrophysics questions,
but I think I can debunk your idea. There is no helium or hydrogen in ther
earth's atmosphere. Why is that? The reason is that, at current
temperatures, the speed of those elements is so high that many have a speed
big enough to escape the earth's gravity and fly out into space; soon they
all leak away. If temperatures were even hotter, as you suggest, hydrogen
would leak into space even faster. So your supposition that "… the
early composition of Earth's atmosphere was mostly Hydrogen… "
could not possibly be true. The earth's gravity is too weak to hold a
hydrogen atmosphere at temperatures which might have been present.

QUESTION:
If all inertial frames of reference are equivalent, and there is no way to identify which of two objects which are becoming further apart is moving, how is it possible to determine which of two objects is accelerating? For example, how do we know that a train is accelerating away from a platform, and not vice-versa, if it is impossible to determine which is moving when the train is traveling at a constant velocity?
If all inertial frames of reference are equivalent, and there is no way to identify which of two objects which are becoming further apart is moving, how is it possible to determine which of two objects is accelerating? For example, how do we know that a train is accelerating away from a platform, and not vice-versa, if it is impossible to determine which is moving when the train is traveling at a constant velocity?

ANSWER: One way you
can identify an inertial frame is if Newton's first law is true (an object
at rest or moving with constant velocity has zero net force on it). For
example, sitting at the train station there is a pendulum bob hanging
straight down; the earth is pulling straight down with a force (called the
weight) and the string is pulling straight up with a force which may be
measured to have the same magnitude as the weight. Now, if somebody is
accelerating in a train going by and also has a pendulum, it will not hang
straight down but at an angle. But there are still only two forces on that
pendulum bob, the weight straight down and the string not straight up. So,
if you are in the train's frame of reference, you conclude that Newton's
first law is not true.

QUESTION:
What is the function of neutron in the nucleus of the atom ?

ANSWER: You might as
well ask what is the function of the proton. Neutrons and protons both feel
the strong interaction (nuclear force) which holds nuclei together, protons
are stable, and neutrons are almost stable but are stable when inside a
nucleus. Although the coulomb force repelling the protons is weak compared
to the nuclear force holding the nucleus together, it still plays an
important role inside the nucleus because the protons are so close together.
There are, in fact, no stable nuclei with only protons and you might say
that adding neutrons helps keep the protons a little further apart reducing
the coulomb repulsion. If a nucleus has too many protons, one of the protons
will turn itself into a neutron by ejecting a positron and a neutrino
(called
β^{+} decay) .

QUESTION:
Why don't iron filings simply align themself due to the magnetic field , why is it necessary to tap the board on which thay lie?

ANSWER: The filings
have friction between them and the surface they sit on. The force from the
field is not big enough to move them. If you tap on the surface, they tend
to "jump up" either eliminating or reducing the frictional force for a short
time, time enough for the filings to align with the field.

QUESTION:
magnetic powered water pumps by cars over road ways, and eme produced from traffic also on road way , based soley upon magnets placed in locations throught the hy system and on vehicels. Is this feasible, Or just another pipe dream??

ANSWER: One thing you
have to remember —there is no free
lunch. If you try use magnets to extract energy from a passing car, eddy
currents in the metal of the car will cause drag on the car which will
require additional fuel to be burned to keep the car moving at a constant
speed.

QUESTION:
I'm having a hard time understanding virtual particles in empty space.
E=mc^2 says that matter and energy are equivalent.
Is there an energy field in empty space? If not, then the math doesn't seem to work out.
If empty space is literally nothing, no energy field, but nothing, zilch, zip, nada, NO THING, then how is it possible for an electron and postitron pair to form?
The problem is, suddenly we now have an electron and positron pair that formed out of nothing, zilch, zip, nada, NO THING and both have mass. Then the electron and positron pair annihilate leaving energy in the form of heat:
E=(e-)c^2 + (e+)c^2
Please make the appropriate corrections to the above equation if I have expressed it incorrectly, but what I am trying to say is that the mass of the electron and the mass of the positron become energy when they annihilate.
This left over energy should be causing the universe to be heating up since now there is energy where there was none before.
Of course if empty space is an energy field and the formation of electron and positron pairs is simply existing energy transforming briefly into matter then annihilating back into the same existing energy, then there is no net gain in energy. The universe will not be heating up.
Even if these are not real electron and positron pairs or they exist so briefly that somehow the law of conservation of energy isn't violated, there is still the problem of Hawking Radiation.
A positron and electron pair form out of nothing, zilch, zip, nada, NO THING near the event horizon of a black hole and the positron falls into the black hole leaving the electron--NOW A REAL ELECTRON--to drift off into space, we now have the creation of new matter--this new electron--where none existed before! Doesn't this violate the law of conservation of energy?

ANSWER: Calm down,
for heaven's sake! The important point you are missing in your rambling
question is the Heisenberg uncertainty principle (HUP). Usually we think of
the momentum-position version of the HUP,
Δx Δp ≈h /2π, you can not simultaneously know
momentum and position to arbitrary precision. But, there is a similar
version involving energy E and time t , ΔE Δt ≈h /2π,
you can not simultaneously know energy and time to arbitrary precision.
What this implies is that energy conservation need not be obeyed provided it
lasts only a very short time. Indeed, creation of the rest masses of an
electron-positron pair violates energy conservation, it is OK as long as
they don't last very long. Because they are virtual, when they recombine
after a short time (annihilate), they do not create photons but simply go
away, pop back out of existence.

FOLLOWUP QUESTION:
I will try to restate the second part of my question more soberly (with less rant) regarding Hawking Radiation. I heard somewhere that in Hawking radiation, the "virtual particles" such as an electron and positron pair pop into existence of nothing--you know the drill--nothing, zilch, zip, nada, no thing near the event horizon of a black hole. As in the previous question we need to consider Einstien's mass energy equivalency, E=mc^2 since now we have an electron and positron pair that popped into existence from nothing, zilch, zip, nada, no thing:
E=(electron)c^2 + (positron)c^2
As in the previous question the positron falls into the black hole leaving the electron to drift off into space. So now we have:
E=(electron)c^2
Doesn't this violate the law of conservation of energy?

ANSWER: No, energy is
still conserved because the black hole provides the energy which the
escaping particle carries away. That is the whole point of Hawking
radiation, it is a mechanism by which black holes may "evaporate". It was
previously thought that a black hole was forever and the inevitable ultimate
fate of the universe was a bunch of black holes or at least a black hole
could never get smaller, only bigger.

QUESTION:
Is the speed of light constant?

ANSWER: The speed of
light in a vacuum is a universal constant, it is the same regardless of the
motion of the observer or the source. To understand why, see the
FAQ page. In a material medium (think
of glass, water, diamond, etc .), light travels more slowly.

QUESTION:
I had a question regarding relativity. If I'm not mistaken, the principle of relativity states that different reference frames are essentially equivalent in that you can't discern a "stationary" reference frame from a "moving" one (that basically there is no such thing as absolute position). So, i was considering a collision of two objects, one of which is much more massive than the other (say, a train and a person). From an observer relative to the person, the train is approaching at a certain speed (say, 5 mph). if the train were to crash into the person, the result would be very bloody and cause severe damage to the person. If, however, this were viewed by an observer moving alongside the train, it would appear as if the person simply walked into the train at 5 mph, and would emerge damaged, but not severely harmed. How is it that this doesn't contradict the principle of relativity? An event's result seems to depend on a "stationary" reference frame (that either the person or the train is moving), when according to relativity the result should be the same regardless of the observer's frame of reference.

ANSWER: I'm sorry,
but your assumption that the 5 mph train will be a violent accident is
simply wrong. If you are hit by a 5 mph train you will suffer the same
injuries as if you run into a train at a speed of 5 mph. Think about it —as
the train approached you could simply grab on to the front and start moving
with it. It would hurt a little bit, but no more than if you ran real hard
into the train at rest. If the train were going 100 mph, it would kill you,
but if you were launched at 100 mph into the train, you would die with
similar injuries.

QUESTION:
Why does speed of light reduce when entering a denser medium e. g. glass, and why does it return to initial speed when reentering original medium.
Why doe different colours(frequencies) of light refract at different angles through a prism.

ANSWER: I am afraid
that there is no simple explanation for this, it is probably best to take it
as an experimental fact. The answer requires consideration of how the
oscillating electric fields interact with the electrons in the medium
through which it is traveling. The reason for the the prism effects is that,
when a more detailed analysis is done, it turns out that light of different
colors travel with different speeds in a material. This is called
dispersion. So, when we say the index of refraction of some medium is, say
n =1.33, this is an average number and there are slightly different
values for colors, that is the index of refraction is actually a function of
the wavelength.

QUESTION:
Bearing in mind that the earth revolves every 24 hours, if I wanted to travel from, say, England to Australia, why cant I float up into the sky and 12 hours later float down into Australia? Thus saving about 12 hours flying time and huge amounts of fuel!

ANSWER: Even if this
scheme could work, you could travel only due west, starting in England and
ending up in Mongolia, maybe. But, it does not work because when you float
you are at rest with respect to the air you are floating in and the earth
drags its atmosphere along with it.

QUESTION:
In E= mc2, c2 cannot merely be velocity squared because a massless photon would then have no energy.

ANSWER: Always check
the FAQ page before asking a
question.

QUESTION:
When we are talking about the spacial dimensions it confuses me how we can talk about the first and second and third dimensions as separate from each other. My question is essentially how can there actually be flat planes in the physical universe; are they only conceptual or do physicists think an object can have length without width or depth?
It seems like common sense to me that it is not possible, but I was watching a Carl Sagan clip in which he talks about flat-landers. If they are flat they would still have a finite three dimensional shape I would think...

ANSWER: The
philosopher/mathematician Ren й
Descartes (1596-1650) was the first person to ask "how many numbers must
I use to unambigously specify where an object is at a given time?" Legend
has it that he asked this question while watching a fly flying around his
room. He realized that the distance from the floor (defining a plane
parallel to the floor), the distance from the north wall (defining a second
plane intersecting with the first in a straight line), and the distance from
the east wall (defining a third plane intersecting the line at a point)
would do it —three lengths need be
specified. That is why we call that coordinate system Cartesian . Now,
of course, these planes are constructs of our imagination, nobody would
think that they actually exist in a real way in the "physical universe". But
they are not unique ways of specifying a point in space, there are dozens of
coordinate systems which have been dreamt up over the years, many of which
are far superior to Cartesian coordinates for solving specific problems. But
Descartes might also have asked how many numbers he needed to specify where
the fly was if walking around the floor; clearly the answer is only two,
this is a two dimensional problem. Or he could have asked how many he needed
if the fly was walking along a clothesline strung across the room, just one
for this one dimensional problem. But the world as we perceive it stops at
three dimensions. A two dimensional problem does not have to be a plane. The
surface of the earth is two dimensional and we need only two numbers to
specify where we are, longitude and latitude. You can imagine how cumbersome
it would be to try to use cartesian coordinates to specify where we are on
the earth. A three dimensional coordinate system often used is spherical
polar coordinates which adds the distance from the center to the longitude
(azimuth) and latitude (polar). Finally, the whole flatlander idea is based
on the premise that the people are truly two-dimensional, like characters
printed on a page, not capable of perceiving a higher dimension than two
they exist in, just as we are incapable of perceiving a higher dimension
than the three we exist in.

QUESTION:
Why
are massless photons stopped by paper while a neutrino (with mass),
easily passes through the earth?

ANSWER: A photon
interacts strongly with electric charges of which all matter is composed.
Neutrinos do not.

QUESTION:
In V=W/Q or W=VQ where V is the potential difference, W is the work done to move a unit negative charge Q, from one point to another, then what is here the force and displacement as we know that work done is force multiplied by displacement ?

ANSWER: The only
thing that the work in moving a charge between two points depends on is the
potential difference between the points. If V =10 V between points A
and B, it makes no difference whether they are 1 cm apart or 1 km apart, the
work necessary to move a 1 C charge from A to B is 10 J. But, it is still
true that work can be computed as force times distance, although the force
might not be a constant so you cannot just write W=Fs . It follows
then, that the average force experienced by the charge in the example
above must be 10,000 times bigger when A and B are separated by 1 cm; the
fields are stronger when A and B are closer together.

QUESTION:
I have a question on Law of Decay.I never heard of it before and I am quite confused with the question asked.Its about a radioactive material that is known to decay at a rate proportional to the amount present. What does the mass of the material initially,got to do with the time taken for example,5 hours, and after this hours ,the material has lost some percentage of its mass.Can you try explain to me what's this whole thing about and how do I even get to start??

ANSWER: This is very
fundamental and can apply either to decay or growth. The idea is that the
rate of decay or growth is proportional to the population. Does it make
sense? Suppose we have 10,000 radioactive nuclei and we measure that they
are decaying at a rate of 50 per second; it is then perfectly reasonable to
assume that if we had 20,000 instead, we would see 100 per second decaying.
It simply reflects the situation that it is equally probable for any of the
population to decay in the next 5 seconds, for example. It also applies to
many growth problems. For example, if you have 10,000 bacteria and they are
growing at the rate of 50 per second, the rate of growth of 20,000 of them
would be 100 per second. The general way to write this down (I assume you
know some calculus?) is dN /dt = ï¿½λN
where N is the number at any time t , the + sign is for
growth, the - sign is for decay, and λ is called the decay constant,
the proportionality constant. You now have to solve this equation for N . It
is a basic differential equation, easily shown to have a solution N=N _{0} e^{ï¿½λt}
where N _{0} is the number at the time t =0. The
graph shows two decay curves with N _{0} =10,000, the black
curve with λ= 1 and the red curve with λ= 2. λ may also
be related to the half life T _{Ѕ} , the time it takes the
initial population to drop to N _{0} /2, by T _{Ѕ} =0.692/λ ;
note that the black curve is at 5000 at about t =0.7.

QUESTION:
Force is a push or pull. Is it just the transfer of energy from one body to another? If yes, why does it mostly result in the acceleration of the body?
If no, then how does force travel?

ANSWER: A force
transfers no energy unless it does work. For a force to do work, it must act
over some distance. So, if you push against a wall, you do no work and the
wall acquires no energy. If you throw a baseball, the force from your hand
on the ball acts over a couple of feet and gives the ball kinetic energy.
So, the answer is either yes or no.

QUESTION:
hey man,
we had a discussion at the bar and we were not sure about this.
here it goes.
when you knock on a vacuum? like, u have a trunk with a vacuum inside.
what will be the sound. is it A. none at all? B. like a solid piece of wood?( in the case of a trunk ) c. something else
It may not be the greatest of questions but i'd apprieciate any answer.

ANSWER: What sound
does something make when you "excite" it? Something has to move to make a
sound. The classic example would be a guitar string. If you just had a
string stretched between two nails in a concrete floor, you would barely be
able to hear it when plucked. Now, if you connect the string to a flexible
piece of wood (like with the bridge of a guitar), the vibrations of the
string will be transferred to the wood which will also start vibrating in
some complex fashion (like the head of a drum makes sound). You may now be
able to hear it a little better. Now put a volume of air behind the wood
(that is, make a guitar). Now the whole volume of air will start vibrating
which will amplify the sound made by the vibrating wood which is amplifying
the vibrating string. There is a lot more to a guitar than a plucked string.
If you now close up the hole and pump out that vibrating air, you will still
be able to hear the guitar but it will not be nearly as loud (or as pretty)
without that vibrating chamber of air.

QUESTION:
Will a stationary wave formed when two wave trains with same speed, frequency but different amplitudes moving in opposite directions overlap? I know if the frequency and amplitude were the same, a stationary wave is formed. But I am not sure when the amplitudes are different. My guess is it will form a stationary wave as long as the frequency is the same. But when tried it in a simulation, it looks as if the resultant wave is not stationary.

ANSWER: No, the
amplitudes must be the same to result in a standing wave. One of the
characteristics of a standing wave is that there are times when the
displacement is everywhere zero and this cannot happen unless you have equal
amplitudes. Imagine one wave, going south, with an amplitude 1000 times
greater than the amplitude of another wave with equal frequency going north;
surely all you will see is a big wave going south.

QUESTION:
I was taught in a meteorology class that the Coriolis force is proportional to the speed of an object. The faster an object is moving, the greater the amount of deflection. The classic example of the Coriolis force is the deflection of a ball when tossing it on a merry-go-round. On a counterclockwise rotating merry-go-round, a person tosses a ball to someone directly across from them. The ball appears to deflect to the right (from a rotating reference point). However, if you toss the ball fast enough it will reach the person on the other side before it has time to deflect too far to the right. So, wouldn't an object with a lower speed have greater deflection?

ANSWER: You confuse
force with deflection. Big force does not mean big deflection. Consider a
bullet fired horizontally from a gun and an identical bullet thrown
horizontally from your hand. Both have the same force on them but the fired
bullet has considerably less "deflection". I hope you keep in mind that the
Coriolis force is a
fictitious force .

QUESTION:
Hey, so it's pretty basic physics that tells us why some planets revolve around our sun, it's because of gravity, but gravity tells me that the Earth should just spiral into the sun, so the question is what is defying gravity and keeping us from colliding and keeping orbit?

ANSWER: " …gravity
tells me…"? Consider a car going around a circular track. Is there a force
needed to keep it going in a circle? Yes, the friction between the wheels
and the track exert a force toward the center of the track. So, here is a
force pointing in toward the center, so why does the car not spiral into the
center? Because the force is changing the direction of the car's velocity,
not dragging it in. The same thing happens with gravity—it causes a closed
orbit, not one that spirals in. Not all forces necessarily work so nicely.
It just happens that a force like gravitational or electrical, forces that
fall off like 1/r ^{2} , give rise to stable orbits.

QUESTION:
Why does liquid flow out of a container faster (or in greater quantity) from a spout after a separate nozzle for air flow is opened?

ANSWER: As the liquid
leaves, the volume of air gets bigger so the pressure gets smaller. So the
atmospheric pressure on the outside is larger than the pressure of the air
inside the container which tries to keep the liquid from leaving. If another
opening is made, the pressure inside will be the same as outside and the
liquid will flow more easily.

QUESTION:
under momentum in physics why is it that when you shoot a gun the gun moves backward

ANSWER: Momentum of
an isolated system (no external forces on it) is conserved. The gun plus
bullet system may be approximated as isolated because the only force on it
is the explosion of the bullet which is internal. The system starts with
zero momentum. After the bullet is fired, the bullet has momentum so the gun
must also have momentum in the opposite direction so that the momentum is
still zero.

QUESTION:
It was explained to me that if I jump off my roof into my pool, in free fall I am at rest and the pool is accelerating up towards me. The exact quote: "if you were to go jump off of your house into your pool, you aren't falling into the earth--you stay perfectly still and the earth comes up to catch you." This doesn't make any sense to me.

ANSWER: The reason it
makes no sense to you is that it is utter nonsense. The earth exerts a force
down on you, called your weight (Newton's universal law of gravitation).
This force causes you to accelerate toward the earth with an acceleration
inversely proportional to your mass (Newton's second law). You exert a force
up on the earth which, in magnitude, is equal to your weight (Newton's third
law). This force causes the earth to accelerate toward the you with an
acceleration inversely proportional to the mass of the earth (Newton's
second law). So, you accelerate toward each other. However, since the mass
of the earth is huge compared to your mass, the earth has a tiny
acceleration, far too small to ever hope to measure. For all intents and
purposes, you fall to the earth, it does not fall to you.

QUESTION:
I refer to one of your previous posting where a school teacher asked if a weight of 1kg is dropped from 1m on to the head of a teacher how much force is exerted on the teacher. Your response was that F=M*A so unless you know how long the weight was in contact with the teachers head you could not calculate the force.
I understand what you are saying that the longer the force is is contact the less the accel, thus the less force is imparted. But I think your answer, respectively is not quite correct.

Instead of calculating the Force in terms of F=MA we should be actually calculating the momentum of the of object in N.m/s ?.

My question is, if we calculate your way (F=M.A) then if the object hits the head of the teacher and instantaneously decelerates to 0 M/s we should consider instantaneous deceleration as A=1?

Because if Acceleration = (Vfinial-Vinitial)/1 which means the total force is imparted, however if the object was in contact for only 0.1 seconds then F=M*((Vfinal ЁC Vintial) / .01s) - F=9.8N *(0 ЁC 9.8M/s) / .01 = 9604Nm

(Can you see my point that 1 sec = full force but 0 .1 is still an incredible amount of force that cant be right.?)

Could you please clarify? And show how to calculate the force in terms of momentum force of the object which must be complexly transferred to the teachers head if the velocity of the 1KG comes instantaneously to a complete rest, and in 0.1 and 1 second intervals?

And also if the object was a ball and spinning as it falls which means we should calculate the impact in terms of Inertia = Energy transfer to the teachers head.

ANSWER: You have
incredible misconceptions. I have converted your question to a numbered list
so I can answer your multiple questions.

I don't get this statement at all. Momentum is
momentum, not a force, and N m/s is the units of neither force (N=kg m/s^{2} )
nor momentum (kg m/s). If you want to bring momentum into the
discussion, then you should write F =

Δp /Δt
where p=mv and Δp is the change in momentum. Δt is
the time over which momentum changes. But, since mass is not changing,
this is exactly the same as F=ma .

No acceleration is instantaneous, and if it
were, it would be a =∞ because instantaneous means
Δt =0. Most certainly, a is not 1 m/s^{2} . It
always takes time to stop, a short time means a big force (a 1 kg brick,
for example) and a long time means a little force (a 1 kg pillow, for
example).

All your numbers are wrong and you do not even
seem to know what mass is. Mass is 1 kg, not 1x9.8 N which is what
weight is. The initial velocity is not 9.8 m/s. If you do the
kinematics, you will find that after an object has dropped 1 m it has a
speed of about 4.4 m/s (neglecting air friction). Hence, if the object
stops in 0.01 s, the force is (1 kg x 4.4 m/s / 0.01 s)=440 N. This
force lasts for 0.01 s and is actually the average force over that time.

You have no point. What does "1 sec =
full force" mean? Why can "an incredible amount of force" not be
"right"? A brick dropped from 3 ft above your head can really hurt.
Probably 0.01 s is a little short, but not out of the ball park.

There is no such thing as "momentum force".
Again, instantaneous stops do not happen.

This is irrelevant here. Furthermore,
"inertia=energy transfer" has no meaning.

Tell me, please, that
you have never had a physics course. You do not seem to have a clue.

QUESTION:
Does gravity have a speed or is it instantaneous? My thought is...if two bodies are orbiting each other (Earth and Moon for instance), is the gravitational attraction between them a vector force that points directly at the center of the other's mass, or does the vector point somewhere behind its mass due to the finite speed of gravity?

ANSWER: Always check
the FAQ page before asking a
question!

QUESTION:
When your eye sees light, are those photons "lost" in the absorption and subsequent conversion to signals to the brain? Is the original source of light diminished by the amount of photons absorbed by the eye (or a measuring device is used in lab)? I am trying to figure out if light that travels through the universe remains whole or are photons lost due to to the simple act of "observation".

ANSWER: Yes, the
photon is destroyed in your eye. The original source lost those photons when
they left, not when they got absorbed. A photon traveling through the
universe continues until it interacts with something.

QUESTION:
If the gas discharge tube contains MOLECULAR hydrogen, why do we only see ATOMIC hydrogen line spectra, not molecular?

ANSWER: Because the
tube ionizes the gas which causes the H_{2} to dissociate into
atomic H.

QUESTION:
Why can electomagnetic radiation travel through the vacuum of space?

ANSWER: Because the
waves are time varying electric and magnetic fields which do not require a
medium to exist. See
earlier answer .

QUESTION:
What is the smallest instance of time?

ANSWER: Nobody really
knows whether time is continuous or discrete. See an
earlier answer .

QUESTION:
Last night I heard an interview on the BBC with a man from Sweden who had tried to build a nuclear reactor in his kitchen. What disturbed me most was that he had bought radium on eBay. Admittedly it was only a small amount, about half a gram. I know that the conversion of mass to energy is normally inefficient, but for a moment, assuming 100% efficiency, then given that the square of the speed of light is an enormous number, at first sight it looks as though half a gram of radium might potentially produce a rather big bang. I wondered quite how big it might be, compared to, say, a firework, a stick of dynamite, or a ton of TNT, whichever is the most appropriate. I am so glad the police stopped him before he got his project underway!

ANSWER: Radium is not
a material suitable for a chain reaction. And, a half gram of it is not
enough, even if it were plutonium or uranium, to make a critical mass for a
chain reaction to occur. And, 100% mass conversion is so far off what is
possible that it does not merit discussion. A completely "efficient" fission
reaction (by which I mean 100% of the atoms undergo fission) of uranium or
plutonium yields only on the order of 0.1% mass conversion. The danger in
this instance is, mainly, to the man himself from the radiation, very little
other danger.

QUESTION:
Baseball pitcher throws 90 mph in 60-dgree ambient air temp. If he threw with the same force in 90-degree ambient air temp, would the pitch be faster or slower. If I remember my physics, shouldn't humidity play a role? If so, make humidity constant. If colder air is denser (irrespective of humidity), then wouldn't the same force produce a faster pitch in warmer ambient air temperatures?

ANSWER: A couple of
problems here. Constant humidity does not mean the same amount of water in
the air at different temperatures; humidity is the fraction of the maximum
water that the air can hold at a given temperature, and cold air can hold
less. Density of a gas is not solely determined by temperature, pressure is
also a factor. So, let's just say the humidity is zero and the atmospheric
pressure is constant. Then, you are right, the 90 mph pitch (leaving the
pitcher's hand) in cold air will have lost more velocity when it reaches the
plate because the density of the air is greater. However, the effect will, I
believe, be rather small; the difference in densities is only about 5%.
Since the
speed lost by a 90 mph fastball is about 10 mph, the difference in final
speeds would be be about 0.5 mph (one arriving with 80 mph, one arriving
with 80.5 mph). There are lots of FAQ
links to questions about air drag.

QUESTION:
I read that atoms come together to form molecules, which combine to form solid matter. I'll site the uranium
atom which has I believe 92 electrons spinning at high velocity in orbit around the nucleus. some may be at lower orbit than others.
So, all these negative charged electrons occupying this "shell" of <---> electric charge. Would that not create some type of a barrier, which
would prevent the atom from approaching another due to this perimeter ?? of negative electric charge ?
I know I'm reaching out too far here as a "novice", but I imagine this outer electron perimeter as having an electron located at any given
point, at any given moment. Or, no electron located at any given point at any given moment. Because of their velocity they would seem to be
everywhere all the time, thus creating this field (barrier) ?? of negative electric force. " LIKES REPEL"
Also may I add another point here, If a single proton reaches out to attract a single electron, why does the electron stop at a great
distance from the proton. What causes it to stop instead of going "all the way", BUMP ????

ANSWER: The simplest way for you to think about this is to think of the electrons as occupying concentric spherical shells around the nucleus. And,
only a limited number of electrons can occupy each shell. So, for example, H
has 1 electron in the first shell, He has 2 electrons in the first shell, Li
has 2 electrons in the first shell and 1 electron in the second shell which
has a larger radius. Hence, the third electron in Li is "shielded" from the
nucleus, sees only approximately one positive charge instead of the three
that are really there. Also, Li looks a lot like H (one net positive charge
inside, one negative charge outside) and, indeed, H and Li are very similar
chemically. Now, you wonder why atoms bond into molecules; you are right
that the tendency is, if brought close together, that two atoms will repel
each other because of the outermost shells of each being negatively charged.
But, if the atoms get close enough together, the whole shell idea starts to
break down because, the outermost electrons (which chemists call valence
electrons) in atom A start feeling an attraction to the positive nucleus of
atom B and vice versa , and so the net result is an attractive force
which may bond the atoms into molecules.

QUESTION:
I have a question which is part of a project I'm working on and thought you might help out :)
It is not homework!
The objective is to calculate the volume and pressure of the air coming out of the following experiment:
Imagine a huge piston and cylinder. The surface of the piston is 10000sq ft and it is moving up in the cylinder at a rate of 1 ft per hour. As a result air is being pressured to exit through a 1sq ft opening at the top.
1- what will be the pressure and airflow volume exiting the opening.
If any one has a basic equation for this it will be wonderful.

ANSWER: I presume the
end with the hole is just open to the atmosphere. Therefore, since the
piston is moving so slowly, the pressure everywhere will be atmospheric
pressure to an excellent approximation. The velocity of the air exiting the
hole will be 10,000 times 1 ft/hr, 2.78 ft/s. The equation you need is that
vA =constant, where A is the cross section and v is the
velocity. Technically, this is true only for an incompressible fluid, but
with such small speeds, air may be approximated as incompressible.

QUESTION:
I know that gravity decreases as you move away from its center, so at the top of a high mountain, you would weigh a small percentage less. Now, what if you had a very tall mountain, say 100 miles tall, how much less would you weigh at that height, and what general formula would one use to calculate the relative weights at various heights from sea level to any given altitude? I am not a mathematician or scientist, but it would be very helpful to have this information for a novel I'm writing.

ANSWER: If you are
outside a spherically symmetric mass M , a distance r from the
center, and have a mass m , the force you feel is called your weight:
W=MmG /r ^{2} where G is the universal constant
of gravitation, 6.67x10^{-11} N m^{2} /kg^{2} . The
mass of the earth is M =6x10^{24} kg, the radius of the earth
is R =6.4x10^{6} m, so the weight at r=R is W =9.77m.
(If m is in kg, W is in Newtons (N) and 1 lb=4.448 N.) If
you go to an altitude of A =100 mi=1.61x10^{5} m, W=MmG /(R+A )^{2} =9.30m ,
about a 5% decrease. That last equation is the one you want to use. A word
of warning —this formula is not correct
if you are inside the earth, for example in a deep hole; the force at the
center of the earth is zero. If you are not comfortable using metric units,
I recommend a little unit
conversion program
you can use to convert, e.g. , miles to meters and others (doesn't
work on Macs, but there are probably similar programs available).

QUESTION:
Why do they say the center of the earth is like a bar magnet when the center is more like a perfect gyro scope? Seriously almost like a mini sun... a uber dwarf sun. That's why the earth plates work in a fluid motion cause fluids are sitting on plasma?

ANSWER: "They" do not
say the center of the earth is like a bar magnet, "they" say that the
earth's magnetic field resembles the field of a bar magnet, has a north and
south pole, for example. But, the field of any simple current loop resembles
a simple bar magnet field also. The earth is thought to have, at its center,
currents in the molten core, which give rise to the field. It does not
really resemble the sun at all, no hydrogen and no fusion going on.
Radioactivity is believed to be a source of energy for heating the core of
the earth.

QUESTION:
If dark matter/dark energy exists, would it be distributed throughout space, interspersed or integrated, so to speak, with observable matter/energy? Or is dark matter/dark energy thought to exist separately, in regions of what appear to be vacuum, apart from ordinary matter/energy?

ANSWER: First of all,
my take on dark matter and dark energy is not mainstream, see the
FAQ page. Also, note that dark matter
and dark energy are quite different things, neither of which are well
understood because they have not been directly observed. Dark energy is
already a part of general relativity, represented by something called the
cosmological constant, again, not well understood what its origin or value
are. Regarding whether dark matter is distributed uniformly throughout
space, the answer has to be no. If it were, there would no effects because
you should interact only gravitationally and, no matter what direction you
looked, you would see the same amount of mass and therefore no net force. I
believe that dark matter advocates believe that it is concentrated in
galaxies and it is "hot", that is it has a high speed so that it does not
just build up around and in stars and planets.

QUESTION:
here's my question which I read in an article from the early days of space flight:
A manned spacecraft is circling the moon at an altitude of 100 miles above the surface, and the astronaut on board noted the "angle of depression of the horizon was 23.8 degrees. With this known data, the astronaut was able to determine the radius (or diameter) of the moon. I drew a diagram of the moon and the orbiting spacecraft with the known height of craft above the moon and the depression angle of 23.8 degrees. Thinking that this would be a simple case of similar triangles, etc. I found that I could not determine the radius of the moon from the given data

ANSWER: The figure to
the right shows what you need. The angle A in this figure is, you will
agree, your 23.8^{0} . You can see that cos23.8^{0} =R /(R +100).
Solving, I find R=1076 mi, pretty close to the known radius of 1079 mi.

QUESTION:
I have been instructed to ask about photons following a long discussion with my teenage sons. Is it not possible that a photon is an energy more akin to a sound wave than a particle? Is it possible that the photon is just an energy moving through an, as yet un-identified, substrate ? I thought possibly the universe could be like a , bad analogy, bowl of thin cornflower mixture or wallpaper paste with the cornflower, say, instantly able to take the property of light and pass it on. Holds us all apart, no vacuum, no photon as such: just a property of energy moving through the substrate.

ANSWER: First, you
might want to read the question and answer just after yours to find out what
photons are. They are indeed particles and not the least bit like sound
waves. Electromagnetic radiation is sort of like somebody with two
personalities, sometimes a particle, sometimes a wave. And, by "substrate"
(cornflower and wallpaper paste?!) I presume you mean like some all
pervading medium which is the medium in which your light travels like sound
travels, for example, through air? This was the assumption made by 19th
century physicists because all waves must travel through a medium, right?
They called it the luminiferous
ж ther and
set out to calculate and measure its properties. This was a fruitless, dead
end idea but occupied a lot of physicists for quite a while before Einstein
finally introduced the theory of special relativity. It turns out that light
(hence photons) does not need a medium through which to travel, it moves
perfectly well through a perfect vacuum. If there were a universal
cornflower paste, that would mean there is a preferred reference frame (the
one in which the cornflower is at rest) and it has been shown with numerous
experiments that this is not the case.

QUESTION:
I do not understand the pictorial representation of an electromagnetic wave. Does the wave symbolize the trajectory of a stream of photons? Do the photons actually travel in an up and down wave motion? Or does the wave just symbolize energy oscillations? Or is it a cross-section of a disturbance in the electromagnetic field? Or heaven forbid, does the wave depict a probability? In other words, if the picture of the wave was on a graph, what would the axes of the graph represent? It's amazing that I'm having such a difficult time finding the answer!
I am already familiar with the double-slit experiment and how light can act as both a stream of particles and a wave. I'm not sure how to interpret that either.

ANSWER: In the 19th
century light came to be understood as being
waves of electric
and magnetic fields. The picture I draw in that earlier answer shows what is
called a one-dimensional, plane-polarized, monochromatic electromagnetic
wave and, though simplified and idealized, is what you should think of when
you think about light as a wave. It would be less confusing for you if you
stopped right there. Alas, nature is not quite so simple. It turns out that
there are experiments which can be done which can only be understood if
light is not a wave but rather a stream of particles (photons). Examples of
such experiments are the photoelectric effect (light knocking electrons out
of a metal), Compton scattering (light scattering from electrons), and any
light emitted from atoms. Light indeed is a wave and it is also a particle,
a situation called wave/particle duality. If you design an experiment to
prove that light is one or the other you will certainly succeed. It is each
and it is both. This way of thinking about something is antithetical to how
we usually think about nature, ambiguity is usually considered unscientific,
and that is why you are having trouble getting your head around this.

QUESTION:
With reference to binding energy and mass defect how do you explain why both fission and fusion releases energy.

ANSWER: See an
earlier answer .

QUESTION:
please settel this argument
if i was in a moving veichal that was moving at 30mph and i got up to walk what speed would i be traveling

ANSWER: You always
have to ask speed with respect to what. If the bus moves relative to the
ground with speed 30 mph and you walk forward relative to the bus with speed
10 mph, someone on the ground will see you moving with a speed of 40 mph
relative to him. If you walk 10 mph toward the rear of the bus, your speed
relative to the ground will be 20 mph (moving in the same direction as the
bus).

QUESTION:
is
the speed of light limited by inertia?

ANSWER: Light, having
no mass, has therefore no inertia. The
speed of light is just what it is and that is determined solely by
electromagnetism. Maybe you meant to ask about the speed limit for material
objects being the speed of light? There you could say that inertia provides
the limit because as you approach the
speed of light the mass (i.e . inertia) approaches infinity.

QUESTION:
When 64Co27 (where 64 is the mass number of cobalt and 27 is the proton number) undergoes gamma decay it produces 64Co27 and energy which is in the form of the gamma ray. I don't get why the binding energy per nucleon has increased?, there has been no change in mass number?

ANSWER: You have to
conserve energy, right? The photon carries off an energy E , the
original nucleus has energy M*c ^{2} , and the final nucleus has energy
Mc ^{2} . Therefore, Mc ^{2} +E =M*c ^{2} .
This means that the final nucleus has less mass than the original nucleus,
M<M* , which means that the final nucleus is more tightly bound, which
means its binding energy per nucleon is greater. The binding energy per
nucleon is determined by the structure of the nucleus, not the number of
nucleons it has.

QUESTION:
I am working on an educational speech about the earth's motion around the sun, and how the primary factor in the seasons is not the distance to the sun, but rather the time the atmosphere in a hemispheres is exposed to the sun's energy in a given day. As I thought about the atmosphere, I had to remind myself that our atmosphere has mass, and thus that mass is attracted to the earth's surface by gravity. There is no invisible wall holding it all in. But, as I thought about it, the density of the atmosphere decreases as the altitude above the earth's surface increases, and it made me wonder -- at the very top-most reaches of the upper atmosphere, is the earth trailing a sort of "soap bubble" of oxygen, hydrogen, and nitrogen atoms? And, are some of those atoms drifting far enough away to "fall off" into space?

ANSWER: You are
right, the reason we have an atmosphere is because of gravity. The reason
the density of the air changes with altitude is because the pressure is
increasing. Think of going deep in the ocean: the deeper you go, the more
water there is above you whose weight is pressing down on you, so the
pressure gets bigger and bigger. But, water is almost incompressible, so the
density hardly changes at all as you go deeper. But the origin of
atmospheric pressure is the weight of the air in the "sea of air" above you,
just like in water. The difference is that air is compressible, so the
higher the pressure, the greater the density. It is not really possible to
find a boundary, it just fades away. If you look at some particular
boundary, say when the density drops to 1/100 of the density at sea level,
it is not some constant altitude. In times of high solar activity the
atmosphere swells. Perhaps the most interesting part of your question is the
"far enough away to 'fall off' into space" part. It is not the distance that
matters for atmosphere to escape into space, but rather the speed which the
molecules have. Temperature is a measure of the average kinetic energy per
molecule in a gas, and kinetic energy is
Ѕmv ^{2} where m is the mass and v is the
speed. So, if you have a mixture of gasses at some temperature, the lighter
ones are going faster, on average. For example, a molecule of oxygen has a
mass 16 times larger than hydrogen molecules, so the speed of H compared to
O is v _{H} =4v _{O} . It turns out that for
hydrogen or helium, many of the molecules have speeds greater than the
escape velocity (the speed necessary to escape the earth's gravity) and
there is therefore virtually no H or He in the atmosphere—it has all escaped
into space. The only source of He is from natural gas wells and H can be
gotten from chemistry (lots of hydrogen in the ocean!). This is also the
reason for there being no atmosphere on the moon—its gravity is so weak that
all gasses have escape velocity. And, by the way, it is not just the length
of day which is responsible for the seasons but also the fact that the sun's
rays hit more obliquely in the winter, thereby delivering less energy per
square meter.

QUESTION:
In a junkyard what is doing the work when an electromagnet picks up the car. Many people say it is the magnetic field but others say it is the electric field? Is there any real way of knowing?

ANSWER: The magnetic
field never does work. I have fully discussed this in a
recent answer .

QUESTION:
If I'm trying to calculate mass of all of the light that comes
out of the light bulb per second, then I first have to find the mass of each
photon which is given by the formula E=hf, then I have to include the
amplitude- which will show me how much of those light quanta (photons) will
come out per second. Which formulas do I have to use to find out the amount
of photons coming out? Which distance are the photons from each other? And
is there some proportionality between frequency and amplitude that is
between mass of each photon and numbers of them coming out? And finally can
you increase frequency without increasing amplitude?

ANSWER: This is easy — light has no mass.

FOLLOWUP QUESTION:
Light has it's mass as long as it travels right?

ANSWER: Wrong, light
never has mass.

FOLLOWUP QUESTION:
if photon is massless then how could it then itraract with normal matter such as electrons... and also how could massless object be travaling at all? it'll be impossible to do work on it (photoelectric effect) electron requires energy to move electron and electron emits photon whenever it is changing it's orbit into lower one. How can you think of photon then if it doesn't has mass? How momentum of photon is measured then if on in P=M/V

ANSWER: Since when is
mass required to interact with electrons? A photon has electromagnetic
fields to which an electron responds. An electron has electric charge and
therefore interacts with photons. A photon has energy and momentum but not
mass. In the theory of special relativity, momentum is no longer mv
and kinetic energy is no longer Ѕmv ^{2} .
Read my earlier answers regarding the mass of a
photon and relativistic
momentum .

QUESTION:
If I have a wire with a current flowing through it, the magnetic field B generated by the current can be found using B= (mu*I)/(2piR), with R being the radius away from the wire. As R approaches zero as you get closer to the wire, does the magnetic field strength approach infinity? Or does it simply max out at the radius of the wire?

ANSWER: Inside the
wire the field is no longer
μ _{0} I /(2πr ). If the current is uniformly
distributed throughout the wire, the field may be shown to be μ _{0} Ir /(2πR ^{2} )
where R is the radius of the wire. Note that the field increases
linearly from zero at the center of the wire to μ _{0} I /(2πR )
at r=R , so the field is continuous at the surface. If you want to
learn how calculate this field, research Ampere's Law.

QUESTION:
Someone told that newton is wrong in his second equation of motion as it gives an impossible situation in the following statement:
1- F = M x a ok now we have a car with 500 KG mass and constant velocity 50 mph the car hit some wall what force will be applied on the wall ??? as the velocity is constant the acceleration would be zero and substituting in the 2nd law F = 500 x 0 =0 ???which is impossible

ANSWER: Let's start
with the car. It experiences a sudden acceleration a and therefore a
large force F ; certainly we can agree that F= 500a . This
force is the force the wall exerts on the car. Now, Newton's third law says
that if the wall exerts a force on the car, the car must exert an equal and
opposite force on the wall. So the wall experiences a force of magnitude
F . Now, the wall's acceleration is certainly zero but its mass is not
500, it is infinite. Of course, it is not really infinite, but it is so
large compared to 500 kg that it might as well be; the wall has to move some
but really a tiny amount. So, we have that 500a = ∞•0.
Mathematically, ∞•0 is whatever it has to be to make the equation correct.

QUESTION:
what makes sunset and sunrise red?

ANSWER: See an
earlier answer .

QUESTION:
can u explain me wat parity is and why it is not conserved...

ANSWER: Parity is a
symmetry having to do with how a function behaves under reflection. In
particular, if f (x,y,z )=f(-x,-y,-z ), f
is said to have even parity and, if f (x,y,z )=-f (-x,-y,-z ),
f is said to have odd parity. An example of a function with even
parity would be f=x ^{2} +y ^{2} +z ^{2} ;
an example of a function with odd parity would be f=x+y+z ; an example of a function with no parity would be f=x ^{2} +x .
Parity nonconservation refers to the wave function of a system having a
different parity after some interaction. Beta decay is an example of a
situation where parity is not conserved. It is much too involved to try to
explain the theory of beta decay here.

QUESTION:
When an airplane tilts, why doesn't water fall out of a full cup . Am i only imagining this and is the tilt less than what i observed out of my window.

ANSWER: A plane
seldom tilts (banks) without turning. You are then moving in a circle and
the (fictitious) centrifugal force keeps the surface of the water from being
horizontal. If you are tilted upwards steeply but going with constant speed
in a straight line, your water will spill if you are not careful to hold the
cup horizontally.

QUESTION:
is it true that the decay of one field causes the build up of another?

ANSWER: Yes. Two of
Maxwell's equations contain the fact that time varying electric fields
cause magnetic fields (Ampere's
Law ) and time varying magnetic fields cause electric fields (Faraday's
Law ).

QUESTION:
how does the electric force changes into magnetic force by giving certain velocity to the observer i.e. how are these 2 forces frame dependent?

ANSWER: Actually,
electric and magnetic fields are not different things, they are
manifestations of a single field, the electromagnetic field. When you
transform from one frame of reference to some other moving frame, you change
the mixture of the electric and magnetic parts of those fields. For example,
a point charge at rest has a pure electric field; but if you view it from a
moving frame of reference, it has both an electric and magnetic field. It is
far too involved to show the details here. The reason that we still talk
about electric and magnetic fields in elementary physics courses is that,
historically, it took a long time to realize their connections. The
electromagnetic field is not a vector field, it is a tensor field, a field
which has more than 3 components.

QUESTION:
I understand why they say the speed limit of a rocket engine is based on the maximum velocity of escaping gasses, but when I consider it this way I get lost. Where does my thinking go off the rails?
In space everything is relative, especially velocity. So let's imagine that we have a spacecraft in open space with no points of reference to determine velocity. The spacecraft may be travelling a million miles per hour, but there would be no way to tell. The only point of reference is the spacecraft itself. So relative to the spacecraft, its engine is at a standstill, and when it is fired the escaping gases must expand and in doing so must provide pressure against the spacecraft, and consequently must provide accelleration.
If not why not? How would the engine "know" it was already at any particular velocity and couldn't go any faster?

ANSWER: No wonder you
are lost —y our
premise is wrong . The only "speed
limit" for a rocket is the speed of light. If you have fuel, you can always
increase the velocity of the rocket. If you are in the rocket, the increase
in speed you will see is not the same as the
increase in speed which will be measured in a frame with respect to
which you are moving at a high velocity compared to the speed of light. All
the rest of your question indicates that you understand what is happening
just fine, not lost at all!

QUESTION:
Many sites and books discuss space-time curvature as the reason for gravitational force. They use an analogy of a bowling ball sitting on a bed and causing a depression due to its mass. Rolling a marble across the bed will roll down the depression accelerating towards to the bowling ball and eventually hitting the bowling ball and coning to rest. I don't under this analogy because the ball accelerating is due to the earth's gravity, but in space, the moon does not roll down the earth's space-time curvature and run into the earth. Also, if gravity is weaker the further out we go, then how is it possible that Sun's gravitational force can keep pluto in orbit when Jupiter's mass should overpower the Sun's gravity effect since it is closer to pluto? Why doesn't pluto orbit Jupiter, Saturn, Uranus or Neptune? Likewise, why doesn't mercury fall into the Sun since it is so close? If gravity from the sun is the reason that earth remains in an elliptical orbit, then what started the initial inertial momentum of earth? Lastly, if gravity was not a force and only space-time curvature, then how do we explain that space is curved all around us and that is the reason that a ball drops to the floor when we let it go and comes to rest on the earth?

ANSWER: First, the
bowling ball on the bed (often a trampoline) is not meant to be literally
true, it is a way to illustrate an idea. See an
earlier answer .
Regarding your question about why Pluto is not in orbit around Jupiter
instead of the sun, it is simply that, although Pluto is certainly closer
to Jupiter sometimes*, the gravitational force from the sun is still much
stronger because the sun has so much more mass; the gravitational force of
the sun is always much larger. In terms of your "bed" model, the sun is a
sumo wrestler and Jupiter is a baby, so whom you gonna roll toward? (*I say
sometimes because there will be times in their orbits when they are on
opposite sides of the sun.)

QUESTION:
I need to keep quarter gallon of water in trunk of my car (asume same temperatures as outside of the car) not frozen at all times (Water is kept in plastic and/or glass bottle. How much electric energy do I need if during any 24H period temperatures are between 10 to-10F. How much less electric energy would I need if water is contained in a thermos?

ANSWER: This will be
a very rough calculation which will show you the principles involved. The
rate R at which heat passes through a material of thickness t
and area A is given by R=kA ΔT /t
where ΔT is the temperature difference and k is a constant
which depends on the material. For glass, k =1.1 watts/m-^{0} C.
I estimate the area enclosing a quart A ≈0.06 m^{2} , the
thickness of a glass bottle t ≈1/8 in≈3 mm=0.003 m, and take the temperature
difference (you would have to keep the temperature of the water above 32
^{0} F) to be ΔT =(32-(-10))^{0} F=42^{0} F=23^{0} C. So, R =1.1x23x.06/.003≈500 watts. This answer is
a bit bigger than I would have expected, but it is only a rough estimate. The thermos answer
would depend on the properties of the thermos, some are better than others.
A good thermos would probably last a day with no heat input if you started
with hot water. The idea of a thermos is that there is a vacuum between two
layers of glass. If it were a perfect vacuum, no heat would be conducted.
So, radiation is the only way to transmit heat across the vacuum and this
happens much more slowly than conduction, particularly since the glass is
mirrored. I suspect most heat loss would be through the stopper.

QUESTION:
Is there an explanation for a light clock for dummies?

ANSWER: I guess you
want to explain it to a friend, right? First, you have to accept a
fundamental premise of the theory of special relativity —the
speed of light is the same no matter
what your motion or the motion of the source of the light is. Then, read an
earlier answer of mine.

QUESTION:
i have heard several times that einstein claims that an object's mass increases as its velocity's magnitude increases. but einstein also says that all motion is relative, and there is no absolute state of rest. what, then, is the frame of reference for the calculation of the increase of mass of an object? we cannot use the earth because the earth is moving; that would be like trying to calculate the speed of a car from a different car moving at an unknown speed.

ANSWER: There is no
"absolute state of rest" but every object has a frame in which it is at
rest. In that frame the mass of the object— which
measures its inertia, its resistance to accelerate if pushed or pulled —is
called the rest mass, m _{0} . If that mass has a speed v
in some other frame, it is found that it has more inertia, it is harder to
accelerate, its mass m is bigger: m=m _{0} /√[1-(v /c )^{2} ].

QUESTION:
Does energy create momentum to create force?

ANSWER: I have no
idea what you are asking! Momentum is caused by impulse, a force exerted
over some time. Energy is created by work, a force exerted over some
distance.

QUESTION:
If I carefully arrange two permanent magnets in such a fashion as to levitate one of the magnets due to mutual repulsion, will the distance between the two eventually go to zero? What is the nature of the energy transfer that occurs when the "experiment" is performed?

ANSWER: If you
"carefully arrange" them, the energy to place the levitated magnet just
right comes from you. Once they get there, they will stay there forever
(assuming the magnets do not become demagnetized somehow). But, the magnetic
fields themselves never do any work. For more discussion on that, see the
following answer.

QUESTION:
two permanent magnets will repel each other indefinitely
often with great force
not hard to make a permanently magnetically levitated platform
Where does the energy to perform the work come from?

ANSWER: One of the
truths of electromagnetism is that a magnetic field never does work, you
never get work (create energy) directly from a magnetic field. The simple
reason is that the force of magnetism on a moving charge is always
perpendicular to the charge's velocity. The explanation is always that an
induced electric field is what actually does the work. The case of permanent
magnets is more complicated than most to try to explain explicitly, so I
will give a similar example, taken from the classic text Introduction to
Electrodynamics , David Griffiths, Prentice Hall Publishers. In the
figure to the right there is a constant current in the loop as
indicated and there is a magnetic field pointing into the page in the
cross-hatched area. The upward force on the loop is just right so that the
weight, mg , levitates. Of course, right now, no work is being done
since the force is not moving the mass. But, suppose that the field is now
increased so the upward force now exceeds mg . Now the mass starts
rising. The magnetic force is doing the work, right? Wrong! Before you
increased the field the direction of the electrons' velocity was to the left
(don't forget the electrons go in the direction opposite the current). But,
when the loop starts moving up, the electrons now also have a component of
their velocity upward and that gives rise to a component of the force on the
electrons which is to the right; but this would reduce the current which we
have stipulated to be constant. So the battery in the circuit (which is not
shown but has to be there if there is a steady current) has to supply more
energy than before and that is where the work done on the rising coil comes
from —the battery. This is really just
Faraday's law, an induced back EMF in the loop arises because of the
changing flux. I know this is not the exact question you asked, but, trust
me, all "where does the energy come from when magnetic fields seem to
do work " questions boil down to similar explanations—an electric
field you forgot to think about.

QUESTION:
I was reading about the Large Hadron Collider and their ALICE experiment with the collison of lead ion particles. In the experiment the collison of these particles creates temperatures of 100,000 times hotter than the center of the sun. How can the collider itself handle such high temperatures? It would seem any metal, or substance the collider would be made of would simply melt (along with any instruments measuring the collison). What is the collider made of - or what condition makes the collider able to withstand such high temperatures?

ANSWER: Whenever you
have a relatively large number of particles, you can talk about their
temperature; temperature is essentially a measure of the average kinetic
energy ( Ѕmv ^{2} )
per particle. But the temperature tells you this average, it does not
tell you the total amount of energy —that
is determined by how many particles there are. When you collide two lead
nuclei, the total number of particles (the protons and neutrons)
participating is about 400. But, that is a really tiny amount compared to
any macroscopic object which might have on the order of 10^{24}
particles. The total energy is tiny even if the temperature is huge. Just as
numerical example, suppose that the average speed were 2x10^{8} m/s
(about 2/3 the speed of light) and the mass per particle (neutron or proton)
is about 2x10^{-27} kg. Then the total energy of 400 particles would
be about 400xЅx2x10^{-27} x(2x10^{8} )^{2} ≈2x10^{-8}
J≈5x10^{-9} calories. Wow, that is enough energy to raise the
temperature of a gram of water 5 nanodegrees C! (This calculation ignores
relativity which would change the kinetic energy per particle somewhat, but
the general idea—that the total energy is tiny—is still true.)

QUESTION:
Why do racecars have bigger tires? My first thought is that a racer would want to reduce the surface area between the tire and the ground to reduce static friciton so that he or she needs less acceleration to overcome the static friction; however, in class today we learned that friction does not depend on surface area with some exception, tires possibly being one of the those exceptions. Why would a racer want to increase static friction when it resists movement of the car in the opposite direction? I think my teacher said that static friction resists motion but also accelerates it. If this is true, how does it does it also accelerate in the object?

ANSWER: It is static
friction which provides the force which drives a car forward. If you were on
a frictionless surface at rest in a car, you would not be able to drive
forward. You do not
want your wheels to spin because you can get more friction from nonslipping
tires. This is one of the tricks to driving on snow or mud or ice —push
the accelerator just enough to not have the wheels slip; slamming your foot
on the accelerator is the worst strategy if you are stuck. Regarding the
efficacy of wide tires increasing friction, this is a very complicated
question not easily addressed by simple introductory textbook physics—see an
earlier answer .

QUESTION:
My teacher did a demonstration in class where there was a mass of 1 kg suspended by a string above it and then he attached a string below the mass. For the first demonstration, he pulled quickly to show that a rapid acceleration on the string below the mass would break that string. For the second demonstration, he showed that if you gradually accelerate the pulling of the string below, the string holding the mass would break instead. Could you please explain why this is the case, including the interaction between the two tensions between the strings? I understand that a greater acceleration means a greater force, but I have difficulty understanding why the gradual pull breaks the upper string and what "breaking the tension of the string" means.

ANSWER: If you pull
up gradually with a force F and the mass M is at rest, then
the forces on the mass are its weight, -Mg , the lower string, -T _{lower} ,
and the upper string T _{upper} =F . Newton's first law states that 0=F-Mg-T _{lower} .
So, the tension in the upper string is greater than the tension in the
lower, F =Mg+T _{lower} . So, as you gradually increase F, the tension
in the upper string will reach the breaking point first. I assume that in
the second case, the lower string starts out slack and the mass is pulled
upward by the upper string. When the lower string suddenly gets taught,
there is suddenly a very large acceleration in the downward direction which
can only be provided by the lower string. Therefore, there is suddenly a
large tension in the lower string which exceeds the breaking strength and
breaks it.

QUESTION:
in photoelectric effect (keeping intensity and frequency constant) as we gradually increase the +ve potential the photoelectric current increases.but after a certain value of current(saturation point) the current does not increses eventhough potential is increased..i want to know why the current is not increasing??if all the electrons reach the positive plate then how does the current still exist without charge flow?

ANSWER: A certain
intensity of photons will give a certain number of electrons per second
ejected. So the current coming off the cathode is some constant. With zero
or small voltage to the cathode, some of the electrons do not make it (they
hit the walls of the enclosure or scatter off the diffuse gas that might be
present). But as you increase the voltage, you capture more and more of them
until finally you get them all and you can't get any more than all of them.

QUESTION:
if a bullet was travelling at 823meters per sec and hit an object that stopped it dead how much force per square cm would be exerted onto the target?

ANSWER: Here is the
question which I get in one form or another which indicates how poorly
understood the concept of force is! First, you want to ask what the force
exerted by the bullet is, not the force per cm^{2} . If you were able
to get the force, you could divide that by the cross sectional area of the
bullet to find the effective force per unit area; but you cannot get the
force because it depends on how quickly the bullet stops. If you mean by
"stopped it dead" that it stops instantaneously, then the force would be
infinite. The average force is the change in momentum (mass times velocity)
divided by the time to stop. So, you need also the mass of the bullet.
Suppose the bullet had a mass 0.02 kg, then the change in momentum
(0.02x823) is about 16 kg m/s. If it stops in 0.01 s the average force is
1600 N=360 lb, if it stops in 0.001 s the average force is 16,000 N=3600 lb.

QUESTION:
I understand the general principles of spacecraft in orbit around the Earth (viz., gravity continues to act on an object in orbit, but because it is in freefall - continually "falling over the horizon" - the object is weightless). I'm confused about one thing though. If terminal velocity in a vacuum (which for all intents and purposes applies to low earth orbit) is 9.8m per second per second, how come the acceleration of the object towards the Earth does not continue to increase such that the object cannot remain in orbit but instead falls to Earth? If terminal velocity is reached or is stabilized, then I understand how orbit is maintained, but there is clearly something I'm missing. Obviously objects do orbit, so I'm unclear on the physics behind the orbital motion vis a vis gravity.

ANSWER: Funny, you
start out saying you "understand" orbiting and end up saying you are
"unclear". Believe me, it is the latter. 9.8 m/s^{2} is
gravitational acceleration near the earth's surface, not "terminal velocity"
which is the speed something moving through the air has when it stops
accelerating (like a parachute, for example). If you drop an object near the
earth's surface it will (neglecting air drag) have a downward acceleration
of 9.8 m/s^{2} which means that after 1 second it has a speed of 9.8
m/s, after 2 seconds it has a speed of 19.6 m/s, after 3 seconds it has a
speed of 29.4 m/s, and so forth. The acceleration of a satellite near the
earth's surface and moving in a circle is called the centripetal
acceleration (its direction is toward the center of the circle) and has, of
course, a value of 9.8 m/s^{2} because the only force on it is
gravity. It may be shown that the acceleration of something moving with
speed v in a circle of radius R with constant speed is v ^{2} /R.
So, the speed of the shuttle, for example, is about v = √(9.8R _{E} )≈7900
m/s≈18,000 mph. Check out
Newton's mountain .

QUESTION:
Is the electric field of any charged particle moving in a straight line dependent in its velocity? If it is, and it is just a constant velocity, would that make the electric field equal to zero?

ANSWER: Yes, but this
does not become noticeable until very high velocities (comparable to the
speed of light). The electric field gets smaller in the direction parallel
or antiparallel to the velocity and stronger perpendicula to it. The picture
to the right shows the field for a negative charge with speed 90% the speed
of light; if the charge were at rest, the field lines would be of equal
intensity in all directions. You cannot transform the field to zero.

QUESTION:
is it possible,in any way,shape, or form that time travel is possible?

ANSWER: From what we
know, it is possible to travel forward in time (e.g. the
twin
paradox ), but not backward.

QUESTION:
I have a question pertaining to a gas in a fixed volume:
If I were to ionize (such as through an air ionization device) the gas, would the density change, and what would be a good equation for calculating the change?
(Would this be a piece of the Ideal Gas Law that I am missing, or something else entirely?) And second, how would this interact witht he Ideal Gas Law if I were to also change the temperature?

ANSWER: The only way
to ionize it is to add energy; if you add energy you increase the mass and
therefore the density (E=mc ^{2} ). That said, the amount of
mass increase would be unmeasureably small. The ideal gas law does not apply
because you have a mixture of ions and electrons of different masses and
both are highly interactive with the walls of the container.

QUESTION:
If we have a long tank filled with water and a motorboat motor attached to the end of the tank with its propeller spinning rapidly under the surface of the water sending a water jet out into the tank, and given a long enough tank, is it possible the water jet’s momentum can dissipate before the jet--or some of the jet--hits the other end of the tank and thus preventing any further push on the wall opposite the motor? Or will there always be momentum of the jet hitting the opposite end of the tank(opposite the motor)? In short, when will there be a push on the wall opposite the motor not due to normal water pressure in the tank but rather stemming from the jet. How much?

ANSWER: If you have a
short tank water will be sloshing back and forth but, if you turn off the
motor, it will eventually calm. The reason is that water is a viscous fluid
and viscosity is friction in a fluid. So energy the motor puts into the
water is eventually taken away by this friction. If the tank is long enough,
any disturbances caused by the motor will eventually be damped out before
reaching the other end. But, then, you didn't really expect to be able to
tell at the end of a 1000 mile long tank whether there was a motor going at
the other end, did you?

QUESTION:
Suppose a deserted boat on the sea advancing in a straight line at a fixed speed due to it's inertia, the motor is shut off and nobody is on the boat to obey commands. Is it possible to "knock" it off course, to change the direction of the boat to modify it's vectors without adding energy to the boat ?
If I ram into the boat with another boat, I would have knocked it off course but I would need to transfer energy from my boat to the boat I wish to knock off course.
My question is that is it possible to modify a system without modifying it's stored kinetic or potential energy ? Is it possible to change the direction of the boat without adding energy to the boat ?

ANSWER: Yes, you can
change the direction of the boat without changing its energy. The idea is
that you must exert a force perpendicular to its direction. The result is
that you cause the direction of its velocity to change but not the magnitude
of its velocity. Examples are a car going around a circular track with
constant speed due to the (centripetal) force exerted by the friction of the
tires, a charged particle in a magnetic field will not change its speed but
will change its direction, etc . If you "ram" the other boat, you lose
energy but it does not show up in the energy of the other boat, it is lost
to frictional forces.

QUESTION:
If gravity affects all matter in the same way, regardless of mass, then why does a beam of light not deflect the same way as other particles?
For example, it is known that light bends due to gravity. Therefore, it is possible to take deep space pictures due to light bending around massive objects such as stars. Traditionally it is said that photons are "massless." However, if this were the case, they would not be able to be bent by gravity in the first place.
So, to formulate my question: take for example a beam of light being projected by a flashlight. It would seem that the light should deflect toward the ground due to gravity. However, this is not the case. Why is it that the beam projects out as a straight line?

ANSWER: First, you
should read the FAQ links on this
question. Regarding your question about the flashlight: the light does bend
down, but the amount is so tiny that you cannot observe it. An
earlier answer can give you a more quantitative idea of how small this
bending is.

QUESTION:
if you were to suddenly put a large object in space then the space would be pushed out to the sides (displacement) . because the object is now in the place were space once occupied ?? so space is just pushing back to its original area ???? is this what gravity is??? or bending of space?

ANSWER: Your basic
premise is wrong. The object occupies the space, does not "push it out".

QUESTION:
Is there a way to vary or regulate the strength being produced by an electromagnet? If i was to run more current would it have a stronger polarity and if so, what would be the minimum amount of current I could use to give the magnet charge?
I also am wondering by running magnets in succession for long periods of time if there is any risk? I have heard that you can over charge your magnets witch makes them like a bomb. I that true? If so, is there a a way to measure the polarity being produced to avoid an incident like this.

ANSWER: The strength
of the field may be increased by increasing the current. However, assuming
this is a magnet with an iron core, the field will eventually reach a
maximum limit called saturation. The field can be increased by increasing
the current, but by a quite small amount for each increase in current. There
is also the problem of hysteresis which means that if you turn the current
back down to zero, the field will not completely go away because of the
residual magnetization of the iron. There is no limit to how long you can
operate a magnet. The only way it would "explode like a bomb" would be if
the wiring overheated or something like that. Maybe you heard something
about magnets at the accelerator at CERN exploding; these are
superconducting magnets which have to be kept at extremely low temperatures
and a malfunction resulted in an explosion of the cooling system.

QUESTION:
i was reading up on Quantum Teleportation and i found the it is descused to sound more like a form of cloning than telrportation.
it made it sound like of insted of you being telaported yourself, there would be a copy of u at the destnation, and not a vary good one.
it also said that when you get copied the first u must no longer exist so that the second will. so is this what Teleportation really is a way to send not u but a copy of u somewhere else, or did i get the idea of how it might work?

ANSWER: Here we have
a relatively concise question which cannot possibly have a concise answer.
For a thorough discussion of the "beam-me-up" problem, I recommend Lawrence
Krauss's book The Physics of Star Trek .

QUESTION:
I am doing an extended essay in physics where I am investigating how the volume of water in a plastic bottle affects the time taken for it to roll down an incline and then after leaving the incline a particular distance along an horizontal plane. My results show me that if you increase the volume inside the plastic bottle, the bottle takes less time in that it is faster. How would you explain this relationship in terms of different volumes of water inside the plastic bottle?

ANSWER: You have
chosen a very difficult project because the water, because it has viscosity
and sloshes around in the bottle, greatly complicates things. Consider the
two extremes —empty and full. If the
bottle is full, it is like a solid cylinder and if it is empty, it is like a
hollow cylinder. Because the two have different moments of inertia (MR ^{2}
for the hollow cylinder and ЅMR ^{2} for the solid cylinder),
the full bottle will reach the bottom more quickly. And, both should go a
good long way on the horizontal surface. But, for the half-filled bottle,
the water will rub on the walls of the bottle as it rolls resulting in the
water getting "stirred". Since it takes work to stir a fluid, energy will be
lost this way. It is very difficult to try to predict or calculate the
part-full cases and the best I think you can do is to measure and graph your
measurements. Just to emphasize how the viscosity and energy loss works, try
a fluid more viscous (say motor oil or honey) for the half-full case and it
should roll very much more slowly than for water.

QUESTION:
The sun sits there... burning. Converting H to He and it's been doing this for millions and millions of years. My question is, why does the sun burn so long instead of just blowing up like a nuclear bomb?

ANSWER: The trick to
making a hydrogen bomb is to hold everything togther long enough for a
significant amount of hydrogen to undergo fusion reactions. This is really
hard because to make fusion you need very high temperatures and it is
difficult to hold a high temperature gas together. And, you do not hold it
together for long but long enough to put a lot of energy into the expanding
explosion afterwards. This is the same reason it is so challenging to make a
fusion reactor, just holding the hot plasma contained is not easy. So, what
makes this not an issue for the sun? What the sun has going for it is its
huge gravity (because it has a huge mass) which holds the whole thing
together, even at a high temperature. The sun is no more likely to explode
than are you able to just jump into outer space because the sun holds
everything down. The reason it lasts a long time is that there is a huge
amount of fuel.

QUESTION:
Electricity is made of electrons.
Atoms have electrons, protons and neutrons.
We split atoms, and fuse atoms to cause a chain reaction, which releases more energy than any other chemical reaction that we can achieve currently. And we do this in order to boil water. To boil water quickly, to spin mechanical turbines, which rotate magnets inside wire coils to produce electromagnetic induction and siphon away electrons from that induction to be used as electricity.
Why can't we just directly access the electrons in each atom. Why can't we simply siphon off all of the electrons from a 5 gram cube of uranium 235.
What stops us from breaking down the atoms and just absorbing the atomic level electrons, even the protons since they have a positive charge.

ANSWER: You have a
mess of misconceptions here. Electricity is not just getting electrons. A
conductor like copper has beaucoup electrons just waiting to be moved
around. You use energy to get these electrons moving so you can use them. So
your scheme of "siphoning off" electrons is unnecessary —all
the electrons you could ever want are just sitting around in conductors.
Furthermore, the energy you would have to expend to strip off all the
electrons in a macroscopic piece of stuff is just astronomical. And, if you
did it, the remaining positive charge could not be held together and would
blow itself up very dramatically. The beauty of how electric power works is
that we move electrons around but the materials all keep a net zero charge
so the Coulomb force does not blow things apart.

QUESTION:
Is it possible to find the density of protons, neutrons and electrons or just the nuclei?

ANSWER: Unlike
electrons in atoms, neutrons and protons in nuclei are pretty much tightly
packed. Therefore, the density of neutrons and protons is about the same as
for nuclei, about 2x10^{17} kg/m^{3} . Electrons really do
not have a meaningful size since they are elementary, so density is rather
meaningless. You could calculate a density using the classical electron
radius , about 3x10^{-15} m, but this number does not really
represent anything physical.

QUESTION:
Is everything moving relative to everything else? Can anything be 'still'?

ANSWER: There is no
such thing as "absolutely still", that is one still is no 'stiller' than any
other still. In classical physics, all inertial frames are
equivalent. An inertial frame is one in which Newton's laws are true. Once
you find one inertial frame, any other frame moving with constant velocity
relative to it are also inertial frames and nothing makes any one of them
special. However, lots of things are still if they happen to have zero
velocity in your frame. I should add that, quantum mechanically, the
Heisenberg uncertainty principle states that you cannot know both the
velocity and position of a particle to arbitrary precision. Therefore, the
only an object can be perfectly still is if you also are totally ignorant of
where it is. But, that is probably not what you were asking about.

QUESTION:
are cell phones harmless to ones health?

ANSWER: I would be
reluctant to say they are "harmless". But the recent hoopla over brain
cancer and cell phones deserves some comment. The recent designation by the
World Health Organization that cell phone use is possibly carcinogenic
(added to a list which includes coffee and pickles), is based on a very weak
study which asked people with brain cancer to recollect how much they used
cell phones. Also, although cell phone use has increased astronomically in
the past 20 years, there has been no corresponding increase in brain cancer
rates. Finally, there is no proposed mechanism for cell phone radiation to
cause cancer because unlike x-rays, for example, they do not have sufficient
energy to alter DNA. A good discussion can be read on the
New York Times . (I personally think excessive cell phone use causes
stupidity just like excessive Facebook and Twitter!)

QUESTION:
if a box drops from a moving plane and there is no wind which way does it drop

ANSWER: There is
never "no wind" because the plane is moving through the air. If you neglect
air drag, the box will drop such that it stays directly under the airplane.
To the airplane pilot the box will appear to drop straight down, always
being directly below. However, air drag will never really be negligible in
this situation. The box will hit the ground behind where the airplane is
when it hits; the details depend on the speed of the airplane, the mass of
the box, the geometry of the box, etc .

QUESTION:
What is the speed of electric current through a copper wire?

ANSWER: Electrons
carry the electric current, so your question could mean either how fast are
the electrons moving in the direction of the current or how fast does
information move down the wire (like how long after the switch is turned on
do I have to wait until the light turns on). The electrons in a conductor
which carry the current are called conduction electrons and are free to move
around; they do so in random directions in an isolated wire of copper; so,
there is no net flow of electrons just as there is no net flow of air
molecules in a room even though they are zipping about very fast. When you
apply a potential difference across the ends of a wire (e.g . connect
a battery or plug them into the wall) an electric field is established in
the wire which pushes the electrons opposite the direction of the field
(because they are negatively charged particles). As they move along, they
bump into atoms in the copper and slow down but then the field speeds them
back up. The net effect is that, on average, the electrons have some average
speed in the wire called the drift velocity . The actual magnitude of
the drift velocity depends on how much current flows, how thick the wire,
etc ., but it is quite slow. A
numerical calculation is shown for a specific case in the Wikepedia
article on drift velocity; the drift velocity for this example is about 1
meter per hour. But, the electrons in the whole wire start moving together
almost instantaneously because the electric field which drives the electrons
is set up at
almost the speed of light.

QUESTION:
Electrons, as negatively charged particles, not only rotate about the core of atom but also about their own axis (spin)
Why this happens ?

ANSWER: Electrons,
indeed, have spin, that is what we call intrinsic angular momentum .
However, it is not a classical idea like the orbital angular momentum is and
it is wrong to visualize it as a little sphere rotating on its axis. It is
an angular momentum which cannot be stopped. It is the natural consequence
of solving the Dirac equation in relativistic quantum mechanics. When you do
this you find, if the orbital angular momentum quantum number for an electron is
L , the the total angular momentum
quantum number is L ï¿½Ѕ, implying a spin quantum number of Ѕ. There is
an earlier answer which
might be of interest to you.

QUESTION:
If I wanted to build a antigravity chamber by using a sphere made of osmium or iridium suspended 1 meter above the Earth's surface to cancel out the Earth's gravity, how massive would the sphere need to be and how large considering that these elements are over 4 times more dense that the Earth's average density? What if I used neutron star matter?

ANSWER: I won't do
your arithmetic for you, but I'll tell you how to do the calculation. The
gravitational force on a mass m near the earth's surface is F _{E} =GmM _{E} /R _{E} ^{2} .
The gravitational force on a mass m near the surface of a sphere of
radius R , mass M , density
ρ
is F =GmM /R ^{2} =4 π ρGR /3.
Setting F=F _{E} , R =3M _{E} /(4πρ R _{E} ^{2} ).
The density of a neutron star is about 2x10^{17} kg/m^{3} .

QUESTION:
If 2 waves interfere destructively, where does the energy go?

ANSWER: This question
was answered, for a double-slit experiment, in an
earlier answer . For a much more detailed
discussion, go here .

QUESTION:
im currently undergoing an Extended Experimental Investigation at school. My friend and i are looking at doing something with saxons bowls. Our variables are to be the hole size and also the viscosity of the substance (oil, water, thinners). As you would know for any experiment there is a main purpose to find something. However, we are unsure what can be our main purpose. We are unsure of what we should be comparing to make this an A standard experiment. We were thinking to possibly look at the different rates at which they sink and compare that between the different substances as it is obvious that they will all be exponential growths.

ANSWER: I did not
know what a Saxon bowl was! A bowl with a hole in the bottom is placed in
water and sinks when it fills; they were used by Saxons to limit the time of
orations. They were also used, reputedly, by ancient Greek prostitutes to
allocate time to customers.

So, your question …The first thing which
strikes me is your statement " it
is obvious that they will all be exponential growths". A good scientist does
not make prejudgments as to what is "obvious"; that is what you do an
experiment for. And, are you sure you know what exponential growth is? To
make your experiment as clean as possible, you should study one variable at
a time. It sounds like size of bowl will not be a variable, only hole size
and fluid properties. What you want to do is record time to total submersion
as a function of hole size for a range of sizes and then plot your data. You
should consider whether hole diameter or hole area is a more appropriate
measure of size. Repeat this experiment for each fluid. From the graphs try
to conclude the functional dependence of time on hole size. For each hole
size, compare and graph the time as a function of viscosity.

QUESTION:
I'm trying to make a simulation game in which the player or players change the angle and velocity of a projectile launched. To make it simple, i'm stating that "easy mode will be in a vaccum, so no air resistance. I've been able to calculate how far away it will hit away from the origin and how long it will take to get there... what I haven't been able to calculate is how you determine the parabolic curve it takes to get there. In other words, so far i have a game where you can't see the projectile, then a random explosion goes off wherever it impacts. i've gone through pages and pages of google searches and a college physics textbook. i wasn't able to find it in either of those. I also forund one formula for finding the X,Y coordinates, but when the angle < 19="" or=""> 69, every coordinate becomes negative, but that can't be...

ANSWER: Well, I
certainly cannot debug your game, but I can give you the equations of motion
for a projectile without air drag (which, despite your search, is in any
introductory physics book). I will write things in terms of x, y, t, v _{0} ,
and
θ assuming that, at t =0, x=y =0. The projectile
starts at the origin at t =0 with a speed of v _{0 } at
an angle of θ relative to the horizontal. Therefore the two
components of the initial velocity are v _{0x} =v _{0} cosθ
and v _{0y} =v _{0} sinθ . The equations of
motion, which give you the position and velocity as functions of time are:

y =v _{0y} t- Ѕgt ^{2}

v _{y} =v _{0y} -gt

x=v _{0x} t

v _{x} =v _{0x}

where g =9.8 m/s^{2} is the accleration due to gravity. If you
want to get the equation for the parabola, put t=x /v _{0x}
into the y equation: y =(v _{0y} x /v _{0y} )-(Ѕgx ^{2} /v _{0x} ^{2} )=x tanθ- Ѕgx ^{2} /v _{0} cosθ ;
this is the equation of a parabola since the general form is y=a +bx +cx ^{2} .
In all these equations, x and y are measured in meters, t is measured in
seconds. If you prefer x and y to be measured in feet, use g =32 ft/s^{2} .

QUESTION:
I was wondering whether objects that bend space-time and move through space-time would produce something like waves of space-time or gravity waves. I looked it up and found out that physicists were looking or trying to find a way by which to detect gravity waves. What exactly are these waves and what is "carrying" this wave. What is the medium?

ANSWER: It is thought
that an accelerating mass radiates energy (gravitational waves) similar to
how accelerating electric charges radiate energy (electromagnetic waves).
Gravity waves are waves which are carried by space-time itself. They have
never been directly observed. They have been indirectly observed by
measuring the rate at which energy is disappearing in binary systems of very
massive stars. However, like all of science, they should be considered a
hypothesis until they are directly observed.

QUESTION:
Say a hole was dug that ran from the surface of the earth, thru the core of the earth and exits on the exact opposite side of the earth... so in a sense, one could see straight thru the earth to the other side. Say, someone fell thru that hole, they reach terminal velosity but since theres no resistance other than air, theres no way to slow down when they get closer to the core center of the earth. What are the likely affects and effects of gravity would that person experience once they reach exact core center of the earth? They wouldn't come to a dead stop once they reach core center would they? Or would they shoot past the core center, slow down then fall back towards the core center again, then repeat until they rest in the middle of the air at core center?

ANSWER: Always read
the FAQ page before asking a
question!

QUESTION:
I am confused about kinetic energy. Newtonian kinetic energy law KE = 1/2MV^2 says that when I drive my car at a certain speed and double it I have 4 times the energy compared to my un-modified speed. Supposed I was floating in space and I fire a gun to go at a certain speed (obviously moving backward away from the bullet), does that mean I need to fire 3 more bullets to double my speed or just 1? Keeping in mind that in this case the bullet's mass is negligible compared to my body so each bullet fired should intuitively create consistent force against my mass. I fail to see why I need to fire 4 bullets in total to go at a speed 2 times that compared to a speed achieved by firing only 1 bullet. But at the same time kinetic energy law says I will possess 4 times the energy is I double my speed. So this screams free energy to me if I fire 2 bullets only and possess 4 bullets worth of energy.

ANSWER: The
experiment you are describing changes your velocity using momentum
conservation in the following way: suppose the bullet has a mass m
and emerges with speed v ; then you, with mass M , recoil with
speed V such that MV=mv . Hence, the speed you acquire is
V=mv /M and every time you fire a bullet you increase your speed
by this amount. So, if you fire a second bullet you quadruple your kinetic
energy because you double your speed. The preceding is really an
approximation because the second time you fire a bullet your mass is less
than the first time since you lost a bullet's worth of mass in the previous
firing. So the second time you fire the bullet, you recoil with an increase
in speed V'=mv' /(M-m ), the third time V"=mv" /(M- 2m ),
etc . where v' is the recoil of the second bullet, etc .
Typical numbers are m =0.02 kg, v =500 m/s, and M =100 kg,
so the loss of mass is indeed neglegible since we're talking about bullets,
not cannonballs; it then works out that V =0.1 m/s. There is no "free
energy" here because the chemistry in the bullet explosion adds the same
amount of energy with each firing. Keep in mind that you and the bullet
share this energy, you do not get it all; although you and the bullet share
the momentum equally, it gets the lion's share of the energy, 2500 J
compared to your
Ѕ J .
(Incidentally, v is not the muzzle velocity of the rifle because the rifle
recoils with you.)

QUESTION:
I cannot find the answer to this on the internet (at least not for a physics novice).
Q: What is the relevance of light speed in the equation E=MC2? How do light speed fit in with Mass and Energy?

ANSWER: What is the
relevance of anything in a physical equation? Like many equations, this one
was found by taking earlier equations and deriving this one. The speed of
light follows along in the mathematics. There is a brief summary of the
derivation in an
earlier answer , but if you do not know anything about the theory of
special relativity, you will probably have trouble following it. The
qualitative idea is that relativity is based on the notion that the speed
of light is a universal constant; this means that all observers measure the
same speed for a light beam regardless
of the observers' speeds. With this assumption we find that linear momentum
must be redefined from what we use in classical physics. But, linear
momentum may be related to kinetic energy of something and it turns out that
the total energy of something is its kinetic energy plus a quantity which
turns out to be its mass times the speed of light squared; we call that
quantity the rest mass energy of the object.

QUESTION:
Is it possible to fire a gun with absence of oxygen ? Why ?

ANSWER: See an
earlier answer .

QUESTION:
We know that a pendulum attached to the ceiling of an accelerated car
will make an angle with the vertical. I wish to know why this only
happens when the car is accelerating and the angle returns to zero when
there is no more acceleration. Some books use the concept of fictitious
force to explain which is difficult to understand.

ANSWER: If you view
this from a coordinate system attached to the earth, the pendulum bob is
accelerating (along with the car). Now, according to Newton's second law, if
an object is accelerating in a particular direction, it must have a force on
it and in that direction. What can possibly exert that force? Only the
string. But, if the string is vertical, it cannot exert a force forward, so
it swings such that the magnitude of the vertical component of the tension
on the string equals the magnitude of the weight of the bob and the
horizontal component is ma where m is the mass of the bob and a
is the acceleration. That is all you need to know. For completeness, let us
ask how we understand things if we use a coordinate system attached to the
car, not the ground. In this coordinate system the ball is not accelerating
even though there is clearly an unbalanced force on it (the horizontal
component of the tension). The key is that Newton's laws of motion are not
true in accelerating frames of reference. You cannot write F=ma
because here F is not zero but a is. However, you can force
Newton's second law to be true by introducing a force which, in this case,
would be ma toward the rear of the car; then, there are two forces, one
forward and one backward, which cancel each other out and the particle is
therefore obeying Newton's first law, F _{net} =0 if the object
is not accelerating. It is important to understand that this force does not really exist,
that's why we call it a fictitious force. The best known example of a
fictitious force is the centrifugal force, the force you think you
feel when you are inside a rotating drum; actually, you feel the drum
pushing you toward the middle (centripetal force).

QUESTION:
I am building a 3D graphical simulator of atoms and sub atomic particles that allow people to view atoms in bhor as well as quantum models.
Could you help me by answering short concise questions as you so prefer every now and then?
At present I need to know how fast an electron travels in a ground state hydrogen atom in the s1 shell in meters per second.

ANSWER: The Bohr
model has the velocity v at distance r given by: v = √[kZe ^{2} /(mr )].
Where Z is the nuclear charge (1), k is the Coulomb constant
(9x10^{9} ), e is the electron charge (1.6x10^{-19} ),
and m is the electron mass (9x10^{-31} ); all are in are SI
units. The radius of the ground state orbit is about r =5.3x10^{-11}
m. So, I find, v =2.2x10^{6} m/s.

QUESTION:
There are all these stories latley of anti-hydrogen and how they got it to last >15min by cooling it down. This doesn't imply a stable atom since obviously the colder it is the longer it stays together before the orbital or force instabilities blow it apart. Since Hydrogen exist in a stable form in nature, have scientists ever created Hydrogen from the collision of protons and neutrons that was stable? It seems any science experiment should be able to reproduce the obserable before trying to understand the unknown.

ANSWER: Antihydrogen
is an antiproton plus a positron. The properties should be similar to
hydrogen because the masses are equal, only the charges are flipped. The
difficulty in trying to make and study it is that if it encounters normal
matter, the positron will annihilate with an electron and the antiproton
will annihilate with a proton. So, the reason for keeping it cold is because
low temperature means that it moves slowly so that it will not bump into a
wall. I am sure it is also kept in as near to perfect vacuum as possible
since you do not it colliding with any ordinary gas atoms. The cold has
nothing to do with the stability of the atom; atomic properties are not
dependent on temperature unless the temperature gets so high that collisions
excite or ionize the atoms. I do not know why you would think that colliding
a neutron and proton would make hydrogen since it is not made from those.

QUESTION:
can we break a neutron??

ANSWER: We don't have
to because it falls apart on its own. The free neutron has a half life of
about 15 minutes and decays into a proton, an electron, and a neutrino.

QUESTION:
Ice sometimes forms in my freezer that has small columns like spikes sticking out of the ice cubes. What causes that to happen?

ANSWER: See an
earlier answer .

QUESTION:
If you place a heat source in a room, and said heat source is independently capable of bringing the temperature to 30 degrees, would adding another identical source make the room hotter than 30 degrees, or simply increase how quickly the room gets to 30 degrees?

ANSWER: I assume that
these heaters do not have thermostats, that is do not shut off when the
temperature reaches 30^{0} . A heater gives off heat at a given rate;
when the rate of heat coming from the heater equals the rate of heat leaking
from the room, equilibrium is reached and the temperature remains constant.
If there are two such heaters, when the temperature reaches 30^{0}
there is more heat flowing into the room than out of it so the temperature
continues to rise. Both your scenarios are right, the temperature will rise
more quickly and reach a higher level with two heaters. Incidentally, I am
assuming that the temperature outside the room is colder than 30^{0} ;
if not, heat will flow into the room even with no heaters.

QUESTION:
my question is regarding the dimensions. physical quantities do have dimension in the form [ LMT ] but what is this actually?

ANSWER: Once you have
operational definitions of length [L] (distance between two points in
space), time [T] ("distance" between two events at the same point in space),
and mass [M] (inertia, resistance to being accelerated), every other
physical thing you might measure or observe may be expressed in terms of
these three. These are usually called dimensions but are not the same things
as when we talk about three dimensions, e.g .; each of the "three
dimensions" are length. Sometimes we talk about space-time or
four-dimensional space which adds time to three dimensions. Scientists
usually prefer units in the SI system to measure [L], meters (m); [M],
kilograms (kg); and [T], seconds (s). Some examples are

area: [L^{2} ] m^{2}

volume: [L^{3} ] m^{3}

speed: [L/T] m/s

force: [ML/T^{2} ] kg
m/s^{2} (called a newton, N)

energy: [ML^{2} /T^{2} ]
kg m^{2} /s^{2} =N m (called a joule, J)

power: [ML^{2} /T^{3} ]
kg m^{2} /s^{3} =J/s (called a watt, W)

Dimensional analysis is often useful in checking
calculations. Supposed you derived that the energy of something was E =5mv ^{3}
where m is mass and v is speed. Then the dimensions of the left side are [ML^{2} /T^{2} ]
and the right side are [ML^{3} /T^{3} ]. Since these are not
the same, your derivation must have been wrong.

QUESTION:
Why does the energy level of shells increase in terms of potential energy...I get that the atomic radius increases, however electric field of the nucleus decreases with distance and further more shielding by other electrons decrease charge felt by the electron from the nucleus...Therefore as distance is increases electric potential is decreasing so how does potential energy increase!
Maybe the answer is that is just the way it is according to experimental testing...the physical world doesn't follow formulas exactly...especially my basic understanding of them...if I am missing something in my thought process please point it out to me.

ANSWER:
The potential energy of an electron near a positive charge is negative,
U (r )=-C /r where C is some positive constant
which need not concern us. Recall that potential energy may be chosen to
be zero wherever we wish, and it is customary, as here, to choose the
U (r= ∞ )=0.
So the electron energy is zero when the electron is infinitely far away
and at rest. If the electron is bound in the atom, its total energy is
negative and negative and its potential energy is negative. As r
gets bigger, U gets less negative so it is getting larger (even
though its magnitude is getting smaller). If you give the electron
enough energy so that it becomes unbound from the atom, its total energy
is positive.

QUESTION:
How does dark matter interact with gamma-ray bursts?

ANSWER:
In my opinion, dark matter is a hypothesis, possible but essentially a
fiction, until it is actually observed experimentally. As advocates of
see it, dark matter interacts only gravitationally, that is it is
perfectly transparent to electromagnetic radiation which is what gamma
rays are.

QUESTION:
I was trying to explain Time Dilation and the Twin Paradox to a friend and I am having trouble explain it to him. I have shown him experiments of precise clocks traveling at high speed being slightly slower than stationary clocks,but he always seems to come back to the same point.
If 2 twins were floating in empty space then suddenly one of the twins was moved at the speed of light for 1/2 a year to the left and then moved at the speed of light for 1/2 a year to the right, the time that has passed is only 1 year yet the moving twin is supposedly younger than the stationary one. Also, if there are no reference points around then neither of them would know whether they are moving or stationary.

ANSWER:
First of all, neither twin can move at
the speed of light , just close to it. To understand time dilation,
see my earlier discussion of the
light clock . Of course, to see how the light clock works you need to
accept that the speed of light is
the same for all observers. Then you can look at my explanation of the
twin paradox . You can also
tell your friend that GPS would not work at all if software corrections
in the GPS were not made for time dilation (special relativity), not to
mention necessary corrections for different clock rates at different
altitudes (general relativity). Regarding which twin is moving, it is
pretty clear who turned his rocket on to depart and who has to turn
around to come back. The easiest way to see why the traveling twin's
clock will read less than one year is to invoke length contraction. If
he is traveling at, say, 80% the speed of light, he sees the distance he
travels shrink by a factor of
√(1-0.8^{2} )=0.6 and so he will be only 0.6 years older when he
returns.

QUESTION:
why does a ball bounce higher if the height in which it's dropped from increases? i know its something to do with the gravitational potential increasing but how does that have an effect?

ANSWER:
When a ball bounces, it loses a particular fraction of its kinetic
energy because of the properties of the material it is made of (and what
it bounces from). Kinetic energy is
Ѕmv ^{2 } where m is the mass and v is the
speed. If you drop something from a height h , its kinetic energy
at the floor will be mgh where g is the acceleration due
to gravity. Suppose the the ball loses half its energy in the
collision. Then, if it is dropped from a height h , it will
rebound to a height Ѕh .

QUESTION:
How many joules of solar energy are consumed in the evaporation of water from the ocean (say 1kg)?

ANSWER:
The quantity you seek, latent heat of evaporation, depends on water
temperature and salinity. It is approximately 2200 kJ/kg. You can get
more specific information at
this link .

QUESTION:
in my school their is a great contrdiction between the motion of rocket in space will tell me just that when a rocket move away from the gravitational field of earth then will it use fuel to go farword because according to 1st law of motion it will not use fuel . so please tell me that which law should be obeyed by the rocket newtons 1st law or newtons 3rd law of motion???

ANSWER:
If there are no forces on the rocket, which is approximately true far
from any planet or the sun, it will continue moving with constant speed
in a straight line without burning its engine; this is Newton's second
law. But, if you are close to a planet then the rocket experiences the
gravitational force and to move with a constant speed in a straight
direction your rocket needs to apply a force opposite to the direction
of the weight of the rocket and equal in magnitude.

QUESTION:
a large number of free electrons are present in metals .Why is there no current in absence of electric potential across it?

ANSWER:
The conduction electrons act like a gas where they move around in random
directions. But there is no preferred direction and so, at any given
time, there are just as many electrons moving in one direction as the
opposite direction, no net flow. Your question is sort of like asking
why there is no breeze in a closed room.

QUESTION:
i understand that gravity is merely the force of 2 masses attracting. They both attract each other in relation to each other's mass. I ALMOST understand the gravitational constant, but just to be clear, is that just a number that is always the same? Like unto Pi? Something I could closely represent with 2.4 or 59.886 (probably both way off) for simplicity sake? Or is it a different value based on the mass of the 2 attracting masses?

ANSWER:
Newton's universal law of gravitation is F=Gm _{1} m _{2} /r ^{2}
where G =6.67x10^{-11} m^{3} kg^{-1} s^{-2} and it is a constant
which never changes. r is the distance between the masses m _{1}
and m _{2} .
So, you see, gravity is incredibly weak unless one of the masses is very
big.

QUESTION:
There are 2 masses floating in an empty universe, they are 1 mile apart. One has a mass of 100 pounds, and the other has a mass of 200 pounds. Where is the point of impact in relation to both of their starting points? I understand it should be closer to the one of more mass, but is it exactly twice as close since it is twice as massive, which would put the point of impact at 2/3 between mass 1 and mass 2? Or is it a more complicated formula? Thanks.

ANSWER:
They collide at their center of mass. The distance of the center of mass
from mass m _{1} may be found from R=m _{2} r _{2} /(m _{1} +m _{2} )
where r _{2} is the distance from m _{1} to
m _{2} . For your example, R =100x1/(200+100)=1/3
mile from the 200 lb object. You realize, however, that the force is
really tiny which means you would wait almost forever for the collision
to occur. See the following question for a little more quantitative
discussion.

QUESTION:
I saw the experiment by an Astronaut in which bits of salt or sugar clumped together while in a bag of water, thus showing gravitational attraction. Made me wonder why we do not see similar occurances on Earth. Eg. Why doesn't a grain of sand on a pier migrate toward an aircraft carrier while docked?

ANSWER:
You must have misunderstood what that experiment was trying to
demonstrate. Gravity is an incredibly weak force. For example, the mass
of a grain of salt is about 10^{-7} kg, so two such grains
separated by 1 mm=10^{-3} m would be about F =10^{-7} x10^{-7} x6.67x10^{-11} /(10^{-3} )^{2} ≈10^{-18}
N. This force would result in an acceleration of about 10^{-7} x10^{-18}
m/s^{2} =10^{-25} m/s^{2} . The time for something
to accelerate to a speed of 1 mm/s with that acceleration would be 10^{22}
s, about 10^{14} years, far longer than the age of the universe.
The mass of an aircraft carrier is about 3x10^{7} kg, a grain of
sand about 10^{-7} kg, so the the force on the sand about 10 m
from the carrier will be about 3x10^{7} x10^{-7} x6.67x10^{-11} /(10)^{2}
N≈10^{-12} N. This force is way smaller than the frictional
force keeping the sand from moving; and besides, the acceleration of
about 10^{-5} m/s^{2} would be too tiny to observe
before the ship left port.

QUESTION:
Assuming everything else is equal (chemistry, length, width, microscopic defects, etc), why specifically is a thinner piece of glass less durable (i.e. more fragile) than a thicker piece?

ANSWER:
Imagine two springs, identical except that one has much thicker coils
than the other. To stretch the thinner spring 1 inch will require a much
smaller force than to stretch the other 1 inch. Glass is like a spring:
you push it and it bends and when you stop it goes back to how it was.
But, if you bend it too far, it breaks. It takes less force to bend thin
glass to the breaking point than a thicker piece.

QUESTION:
How old would someone be to themselves if they traveled 10 light years in 20 earth years? How old would they be to us back on earth?
OR in other words -
How long would a 10 lightyear trip feel to the occupant, and how long would it be for mission control at with the vessel going 1/2 the speed of light?

ANSWER:
If you are traveling at half the speed of light, the distance you go shrinks
by a factor
√(1-.5^{2} )=0.866, so the distance you have to travel is
20x0.866=17.3 light years. So your clock will tick forward 17.3 years.
Mission control would say it took 20 years.

QUESTION:
i am curious about the nature of electromagnetic waves...looking at the diagram of an em wave from maxwell's equation, i can't figure out...do the electric and magnetic fields (represented by vectors) continually rise and fall over time like a wave, or does the entire waveform (all vectors) occur at a single instant? It's probably a misconception, but i'd like any advice to help me visualize an electromagnetic wave. ty for your help!

ANSWER:
Look at an earlier
answer .

QUESTION:
According to my understanding about the force acting between two
parallel straight conductors, i have a feeling that two overhead power
cables must undergo attraction or repulsion. Then why dont we notice
this attraction or repulsion?

ANSWER:
Suppose that the wires are separated by 1 m and each carries a current
of 100 A. The force each meter of each wire would feel is about 0.002 N,
about 0.01 oz. Magnetism is a pretty weak force.

QUESTION:
Imagine, Earth like a big round stone bead.
means a hole in the earth from one side to another, passing thru the center crust of the earth, the game is ....what will happen when one jumps in that hole ???
will it hurt him ? will he be able to reach the other end ? or will the hole act like a spring ? what will happen ???

ANSWER:
Always look at the FAQ page
before asking a question!

QUESTION:
Consider a ball with mass M attached to a very long cord with linear density
λ . If you calculate the ball's maximum height when you throw the ball up with initial velocity
v, using F = dp / dt, F = - (M +
λ y) g, and p = (M +
λ y) dy / dt, you will obtain something different of what you obtain trying to calculate the maximum height using conservation of energy. How energy is dissipated in this case?
The cord is straight below M and it is longer than the maximum height.
In the beginning of the movement, only the ball moves, and the cord is on the ground. As the ball goes up, it pulls the cord.
[I have changed your notation slightly.]

MY FIRST (INCORRECT!) ANSWER:
Let's first do energy conservation. Energy is conserved because the only
force on the ball + cord is gravity. (There is also the force of the
table on the stationary part of the string, but it does no work. E _{1} = ЅMv ^{2} ,
E _{2} =Mgh + λhg (Ѕh )=Mgh +Ѕ λg h ^{2}
where h is the maximum value of y . Setting
E _{1} = E _{2} ,
you must solve the quadratic equation Ѕλg h ^{2} +Mgh- ЅMv ^{2} =0.
You get two solutions, one for which h <0 which is discarded. The
final solution is h =(M /λ )(-1+√{1+[ λv ^{2} /(Mg )]}).
As a check, if
λ= 0 this may be shown to be h=v ^{2} /(2g ),
the well-known answer for no hanging cord. Now, you say you solved it by
solving F =dp /dt ? I
was unable to do so. I can get the differential equation for Newton's
second law, but it is a nonlinear, second-order differential equation:
0 = (M+ λ y )(g +d^{2} y /dt ^{2} )+λ (dy /dt )^{2} .
Maybe I could solve this if I brushed up on differential equations, but
I am not willing to do that!

MY SECOND ANSWER:
The questioner, Emilio Matos of Brazil, sent me his solution solving
the
F =dp /dt equation. He was much more clever
than I and achieved a solution of h= (M / λ )(-1+(^{3} √{1+[3λv ^{2} /(2Mg )]}).
A copy of his solution (in Portuguese) is
attached ; note his slightly
different notation: h for y , m for λ , v _{0}
for v , h _{max} for h . The graph shown above
compares the correct solution with my incorrect solution assuming energy
conservation; the correct solution has a much lower maximum height
indicating substantial loss of energy. My statement that " the
only force on the ball + cord is gravity" is incorrect because the mass
is increasing. In order for the "next piece" of cord to be moved from
rest to moving up, a force must be exerted on it; that means that the
next piece of cord exerts a downward force on the end of the moving cord
which does negative work on the system. This is most easily
seen by imagining a mass m sliding on a horizontal frictionless surface
(avoiding the complication of gravity) with speed v and picking up a mass dm at rest. An impulse
of F dt =dp =(m +dm )(v +dv )-mv =v dm +m dv
must be delivered to dm to speed it up, and so m
experiences an impulse which slows it down. Of course, this is nothing
more than F= dp /dt , but with a focus on the
nature of F . In the case of interest, there is the additional
force of gravity; the important thing to realize is that this force is
not the only force because if it were the acceleration would simply be -g
which it clearly is not. Hence, energy is not conserved. Incidentally,
the correct solution for h also reduces to
h=v ^{2} /(2g ) for a weightless cord showing that,
just because an answer reduces to a known result in a particular limit,
the answer is not necessarily correct! Now that we understand the
problem, I am not going through the calculations necessary to use
energy to solve this problem (by calculating the work done by F )
because it would simply go over the same ground already covered by Mr.
Matos' solution.

QUESTION:
let's say there is a star 7 billion light years away. we can see the star that was there 7 billion years ago. how long will it take for the photons to stop reaching us and make the star 'disappear'? once those photons arrive, shouldn't the star disappear or is there a constant stream of photons? shouldn't the star only be visible for a short periof of time?

ANSWER:
As long as a star lives it will shine. Suppose that the star in question
had a lifetime of 5 billion years and we are seeing it right now at the
halfway point of its lifetime. That means we can continue looking at
that star for 2.5 billion more years before it burns out.

QUESTION:
I'm not sure how to word this properly but here it goes. Radio waves can generally be assumed to travel at the speed of light. My question however, is if the distance from point a to be is a straight line and a wave doesn't follow a straight line (sine curve for simplicities sake), with changing frequencies and amplitudes isn't the "distance" traveled greater than the straight line distance?

ANSWER:
The wave travels in a straight line. The sine curve describes the shape
of "what is waving", electric and magnetic fields for radio waves. Think
of a sine-shaped stick which you move in a straight line. You should
read my earlier answer on
electromagnetic
waves .

QUESTION:
T Jefferson used his famous example of the "candle vs the flame" to describe what is and what is not "property".
Here is the Set up:
You have a candle with a burning flame on it. I have an unlit candle with no flame. If I extend my unlit candle into your candle's burning flame in order to light my candle and withdraw my candle at the moment it ignites, .....
Here is the Question: Have I slightly taken any energy away from your candle & flame, slightly added to the total overall energy, or is it all equal or balanced out "even-stevens" ?

ANSWER:
I don't understand what this has to do with property. Energy
conservation says that the total energy of the two candles and all the
gases they exhaust remains constant. This is very complicated, of
course, because a candle burning is the conversion of chemical potential
energy into heat, light, and sound energies. Until the second candle is
lit, its energy is simply stored chemical energy. To light it requires
adding some energy, it will not spontaneously start burning. If you
light it, that ignition energy must come from somewhere, obviously the
released heat energy from the first candle. I guess you could say that
it is taking energy from the hot gases in the flame, not from the
candle, but that would be splitting hairs. Imagining the two candles to
be inside a box which isolates them from the rest of the universe, the
total energy in the box never changes.

QUESTION:
Two trains 200 km apart are moving at a relative speed of 50 km/h. A fly takes off from one train, flying straight to the other one at the speed of 75 km/h, touches and flies back to the first train. How far will the fly have gone when the trains collide?

ANSWER:
I suppose this might be somebody's homework, but I have waited long enough
that the homework was probably due long ago! It takes the trains 4 hours
to collide. Something going 75 km/hr will travel 300 km in that time.

QUESTION:
Suppose a space ship traveling at relativistic velocities, say 0.9c, could overcome inertia and stop on a dime, how many g-forces would the ship and its crew feel? What is the equation for calculating this?

ANSWER:
I cannot do a calculation given only "stops on a dime". You could say it
stops in 2 minutes having started with a speed of 90% the speed of light.
The average force is given by F= Δ p/ Δ t .
The initial momentum of a mass m is p=mv √(1-(v /c )^{2} )= 0.9x3x10^{8} m √(1-0.9^{2} )=1.2x10^{8} m
which is the change in momentum Δ p
since it ends up at rest. So, F =1.2x10^{8} m /120=10^{6} m .
The weight of m is mg , so, since g ≈10 m/s^{2} ,
m will experience a force about 10^{5} times its weight,
100,000 g 's. If m is you, you're dead.

QUESTION:
I know that the full Law of Conservation of Energy can be shown as:
"KE{gained} + PE{gained} = KE{lost} + PE{lost} + (F{friction} * {distance})"
The "F{f}d" part is equal to the energy lost/gained by thermal energy due to friction. This would suggest that thermal energy due to friction is equal to the force of friction times the distance. However, I cannot find anyone that says this, or even backs this up! Most of the answers I find (of which their are very few) are differential calculus equations, none of which agree with each other. What is the actual formula, and why isn't it "F{f}d?"

ANSWER:
First, that is not at all how I would write the energy equation. I
prefer E _{2} =E _{1} +W where E _{2}
is the energy at the end, E _{1} is the energy at the
beginning, and W is the work done by all external agents. When I
write E this means the sum of kinetic energy and potential
energy. All external agents mean all forces for which we have not
introduced a potential energy function. I tell my students this is easy
to remember because it says "what you end up with is what you started
with plus what you added". If the external agent is friction, the work
it does is negative because it takes energy away. This is not what you
should call "Law of Conservation of
Energy" because energy is not conserved but changes by W . So,
the work done by friction is W=-Fd. The law of energy
conservation states that the energy of an isolated system never
changes. So, where did the energy which friction "took away" go if we
imagine doing the whole experiment in some enclosed box to isolate it
from the rest of the universe? Friction must be converting some of the
energy we started with into some other form of energy (not showing up as
obvious kinetic or potential energy), as you correctly feel. Most of the
energy will end up as thermal energy, everything gets hotter. Also, a
sliding object makes noise and sound waves carry energy. Thermal energy
and sound waves are just microscopically another form of kinetic energy.

QUESTION:
Is it possible to find the exact a volume of a proton neutron or electron?

ANSWER:
Basically, no, because there is no well-defined surface. This is most
clearly illustrated by looking at the density distribution of nuclei
which are spheres similar but bigger than a proton (actually collections
of protons and neutrons). Here we see the density of several nuclei
plotted as a function of how far we are from their centers. Inside they
are pretty constant, like a billiard ball, for example. But at the
"surface" they are diffuse, not sharply dropping off like a billiard
ball. (Incidentally, fm means 10^{-15} m.)

QUESTION:
If a photon has no mass how can it have energy? (e=mc^2). I've read about resting mass being 0 and relativistic mass ( in motion) being different but it seems like semantics to me. Is there mathematical proof of a photon having mass in motion but having zero mass at rest? Or is it inferred. Or is this something we have to accept?

ANSWER:
Always look at the FAQ page
before asking a question!

QUESTION:
A particle, of mass m, on a hill of height h, moves down the hill and obtains a velocity, v, after which it moves on flat ground. All surfaces are frictionless. Suppose I observe this event from an inertial reference frame that is moving at velocity v. At first, the total energy of the ball will be:
1/2 mv^2+mgh
However, after the ball has moved down the hill, I will observe its velocity to be zero. Therefore, the kinetic energy is zero and the gravitational potential energy is also zero.

ANSWER:
First consider the simpler case of m falling from rest straight
down a distance h and acquiring a speed v . Then E _{1} =E _{2} ,
mgh = Ѕmv ^{2} .
If we look at it from a frame moving to the left with speed v ,
the object, at the bottom, will have a horizontal component of velocity
v _{x} =v and a vertical component of velocity v _{y} =v .
So, the speed at the bottom from this frame is √(v _{x} ^{2} +v _{y} ^{2} )=v √2,
so kinetic energy =Ѕm (v √2)^{2} =mv ^{2} .
So, energy is conserved,
mgh + Ѕmv ^{2} =mv ^{2} which
gives the same result as observed from the frame where m is at
rest at the start. Now, the case you give apparently does not have
energy conserved in the moving frame but it is conserved in the rest
frame. How can that be? You have to ask the question: when is energy
conserved? Energy is conserved if there are no external forces doing
work. What forces are there in the rest frame (not including gravity
which is taken care of by the potential energy)? Only the force the hill
exerts on the mass which is normal to the surface. But, since m
is moving parallel to the surface, that force does no work and energy is
conserved. In the moving frame, however, m is not moving parallel
to the surface because of its constant horizontal velocity of v .
Therefore the hill does work on m in the moving frame and energy
is not conserved. It is easy to calculate how much work the hill does on
m : E2=E1+W , 0= mgh +Ѕmv ^{2} +W ,
so W =- mgh -Ѕmv ^{2} =-2mgh.

FOLLOWUP QUESTION:
I would also like to know where the energy goes.

ANSWER:
This is a little tricky. We assume that the hill cannot move, that is,
it has infinite mass. Therefore we can say that the lost energy is in
the kinetic energy of the hill's recoil (even though the recoil velocity
is zero!). With M _{hill} = ∞
and v _{hill} =0, it is not wrong to say that Ѕ M _{hill} v _{hill} ^{2} =2mgh.

QUESTION:
Can we literally see an atom given today's technology ?

ANSWER:
Depends on what you mean by "see". If you mean seeing light from the
atom forming an image of it in a microscope, like we observe bacteria
for example, then no. The reason is that an atom is much smaller than
the wavelength of visible light and geometrical optics does not allow
that. But individual atoms can be imaged with electron microscopes or
things called atomic force microscopes and scanning tunneling
microscopes. Google those to see how they work. Atoms can be manipulated
too. The picture shows a famous IBM image of individual atoms pushed
around to spell IBM.

QUESTION:
Since the Milky Way is a spiraling galaxy, can Hubble or perhaps a stronger telescope in the future, be able to look back and see Earth when it was positioned on the opposite side of the galaxy from it's present location?

ANSWER:
Think about it: if nothing can travel faster than (or as fast as) the
speed of light, how could the earth, having to travel a longer distance
than the light would, arrive on the other side of the galaxy before the
light?

QUESTION:
In quantum mechanics a free particle has a wave associated with it.The wave is a wave packet localized in space.It is supposed to be formed by supeposition of many plane waves with value of k quite close to each other.De Broglie says the wavelength of the matter wave associated with the material particle is h/p.If the wave packet is localized in space (as it represents a particle which is always localized in space) it lacks spatial periodicity.Then how can we assign a wavelength to it?????

ANSWER:
You are getting into some pretty subtle stuff here. At the root of your
question is the
Heisenberg
Uncertainty Principle (HUP) which states that you cannot
simultaneously know both the position and momentum of a particle to
arbitrary precision. If you think of a particle as a wave packet with
some finite extent (some uncertainty in position), then, as you seem to appreciate, it does not have a precisely observable wavelength
but rather a distribution of wavelengths (some uncertainty in momentum).
If you state that a particle has a specific momentum (wavelength), the
particle wave is one of infinite extent, that is you have total
ignorance of the particle's position. The first statement of deBroglie
wavelength assumed an infinite wave train for the particle.

QUESTION:
if person is standing on earth exerts force equal to 'mg' ,then the same person falling from height when reaches to land exerts force 'mg' as in the both cases the distance of the person from earth's gravitational center is same.But in practical , in second case the person experiences more force of reaction than first case when he was stable ,so why this happens? in the both cases the person was at the same distance ,only , in first case he was stable and in second, in motion then how the motion affect the value of force? is the value F= G.M.m/square of dist. for only stable body ? then what about the body in motion?

ANSWER:
You must consider all the forces on an object. When standing on the
ground, the earth exerts a downward force mg . If you are at rest,
the sum of all the forces on you must add to zero, so the ground exerts
an upward force on you, call it N , which must equal mg in
magnitude; this is Newton's first law. If you hit the ground after
falling, you are not in equilibrium when you are stopping, instead you
have a large upward acceleration. Therefore there must be a net upward
force on you to stop you, so the upward force N the floor exerts
on you must be bigger than your weight mg but the force of
gravity on you is still mg . Specifically, ma=N-mg . This is
Newton's second law. If the floor exerts on you, you exert an equal and
opposite force on the floor. Therefore you exert a downward force
of magnitude N on the floor. This is Newton's third law.

QUESTION:
My question is regarding the dimensional formula of trogue. Dimensional formula of trogue & energy (potential, kinetic & also work) are same but both are different physical quantities. How can different physical quantities have same dimensional formula? If they have same dimensional formula then they should have some relation in between them? how is it possible? please explain it..

ANSWER:
Just because two quantities have the same dimensions does not mean that
they are related in a fundamental way. Energy has its roots in the
concept of work which is the component of the force parallel to
the distance over which it acts times that distance, F _{parallel} d .
Torque about some axis is the component of the force perpendicular
to the distance from the axis times that distance, F _{perpendicular} d .

QUESTION:
i was reading some ideas about Einstein's theory of relativity and it might be too difficult to explain but the main part that i didnt understand was that the speed if light remains constant regardless of the frame of reference. This is one of the major ideas about the topic that im struggling with and i was hoping you could explain it to me. Because it maths at etc, we learnt about relative velocity and such ideas that make it difficult for me to grasp the concept.

ANSWER:
I think that two of my FAQs might help you. The
first discusses why the speed of
light is a universal constant. The
second gives information about how velocity addition differs in
relativity from classical physics.

QUESTION:
Imagine I have a million-mile long, unstretchable wire out in space. If I pull on one end of the wire, will the other end move at the exact same instant? If not, then how fast will it be?

ANSWER:
Always read the FAQ before
asking a question!

QUESTION:
If we make a hole in the earth & throw a ball in it then what will happen

ANSWER:
Always read the FAQ before
asking a question!

QUESTION:
It is known as light particles has no mass.
Therefore mass=0
But F=ma > F=0
But what about laser which exerts a force on the everything and penetrate?

ANSWER:
Newton's second law, in the form F=ma , is not correct for
relativistic situations, situations where the speed is comparable to the
speed of light. If anything is relativistic, it is light itself! The
correct form is F =dp /dt , time rate of change of
linear momentum. A photon has linear momentum even though it does not
have mass.

QUESTION:
Is energy required to exert a force? (not talking about doing work, just exerting the force itself...)

ANSWER:
Consider a 10 lb weight hanging from a rope. The rope exerts a force and
no energy is required. Now consider you holding that 10 lb weight. In
this case, although no work is being done on the 10 lb weight, energy is being consumed
by your muscles. It
turns out that when a muscle in your body is exerting a force, the cells
are continuously slipping and retightening on the microscopic level, so
this "microwork" being done internally requires energy.

QUESTION:
you know, how if you give a hanging string, like from a wall or something, a small displacement, the subsequent wave takes some time to go all the way to the top of the string. What does the time this displacement take to reach the top of the string depend on? Is it on the mass and the length. If so, how?

ANSWER:
I had never worked this out before and it is kind of interesting. The
speed v of a wave in a string of length L and mass M
with a tension T is v = √(TL /M ).
But, in your case, the tension is dependent on where you are; think
about it—the tension is biggest at the top and smallest at the bottom
and so the pulse speeds up as it goes up the string. The tension can
easily be shown to be T (y )=yMg /L where g
is the acceleration due to gravity and y is the distance from the
bottom end. If you put that into the expression for speed, v (y )=√(yMg /L )(L /M )=√(yg ).
So, the speed (as a function of distance from the bottom) is independent
of mass or length; I would not have guessed it! However, to get the time
t you need to integrate over the length of the string. Doing that
integration, I find t =2√(L /g ). So the time depends
on length but not mass.

QUESTION:
Although it is not possible to travel at the speed of light, unless I am wrong with enough energy, you could accelerate yourself to nearly the speed of light, and not violate the laws of physics. My question is this: Say you wanted to cross a galaxy, and it is 50,000 light years across. If, with some future technology, you could achieve 99.99% the speed of light how long would you (in your ship) have to wait to get across? Although someone on earth would have to wait approximately 50,000 years for you to get there (plus the time for the light to get back), wouldn't the fact that time for you is passing so slowly, leave you with like a 50 year wait? And would you be able to watch say a planet you are aiming for, have 50,000 years of history play out in what looks to you like someone hit a fast forward button?

ANSWER:
It would certainly take less time in your frame of reference. The
formula for time dilation is, for your example, T =50,000 √(1-.9999^{2} )≈707
years. Yes, you would see a "fast-forward history" of the approached
planet; it would start at a time 50,000 years before you left and end
707 years after you left.

QUESTION:
It seems possible to me that the tools we have to measure time are electromagnetic in nature and, therefore, lead us to define the speed of light the speed limit of the universe. After all, if the tools cannot read anything faster than the speed of light, how would we find anything faster? I don't see this possibility considered by reputable scientists. Why are we so confident in light being the absolute speed?

ANSWER:
The speed of light being the fastest possible speed has nothing to do
with measurement. We do not " …define
the speed of light to be the speed limit…", it is a necessary
consequence of electromagnetism and the theory of special relativity.
You really need to read the related FAQ .

QUESTION:
If all of the elementary particles in the atoms of the sun (you can assume pure hydrogen for simplicity if you like) were to condense in on themselves such that all electrons, protons and neutrons were in intimate contact with one another, how large would the sun be? What would its (approximate) density be? How would this compare to the density of a novaed star? I know this reads like a textbook question, but I'm an engineer and I cant think of any less-textbooky way of asking.

ANSWER:
What you are describing, essentially, is a neutron star. It results from
gravitational collapsed of an exhausted star. The density is comparable
to that of a nucleus, about 2x10^{17} kg/m^{3} .

QUESTION:
My question concerns positronium.
Literature speaks of a bound state of electron and positron, hydrogen-like spectrum, orthopositronium and parapositronium and annihilation.
I have made many searches but I did not find any answer to the following question :
Which, for sure, of the positron or the electron is in the place of the nucleus ? I don't want an answer like : it goes without saying that the positron replaces the proton of a hydrogen atom. I need experimental evidence, not theoretical assumptions of whatever theory.

ANSWER:
You do not need "experimental evidence" to answer your question. The
two-body system will always rotate about the center of mass which, for
positronium, will be halfway between the two. So neither should be
thought of as the center. Experimental proof would be that, assuming the
two move around the center of mass, you can predict energy levels which
would differ if you held one of the two fixed; this prediction matches
experimental measurements of the spectrum. It is no different for
macroscopic systems; if the earth and moon had the same mass, they would
both orbit around a point halfway between them. In fact, since the earth
does not have infinite mass, the earth actually "wobbles" as the moon
orbits because the moon does not orbit the center of the earth but a
point away from the center. You can calculate the position of the center
of mass from one of the masses (m _{1} ) as R _{cm} =r _{2} m _{2} /(m _{1} +m _{2} )
where r _{2} is the distance of m _{2} from
m _{1} .

QUESTION:
40 some odd years ago when i was in grammar school the science class had this devise that was shaped like clear light bulb at the base was a day glo reflector in the center of this bulb was a sort of turbine with a vertical axis almost resembling a wind measuring devise when exposed to sun light the turbine would spin my question is what is the name of this thing and how does it work

ANSWER:
It is called a Crookes radiometer. See an
earlier answer and follow
the link to Wikepedia for an explanation.

QUESTION:
So if the earth is spinning so fast that it rotates completley in 24 hours, then why cant you get into a helicopter, and just stay in place for 12 hours and end up in China???

ANSWER:
The helicopter flies relative to the air which means that, when it
hovers, it is at rest in the air. But the air moves with the spinning
earth, that is the world drags its atmosphere with it as it moves. So
the helicopter stays in the same place relative to the ground.

QUESTION:
how can we tell the difference from large bright stars at the edge of the universe(the farthest we can see) and small less intense stars relatively closer? wouldn't they appear to have the same luminosity or brightness?

ANSWER:
You are right, you cannot tell by how bright a star is how far away it
is. Distance is usually determined by measuring the red shift of light
observed. Another important method for measuring distance is to use
"standard candles", a particular type of supernova whose absolute
brightness is known. Since (as stated on the site) I do not answer most
astrophysics questions, this is as much detail as I will give here.

QUESTION:
one of my co workers came with a challenge and I was wondering if it was correct in his assumption.
He said that if you have 2 cars coming in opposite direction and both cars have the same mass running at the same speed 50 miles per hour, when they collide head on the point of impact will be the same if one of the Cars collide head on with an unmovable wall.
Is this correct? Would the forces cancel each other?

ANSWER:
The only thing that matters is how much the momentum (mass times
velocity) changes and how long the collision lasts. I assume we are
talking about a perfectly inelastic collision (the two cars stick
together or the car sticks to the wall). In both cases, the momentum
change is the same since the car started with the same speed and
stopped. If the time to stop was the same, the collisions are
indistinguishable from the perspective of one car. It has nothing to do
with "forces cancelling each other".

QUESTION:
if attraction between two object is called gravity then why there is 38% less gravity in mars or any other planet in the universe?

ANSWER:
Gravitational force at the surface of a planet depends on the mass (M )
of the planet and on its radius (R ). The dependence is M /R ^{2} .
So, for example, if a planet has a mass twice as big as the earth and
its radius is twice the earth's, then the gravity at the surface is 2/2^{2} =1/2
as big as the earth's.

QUESTION:
Faster than light travel? If your traveling in a car at 30mph and throw a ball forward at 30 mph the ball will then be traveling at 60 mph.
So what happens when you turn on your headlights?
Would that accelerate, the light coming from the car, by 30 mph?

ANSWER:
Always look on the FAQ page
before asking a question!

QUESTION:
In a neutral atom, there is an equal number of protons and electrons, right?
But how could the charges of the protons and electrons cancel out if one proton is so much more massive than one electron? Are charges not linked to the mass of the particles? What does charge depend on if not mass?

ANSWER:
Mass and charge are not related.

QUESTION:
Some young earth creationists have suggested that the speed of light was orders of magnitude higher in the past. It seems to me that if this were true, electromagnetism would have been stronger, since photons carry the electromagnetic force. Thus, atoms would either be unstable or very different than they are today. Do I have a valid argument?

ANSWER:
The speed of light is 1/ √(ε _{0} μ _{0} )
where ε _{0 } and μ _{0 } are the constants
which tell the strength of electric and magnetic forces, respectively;
larger values of the constants mean weaker electromagnetic forces.
Hence, if the speed of light is larger, electromagnetism would be
stronger as you suggest. I prefer that argument to your photon argument.
In any case, if the constants were significantly different, so would
atomic structure be.

QUESTION:
There was a demonstration in our school's physics club about a spring with two masses attached to it. The guy showing this mentioned that if you apply an instantaneous impulse on the lighter mass, the total energy of the oscillation can be found out by first squaring the magnitude of the impulse, then multiplying it by the ratio of the big mass to the smaller mass, and then dividing it by the sum of the masses.
Just HOW does he claim that? I'm just curious.. because I've no idea of how to prove/refute him...

ANSWER:
Well, if I understand the situation, I get almost that answer —I
get half what you gave me. Here is how I got it. If the impulse I
is indeed instantaneous, then immediately after the impulse the small
mass m has a speed v given by v=I /m because
the change in linear momentum (mv ) equals impulse. Hence, the
total energy transferred to m is E _{total} =Ѕmv ^{2} =ЅI ^{2} /m.
But, because m is coupled to the larger mass M , the
two of them as a system must have the same amount of linear momentum,
that is, their center of mass must have momentum mv , so (M+m )u=mv ,
where u is the speed of the center of mass. So the
translational kinetic energy of the whole system is E _{trans} =Ѕ(M+m )u ^{2} =ЅI ^{2} /(m+M ).
So, the the translational kinetic energy of the whole system is less
than the total energy of the whole system. This makes sense since it is
vibrating as it moves along and some of the total energy resides in that
vibrational motion. E _{vib} =E _{total} -E _{trans} =ЅI ^{2} [(1/m )-(1/(m+M ))]=ЅI ^{2} [(m+M-m )/(m (m+M ))]=ЅI ^{2} [(M /m )/(m+M )].
So, you see, I get half of what he says it is. (I have been known to be
wrong!)

QUESTION:
Inspite of the fact that is generally beleived that like charges repels each other while unlike charges.why is it possible to que all the protons in the nucleus of the atom together.

ANSWER:
A proton does not interact only with the electromagnetic force. It has
mass and therefore interacts gravitationally with other masses. This,
however, is negligibly small. It also interacts via the strong or
nuclear interaction. This is a force felt by protons and neutrons, is
much more attractive than the coulomb is repulsive at short distances.
It is the strong interaction which holds nuclei together.

QUESTION:
If you had a pole that somehow extended from here to a planet in another solar system, many lightyears away, would you be able to communicate a signal faster than the speed of light?

ANSWER:
Always look at the FAQ page before
asking a question!

QUESTION:
If a ramp(20 ft by 40in. approximately 500 lbs) is attached to a floating dock (8ft by 14.5 ft), is there a difference in the amount of pressure on the dock if the ramp is at a steeper angle when the water level goes down. We are trying to figure out if one end of the dock needs more styrofoam where to ramp attaches.

ANSWER:
As long as the dock is free to move horizontally, the downward force
from the ramp will be independent of the angle. If the ramp is uniform,
i.e. would balance on its center, the force down on the dock is
half the weight of the ramp.

QUESTION:
A beer is placed in the freezer for a little while and then removed. It's obvious that the liquid in the bottle is not frozen at this time, however when I pop the top the liquid begins to freeze. I'm not a science student, I just happened upon your site and really enjoy reading all the cool stuff so I decided to ask you about my slushy beer.

ANSWER:
The essence of the answer is that the disolved CO_{2} in the
beer lowers the freezing point of the liquid; when the beer is opened,
the pressure decrease causes the disolved gas to leave and the
supercooled liquid freezes. A more detailed explanation can be found at
this link .

QUESTION:
I am trying to find out the total weight of impact. 3 6 lb fire logs fell on top of my foot from approximately 4 feet high on a shelf and hit about the same time on the top of my foot where the toes start to form. So with the height and velocity what would the impact weight be?

ANSWER:
There is no such thing as "weight of impact". You can estimate the force
which you felt during the time the log was in contact with you. If you
drop an 18 lb weight from 4 ft it will have a speed of about 16 ft/s
(about 11 mph) when it hits your foot. Now you have to make an estimate
to determine the most important quantity, how long it takes this moving
object to stop. I assumed that it traveled about 0.2 in, crushing your
foot that much and that its acceleration was uniform. Under those
assumptions, the time to stop would be about 0.002 s. The average force
to stop it would be exerted by your foot and would have been about 4500
lb over the 2 thousandths of a second. Since your foot exerts that force
on the log, the log exerted an equal and opposite force down on your
foot.

QUESTION:
Why electric field lines are taken to be coming out of positive charges, and going in negative charges?

ANSWER:
Electric field is defined as E =F /Q
where F is the force felt by a charge Q ; if Q
is positive, E is in the same direction as F .
The force felt by a positive charge Q near another positive
charge Q' is repulsive and therefore the field due to Q'
must be away from Q' . Similarly, the force felt by a positive
charge Q near a negative charge Q' is attractive and
therefore the field due to Q' must be into Q' .

QUESTION:
Im wondering if work can happen in 0 gravity environments like outer
space. I know work involves the exertion of a force and the movement of
something by the force in a certain direction where a force would be
applied, but since there is no real direction in space ( up and down )
and no gravity for much energy to be used how could work in the physics
term be used in space? simple question: Is work possible in space
according to physics?

ANSWER:
Well, you never find zero gravity in this universe but I guess you could
get pretty close away from galaxies. There is no reason why, if you had
a way to exert a force you couldn't. If you could not exert a force you
could not steer a spaceship or probe.

QUESTION:
If the light from a star 26,000 light years away takes 26,000 years to reach our eyes, is that time diminished for every X power of the telescope used to observe it? And if that same star went supernova today, how long would it take us to be able to see it? Is it possible some of the things we are looking at don't even exist anymore but the light (image) is still travelling toward us?

ANSWER:
No matter what kind of telescope you use, you are seeing light which
originated at the star 26,000 years ago. Whatever happens right now will
be unknown to us for 26,000 years. Certainly, anything we see now might
not exist now.

QUESTION:
One student asked me the following question he saw in a book-
The displacement of a linear damped oscillator is given, for weak damping by
x=Ke^(-bΩt) sin(βt+ ε)
Show x has a maximum at times given by βt = 2nπ + d where n is an integer and d is gives the position of the first maximum, because although they stated this in the book without proof!
Show also that the ratio of successive maxima is e^(2bΩπ/β). I think this is the logarithmic decrement?
I could not answer his questions... Could you please help?

ANSWER:
Normally I do not answer homework-like questions, but since this is from
a teacher (in Singapore) I will make an exception. To find the times
where maxima occur, set dx /dt =0. If you do this you find
tan( βt+ε )=β /(bΩ ).
The constant values of a tangent function are spaced by 2π
and so
βt _{n+1} =βt _{n} + 2 π.
So, the maxima at times t _{n+1}
and t _{n}
are related as x (t _{n} )/x (t _{n+1} )=exp{-[bΩ (t _{n} -t _{n+1} )]}=exp[2πbΩ / β ].

FOLLOWUP QUESTION:
Thank you for your answer to my question. But what about the answer to the first part of the question, where it has to be shown that x occurs at maxima at times given by βt = 2nπ + d where n is an integer and d is gives the position of the first maximum?

ANSWER:
You did not think carefully about my original answer. I did answer that
question by stating that " the
constant values of a tangent function are spaced by 2 π ".
The first maximum is
β t _{0} =d =tan^{-1} (β /(bΩ ))-ε
(the smallest positive value); the next is β t _{1} = d +2π ;
the next is β t _{2} = d +4π ;
the next is β t _{3} = d +6 π ;
etc . so that
β t _{n} = d +2n π.

QUESTION:
If You Were To Fall In A Hole Through The Center Of The Earth, How Long Would It Take You To Land In Beijing?

ANSWER:
If the earth is modeled as a uniform sphere (same density throughout its
volume), it turns out that a particle of mass m behaves as if it were on
a spring with spring constant k=mMG /R ^{3} where
M is the mass of the earth, G is Newton's universal constant
of gravitation, and R is the radius of the earth. Putting in the
numbers and using
ω =2πf =2π /T =√(k /m ), I find that
half a period is T /2=42 min.

QUESTION:
I know the phenomena is well understood but I have been unable to find a good answer for my question anywhere.
We know that entangled particles are dependent upon one another and for two particles we know that, amongst them, are two sides of the same property such as spin including up and down. But we must have both, and after the measurement each particle will hold one or the other and they will be opposites.
I understand not knowing initially which particle is which but where does the idea come from, and in what way is it confirmed, that neither particle has either state until it measured or disturbed in some way?
I mean, logic tells me that if you find one is up then of course the other is down! What am I missing? How do physicists know that neither is determined beforehand thereby implying the instantaneous interaction at a distance?

ANSWER:
Here's what you are missing: logic learned from everyday life is likely
to lead you down the wrong path if you are in a realm where quantum
mechanics governs how things behave. In quantum mechanics, the
properties of a particle are determined by something called the wave
function. A particle might have a spin wave function of either
Ψ = ψ _{up}
or
Ψ = ψ _{down}
for its spin being either up or down, respectively. Entangled particles
must have total spin zero, and you surmise that this means one must be
up and the other down. But, in quantum mechanics, a particle might have
a wave function which is part up and part down, that is
Ψ =A ψ _{up} +B ψ _{down}
where A and B are constants which tell you how likely it
is that the particle would be observed in the up or down state. If
A=B , it is equally probable to be observed in either state. When you
make a measurement on a particle, you "put" it into one state or the
other (called collapsing the wave function). So, if you observe one of
the entangled particles, you force either A or B to be
zero. But, since the two particles must have zero total spin, that means
you have simultaneously put the wave function of the other particle into
the opposite state. Skeptics originally said that this must simply mean
that the two particles were already in that state, you just found it out
when you observed one; experiments have been designed to rule out this
possibility, but they are a little too complicated to go into here.

QUESTION:
Why are electrons unimportant to nuclear chemistry?
or my question would be; Wiki says that electrons are not importan when it comes to nuclear chemistry, so Does creating nuclear reactors and bombs focus mainly on the nucleus, (or the protons and neutrons)?

ANSWER:
Electrons, for all intents and purposes, do not participate in nuclear
reactions. The reason is that they interact only via the Coulomb
interaction (electromagnetic) and the strong interaction dominates the
behavior of nuclear reactions like fission and fusion (bombs and
reactors).

QUESTION:
Why does a fan have a cooling effect, when it seems that it would only speed up the particles, thereby increasing the temperature.

ANSWER:
Average molecular speeds are hundreds of meters/second, so the added
speed caused by the fan causes a negligible change in the overall
temperature. The cooling effect is that heat from your body is more
efficiently carried off by a breeze than by still air. It also expedites
evaporation from your skin which is the main way of the body cooling
itself.

QUESTION:
if it takes an infinite amount of energy to travel at the speed of light, how can the photon's energy be measured?

ANSWER:
It does not take "an infinite amount of
energy to travel at the speed of light". It takes "an infinite amount of
energy " to accelerate an object with mass to the speed of
light. A photon has no mass, and so it can (and must only) travel with
the speed of light in vacuum.

QUESTION:
When an atom is excited the electrons go
up in energy level and then soon return to ground state emitting a
photon. If energy cannot be created, where, physically, does this photon
come from?

ANSWER:
The excited atom has more energy than the ground state by exactly the
amount of the photon's energy, so energy is conserved. In order to
excite the atom you must do work on it, so the increased energy of the
excited atom came from whatever agent excited it. Another way to look at
the situation is to say that the excited atom has a tiny bit more mass,
Δm than the atom in its ground state and that is where the photon
gets its energy: E _{photon} =Δmc ^{2} .

QUESTION:
If kinetic energy is conserve in compton effect?
I know that total energy and momentum is conserve, so just want to know about Ek.

ANSWER:
Kinetic energy is total energy minus rest mass energy. Since the rest
mass energy does not change, the kinetic energy does not change.

QUESTION:
If an object in space changes its mass ( don't worry about how for the moment) how fast is that change noticed elsewhere. My question arises from the fact that I cannot find any source for the speed of gravity, by this I mean the speed with which changes in mass are apparent from a distance. Is it instantaneous, if not what has to move

ANSWER:
Always read the FAQ page
before asking a question!

QUESTION:
how do we know that distant galaxy, or star, or exoplanet is really regular matter at all? How do we know the universe is made up of matter, and its not a patch quilt of antimatter galaxies and matter galaxies? Or would this be the likely conclusion if it were determined that neither particle holds an advantage over the other?

ANSWER:
The thing is that all the empty space between stars and galaxies and
planets and so forth is not empty space. There is still a lot of
intellerstellar gas and dust and therefore you could not have, for
example, an antimatter galaxy because we would see it violently
interacting with the matter in the space near it.

QUESTION:
How much force does it take to move a 3600 lb object?

ANSWER:
Again we have that question "how much force does it take …"
when there is no answer. Suppose the object were on a very smooth level
surface (virtually firctionless). Then it would take maybe 1/10 oz of
force to get it moving 1 ft/s if you pushed for a second. But suppose
the object were on a rough surface. It might take a million pounds of
force just to break it away from the friction. Suppose that the object
were welded to the floor. Then maybe it would take a trillion tons of
force to get it moving. The only sensible question you can ask without
giving more information (and a definition of what is meant by "to move")
is how much force does it take to lift a 3600 lb object? That would be
3600 lb.

QUESTION:
I think that if something were to reach the speed of light (or close enough) would it collapse into a singularity? My thought is that if when things go faster they gain mass and shorten then they must become denser and denser. So if it gets to the speed of light (or close enough) than it would turn into a singularity.

ANSWER:
As I have said more times than I can count, no material object can reach
the speed of light since it requires infinite energy to achieve that.
(Also, site groundrules forbid questions which assume that something
goes the speed of light.) You are right, as the volume gets smaller and
the mass gets larger, certainly the density gets larger. In the object's
own rest frame, however, everything is absolutely normal. Singularity?
Never!

QUESTION:
If light/electromagnetism can function as both a wave and a particle, what is the medium in which the wave travels?

ANSWER:
Among all waves we know, light is the only wave which does not require a
medium to propogate. It will propogate through completely empty space.

QUESTION:
I have a couple questions relating to the same thing. My first question is this, if the gunpowder in a bullet is sealed air tight and you fire it in space, what would happen? I would figure that the air in the shell would help ignight the powder but once that explosion hits the vacume of space would the bullet still fire?

ANSWER:
Gunpowder has, as one of its ingredients, a nitrate (usually potassium
nitrate) which supplies the oxygen for the chemical reaction (burning)
and therefore should work just the same in a vacuum as in air.

QUESTION:
As I understand it, [simplistically] every atom beyond hydrogen, or maybe helium too, was forged in the heart of a star or from the death throws of a star, then distributed violently about the universe to coalesce with other things and make rocks and planets and such. Earth was formed this way, and the various elements and chemicals somehow collected together in little "clumps" (such as veins of gold or copper, concentrations of carbon, oxygen, hydrogen, or combinations of them helping bring about life) rather than being randomly distributed. At what point(s) did each clump form? Are like elements and chemicals somehow attracted to each other while zooming through space, like a gravitational gradient automagically sorting atoms, or is there something about the planetary environment that pools and sorts the elements? Or is it more that the stage of the star or the type of star determines what elements are being forged at any particular time, so they are already pretty much clumped together when they, um, exit the star in great haste. I hope you understand what I'm getting at and my understanding isn't too far off base :-).
I think it boils down to, "how did the elements necessary for life manage to be in the right place at the right time for life on Earth to come about, and even be sustained?" To make carbon based life forms, you kinda have to have a bunch of carbon handy in a readily accessible form, along with all of the other elements necessary, right? How did it get there?
I also wonder "how many of the current collection of atoms that make me are the original ones I was born with, or are those long gone and replaced by natural processes?" but that's probably beyond the scope of what I can hope to get answers for, if I get any answers for this

ANSWER:
This is not a concise, well-focused single question as required by the site
groundrules , is it? Interesting questions, though. Yes, all atoms come from dead stars. Elements up to iron can be made in any star, beyond that extra energy must be added in a supernova explosion;
see my answer on
fission and fusion . Questions about how veins
of gold, e.g. , come to be should be referred to a geologist, not physicist. I suspect that it is done by catastrophic events and early high temperatures. The earth's core is mainly iron. The question of how much of you is original is for biology to answer. I suspect that most cells completely turn over; that is certainly true for skin and bones. On the other hand, there are surely many atoms in your body which are "recycled", that is they used to be in your body, were sloughed off, and came back. This is because there are such vast numbers of atoms in your body; see an
earlier answer . (In
future, please ask one question at a time!)

QUESTION:
How large does a particle need to be in order to physically (and immediately) damage the human body as it passes through? For example, we read about neutrinos passing through us and the Earth constantly, without doing damage. Then I read about the astronaut who got something in his eye during a space walk. So, what size particle would need to hit the astronaut to create a truly dangerous situation?

ANSWER:
This question is impossible to answer because the questioner does not
specify what is meant by "damage". The neutrino is probably the only
thing you could say does no damage because it passes through without
interacting at all (that statement is not quite true, but the
overwhelming majority do not). Any charged particle is what is called
ionizing radiation, and provided it has sufficient energy to penetrate
it will ionize atoms and molecules in its path. A single energetic
electron will not do noticable damage but a billion of them might.
Similarly, photons are ionizing radiation so, x-rays or gamma-rays in
sufficient numbers will cause damage. Fast neutrons are not charged, but
because their mass is close to that of a proton and there is really a
lot of hydrogen in our bodies, they are fairly efficient at knocking
protons out of the molecules they might be residing in. Slow neutrons
might be absorbed by a nucleus and result in radioactivity in the body.
Regarding the astronauts, in addition to ionizing radiation they
encounter, need to worry about micrometeorites or debris which would be
big enough to create an air leak in their space suit; I don't really
know how tough those suits are.

QUESTION:
Hi, how do physicists or mathemeticians transform a complex equation into a concretete picture of reality. We have all seen these (to a layman,unintelligible) equations on a blackboard but I don't understand how the leap is made to then describe the world around us.

ANSWER:
This is not a well-focused question as stipulated in the groundrules of the site. How could you possibly expect a succinct answer to such a broad question? It is sort of akin to asking "how can those Chinese people understand all that gibberish" without bothering to learn Chinese.

QUESTION:
If an object were rotating near the spead of light, would it undergo circular lorenz contraction so that it would not be visible when looking at the poles? What if it were also traveling near the speed of light in a linear fashion, would it then become almost completely invisible? (nearly pointlike?)

ANSWER:
The first question cannot be addressed by just thinking about length
contraction. Think about it: if we look at the "equator" of your
rotating object and assume that the circumference is shortened by length
contraction, what about the radius of that circle? It is not contracted
because its motion is perpendicular to its length. So that means that
π ≠3.14159… That would mean that geometry is not Euclidean.
Objects which are accelerating are not very conveniently understood
in special relativity. The second part of your
question, though, can be understood with special relativity. If you are
thinking about a spherical object, it is actually only shortened along the
direction of its velocity so it is liked a squashed ball, becoming
pancake-like at very high speeds (not point-like). But here is the
really cool thing—it does not look like a pancake, but still like a
sphere. For my earlier discussions of how things look as opposed
to how things are , see an
earlier answer and links there.

QUESTION:
If it is possible to take out all the electrons from a metal one by one what will happen to it? Is it possible to remove the core electrons?

ANSWER:
Sure. Just add enough energy to ionize the atom and keep any electrons
from hopping back on. Inside stars, many of the atoms are fully ionized.

QUESTION:
Is there anyway to hold a plasma within a metal tube without it touching the sides?

ANSWER:
In principle, you can confine a plasma with a magnetic field, called a
magnetic bottle, but it is extremely difficult and many plasma
physicists have devoted their entire careers to being able to do so for
a reasonably long time and it is still not an easy task.

QUESTION:
For "special relativity" Newton's 2nd law must be expressed as F = dp/dt, but do his 1st & 3rd laws still apply?
I've never seen these laws mentioned in any text I've read, hence they must not be relevant?

ANSWER:
Force and acceleration both play a much smaller role in relativity and
quantum mechanics than in classical physics. The notion of an isolated
system having never-changing linear momentum (p =mv )
is, as anyone who has studied classical mechanics knows, extremely
powerful; this is called momentum conservation. One of the first things
we find when studying special relativity is that momentum conservation,
in the classical sense, is not true. But we like it so much that we seek
a quantity which is approximately mv at low speeds but is
conserved in an isolated system. This leads to a new definition of
linear momentum which you can read about in an
earlier answer . This allows us to recover the no-longer-true
Newton's first law which states that for an isolated system (that is, in
classical terms, no forces from outside the system) dp /dt =0.
(Some people consider the "meat" of Newton's first law to be the
definition of what an inertial frame of reference is, rather than a
"special case" of Newton's second law. In special relativity we start
with a definition of inertial frames as those in which all laws of
physics are true and from that we find Newton's first law in the form of
momentum conservation.) Newton's second law could be thought of as
F =dp /dt although force is just not used
that much in relativity. An
interesting exercise is to calculate the motion of a point mass
under the conditions that the "force" is constant; I define a constant
force to be one which results in a constant rate of change of
(relativistic) momentum. Newton's third law is actually very useful in
simple introductory physics but not really true under many
circumstances. Magnetic forces, for example, often do not obey Newton's
third law. The problem with the third law is that elementary physics
presents it in too constrained a way — the
equal and opposite forces between interacting particles; it is more
general to say that the third law is, in essence, a statement of
momentum conservation. Then, in electromagnetism, the fields carry also
energy and momentum and the momentum is still conserved even though if
you just look at the charged masses, it appears not to be. In that
sense, the third law is still obeyed but really does not carry any
really new information.

QUESTION:
Is there a way to derive conservation of angular momentum solely from linear momentum concepts? I am asking because it seems like defining L=mvr was pulled from nowhere. Out of couse, it works very nicely as a conserved quantity if no external torques act...but what would give someone the idea to do this in the first place?

ANSWER:
Rotational dynamics, at least in elementary physics, comes entirely from
translational dynamics. The question you ask, though, is the essence of
a whole big part of an elementary physics course, really involved. I
will provide an answer assuming you know a little physics, but get a
good textbook and study it. First of all, L=mvr is not a
definition, it follows from what angular momentum has to be.
Furthermore, this is true only for a point mass. If you accept that
L=I ω, then I=mr ^{2}
for a point mass and ω=v /r , so L =mr ^{2} v /r=mvr.
Now, you talk about angular momentum conservation. Where does this
come from? From Newton's first and second laws. Imagine a point mass
m with velocity v a distance r from an axis.
we can write Newton's second law: F =ma =m( dv /dt)= d(mv )/dt
and so there are two components of F , F _{p}
parallel to r and F _{t} perpendicular to it. F _{p} =ma _{p} =mv _{p} ^{2} /r
is just the centripetal force and acceleration. More interesting is
F _{t} =ma _{t} =mr α= mr dω /dt
where
α is the angular acceleration.
Multiply this equation by r and get, recognizing rF _{t
} as torque,
τ =mr ^{2} dω /dt= d(Iω )/dt= d(L )/dt.
This shows that L =Iω and that torque is time rate of
change of angular momentum. If torque is zero, angular momentum is
conserved. Thus you have the rotational equivalents of Newton's first
and second laws. (By the way, be careful with L=mvr . It is true
only if the velocity v is perpendicular to the distance
r .)

QUESTION:
why a running cycle is more balanced than a stationary cycle?

ANSWER:
See an earlier answer
on bicycle stability.

QUESTION:
How come the "Absolute Hot," or hottest possible temperature is not equivalent to the to temperature at which the average translational velocity if the atoms is the speed of light. I know that mass plays a part too (1/2m(v^2)=3/2kT). It that the reason?

ANSWER:
No object with mass can have the speed of light (read
earlier answer ), so certainly the
average speed would not be the speed of light since this would
correspond to all particles having an impossible speed. See an
earlier answer .
There is no upper limit to temperature. Also, note that
Ѕmv ^{2} is
kinetic energy only for speeds much less than the speed of light; again,
see an earlier answer .

QUESTION:
Why you can see dust particles through a beam of light coming from the wall in a dark room ,but can't see in an open field ?

ANSWER:
Each dust particle reflects a tiny amount of light which is why you can
see it. In the open field, there is so much light from other sources
that the light from the dust has too little intensity to see it from all
the brighter sources behind it.

QUESTION:
I wear glasses to see distant objects clearly, when I don't wear my glasses distant objects look blurry. My glasses broke and left me unable to see my television clearly, so I came up with an idea, if I hold a mirror close to my face and tilted towards the television I should be able to see the television clearly. To my surprise this was not the case, the television still looks blurry in the mirror even though the mirror was close enough to be seen clearly. Normally with my glasses on the television would have appeared clearly in the mirror. Assuming the image of the television is a perfect representation of the television itself why does it still look blurry even though the mirror reflecting the image is close to my face?

ANSWER:
The reason is very simple. A mirror forms an image which is just as far
behind the mirror as the object is in front of the mirror. Your eye
attempts to look at the image, not the mirror , and your eye is
unable to focus on that distant image.

QUESTION:
the object is a stationary rock and is being pushed from the left hand side but doesnt move what is the opposing (reaction) force?

ANSWER:
First of all, it is not the "reaction" force you want. If you push on
the rock, the rock pushes back on you; that is the reaction force and it
is a force on you , not on the rock. The rock does not move
because there is friction between it and the ground. That is the force
which keeps it from moving.

QUESTION:
If the Earth was stationary, would the moon still orbit the Earth?

ANSWER:
Yes.

QUESTION:
if a hole is made from north pole to south pole of earth and a ball is thrown into it. ball will vibrate or not?

ANSWER:
Always read the FAQ page
before asking a question!

QUESTION:
Say two trains went past each over at three-quarters the speed of light. Out of the window, would it look as if the over train was going one and a half times the speed of light.

ANSWER:
Always read the FAQ page
before asking a question!

QUESTION:
If the constituent parts of all matter are so small and (quantumly speaking) very far apart from one another, why do we not "see through" reality? I've seen pictures and videos of atoms through an electron microscope and they appear to be solid pieces, and yet I know they are built of many smaller particles that are quite distant from one another. Why does mostly empty space seem to be solid matter?

ANSWER:
Always read the FAQ page
before asking a question!

QUESTION:
Can a microwave oven reverse the spinning of an atom electron? Does a microwave oven have any effect whatsoever on an atom? If so what would be the effect be on an atom?

ANSWER:
No, because this can only be done if there is a magnetic field so that
an electron spinning in one direction has a different energy from the
other direction. And then, the radiation must be very carefully tuned to
be just right; this is called electron paramagnetic resonance. Of
course, the microwave has an effect on atoms (water, in particular),
heating it. When you heat something hot enough you change the molecular
structure of it; this kind of chemical reaction is called cooking!

QUESTION:
In things like diffraction gratings and slit experiments light is said to interfere destructively to produce interference patterns of light and dark bands, but where does the missing light go. I understand that the photons that make up light have energy but energy is conserved in the universe and can't just disappear. It can only be shuffled around. Is this just a discrepancy arising from the differences in the quantum theory of light and the classical or is there something I'm missing?

ANSWER:
Energy apparently "missing" in the dark bands shows up in the bright
bands. If you measure the total energy at the screen it will equal the
total energy passing through the slits.

QUESTION:
This is a follow-up to a prior question regarding linear accelerators vs. colliders. I understand that both energy and momentum must be conserved. How does that relate to a situation involving automobile collisions? If a car is traveling at 55 mph is there an "advantage" or disadvantage to colliding either directly with a solid, immovable wall vs. colliding directly, head-on with an identical car going the same speed but in the opposite direction? Is there a difference in the apparent collison speed (55 mph vs. 110 mph for head-on?) Is the driver less likely to be dead in either case? Thank you for your previous explanation.

ANSWER:
OK, now we are talking classically, not about particle collisions in
accelerators but about collisions of things in everyday life. The force
experienced by anything is the time rate of change of linear momentum
(which, classically, is mv ); "apparent collision speed", as you
call it, is irrelevant. So, the question is, how much did your momentum
change and how long did it take it to do so. In both cases (another car
or brick wall), you lost all your momentum; if the time to stop was the
same in each case, the force was the same and force is what hurts or
kills you. That is why, incidentally, airbags are so cool —they
lengthen the time of the collision thereby reducing the force you feel.

QUESTION:
Since an electric current is a flow of electrons is it possible to create proton electricity and even neutron electricity?

ANSWER:
Electric current is net flow of electric charge, period. Anything which
is a moving charge constitutes a current. In some batteries, some of the
current is carried by ions; electrical currents in the body are usually
carried by sodium or potasium ions. Neutrons will never work, though,
because they have no electric charge.

QUESTION:
To get into orbit, a horizontal speed of some 8000 m/s is required, so why is the system launched vertically?

ANSWER:
Because, at such a high speed, air drag is very serious (like burns
things up), the vehicle is first lifted out of the atmosphere. The
shuttle may be able to withstand this heating (designed to do so during
reentry), the rest of the launch vehicle isn't.

QUESTION:
Do radio waves travel at the speed of light? Do the radio signals sent to the Mars Rover travel by short wave, long wave, or someother method?

ANSWER:
Yes, radio waves are electromagnetic radiation, all forms of which travel
with the same speed. The Mars Rover uses X-band radio frequencies, about
8-12 GHz.

QUESTION:
I understand tilting solar cells to be normal to the suns rays optimizes their efficiency. Why is this? Please correct me on my following assumptions:
1. exposure of surface area is not a factor (photon saturation will occur no matter what the angle.)
2. angle of incidence has an effect on the ability to energize an electron thereby ultimately creating current more efficiently.

ANSWER:
You are making this way to complicated! The reason is simply that you
maximize the amount of light which hits the cell by orienting it normal
to the rays.

QUESTION:
The following is a practice problem for a qualifying exam in India: Two
spaceships, A and B, with rest lengths of 50 m have speeds of 0.8c
and 0.6c , respectively, relative to an observer S. At t =0
(measured by all three observers) A has just caught up with B as shown.
Find the time for A to overtake B (back of A is at the front of B) as
measured by S and by B.

ANSWER:
Observer S measures the two ships to be length-contracted by the factors
√(1-0.8^{2} )=0.6 for A and √(1-0.6^{2} )=0.8 for B.
Hence, the position of the back of A is x _{A} =0.8ct -50x0.6=0.8ct -30
and the position of the front of B is x _{B} =0.6ct +50x0.8=0.6ct +40.
Setting x _{A} =x _{B} yields t =350/c.
Observer B sees ship A approaching with a speed v given by
the velocity addition formula, v =(0.8c -0.6c )/(1-0.8x0.6)=0.385c
and so the length he measures for A is 50x√(1-0.385^{2} )=46.15
m. Therefore, a distance of 96.15 m must be traversed by a ship with
speed 0.385c , so t =96.15/0.385c =250/c .

QUESTION:
I found what I consider in my partial ignorance to be a discrepancy:
If E = hf (the Planck equation) and v = f (lambda) as well as Kinetic energy = 0.5 mv^2, when I work on these 3 equations I get
lambda = 2h/mv
and not the actual de Broglie equation : lamda = h/mv
Why is that 2 missing in the actual de Broglie equation ?

ANSWER:
E=hf for a photon only, and a photon has no mass. Therefore, you
can calculate neither the kinetic energy nor the momentum (mv ) as
you suggest. Further, the kinetic energy is not
Ѕmv ^{2} and the momentum is not mv , although that
is approximately true for v <<c. (See the question after
yours for the relativistically correct form of kinetic energy and of
linear momentum.)

QUESTION:
I was asking the question about how particle physicists compute the maximum 1% error when the following conditions exist: KE/Eo <1 00="" and="" v/c="">< 1/10...this="" is="" the="" classical="" region="" where="" ke~1/2="" mov^2="" and="" momentum="" (p)~mov.........all="" are="" rest="" masses.="" i="" was="" curious="" about="" how="" the="" 1%="" is="" computed?="" in="" other="" words,="" how="" do="" they="" solve="" for="" the="" vmax="" such="" that="" the="" 1%="" error="" applies?="">

ANSWER:
Click here to see where the v /c ~1/10
comes from. My notation is m for rest mass.

QUESTION:
Is there any process for counting the total number of photons emitting from a source?

ANSWER:
I am not sure what you want, to actually count photons or deduce their
number. If it is the former, there are photon detectors. They do not
detect all the photons but have a known efficiency so that you can infer
the total. If it is the latter, then if you know the luminosity of the
source in watts (W=J/s), you can relate that to the number of photons
per second if the source is monochromatic (because you know the energy
of a single photon); if it is not monochromatic, you need to know the
luminosity spectrum.

QUESTION:
Why does the density of a rock not change when u put it in water but when im in water my density changes when im submerged in water

ANSWER:
Because a rock does not breathe. When you breathe in, you increase your
volume without changing your mass, so your density becomes smaller, you
become more buoyant. When you breathe out, your density increases and
you become less buoyant, maybe even sinking.

QUESTION:
Can I make the mass of a'n atom more by moving electrons further away from center of the atom?
So I want to make the atom heaver by moving electrons to the atoms outer valence rings?
Is that valid?

ANSWER:
In order to excite an atom you need to add an amount of energy Δ ε.
Now, where does this energy show up? Quite simply, the atom has an
increased mass Δm =Δε /c ^{2} .
The thing is, this change in mass is just about unmeasurably small for
an atom which is why you learn in a chemistry class that you end up with
the same amount of mass after some chemical reaction; however, that is
not really true, but it is close enough to do chemistry. If you do the
same thing in a nucleus, excite the nucleus to one of its excited
states, the change in mass can be readily observed.

QUESTION:
If u fall on sand you don't get hurt but On the other hand, if you fall onto a slab of concrete you get hurt, why?

ANSWER:
Because the force is time rate of change of momentum, F=m Δ v /t ,
where
Δ v is
the change of speed, m is the mass, and t is the time for
the speed to change. When you land on something, your speed drops to
zero in some time t . The longer t is, the smaller the
force is. The collision with sand lasts longer than the collision with
concrete.

QUESTION:
How much force would be required to throw a five pound object laterally (east to west) from a vehicle moving 60mph (north to south) 40 feet (east/west)? Assuming the vehicle window is about 4 feet above the ground, and the car is a typical sedan shape?

ANSWER:
This is an often-asked question type —
" …how
much force is required to…" There is no answer possible because just
specifying a force does not give you enough information to know
velocity. Of course, if you think about it, it is pretty obvious: push
with the same force twice as long and you will get a different velocity.
Read an earlier answer .

FOLLOWUP QUESTION:
What I'm really questioning is how hard it would be to throw the object out of a moving vehicle
(60mph) forty feet laterally. That is, if it required someone to be able to begin with an initial throw of 120 mph when throwing the ball out of the window - I doubt anyone could realistically throw the object that far (maybe its velocity I'm worried about, and not force?)
if air drag being worked in makes it a lot more complicated (we're already way beyond my level of understanding) maybe you could help me by indicating what you might expect this difference to make? I imagine if its substantial it could potentially push the requisite arm strength into the improbable, which is the result I usually got side-arming an object out of my vehicle (in a closed area of course :) ).

ANSWER:
Suppose we neglect air drag (not a very good approximation which you can
tell by just feeling the drag on you hand stuck out the window). In that
case, the speed of the car is irrelevant. I calculate that if you throw
the ball horizontally, the required speed would be about 55 mph; if you
threw at 45^{0} angle, the required speed would be 23 mph. The
effect of air drag would be that you have to throw harder, that is the
drag would reduce the range. For a sidearm toss, the horizontal toss
would probably not be possible; for the 45^{0} toss, it would
probably be possible.

QUESTION:
If a photon doesn't have mass how does it produce Force?

ANSWER:
It has been known for a long time that electromagnetic (EM) radiation
carries both energy and momentum. This carries over to a photon which is
the smallest possible amount of EM radiation, it has energy E=hf
and momentum p=hf /c where h is Planck's constant,
f is the frequency of the EM wave, and c is the speed of
light. So, you should not think of momentum as mv which is only
approximately true for v<<c . Now, force is the time rate of
change of momentum; so, any time the momentum of something changes
because of its interaction with something else, a force is exerted. A
photon can be simply absorbed, in which case its momentum changes by the
amount hf /c . Or, a photon may be reflected so that its
momentum changes by 2hf /c . So, if you are designing a
"solar sail" (which has been proposed for propulsion of spaceships in
the solar system), it should be reflective, not black, to get the
maximum force. So, if you are interested in more details of how a
massless particle can have momentum, read my earlier answers regarding
relativistic momentum and the
energy of a photon .

QUESTION:
Can you compress steam back into liquid water without removing heat energy?

ANSWER:
Yes. If you look at the
phase diagram of water, choose a temperature
greater than 0^{0} C. If you keep the temperature constant and
increase the pressure, you find a pressure where the vapor condenses to
liquid.

QUESTION:
if the universe is ever expanding and accelerating, why are galaxies bumping into each other?

ANSWER:
The universe expanding is an average thing. Locally, it need not be the
case. The closest major galaxy to us, Andromeda, is moving toward us.
That is actually not so surprising since nearby galaxies, with enormous
mass, exert attractive forces on each other which can dominate over the
tendency for expansion.

QUESTION:
my question is about uncertainty principle.i was reading Feynman lecture where he had described(in double experiment with electrons) that it is impossible to determine through which hole did electron pass,without disturbing the interference pattern.When we pass one electron at a time,let us say it hits the wall of a particular hole and gets caught by a detector.If we use some kind of force detector on the holes which detects the force when the electron hits it,which doesn't use light which may alter the electron path,cant we determine through which hole it passed without disturbing the pattern?If we could connect the detector on the hole to some kind of a meter which can measure the force or momentum,then cant we determine momentum and position simultaneously?

ANSWER:
Any way you detect the electron will affect it. You want to detect its
electric field by detecting the force on some charge in your detector.
But, if the electron exerts a force on your detector, your detector
exerts an equal and opposite force on the electron (Newton's third law).
The reason the interference pattern is observed is that the electron,
acting like a wave, passes through both slits.

QUESTION:
If light emanates from a point or sphere in straight lines, would there not be a distance so far from the point or sphere at which an observer would be between the emanated light and, thus not see anything?

ANSWER:
In optics, rays are drawn for convenience and visualization, they are
not real. Their purpose is to show the direction in which the light
travels and they are always normal to the wavefronts (spheres for your
case) which represent the loci of points of constant phase and, if you
consider that they are moving out, are more real than the rays. If you
get far enough away that the intensity gets too small, you will only
"see" the object if you wait a long time because you then look at
individual photons and have to wait for enough of them to accumulate to
form an image.

QUESTION:
I am having a discussion with a colleague at the school where I teach. I believe that opposing beams are used in devices like the Large Hadron Collider because the collison of opposing beams (Beam A going clockwise and beam B going counterclockwise) have the additive energy of the the two beams, so the opposing beams give the apparent effect of approaching the speed of light without either particle having to be accelerated to such a great degree. My friend argues that if the two beams have the same energy it is no different than a single beam hitting a solid wall - beam A will be stopped dead in its tracks in either case - so there is no greater energy in either case. What is the real explanation?

ANSWER:
There are lots of misconceptions in what you both believe. Most
importantly, experiments are not done by "hitting a solid wall", they
are done by the beam hitting some other small particle, either some atom
or nucleus; the most favored target is hydrogen which provides a proton
target for the proton beam. This is the way particle physics used to be
done, a beam of energetic protons hitting a target of stationary
protons. But by making two proton beams collide, you gain much more than
the added energy of the other beam (as I will explain below); if this
were the case, it would be much cheaper to build an accelerator of twice
the energy than to make a collider. And, the purpose of having two beams
is not so that neither needs "to be accelerated to such a degree"; both
beams are accelerated to as high as we can do it. The idea of many
experiments is that we want to take the kinetic energy before the
collision and have it available to create the masses of new particles.
The classic example was the design of the first accelerator to make an
antiproton (which has the same mass as a proton). If you collide an
energetic proton with one at rest, you start with two protons and end up
with three protons and one antiproton. Hence, you need to supply twice
the rest mass energy of a proton for this reaction to happen, 2M _{p} c ^{2} .
So, you would just think you need to have the beam have an energy equal
to 2M _{p} c ^{2} , right? Well, it turns out
that all the energy from a single beam is not available for making mass
because, in addition to energy being conserved in the collision, linear
momentum must also be conserved. Linear momentum is the mass times the
velocity of the particle and, in the case of a single beam and a
stationary target, there is a large linear momentum in the direction of
the incoming beam; this same amount of linear momentum must be present
after the collision. It turns out, if you conserve both energy and
momentum, the incoming beam needs about 6M _{p} c ^{2} !
About ^{2} /_{3} of the energy you put into the beam is
"wasted" conserving momentum, just assuring that the system moves in the
direction of the beam after the collision. But now look at the situation
for a collider. Here, the linear momentum before the collision is zero
because the beams move in opposite directions. Therefore all the energy
of the beams is available to create particles. So a collider needs only
M _{p} c ^{2} per beam to create a
proton-antiproton pair.

QUESTION:
if the marble were dropped from a very great height like kilometer instead of meter, why would the final kinetic energy before impact be less than initial potential energy

ANSWER:
As velocities get large, air resistance become large and so energy is
taken from the marble. Dropping anything through air, it eventually
reaches a terminal velocity and never goes any faster. The result of air
resistance is that energy is lost.

QUESTION:
In a 3d game, a 4kg object slides down a frictionless curved path from a height of 3.5 m.
What is the objects speed at the bottom of the track?

ANSWER:
If the track is frictionless, neither the mass nor the shape of the path
matter. You simply conserve energy:
Ѕmv ^{2} =mgh , v ^{2} =2gh , v =√(2gh )=√(2x9.8x3.5)=8.28
m/s. This assumes that it is at rest at the top of the track.

QUESTION:
according to classic mechanics, a negatively charge electron revolving round a positively charge nucleus should radiate energy and eventually collapse into the nucleus. however, the electron does not collapse into the atom. nhow was this discrepancy resolved

ANSWER:
Because things as small as an atom do not behave classically. See an
earlier answer .

QUESTION:
Why red light is used for signals?

ANSWER:
It has nothing to do with physics. It is based on earlier uses of colors
for
railroad signals .

QUESTION:
I want to know if there is a currently conclusive answer to the reason why virtually every shot of the sun you see on tv, has a six sided form. more often than not, its a central point with six lines of light with mathematically precise geometric spacing between them, or if the focus is drawn in and out, then it looks like a six sided star

ANSWER:
Telescopes used are usually reflecting telescopes. They have a large
primary mirror at the end of a big tube; that mirror forms an image of
the object in front of itself, and so either an instrument or another
smaller mirror (secondary mirror) must be suspended inside the tube. It
is supported by wires or thin plates (called a spider), often six of
them, and diffraction around the spider are the things you see.

QUESTION:
In addition to studying cosmic rays, can we use detectors placed in space that utilize these same cosmic rays to create high energy collisions in order to search for fundimental particles in the same way the LHC is searching for these particles (ie: an LHC in space, without the need of the huge magnetic ring)?

ANSWER:
True, there are cosmic rays which have energies as high as achieved by
the LHC, but to try to do serious particle physics with these particles
is simply not practical. The two conditions imperative to perform useful
experiments are high intensity beams of particles and control on the
energy of the particles. You might get one energetic cosmic ray to hit
your detector in a minute in space (I am really just guessing, but the
point is that there are comparatively few) whereas you might get a
billion per second in the LHC; because the events you seek to study
happen with very low probability, you might have to wait a lifetime to
get enough cosmic ray events to be statistically significant, probably
longer. One of the most valuable ways to study particles and their
properties is to see how their production depends on the energy of the
collision; cosmic rays have a whole spectrum of different energies
whereas the beam energy at the LHC is determined by how you set it.

QUESTION:
I'm a writer, working on a lyric essay about the space between things (kindof a lofty idea, I know... which is why it's a lyric essay). I've been trying to investigate what's in between the smallest particles, and I was hoping you could help me.
I understand basic atomic structure, so you don't have to start from there. I'm wondering more (and forgive me if this is obnoxiously unscientific in its phrasing), what exists between an atom's nucleus and its surrounding electron cloud? Is it just emptiness? Am I riddled with little pockets of nothing? How can that be?
Beyond that - what about the quanta? The quarks, up and down and so forth? Is there a space between these things (though, if I und erstand them properly... which I probably don't... they're energy, not physical structures)?
I'm looking to investigate the scientific aspects of distance - large and small. Large, I get. It's just the small... the super tiny small. It's mind-boggling.

ANSWER:
Well, you are thinking too classically. At atomic scales, particles,
like electrons and nuclei in atoms, are not discrete things which are
localized. Rather, they are continuous distributions of what is referred
to as "probability density", like a fog where the density of the fog is
proportional to the probability of finding the object at that point in
space. You are on the right track when you refer to an electron cloud,
but the cloud is not like some shell, it is continuous like a fog and
extends all the way in to the center of the nucleus; yes, the cloud is
very diffuse inside the nucleus, but there is still a probability of
finding an electron inside the nucleus. The probability density for the
ground state of a hydrogen atom can be seen in the figure on the right
(the relative size of the nucleus here is shown way too big). But, you
can see that there is a region of higher density where we usually think
of the "orbit" being but the electron is actually everywhere, just more
likely to be in some places than others. In addition, we think of there
being swarms of virtual photons. Virtual means that they appear and
disappear very quickly; they may be thought of as the "messengers" of
the electric force, called the quanta of the electromagnetic field. A
similar thing happens inside the nucleus. There the density is much
larger and, unlike the atom, pretty uniform, that is, the neutrons and
protons (generically called nucleons) are almost equally likely to be
found anywhere inside the approximately spherical volume and the density
falls off very sharply at some radius so that the particles are much
less likely to be found at, say, twice the average radius of the
distribution than are the electrons in an atom. The quanta of the
nuclear force are called mesons, and so these are popping into and out
of existence inside the nuceus. But, unlike electrons, the nucleons and
mesons are made of quarks, so the nucleus may be thought of as a smear
of quarks and the quanta of their fields called gluons. So, you see,
there is never anything which could be called empty space in matter.
Even if you go into so-called empty space, what we might think was a
vacuum containing no matter, there are constantly particles and their
antiparticles popping into and out of existence, most commonly
electron-positron pairs. This is called vacuum polarization and it is
also going on inside the atom. Vacuum polarization has very minute but
observable effects on the energies of atoms.

QUESTION:
If you build a solid concrete room tottaly sealed from exterior light (no windows, no gaps at the edge of doors.) Would there be absolute darkness in side, zero light? Or would some light penetrate the "opaque" concrete? Is there such thing as absolute absence of light on earth?

ANSWER:
I would say, in the real world, there would never be a situation with no
"light". Every object emits electromagnetic radiation. The concrete
walls would emit radiation which had its greatest intensity in the
infrared region, not visible to your eyes. However, the emitted
radiation has a full spectrum and a very small fraction of it would be
in the visible region. The probability of light from outside actually
penetrating the walls is vanishingly small, but such things are never
zero. There is also cosmic radiation which could penetrate the walls and
cause photons inside your box.

QUESTION:
Atoms have a positively charged nucleus surrounded by a cloud of electrons. If I remember correctly the nucleus is held together by the strong force which is is in general stronger than the repulsive force among the protons.
What explains why the electron cloud does not collapse from the electromagnetic attraction from the nucleus?

ANSWER:
If you accept the common picture of electrons orbiting the nucleus (not
literally true, but a useful picture in this context), you may as well
ask why the planets do not collapse into the sun. The planets
(electrons) are in stable orbits and held that way by the gravitational
(electrostatic) force. You need to study the
Bohr model of the
hydrogen atom to understand why there is a limit to how small these
orbits can be for electrons in atoms.

QUESTION:
I read about atoms "bouncing" off one another, getting "excited", "vibrating" etc. And then I read that a typical atom model would be: if a nucleus were the size of a nickel, then nearest electron would be 2 miles distant.
Given the enormous relative distance between the nucleus and electron and the consequent enormous amount of empty space, how is it possible for the nucleus of atoms to collide, etc.?

ANSWER:
Usually atoms bouncing off each other refers to the whole atoms, not the
nuclei, interacting. The picture you should have is the tiny nucleus
surrounded by a diffuse cloud of negative charge. When two atoms get
very close, the electron clouds repel each other even though the atoms,
on the whole, are electrically neutral. This is the way that atoms in a
gas at normal room temperatures interact with each other, never the
nuclei. However, if you make the atoms move fast enough, the nuclei
might interact but collisions of this sort result in the electrons being
torn out of the atoms and you end up with a "soup" of nuclei and free
electrons called a plasma; this is the state of things in a star. If you
shoot atoms at atoms, the probability of there being a collision is much
bigger than if you shoot nuclei at nuclei, as I can see you understand
by the way you framed your question.

QUESTION:
When one looks at a reflexion of an object on a mirror, hee sees the object in "3D". But if one looks at a photograph of the same object, it does not appear in "3D". So what is the difference between the light comming from a mirror and the light comming from a photograph, that let our brain add a deepness to a seen object?

ANSWER:
Because you are looking at the image with two eyes and so you see two
slightly different perspectives; your brain interprets this information
as three dimensional. The camera has only one "eye" and hence does not
carry the three dimensional information. When you look at the
photograph, you are only looking at part of the information you receive
when you look at the actual mirror. If you close one eye and look in the
mirror you will see only a two dimensional image. (But you must remain
still, because you can get three dimensional information, via
parallax, to your brain by moving your eye around.)

QUESTION:
Itґs not clear to me why atoms with even particles (C and N, for example) donґt have spin values. More precisely, why hydrogen (spin 1/2) loses this property after a neutron is "added" into its nucleus (deuterium)?

ANSWER:
Spin values of particles are intrinsic and they are not changed by
interactions with other particles. Neither the neutron nor the proton
"loses" its spin when they bind together; the spin of a deuteron is 1,
that is the two
Ѕ spins couple to make 1. However, Ѕ and Ѕ could also couple to 0; this
does not imply that those spins are not there, it is just that they are
aligned oppositely (think of one going clockwise, the other
counterclockwise). This is what usually happens in heavier nuclei—pairs
of protons and pairs of neutrons couple to 0. All nuclei with even
numbers of protons and neutrons (called even-even nuclei) have zero
spin; ^{12} C_{6} ^{6} and ^{16} O_{8} ^{8}
are both spin zero nuclei. However, odd-A nuclei never have zero spin
since there is always one unpaired nucleon; for example, ^{13} C
has a spin of Ѕ. The same thing happens with the electrons in the atom:
electrons, also with spin Ѕ, pair to zero, so both C and O have zero
spin of electrons. ^{14} N_{7} ^{7} is an odd-odd
nucleus and has, like the deuteron, a spin of 1. One more detail: the
intrinsic spins are not the only kinds of "spin" a system of particles
can have. In a bound system like a nucleus or an atom, each particle
also has "orbital spin" which results from its motion in the atom or
nucleus. So, a nucleus with one unpaired particle might have a spin of
7/2 resulting from coupling of its intrinsic spin of Ѕ and its orbital
spin of 3.

FOLLOWUP QUESTION:
itґs still not clear to me why deuterium do not ressonate into NMR
spectrometer, once it has spin value +1. I thought that only atoms with
spin 0 (by coupling of opposite spin values) were "invisible" to NMR.

ANSWER:
NMR has a radio frequency perfectly tuned to "flip" the spin of a proton
(from m _{s} = Ѕ to
m _{s} =- Ѕ)
in a particular magnetic field. This tuning is such that the photon's
energy is the same as the energy necessary to cause that energy
difference and that energy difference is determined by the magnitude of
the magnetic field and by the magnetic moment of the proton. The
magnetic moment of the deuteron and the resulting energy differences of
its m _{s} substates simply do not match the frequency for
protons.

QUESTION: ;;
A 100m train travelling at 90% the speed of light enters a tunnel, which is also 100m long Person A is inside the train, person B is outside watching the train enter the tunnel. What do the two people observe?

ANSWER:
B would measure a train 44 m long and a tunnel 100 m long. A would
measure a tunnel 44 m long and a train 100 m long. (Be sure to make the
distinction between
measurement and appearance.) Here is a more interesting question for
you : when the train is halfway through, B quickly and simultaneously
closes and then reopens doors at the ends of the tunnel, having no
effect on the train inside because it is 33 m from either door. So A
must agree that neither door hit or was hit by the train; how can that
be?

QUESTION:
if you dropped a car from a plane and you were tied to it with a parachute what happens when the chute opens my quess it would be bad

ANSWER:
So, I guess you have the picture of the parachute above you and the car
below you? If the parachute were big enough to be able to gently deliver
a car to the ground, than you are dead because when you reached terminal
velocity you would have a car pulling down on you with a force equal to
its weight and the parachute
pulling up with a force equal to the car's weight plus your weight. If it is a little parachute
which isn't big enough for something as big as a car, you would end up
going so fast that you would die when you hit the ground. I can't
recommend it.

QUESTION:
My question is does a hydrogen bomb do other types of fusion like helium to the next heaviest element like our sun will do when it becomes a red giant or even materials past iron like only large stars do in a supernova. If so could a supernova style hydrogen bomb be made wher the initial atomic explosion starst fusion inhydrogen to helium then helium to denser elements till finally it is directing all its force imploding an iron core to create a small scale version of a supernova in an nuclear weapon form and would it be weaker than a standard hydrogen bomb. Also can a nuclear weapon the fusion / fisssion material be formed into a shaped charge like conventional explosive so its energy could be focused in a single direction. Which would be usefull for deflecting meteors or comets that might hit the earth.
I guess this is a couple of questions but its all on the same kind of thing ideas to stop armageddon. i guess the next question would have been if a supernova bomb was possible could it be shaped charge

ANSWER:
Stars, via fusion, make elements up to iron. But the most energy
efficient fusion is with the lightest elements (^{4} He is
extremely tightly bound). As you make heavier and heavier elements, you
get less and less energy out. The reason that it stops at iron is that,
beyond iron, you have to put energy in to make heavier elements.
Finally, a supernova results from gravitational collapse which will
never happen on a bomb-sized device (you need a star-sized device).
There will never be such a thing as your "supernova bomb" and the most
efficient use of fusion is the good old H bomb which takes H to He. You
should read my
earlier answer on fusion and fission.

QUESTION:
we know that a person siting on a single wire does not experiences a shock.....
but how does an electric train move just touching a single wire .....?????

ANSWER:
But, you cannot sit on the wire with your feet on the ground, can you?
There are brushes on the axels of the locomotive which complete the
circuit from the high-voltage wire to the ground.

QUESTION:
Under what conditions could a solid body with rotational and translational kinetic energy transfer some of its rotational energy to its translational energy?

ANSWER:
How many ways are there? Let me count the ways! Here is one example.
Take a basketball and drop it but give it some backspin; it will rebound
not straight up but back toward you. Some of the rotational kinetic
energy has been changed into translational kinetic energy.

QUESTION:
Why do spacecraft such as Mercury, Gemini, Apollo, capsules and the Space Shuttle need ablative or ceramic tile to prevent burnup on re-enter but not on achiveing orbit. The speed to achieve orbit is roughly 16000 m.p.h and the forward surfaces on these spacecraft have no heat protection and in many cases are only about a quarter inch of aluminum. The SR-71 at speeds of supposedly 2200 m.p.h. needed thermal protection on it leading surfaces.

ANSWER:
The problem, of course, is that the atmosphere drag friction is heating
it up if you are at high speeds, but when launched, the very high speeds
are not achieved until the craft is nearly free of atmosphere. The SR-71
had relatively lower speeds, but was in the atmosphere and for sustained
times.

QUESTION:
During alpha decay, it is said that the difference in mass is responsible for the release of energy (kinetic+gamma). But if you compare the mass+binding of the parent with the mass+binding of the products, it is the same. So where is this energy come from? What am I missing here? The loss in mass seems to be in the increased binding of the products.

ANSWER:
You do not include both mass energy and binding energy. You compare the
rest mass energy before the decay with the rest mass energy after the
decay. You will find less mass after the decay. The "missing" energy is
in the kinetic energies of the byproducts (mostly the alpha particle).

QUESTION:
I have a question about light which I'm finding difficult to express, I'll do my best.
If two people look at the same cinema screen, the light is travelling from the square screen to each persons eyes and must presumably do so in straight lines so that the image can be recognised by their brains. My question is how does light reflect/emit from a single source in two directions without interfering with itself. i.e. how can light bounce off of a single object in two different directions so as to create a neat viewable image in two places? This suggest that light travels in every direction possible just in case something needs to be observed.
I was looking at the two slit experiment and wondered if the fact of observing something with the naked eye could collapse a light wave travelling in many directions and interfering with itself, (i.e. not capable of creating a viewable image), into particles travelling in one direction, i.e. to your eye.

ANSWER:
There is no interference because the light going in different directions
are incoherent; this means that they are not all "waving" in phase with
other. Furthermore, for two rays of light to interfere, they must come
together at the same place. Looking at images on a screen is really no
different from looking at some real object. The image of interest is the
one formed on the retina or your eye and this takes the whole of light
coming to your eye from the screen and forming an image of the image you
see on the screen. There is no relation between watching a movie and the
double slit experiment.

QUESTION:
If I take a lawnmower and tie it to a stake in the center of my yard with a rope long enough for it to make a circle 1 mile in circumference, and it takes 1 hour for the lawn mower to make the complete circle, it's going 1 mph. (right?) Now, as the rope wraps itself around the stake, does the lawnmower go faster in mhp, or does it stay the same speed, even though the radius is getting shorter?

ANSWER:
The question you ask has a nonphysics answer: since the motor drives the
lawnmower with constant speed, it will go with constant speed as it
moves inward. However, you have touched on a classic introductory
physics problem —the tetherball
problem where a ball wraps around a pole; there is no friction (and no
motor as in your case). So, let's reframe your problem such that the
lawn mower has no motor but slides or rolls along the ground with no
friction. The only force on the lawn mower (apart from its weight
straight down and the normal force up from the ground) is the tension in
the rope. But the rope always pulls in a direction perpendicular to the
direction the mower is moving at that instant and so it cannot change
the magnitude of the velocity, only its direction. So, even without a
motor, if there were no retarding forces to slow it down it would move
with constant speed all the way in. There is a similar problem often
given as an example paired with the tetherball: imagine that there were
a hole in the ground and somebody underground pulled the rope into the
hole resulting in the lawnmower moving inward again. In this case, the
tension in the rope does change the speed of the lawnmower, and you can
write that v _{2} =v _{1} (r _{1} /r _{2} )
where v _{i } is the speed at radius r _{i} .
So, when the radius halves, the speed doubles. In physics parlance, the
first case has energy conserved because there are no external forces
doing work and the second case has angular momentum conserved because
there are no external forces exerting torques.

QUESTION:
could you explain to me on an atomic scale why water does not burn, please?

ANSWER:
If we define "burning" as oxydizing something, water is already burnt:
if you burn hydrogen you get water by adding an oxygen atom to two
hydrogen atoms (H_{2} O). You cannot add any more atoms to a
water molecule, there is no such thing as H_{2} O_{2} ,
for example.

NOTE ADDED:
Whoops, my mistake! H_{2} O_{2} is hydrogen peroxide. But
you certainly don't get there by burning water. I guess I don't know
enough chemistry to answer your question!

QUESTION:
In your opinion is it possible to measure the position of a billiard ball sitting at rest on a billiard table, such that there is in theoretical principle no uncertainty regarding the billiard ball's actually having a position, as opposed to the billiard ball's position coming down to being fundamentally probabalistic or uncertain in nature. I ask this because it seems to me that we cannot know enough about what the billiard ball is by way of exacting description, so as to make an exact measurement of its position by way of any frame of reference.

ANSWER:
There are truths in physics and there are truths which, when
extrapolated to some situation, are irrelevant and untestable. For one
example , see a recent answer. Your question is
another example. Yes, we know that the uncertainty principle is a truth,
but if extrapolated to a billiard ball, the accuracy to which you would
have to be able to measure its position to observe the fact that it does
not have a definite position would be impossible. Similarly, to prove
that it was "really" at rest you would have to wait like the age of the
universe to see it change its position.

QUESTION:
If Gravity is a by-product of mass, is there a formula one could use to calculate the Gravitational force (in G's/or Earth Gravity) of Sol, or the Galactic Core? In other words, The Earth = 1G, The Sun = X G's and so on?

ANSWER:
The gravitational force on a mass m a distance R from a
mass M has a magnitude of F _{g} =GMm /R ^{2
} where G =6.67x10^{-11} N ∙ m^{2} /kg^{2} .
So, you see, it does not just depend on M , it also depends on
R . You could measure the force in units of 1 earth-surface-unit
(call it a
Ğ ), so
there would be some distance from the sun where the force was 1
Ğ .

QUESTION:
Can mass slow time?

ANSWER:
Clocks run slower in a gravitational field, the stronger the field the
greater the effect.

QUESTION:
My friend and I disagree on what I think is obviously not true. He states that a potatoe in a potatoe gun continues to accelerate once it leaves the barrel. I told him it decelerates because of air residence. But not just that, he says that the fastest a potatoe was able to go was 100 mph. I couldn't believe a potatoe could be accelerated to 100 mph as soon as it left a barrel only a few feet long.

ANSWER:
I think your friend and you have to "split" this one. The potato stops
having a force on it when it leaves the barrel and so it no longer
accelerates. You are right on that half of your question. However, I
suspect he is right about the speed —it
could very well be over 100 mph. After all, a talented pitcher can throw
a baseball this fast. A friend of mine once, demonstrating his potato
gun to my son, had it go off prematurely and it splintered a really
substantial wooden fence. These are not toys to be casual with! (Are you
Dan Quayle, by any chance?)

QUESTION:
How does increasing the mass of a car rolling down an inclined plane affect its averaged velocity?

when I added 400 grams of added weight, the car went faster

but when I added 1000 grams of added weight, the car went slower than the car with 400 grams of added weight

ANSWER::
I have often addressed questions like this. There is no clear answer and
your dilemma is a good example of how the simple physics we learn in a
class sometimes does not work so well for the real world. See
this link and the links it leads
to.

QUESTION:
My son was doing a science fair experiment on friction to see which surface a toy car would go farthest on. We found that when we sent the car down the ramp, on the surfaces with less friction, the car did not keep going straight as we thought it would, it usually veered off, usually to the right. Why?

ANSWER:
If the wheels are rolling and they are well-aligned with the car, the
motion of the car should not much depend on the friction between the
wheels and the ramp. But, when it gets slippery enough that a wheel may
start slipping instead of rolling without slipping, the friction force
changes. There are two kinds of friction, static and kinetic; if a wheel
is just about to slip the frictional force keeping it from slipping is
greater than the frictional force if it were slipping. So, if one wheel
starts to slip before the others, the car will veer.

QUESTION:
Is it true that if you were to hit a 1 mile long piece of metal with a hammer on one side, would the person on the other side simultaneously feel it, or would there be a delay?

ANSWER:
It is certainly not instantaneous. No information may travel faster than the speed of light.
The time your hit would take to get to the other end would be much
slower than that, the hit would travel at the speed of sound in the
metal; that's much faster than the speed of sound in air but a snail's
pace compared to light.

QUESTION:
If even light can't escape a black hole, how can a graviton traveling at the speed of light escape, so that a black hole impacts the bodies around it gravitationally?

ANSWER:
As I have said in many earlier answers, there is no successful theory of
quantum gravity. Therefore the very existence of something called a
graviton has to be regarded as hypothetical. And, if there is such a
thing as graviton, there is no reason to expect it to have the same
constraints on it as a photon does. In fact, if it is the quantum of the
gravitational field, the "carrier" of the force, then there would be
every reason to expect it to not be unable to escape a black hole.

QUESTION:
Why does a rugby
ball dropped vertically from the ground with it's long axis at an angle
to the ground bounce backwards? What provides the force in the
horizontal direction and in which way does it act?Plus if the Contact
force on the ball is resolved and it does not pass through the ball's
centre of mass in which way will the centre of mass of the ball move and
how does this motion relate to the line of action of the contact force ?

ANSWER:
You are certainly right, if there were no force with a horizontal
component, the ball would have to rebound straight back up although with
some spin about the center of mass. However, what you refer to as the
"contact force" has two components, normally called the normal force and
the frictional force. If the surface were frictionless, the ball would
come straight back up. But there is friction, a horizontal force, and
the direction it points in is the direction which the ball will
accelerate while there is contact. If there were no friction (f
in the picture), the normal force (N ) would exert a torque
about the center of mass which would make the ball rotate clockwise. So,
the ball would want to slide to the left during the collision and so the
friction will point to the right and that is the horizontal direction
the ball will move after bouncing. The center of mass will accelerate in
the direction of the sum of all three forces on it (including the weight
W ). The torque about the center of mass will determine how
the ball rotates after bouncing:

if N +f passes
through the center of mass, it will not rotate;

if N +f passes
above the center of mass, it will rotate clockwise;

if N +f passes
below the center of mass, it will rotate counterclockwise.

When I bounce an American football
(I don't have a rugby ball), I find it goes clockwise.

QUESTION:
How is it possible to prove that the Earth moves when you jump according to newton's 3rd law of motion. What if someone was jumping at the exact same time as you on the opposite side of the planet, would the forces cancel out each other?

ANSWER:
It is not possible to prove because the earth moves so little. If you
jump 1 m the earth will move far less than the size of a proton. At
distances this small, you can forget Newton's laws being valid. And, the
thermal motions of a solid are hugely bigger than the recoil. Plus, as
you note, there are many other disturbances pounding on the earth at the
same time and you would never sort them all out even if they were not so
impossibly tiny.

QUESTION:
What's the benefit of colliding proton and anti-protron beams, like at Fermilab, versus proton beams at the LHC? Why didn't the LHC choose proton and anti-proton beams?

ANSWER:
Suppose the storage ring/accelerator is designed to move protons
clockwise as seen from above. Then antiprotons (having negative charge)
will go around that same ring counterclockwise, just like you would wish
if you want to collide them. Therefore you need to build only one ring
to make a collider. The disadvantage is that you have to make the
antiprotons which is not an easy task particularly when they have to
have just the right energy to go round. The result is that you do not
get a very intense beam of antiprotons and so, if you are looking for
low-probability events, you have to wait a really long time to
accumulate enough data. Proton beams, however, can be made in copious
amounts so you can have collisions of two very high-current beams. The
disadvantage, of course, is that you essentially have to build two
concentric accelerator/rings.

QUESTION:
is the angular velocity of a rotating object independent of the radius? also does not change with a changing radius and same for the period. If the radius increases should the period change?

ANSWER:
Angular velocity is constant for a rigid body. If you have a rotating
object for which the distances between its pieces is not constant, the
angular velocity can change with radius. A rotating wheel has the same
angular velocity anywhere. A galaxy has a different angular velocity at
the outer reaches than near the center. The period of rotation T
and angular velocity
ω
are related by
T =2π / ω
so, if one is constant, so is the other.

QUESTION:
If there are 2 identical objects being launched at the same time with the same force, but one is launched straight up, and another at an angle, say 45 degrees--will the two objects hit the ground at the same time? I am having a discussion with a roommate and my argument is that since gravity is the only force acting on the two objects, they would hit the ground at the same time (even though they won't reach the same vertical height). Can you expand on this b/c my roommate believes that the object launched at 45 degrees will hit the ground first.

ANSWER:
Well, prepare to "eat some crow"! The equation for velocity in the y
direction is v _{y} =v _{0y} -gt where
v _{0y} =v _{0} sin θ
is the initial velocity in the y direction, v _{0} is the
initial velocity with which the ball is thrown, θ is the angle
the velocity makes with the horizontal, g is the acceleration due
to gravity (9.8 m/s^{2} ), and t is the time. So, the time
it gets to the top where
v _{y} =0 (half the time to get to the ground) is t =(v _{0} sin θ )/g .
So, for θ= 90^{0} ,
t =v _{0} /g
and for θ= 45^{0} ,
t =0.707v _{0} /g.

QUESTION:
If I propel an object straight downward, in line with Earth's gravitational pull and at high rate of speed; the initial speed being much faster than if I just dropped the object. Is the object just going to continuously go faster?

ANSWER:
The answer, like all answers to this kind of question, simply depends on
what forces act on the object. If gravity is the only force on the
object, it accelerates with an acceleration 9.8 m/s^{2} , just
going faster and faster. But if this object is moving through the air,
air drag can be important, particularly for very high speeds; this force
is opposite gravity and, if its magnitude is greater than the weight,
the object will slow down rather than speed up. When the speed is just
right, the weight and drag will be equal and the object will fall with
constant speed (which is how a parachute works).

QUESTION:
why an electron does't radiating light when it revolves in an orbit?

ANSWER:
I presume you mean orbits in an atom. This is why early models of atoms
were not dynamic, because it was assumed that an electron accelerating
around inside the volume of an atom would radiate away its energy. Niels
Bohr had the insight that things very small must behave differently from
things we see in our macroscopic world. This was one of the origins of
quantum mechanics which is why the orbiting electron does not radiate.

QUESTION:
i wanna ask about nucleus im in 9 grade and we learn about the nucleus my teacher ask me do you know why the nucleus is in center and in it is a proton and neutron and do not a electron?then he say to us this is because the mass of proton and neutron wich are heavier then the electron.But in really i wanna now why if a tiny piece like neutron is without an electricity and it produced the nucleus force?and the second question is do the nucleus move around her axis?

ANSWER:
There are different forces and they do not all depend on electric
charge. For example, gravity is one of the fundamental forces and an
object needs to have mass to feel that force, electric charge has
nothing to do with it. Another force is the nuclear force (also called
the strong interaction) and a particle must be a hadron to feel it;
protons and neutrons are hadrons, electrons are not. So, this force can
attract protons and neutrons together and "glue" them together to make a
nucleus. It must be a much stronger force than the electromagnetic force
(another one of nature's fundamental forces) because the nucleus has
just positively charged protons in it and why doesn't the repulsive
force between positive charges blow it apart? Because the nuclear force
"wins the tug of war" with the electric force. Now, electrons have
negative charge, so they are attracted by the positive nucleus and they
make a swarm of electrons around the nucleus until it has gotten just as
many electrons as protons in the nucleus and the resulting thing is
called an atom. Of course, the nucleus has to be at the center; if it
weren't, there would be more electrons on one side than the other and
they would pull it back. The atom is hugely bigger than the nucleus
because the electric force binding it together is much weaker than the
nuclear force binding the nucleus. Do nuclei "spin"? Yes.

QUESTION:
I have been told that as an object accelerates to the speed of light that it becomes more 'massive'. I have also checked some other answers on the site and they use the term 'increases in mass'
My question is how can something increase in mass? it can't right, there are only some many atoms in the object? Perhaps they are confussing weight with mass?
Also Force = M*A so M=F/A therefore if anything as A increase M decreases?

ANSWER:
Mass is inertia, the resistance to acceleration when you push on it. An
increase in inertia is exactly what we see when an object has a very
large speed. If you are thinking of mass as the count of how many atoms
there are, this assumes that the mass of an individual atom is a
constant, doesn't it? As the speed approaches the speed of light, the
mass approaches infinity which means it would take an infinite force to
accelerate it any faster. Regarding your F=ma argument, you are
confusing acceleration with velocity —you
can have a large acceleration without having a large velocity. Anyway,
Newton's second law, in the form
F=ma , turns out to be wrong
at high speeds. If you want to read an answer which gives the big
picture leading to why inertia increases with speed, my best answer is
probably
this one .

QUESTION:
It is known that in quantum mechanics a wavefunction describing two indistinguishable particles (ie. two electrons) must account for the exchange of the particles. What is the current view about the mechanism of exchange? What physical process do these particles undergo to make this phenomenon a reality?

ANSWER:
There is no mechanism, there is no "physical process". "Exchange" is a
term in quantum mechanics which recognizes that, in a quantum mechanical
system, identical particles (electrons in an atom, for example) are not
distinguishable, cannot be meaningfully labelled. So, if you think of
one electron as being in one state and another being in a different
state, if you interchange them there is no difference in the physics.
The wave function of the system must be constructed to include this
property. This is done by constructing the wave function of a
multiparticle system to have a particular symmetry, either symmetric or
antisymmetric. To keep things comprehensible, consider a two electron
system: suppose we call the wave function of particle "a"
ψ _{1} (a) and the wave function of particle "b" ψ _{2} (b);
but we do not know which is which, so the wave function of the whole
system must be of the form
Ψ _{ab} = ψ _{1} (a)ψ _{2} (b)ï¿½ψ _{1} (b)ψ _{2} (a).
The plus sign is called a symmetric wave function, the minus sign an
antisymmetric wave function; I will not get into which is appropriate,
but note that
Ψ _{ab} = ï¿½ Ψ _{ba} _{
} which means it does not matter which which is
"a" and which is "b" because the
properties of the system are determined by
Ψ ^{2} , not
Ψ
itself.

QUESTION:
Due to the repulsion of the electrons surrounding each atom, is it true that two physical objects never actually touch one another? That is to say, a very small, but finite distance, exists between any objects said to be "touching" one another?

ANSWER:
Well, we get into trouble because —what
does it mean to "…actually touch on another…"? Atoms are not little hard
balls, but more like fuzzy balls and the extent of the "fuzz" goes way
beyond what we usually think about as the "surface" of an atom. So this
fuzz might very well overlap between the objects "touching (or not)"
each other.

QUESTION:
ive already read your other answer on centripetal force so please dont refer to that, my question is: how can a rotating object force things towards its center, I just dont understand it, i know centrifugal force is ficticious. But i dont understand centripetal force, again, pls dont refer me to your other answer because i dont understand it.

ANSWER:
Well, there are lots of my answers dealing with centripetal force, but I
guess you read the most recent. Here is a quick lesson on centripetal
force. Newton's second law says F =ma . Force F
and acceleration a are vectors, that is what their
being boldfaced means; m is mass, not a vector. Acceleration is
the rate of change of velocity v and there are two ways velocity
can change —by changing its
magnitude, i.e . speeding up or slowing down, what most people
think of when we think of acceleration ("deceleration" is essentially
just negative acceleration, an acceleration vector opposite the velocity
vector); but a vector can also change by changing its direction while
its magnitude remains constant. The direction of the acceleration of a
particle moving in a circle with constant speed is a vector pointing
toward the center of the circle and having a magnitude of v ^{2} /R
where R is the radius of the circle. Thus, if an object is to
move in a circle, something must exert a force on it of magnitude mv ^{2} /R
and pointing toward the center; this is called the centripetal
force. How can it happen? Lots of ways. For example, a penny sitting on
a turntable moves in a circle (provided the turntable is not spinning
too fast); the turntable is able to exert a force on the penny because
of friction. For example, an object tied to a string and going round and
round (no gravity); the string exerts a force on the object (called the
tension). For example, the moon circling the earth; the earth exerts a
force on the moon (called gravity). For example, an object inside a
spinning cylinder; the cylinder wall exerts a force (called a contact or
normal force) which pushes toward the axis or rotation.

QUESTION:
As a space ship is accelerated toward the speed of light the more massive it becomes, becoming infinely massive. How can a particle be accelerated to 99% the speed of light and not become so massive that it damages the acclerator?

ANSWER:
The "accelerator" is simply something which exerts a force on something.
If it is in the spaceship's rest frame, it does not see any more mass
than usual and just pushes as it would at any other speed. If it is
external, that is in our rest frame, it just finds that its effect is
smaller and smaller as the object goes faster and faster because its
mass gets bigger and bigger. I have always thought that particle
accelerators should be called particle energizers because they add
energy to the very fast particles without really changing their speeds
significantly.

QUESTION:
If energy of photon, bombarded on electron is more than the minimum amount of energy required to knock that electron off the metal, then would rate of photoelectric effect increase, or what would happen?

ANSWER:
Increase relative to what? If you have a certain rate of photons, then
when their energy is increased across the threshhold the rate will
suddenly increase. As you continue to increase energy, the rate will not
change but the kinetic energy of the ejected electrons will increase.

QUESTION:
Does an object immediately start to lose speed once an unbalanced force stops acting on it? For example, does a baseball achieve its highest speed at the exact moment it leaves the pitcher's hand and then begin to slow down due to air resistance or is there a period of time after it leaves the pitcher's hand when it might actually pick up speed before beginning to slow?

ANSWER:
You have to look at all the forces in a given problem. The mass will
always respond instantaneously to whatever forces are on it at the
moment. After a pitched ball leaves the pitcher's hand, the hand has
nothing to do with what happens thereafter. The important forces on it
are its own weight (gravity) straight down, and whatever air drag does
to it. Air drag is
pretty significant on a baseball —a
95 mph fastball loses about 10-15 mph by the time (about Ѕ second) it
gets to the plate. So the instant it leaves the pitcher's hand, it
starts speeding up in the down direction (due to weight) and slowing
down in the direction it is moving. The air drag always slows it down,
the gravity can either speed it up or slow it down depending on whether
it is moving up (slows it) or down (speeds it). The net effect is
usually to slow it because the pitcher has to loft the ball a bit for it
to reach the strike zone high enough, so there is little net change in
the vertical component of the velocity. Of course, the preceeding does
not apply if you are interested in a breaking (curve, slider, etc .)
ball. In that case there will be air friction forces not strictly
opposite the direction of the velocity; I believe that a curve ball will
sometime "break" at the last moment due to a downward force due to the
spin on the ball which adds to gravity and makes the ball fall faster
than it would just because of gravity.

QUESTION:
Is kinetic energy calculated as kg .ms-1 or g.ms-1

ANSWER:
Since neither of these is dimensionally correct, the answer is neither.
If they were dimensionally correct, the first would be preferred because
we like to stay in a system called SI, where mass is kilograms (kg),
length is meters (m), and time is seconds (s), but the other would be
possible. Since kinetic energy is
Ѕmv ^{2} , the units in SI units would be kg(m/s)^{2} =kg∙m^{2} /s^{2} .
This unit is called a Joule (J).

QUESTION:
Dose the shape a a football interfere with the distance it is kicked?

ANSWER:
The shape of anything affects the distance that it will go in air when
given an initial velocity. The thing which impedes the distance is air
drag. The two most important factors for a projectile like a football
are the cross sectional area presented to the air and the speed it is
going. There are lots of earlier
answers which pertain to air drag effects which you can read if you
are interested. If the ball is traveling with its point toward the
direction it is traveling, it has a smaller cross sectional area than if
it is crosswise or tumbling. That is why kickers try to get the ball
"spiraling" which keeps the minimum area pointed into the air.

QUESTION:
Can an electromagnetic field be created that is completely positive or completely negative? Or will the field always contain positive and negative charges?

ANSWER:
Fields themselves have no electric charge —they
are neither positively nor negatively charged. Charges and currents are
the sources of electric and magnetic fields. For example, a single
electric charge fills the space around it with an electric field.

QUESTION:
Is a finite empty space infinitely divisible? For example in a cubic meter of empty space, how many ways could some piece of matter be positioned? Could I put my hand in an infinite amount of places within a box, or do the physical properties of space and matter allow for only a finite amount of possibilities. Is it anything like a convergent series, infinite yet creating a tangible whole? I also wanted to ask the same about time. Is every second also an an infinity? or does time have a "frame rate?"

ANSWER:
Whether time and space are continuous or discrete is an open question.
See an earlier answer on
the Planck time and the Planck length.

QUESTION:
If the earth's core was not molten and a hole was drilled directly to the other side, could a person fall straight through or would he become stuck in the center due to gravity?

ANSWER:
Always read the FAQ page
before asking a question!

QUESTION:
An object is connected to a conveyor belt traveling at a constant speed. When the conveyor belt and the object reach the end, the object begins to travel in a radial motion until it travels in the opposite direction. Does the oject have the same speed when it is traveling linearly as it does when traveling radialy? If the conveyor belt is traveling at a constant speed, then it would make sense that the object is traveling at a constgant speed, even around the end of the conveyor. However, if you change the radius of the conveyor end roller, the speed or distance traveled by the object will change. Can you simplify this expalination of linear speed versus radial speed?

ANSWER:
I hope I have the picture here: it sounds like the object is stuck to
the belt and swings around and returns upside down. First of all, there
is never what could be called radial speed when the object is moving on
the circle. When the object traverses the circle at the end, its speed
is always tangential, that is it has no component toward the center of
the circle (which is what radial speed would be). This tangential speed
is always equal to the speed v of the belt. The speed of the
object never changes but when it is going around the end it has an
acceleration toward the center of the circle (centripetal acceleration)
of magnitude v ^{2} /R ; if R is changed, this
acceleration is changed.

QUESTION:
Does the tidal interaction between the earth and sun affect the earths rotation or orbit like the interaction between the earth and moon? Specifically, does it affect the length of a year or day or the size of the orbit, if so, how much was a year or day or the distance from the sun millions or billions of years ago?

ANSWER:
I am sure the effect would be negligible.

QUESTION:
if you drilled a hole through the earth from one end to the other and jumped in. would you fall all the way through to the other side or would you fall down half and climb up other half? wat will be effect of gravity?

QUESTION:
Suppose a hole is made inside the earth from north pole to southpole and a ball is dropped inside the hole what happens to it whether it reaches the centre of earth or oscillates between the poles?

ANSWER:
Always read the FAQ page
before asking a question!

QUESTION:
So one proton is traveling one direction at almost the speed of light, another is traveling at the same speed and they are going to hit head on. What relative speed will they meet at?

ANSWER:
Since you ask your question qualitatively, qualitatively is the best
answer I can give you. Each will see the other approaching at a speed
closer still to the speed of light. If you want to get more quantitative
(e.g. each has a speed 99.9% the speed of light), see an
earlier answer ).

QUESTION:
why are all the celestial bodies round in shape?

ANSWER:
Mainly because the gravitational force which holds them together is
spherically symmetric, that is, it pulls with equal force toward the
center no matter which direction you look, for all points the same
distance from the center. Imagine that you have a gully with rocks along
the top edge. The nature of gravity is to pull them down, hence tending
to fill in the gulley and make the object more spherical. The same idea
of why you cannot build a mountain out of water.

QUESTION:
I would like to know about, what I call, "The moon paradox"
It goes like this: the moon moves farther away from the earth each year by about a quarter inch, even though the earth gains about three tons of material each day from space, increasing the gravity of the earth over time.
The paradox is this: Why would the moon move away from the earth, even though the earth is gaining mass and increasing in gravity over time?

ANSWER:
I think you can forget about the added mass having any significant
effect since the mass added per year is like 20 orders of magnitude
smaller than the mass of the earth. The reason for the moon's slowly
moving away is that the the tides cause the earth to transfer angular
momentum to the moon; similarly, the earth is slowing its rotation as
the moon moves away in order to conserve total angular momentum. You can
read about this in
Wikepedia .

QUESTION:
Why does an alive person sink in water but when he is dead he usually floats on water?

ANSWER:
This is not really physics. The physics part is that if an object has a
density greater than water, it sinks; if it has a density less than
water, it floats. If your lungs are full of air, you usually float and
if not you usually sink. When somebody drowns, their lungs are normally
either full of water or at least empty of air; they end up at the
bottom. I presume that decomposition, which is due to bacteria, results
in gas which lightens the body.

QUESTION:
latent heat is the heat that is used to convert the state of matter without increasing the temperature.PLEASE TELL ME WHAT HAPPENS AT THE ATOMIC OR MOLECULAR LEVEL THAT THE TEMPERATURE DOES NOT INCREASE EVEN ON HEATING AND THE STATE CHANGES?

ANSWER:
What does it mean to heat something? It means to add energy. There are
two ways that a material can respond to added energy:

The kinetic energy of the atoms or molecules
can increase. This results in the temperature increasing.

The potential energy of the atoms or
molecules can increase. For example, if there is an atom in a liquid
whose temperature is at the boiling point, it takes energy to remove it
from the surface of the liquid but the heat which supplies this energy
does not increase the kinetic energy and therefore does not change the
temperature.

QUESTION:
In an experiment examining centripetal force where we tied a ball to a line and connected it to a mass and spun it round, if it had been performed in a vacuum with no air resistance would the ball keep spinning forever assuming that gravity still provided a force to a mass indirectly providing the centripetal force?
To put it more simply would the ball never cease to stop in a vacuum but continue in a circular motion?
Also could you please explain how satelites orbit the earth, and the notion of an exact balance of centripetal force and velocity which keeps the satelite in orbit.

ANSWER:
Air drag is not the only dissipative force here, the string flexes and
however the string is attached to the ceiling will contribute friction.
But, what you are really asking is if all friction forces were
eliminated, would the ball spin on forever? The answer is yes. This,
incidentally, is called a spherical pendulum. The centripetal force is
not due to gravity, even as you say indirectly. The centripetal force is
provided by the horizontal component of the tension in the string. The
vertical component balances the weight (gravity). If you took away
gravity, it would still work fine but the string would be horizontal.
Regarding your second question, my favorite way to understand orbital
motion is
Newton's mountain .

QUESTION:
I have seen at MIT site that "A ping-pong ball moving near the speed of light still looks spherical." I thought that the ball would contract in the direction of movement. Why would it still look spherical?

ANSWER:
As my loyal readers know, one of my favoite points to harp on is that
relativity is about how thing are , not how they look . But
it is, nonetheless, interesting to ask how things look. I tried to
illustrate that things look different than they are in a simple example
of a stick moving at high speed directly at or away from you in an
earlier answer .
The key is that when you look at the object, you simultaneously see
light from different parts of the object which were emitted at different
times; hence, if it is moving at a speed comparable to the speed of
light you are not seeing the object as it is at some time but how
different parts of it were at different times. I found a
wonderful movie on the web which illustrates this for a moving
object which would be a sphere when at rest (like your ping pong ball).

QUESTION:
I study chemistry and i was looking over a part about thermodynamics and if 2 atoms collide they exchange energy but as Einstein once said that e=mcІ so when they do collide they exchange a little bit of mass. the one atom gives a bit of energy to the other and thus a bit of his mass. my question is what is this little part they exchange it can't be a neutron , proton or electron (cause you have the same amount in every atom before and after the collision) and thats what atoms are made off so where does this mass come from.
you could consider this as the same question:
if you take the atomic mass of a Carbon 12 atom and you combine the masses of 6 protons 6 electrons and 6 neutrons(what a C12 is made off) you would end up with a bigger mass cause some of the mass escaped as energy but what is that mass that changes in relation to the energy the atom has, cause the neutrons protons and electrons stay the same and there all the atom is made off.

ANSWER:
Actually, your two questions are not "the same question". If two atoms
collide elastically, they may exchange kinetic energy but neither atom
has its rest mass changed. For example, if you have a head-on elastic
collision between two atoms of equal mass, one of them initially at
rest, the one which came in comes to rest and the one which was at rest
goes out with the same speed the other came in with. However, suppose
that one of the atoms gets excited to a state which has energy E
above the ground state of that atom. This atom now will have a mass
larger by some amount
δ
and the energy of this mass,
E=δc ^{2} ,
will be taken from the total kinetic energy before the collision. When
this excited atom decays back to the ground state, a photon of energy
δc ^{2}
will pop out. The total energy of the system will always be the same. In
your second example, consider the ^{12} C atom. Suppose you
remove an electron or a proton or a neutron from it. Do you have to use
any energy to do that? Of course you do because those things are all
bound together in the atom. Therefore if you disassemble an atom you do
a bunch of work W . If all the pieces are at rest after you have
taken them apart, the total mass of all those protons, electrons, and
neutrons will be greater by an amount
Δ than the mass of the ^{12} C atom; and, guess what—Δc ^{2} =W .
The energy added to the system in this case shows up as mass.

QUESTION:
I recently received a copy of the Particle Physics Booklet published by the Particle Data Group at Lawrence Berkeley National Lab. The booklet seems to contain almost 160 pages of distinct particles: Bosons, Leptons, Quarks, Mesons, and Baryons so my question is: with so many particles, why are there so few elements in the periodic table? Besides being derived from elements in the table, are there additional reasons why so many distinct pieces only arrange themselves into neutrons, protons and electrons, which in turn arrange themselves into the hundred or so elements of the periodic table?

ANSWER:
Nearly every one of those particles is unstable, that is it decays into
other particles in a very short time. Of the particles with mass, only
the electron and proton are stable; a neutron, which is unstable if
alone, can be stable inside a nucleus. Thus, there only three kinds of
particles which are available to make stable matter.

QUESTION:
how can a light contain electric or magnetic field and how it can be polarised

ANSWER:
See an earlier
answer to learn about electromagnetic waves. Polarized means that
all the electric fields in a beam of light are parallel.

QUESTION:
Is there a limit to how much light you can focus in a point of space?

ANSWER:
Actually, you cannot focus any light on a point. You can focus any
amount of light through an area arbitrarily small, but never zero area
(a point).

QUESTION:
How long would it take for every man made satilite to fall back to earth if we add non and take no action to keep then up there. THe question arrises from the movie Wall-E where 700 years after leaving earth the satalites are all still there.

ANSWER:
It depends a great deal on what the orbit is. Some will stay up for tens
of thousands of years. Those most likely to fall pretty soon are those
close to the earth where there is still a little bit of atmosphere. I
wouldn't spend too much time worrying about the literal accuracy of
Wall-E!

QUESTION:
Is DARK MATTER and Dark Energy special kind of matter or they are not Matter at all?

ANSWER:
Always look at the FAQ page
before you ask a question!

QUESTION:
what force is required to accelerate a 150 gram stationery object to 150 kmph velocity in 0.1 seconds? A spring of what tension would be appropriate for the same ? I am trying to design a ball throwing machine for poor kids.

ANSWER:
The average force F exerted over a time t to accelerate a
mass m from rest to speed v is F=mv /t= (0.15
kg)(41.7 m/s)/(0.1 s)=62.6 N. But, if you want to know what spring will
give m this v , you are better off thinking about how far
the force acts rather than how long it acts. A spring is characterized
by a spring constant k which has units of N/m; the force a spring
exerts is proportional to how far it is compressed, x , and k
is the proportionality constant, e.g ., if k =100 N/m and
x =10 cm=0.1 m, then the spring exerts a force of 10 N. To calculate
what k you want for a given x , you need the equation
Ѕkx ^{2} =Ѕmv ^{2} . (This comes from energy
conservation, if you care.) So, suppose you want to compress the spring
by 10 cm. If you put in your numbers for m and v you will
find that k =26,000 N/m=260 N/cm. It will take 260 N to compress
this spring by 1 cm. That is a pretty stiff spring (260 N≈58 lb) so you
will have some engineering issues trying to design this machine.

QUESTION:
I am a kindergarten teacher. I was hit on the head by a portable huge basketball hoop, the kind people stick outside their garage. Someone had emptied the weight reservoir. My school ordered a new one and I figured out from the box - unassembled - it weighs a 100 lbs. It struck me to the ground. It stood 10 feet tall. I am wondering with what force it hit me?

ANSWER:
This is a question similar to many others I get. The answer is that
there is no way to calculate the force you experienced without more
information. The crucial piece of missing information is how long the
collision between your head and the hoop lasted. Let me give you a
couple of examples to illustrate how important the time of collision is.

Somebody throws a ball to you, fast. If you
just hold your arm stiffly out and catch it, it stings pretty badly (big
force). But, if you make your hand move the direction the ball is going
you will lengthen the time you are stopping the ball and it hurts much
less (little force).

You jump off a ladder. You hold your legs
real stiff so you stop real quickly; ouch! Now, use your legs like
compressible springs so it takes longer to stop; the average force you
experience is much smaller.

Instead of hoop hitting you right on your
head, you put a thick piece of rubber foam on your head. It would have
made the collision last much longer and hurt much less!

QUESTION:
Assume that a box is floating in the middle of the fluid without moving up or down. Explain how this is possible? Is it that the density of the box is half the density of the fluid so it experiences neutral bouyancy?

ANSWER:
An object with the same density as the fluid will experience a buoyant
force equal to its weight and therefore experience zero net force.

QUESTION:
I know that Electrons are mobile and that They constitute Current but are Protons mobile to some extent or they are completely at Rest.Our Teacher told us that Protons do not move but there are actually 'HOLES' that move. I could not understand?? Please explain.

ANSWER:
No, the protons in a solid are never mobile. For one thing, they are very
strongly bound in the atomic nuclei. For another, a proton has a mass
about 2000 times greater than the electron; if there were mobile
protons, the force on it would be the same magnitude as on an electron
but the acceleration would be 2000 times smaller. The reason that
electrons in conductors are mobile is that the outermost electrons are
essentially unbound, that is it takes almost no force to move them
around in the material. In fact, one of the simplest but nevertheless
accurate models of a conductor describes the conductions electrons
simply as a gas inside the metal. When a voltage is applied, the
electric field causes this gas to "blow" like a wind in the opposite
direction from the field (because electrons are negative). In some
materials, called p-type semiconductors, there are vacancies in some
atomic orbitals called holes. The electrons are not really free to move
as easily as they do in a conductor, but when a voltage is applied, an
electron may jump from a neutral atom into a nearby hole; the net effect
is that it looks like the hole is moving in the opposite direction as
the electrons and, in this case, the current really is positive, that is
the current is carried by the holes. This may be experimentally proven
by using the Hall
effect .

QUESTION:
If you are in a driving and you pass a stationary car, do you hear the same noise as if you were in a stationary car and a car passes you?

ANSWER:
No, the doppler effect is different for a moving source than for a
moving observer. For electromagnetic waves (light, radio, etc .),
though, the doppler effect is symmetric, i.e . independent of
which is moving.

QUESTION:
What happens to matter when it uses up all it's energy?

ANSWER:
This, of course, rarely happens. However, any mass m which
disappears must result in mc ^{2} units of energy
appearing elsewhere. For example, when an electron positron pair
annihilate, two massless photons appear with total energy 1.02 MeV, the
energy of twice the rest mass energy of an electron.

QUESTION:
How is EM radiation created by the random motions/jiggling/kinetic energy of atoms and molecules? For example: the motions of the atoms and molecules of a warm human body creates EM frequencies up to infra-red radiation. I know that accelerating an electrical charge can create EM radiation (such as in radio towers and microwaves), but aren't atoms and many molecules electrically neutral overall? Is it from electrons being bumped into higher energy orbitals then falling back down again?

ANSWER:
An antenna is electrically neutral overall too, it is just that the
conduction electrons are accelerating back and forth. The simplest kind
of radiation is called dipole radiation and results when a positive
charge and a nearby negative charge oscillate but in opposite
directions, as like when a spring connects them. This is an electrically
neutral system and radiates. Think of atoms as little dipoles which
oscillate and radiate, for example the electron cloud might not have the
nucleus at its center at a particular time. This has nothing to do with
the energy orbitals of the atoms.

QUESTION:
when a highly charged metal cube is placed in vacuum, will it
exhibit corona dischargs???

ANSWER:
Corona discharge normally refers to the glow around the area of
discharge. In a gas, either a positively-charged or negatively-charged
object can exhibit corona discharge; the two are quite different
mechanisms —see the
Wikepedia
article .
This, of course, would not occur in a vacuum since it is the gas atoms
which glow. In a vacuum, if the object is negatively charged, electrons
can leak off if the charge gets big enough; you could call this
"discharge" without the corona, I guess.
Any time the electric field at the surface of a negatively charged
object is large enough to remove an electron from the surface, electron discharge will occur. So,
if the charge is
large enough it will "leak" off. You are wanting a more quantitative
answer, though. A cube is a very hard shape to do an analytical solution for. I can only give you a qualitative overview of the problem. When you add electrons they will arrange themselves on the surface in such a way so that the electric field inside the cube is zero. The surface charge density will be largest near the corners. Even if the charge density were uniform, the field outside the cube would be largest at the corners, so you can be sure that, as you add charge,
discharge will occur first at the corners of the cube. Discharge will begin when the electric field at the surface is large
enough to remove an electron from the surface. I suspect that there is no analytical solution to this problem; you would probably have to do a numerical calculation on a computer. Note that the crucial quantity for discharge is the surface charge density, so the answer to your question depends on the size of the cube; the quantity you should seek is (total charge)/(area of cube).

In a vacuum there is no discharge if the net charge on the cube is
positive. In a gas, the cube will capture electrons from the gas to
reduce the positive charge, but that cannot happen in a vacuum, of
course.

QUESTION:

Is it possible to orbit the earth fast enough to produce 1G
of centrifugal force?

If so how fast would a spacecraft need to orbit to attain 1G?

Can you give me a formula that will calculate this velocity for different
orbital altitudes and planets of different masses?

ANSWER:
First you should read my answer on
centrifugal
and fictitious forces. Now you realize that there is really no such
thing as a centrifugal force, it is what appears like a force when you
are in an acceleratring frame. Now, what is meant by a 1G force? If you
mean that it is the magnitude of your weight where you are, then any
orbit corresponds to a 1G force because you feel like you are
weightless, right? So the centrifugal force must be equal to your weight
right there and pointing out. If you mean 1G is the force you feel on
earth, then the centrifugal force is never equal to your weight on earth
because your weight away from the earth is less. Incidentally, these
answers have nothing to do with altitude or mass; the shuttle or a BB
will orbit exactly the same. As the radius of the orbit gets larger, the
period of the orbit gets longer (the moon is a month, the shuttle is 1 Ѕ
hours) but you will feel weightless at any orbit.

QUESTION:
why do the electric line of forces pass through an insulator but not a conductor?

ANSWER:
The definition of a conductor is that the electrons are perfectly free to
move around inside. Therefore if there were an electric field inside the
conductor, the electrons would be accelerating because of the force they
feel. All electrons, for a static field, end up on the surface and
create a field of their own which exactly cancels out the applied field.
It should be noted that an electric field can exist inside a conductor
if it is changing with time, you just cannot have a constant electric
field inside.

QUESTION:
How does one calculate the acceleration a mass would have to have to keep it at a certain angle on a pendulum moving on a x axis, when given the weight and the angle.
Baisically i'm trying to figure out, the accelration a trolly on the top of a crane moving to the right would have to have, to maintain an angle of 5 degrees on a chain of unknown length baring a load.

ANSWER:
The hanging mass has two forces on it, its own weight mg straight
down and the tension in the chain pulling at some angle
θ relative to the horizontal. In the vertical direction, the mass
is in equilibrium and so -mg +T cosθ= 0; hence T=mg /cosθ.
In the horizontal direction, m has an acceleration a
which is caused by the horizontal component of T, T sinθ ;
ma=T sinθ=mg sinθ/ cosθ. Solving, a=g tanθ.
For θ =5^{0} and g =9.8 m/s^{2} , a =0.86
m/s^{2} . All this assumes that the mass of the chain is very
small compared to the mass at the end.

QUESTION:
What is Dark Energy and Dark Matter? What is the Difference between the two? Is their existence proven yet ?

ANSWER:
Always read the FAQ page before
asking a question!

QUESTION:
My question is about the twin paradox. My understanding has been that the twin paradox is not a paradox at all because it can be explained by general relativity, and because their is a lack of symmetry between what the two twins experience(i.e. One accelerates and the other doesn't). My question is what if their was symmetry? What if both twins start out in identical spacecrafts, and both of those spacecrafts did the exact same thing?

ANSWER:
You are right that it is not a paradox at all. You are wrong when you
say that the solution is in general relativity and depends somehow on
the acceleration; periods of acceleration may be made infinitesimally
short and the same results are obtained. The asymmetry you note is not
in the acceleration but in who is actually in motion toward some distant
goal: the traveling twin sees that distance shortened by length
contraction, the stationary twin does not. You can see a full
explanation in an
earlier answer . There you will also find an explanation of the case
where each twin takes off from earth (in opposite directions) and come
back. If that was your question, there is your answer. If you really
meant the two "do the exact same thing", it seems the answer is obvious —their
clocks "do the exact same thing".

QUESTION:
If I have a 150 long cable attached to two poles (100 feet high), and there is a 25 foot sag, how far apart do the poles have to be?

ANSWER:
This is a classic problem in the calculus of variations. The shape the
curve takes is called a catenary and the equation is y=k (cosh(x /k )-cosh(a /k ))
where a is the distance from each suspension point to the center
(essentially what you are asking for) and k is a constant
determined by the conditions (length of the cable, 150 ft, and the
amount of sag, 25 ft). Also, cosh is a mathematical function called the
hyperbolic cosine. When I put in your conditions, I find that k =100 ft
and a =69.3 ft. Hence, the answer to your question is 2x69.3=138.6
ft. (Incidentally, there is a second equation you need which relates
k and a to the length L of the cable,
ЅL =k sinh( a /k ).
Here sinh is the hyperbolic sine.) If I graph y , I get the
picture below.

QUESTION:
what the effective weight/force of a 4 pound object would be, if it were traveling @ 50/mph. As in, how much force would the momentum be if it suddenly stopped.
My theory is if someone were riding a motorcycle, and were thrown off, and they hit an object in their path with their body (like a telephone pole), it would stop them suddenly and likely break some bones, but they might live. However, if they were wearing a helmet (4 pounds), and the pole stopped their body, but their head continued moving due to momentum, how much extra force is that helmet going to exert on that person's head/neck? My guess, is weight might snap their neck.
The context of this is I have to argue against a bill requiring motorcycle helmets at a legislative hearing. Part of my argument is that while helmets may help some situations, they can also hurt in others (therefore, it should be left to the individual to decide whether to use or not).

ANSWER:
In my opinion, this is a feeble argument. If the trunk of your body hits the telephone pole at 50 mph, I don't think you will have to worry much about a broken neck. Your head already weighs about 10 lb, so it already has enough inertia, I would think, to break your neck by not stopping. I believe the only sensible argument is individual liberty. My feeling is that you should be allowed to ride without a helmet but you should pay higher insurance rates
(including your health insurance) so that I don't have to bear the cost of your higher injury rate with my premiums.

QUESTION:
Planck time is the smallest measurable unit of time (theoretically). Does this mean that time is granular or that smaller increments are unobservable?

ANSWER:
This is completely speculative. There is no evidence that the Planck
time is indeed the smallest measureable amount of time. Whether time (or
space, for that matter) is continuous or discrete is an open question.

QUESTION:
Regarding a nuclear reactor. I know neutrons are used to split the nucleus of a uranium atom (fuel rods), causing a chain reaction. However, how does the reactor create the neutrons? How exactly does this process work.

ANSWER:
When a nucleus splits in two (called fission) there are several neutrons
released; i.e. , it really splits into several pieces —two
lighter nuclei and a few neutrons. One way to induce an nucleus to split
is for it to absorb a neutron. If one nucleus splits and then all the
neutrons are absorbed by other nuclei, etc . we have what is
called a chain reaction. Obviously, this process very quickly gets out
of hand and you have what is called a bomb or a reactor meltdown.
However, the neutrons are very fast so they tend to leave the fuel rod
without causing any more fissions. However, if you can slow them down
(called "thermalizing"), you can improve the chance that they will cause
additional fissions. Slowing down the neutrons is the job of the
moderator in a reactor and it is often just water which is efficient at
slowing down neutrons. But, you do not want a runaway situation, you
would rather have each fission to cause one other fission so the energy
would be produced at a constant rate. The control rods are made of a
material which very efficiently absorbs neutrons (cadmium, for example).
When you start the reactor you pull the control rods out and the rate of
fissions increases. When you get to the rate you want, you push them
partly back in so that you maintain a constant rate. In case of an
emergency, just drop the control rods all the way back in and the chain
reaction stops.

QUESTION:
Do objects of different masses traveling at the same speeds viewed by an outside observer appear to be compressed in the same ways? For example if we could get a board 5m long to travel fast enough to appear 50% shorter in length by an outside observer, would a board 10m long traveling the same speed also appear 50% shorter in length?

ANSWER:
I am always answering this kind of question by saying that relativity
does not address how things appear , rather how things are .
If you are asking if the length contraction for two boards moving the
same speed is the same, that is are they shortened by the same amount
along the direction of their velocity, the answer is yes. If you are
asking if they look like they are shortened differently, the answer is
maybe, depending on how they are moving relative to the observer; in
fact one might look longer even though it is actually shorter. See an
earlier answer for
a detailed analysis and explanation.

QUESTION:
it's a question the answer to which I wish to use as an analog when I make (always free) talks to citizen (and other) groups regarding homelessness; and specifically in response to the complaint by some in the audience that the homeless need to just pick themselves up by their own bootstraps and stop being a burden on society. I keep trying to explain to them that once one has fallen all the way down (as opposed to just tipping over a little, or even falling to one's knees; and especially once they've slipped so through certain kinds of society's cracks), it actually takes more effort to get back up again than it took to knock the person down. (And, trust me, it does.)
[The Physicist : The questioner wishes to compare the energy
necessary to tip over a cylinder of radius R whose center of
gravity is a distance h above the floor to the energy required to
lift it back up.] What is the amount of
energy needed to tip it over from vertical to horizontal compared with
(versus) the amount of energy needed to pick/tip it back up and make it
vertical again? I'm looking for a ratio.

ANSWER:
To tip it over, you have to move the center of gravity (COG) so it is above
the point on the floor where the cylinder touches the floor; to do this
you must raise the COG a distance
h [√(1+(R /h )^{2} )-1]. The work necessary to
do this is W _{fall} =mgh [√(1+(R /h )^{2} )-1].
If R is much smaller than h , this may be approximated as
W _{fall} ≈Ѕmgh (R /h )^{2} . The
work necessary to lift it back up is W _{lift} =mg (h-R ).
Again, if R is much smaller than h , W _{lift} ≈mgh.
So, the ratio is W _{lift} /W _{fall} ≈2(h /R )^{2} .
For example, if h =5R , W _{lift} /W _{fall} ≈50;
it takes 50 times the work to lift as to push over!

QUESTION: ;
At what speed would you have to reach in order to have a runaway
condition on a train with 10,000 tons as momentium continues to build on a
1.3% grade and all brakes are operating properly?

ANSWER:
There is not enough information given. You need to know how much frictional
force you can get from the braking system.

FOLLOWUP QUESTION:
How do I determine the frictional force per RR car having four brakes, one per wheel?

ANSWER:
You cannot figure it out, you would have to look it up. However, you
have stated the problem incorrectly. The speed is irrelevant — there
is a certain amount of maximum braking you can get and if that is
sufficient to keep the train from having an acceleration down the track,
the train will not "run away". Furthermore, the mass of the train is not
relevant either, as you will see. I assume the brakes are capable of
locking so that the train just slides on the track. Since static
friction is greater than kinetic friction, maximum braking is achieved
by applying just the right braking so that the wheels are not sliding
but are just about to slide (which is how antilock brakes on a car
work). Hence, it is a simple problem of a mass m just about to slide down an
incline of angle
θ where the coefficient of static friction is μ . The
maximum that frictional force up the incline can be is f=μN=μmg cosθ.
The force of gravity trying to push the train down the incline is
F =mg sinθ. If these are equal, the train will not slide
but be just about to slide. Solving, θ= arctan(μ ); this is
the maximum grade this train can go down. (Note that m and g
have cancelled out.) The coefficient of static friction for steel on
steel (dry) is about 0.80, and so θ= arctan(0.80)=39^{0} .
The angle of a 1.3% grade is less than 1^{0} , so I expect that
there would never be a problem of a "runaway train". That is the
physics—now, the real world! The reason for "runaway trains" is that the
way the brakes usually work is that we have brake shoes rubbing on drums
or disks and this generates a lot of heat. After a while, the heat makes
the brakes work less efficiently or not at all so they become unable to
exert enough force to balance the component of the weight down the
incline; this is sometimes referred to as the brakes "burning out".

QUESTION: ;
Why do we always hear about critical mass in nuclear explosions. Shouldn't it really be critical density? It seems to me that the mass of fissionable material in a bomb is always there; it's when it gets compressed to some high density that the chain reaction occurs.

ANSWER:
Critical mass is one of those unfortunate misnomers (because it
is not a mass) which has been passed down by common usage. (Another
example would be electromotive force which is not a force.) Your
suggestion that we call it critical density is not really very good
either. I would call it critical geometry. For example, a hemisphere of
uranium might not be a critical mass (that is, less than one neutron
from a fission causes another nucleus to fission); but, if you put two
such hemispheres together, they might be critical. When the material in
a bomb is "compressed", it is brought into proximity with more material,
it is not physically compressed to higher density.

QUESTION:
Is it possible to repel rain with electromagnetism? I'm thinking of a science-fictiony device worn on your belt that produces an EM field—either constant or flickering—strong enough to repel rain so that the wearer would remain dry. An EMbrella, as it were. Is this theoretically possible at all?

ANSWER:
It is easy to attract, rather than repel, water with a static charge.
Look at this video .

QUESTION:
When the size of a soap bubble is increased by pushing more air in it, the surface area increases.Does it mean that the average separation between the surface molecules is increased?

ANSWER:
No, the thickness of the film decreases but the average separation
remains the same. If the bubble were a single layer of atoms (which it
is not), the average distance between atoms would increase.

QUESTION:
The force of surface tension acts tangentially to the surface whereas the force due to air pressure acts perpendicularly to the surface.How is it then the force due to excess pressure inside a bubble balanced by the force due to the surface tension?

ANSWER:
Think of a small area on the surface of the bubble. The forces of
surface tension on this small area are at its perimeter and everywhere
tangent to the surface (as you note). But the vector sum of all those
forces will have a component toward the center of the bubble.

QUESTION:
what is meant by saying that a collision between an electron and a gas atom is elastic? And what is an example of an inelastic collision?

ANSWER:
An elastic collision is one in which the total kinetic energies of the
atom and electron are the same after the collision as before. If the
collision causes the atom to become excited to a state higher than the
ground state, this is called an inelastic collsion.

QUESTION:
if E=mc^2 is right and light has no mass then why does light have energy because the equation states that mass = energy so if mass is 0, then shouldnt energy also be 0, I love physics and technology but this eludes me

ANSWER:
You should always check the FAQ
page before asking a question!

QUESTION:
Considering that matter is mostly empty space, with a few small particles separated by large gaps, why don't solid things just fall through each other?

ANSWER:
You should always check the FAQ
page before asking a question!

QUESTION:
when a glass capillary tube is dipped at one end in water,water rises in the tube.The gravitational P.E is thus increased.Is it a violation of conservation of energy?

ANSWER:
Energy conservation only applies to isolated systems, a system which has
no external forces doing work on it. In a tube the molecules of water
experience an attractive force to the molecules in the tube. This
force pulls up the surface and, since there is surface tension
(originating in forces between water molecules), a small volume of the
water is dragged up. The energy of the water increases because the tube
does work on it.

QUESTION:
I am a "Contact Staff Juggler", it is a certain art where one controls and spins a "Contact Staff" throught ones body in a gracefull maner, I am sure that if you write "Contact Staff" on youtube many examples will be found.
The tool of our trade, the "Contact Staff" is a light staff with Big Kevlar Burners in the ends that at night are lighted for asthetic reasons, unfortunately during the day these Big Kevlar Burners are not very pleasing to the eyes but are necessary due to the weight of them at each end, it is because of this concentrated weight at the ends that we can control them and do tricks.
Once that explained here is what want to.
I want to fill the ends of a Titanium Pipe with a dense metal such as Lead or Osmium to act as the weight the Big Kevlar Burners would provide, just that instead of being outside, they would be on the inside of the pipe.
My question is, if theoretically, the big kevlar burners and the dense metal matched on both covered surface of the staff and weight, the staffs would behave the same way, regardless of whether the weight was inside or out.

ANSWER:
What matters, if these staffs have rotational motion which I presume
they do, is not just how much mass the staff has but how it is
distributed. In the case of a staff with burners, much of the mass is at
the ends of the staff making its moment of inertia large. If you
take that mass and distribute it uniformly along the length of the
staff, you decrease its moment of inertia and its response to the
torques you exert on it will be different. To make it behave similarly
to the original staff with burners, you would have to add as much mass
as the burners have but just at the ends. (I cannot really tell from
your question whether you plan to put the mass at the ends as I suggest
or all along the length.) One other thing occurs to me is that the
diameter of a burner is considerably larger than that of the staff and
they are located at the place where the greatest speeds will be if the
staff is rotating; what this means is that the air drag likely plays an
important role in the behavior of the staff so moving all the mass
inside the staff would again make it have different behavior.

QUESTION:
How much force is used to move a 70lbs object (in this case, luggage) first pulling off the bag well and then across 3 feet of rollers and pushed onto a moving conveyor belt? I would like to know the approximate pounds of force used and eliminating variables such as friction from the bag, rollers and any incline.

ANSWER:
If you eliminate friction, move it only horizontally, and do not ever
hold it up without support underneath, no force at all is required. Any
time you are holding it or moving it horizontally (without anything
under it), you are exerting a force (upward) of 70 lb. If all you do is
move it horizontally, the force you exert is equal to (if you move it
with constant speed) or greater than (if you are getting it moving,
accelerating it) the frictional forces involved.

QUESTION:
The surface of a liquid placed in a cylindrical container will form a parabola if the container is rotated about its central axis.
In the case of a non-cylindrical container, such as one with a diameter that increases with height (wider at the top than the bottom), will the surface of the liquid still be a parabola? If not, what will it be?
I would be interested in any online documentation that you can refer me to.

ANSWER:
The container should be cylindrically symmetric (that is, the side walls
should all be equidistant from the axis of rotation for any elevation);
a square box would not work, for example although it seems to me that
any particular cross section would be parabolic, just not a paraboloid
of revolution. It need not be a right cylinder, that is, it could be a
cone for example. The only documentation you need is the
derivation of the surface shape which makes no reference to the
shape of the container.

QUESTION:
Is the ratio of final kinetic energy and initial kinetic energy of a collision affected by the ratio of the masses of the two colliding objects? In other words, do the relative masses affect how much energy is dissipated?

ANSWER:
Let's do the simplest possible case, a perfectly inelastic collision
between m _{1} with speed v and m _{2}
at rest (they stick together). After the collision, conserving linear
momentum, they have a speed u=v (m _{1} /(m _{1} +m _{2} )).
The energy before the collision is E _{1} = Ѕm _{1} v ^{2}
and after the collision is
E _{2} = Ѕ(m _{1} +m _{2} )u ^{2} = E _{1} (m _{1} /(m _{1} +m _{2} )).
So, E _{2} /E _{1} =(m _{1} /(m _{1} +m _{2} ));
the ratio does depend on the masses.

QUESTION:
Can you please explain Uncertainty Principle with examples?

ANSWER:
It says, essentially, that there are certain pairs of observables
(conjugate variables) in nature which cannot be known simultaneously
with abitrary precision. We are not aware of this impossibility in
everyday life because the uncertainty is extraordinarily small and not
noticable. The most often cited uncertainty is that of the position (x )
and momentum (p=mv ) of a particle,
Δ p Δ x> ħ /2
where
Δq indicates uncertainty of q and
ħ is the rationalized Planck
constant, about 10^{-34} J-s. For example, if you have 1 kg ball
moving at a speed of 1 m/s,
and you are uncertain by 1%,
Δ p= 0.01
kg-m/s so you cannot know x to an accuracy better than 0.5x10^{-32}
m; you can see why we never notice the uncertainty principle for
macroscopic objects. Another pair of conjugate variables are energy (E )
and time (t ), so,
Δ E Δ t> ħ /2.
The mass of an atom, ( E=mc ^{2} )
cannot be known precisely unless it is stable, that is if it lasts an
infinitely long time. Energies of excited states of atoms or nuclei are
not precisely knowable, they have inherent "widths" because they only
last a short time before decaying.

QUESTION:
I heard a joke recently on the TV show Big Bang Theory. Usually I get them but not this time.
It was about a farmer that was having trouble getting his chickens to lay eggs so he asked a physisist friend for help. The physisist said he had a solution but it only worked for "sphereical chickens in a vacuum."

ANSWER:
You wouldn't know it from taking a physics class where the impression is
that, if you are clever enough, any problem may be solved on a
blackboard. But real world problems are seldom amenable to simple,
closed-form solutions. To be a good physicist, you need to make
reasonable approximations which, while not leading to a perfectly
correct solution to a problem, provide both an approximate solution
and a better understanding of the problem and its possible solutions.
For example, the earth is almost always considered, in elementary
physics courses, to be a sphere; of course, it is not a perfect sphere,
but you can sure learn a lot if you assume that it is. There is a
classic "physics joke" for which the punch line is
"… consider
a spherical cow … "
to which your tv show is paying homage.

QUESTION:
like trucks have chains for earthing to protect it from damage but if aeroplane is flying then how we protect it from sparking or any damage etc.

ANSWER:
Obviously, airplanes cannot have grounding wires. They are periodically
struck by lightening and the charge just comes in at one point and exits
at another, harmlessly flowing through the aluminium skin of the
airplane. There is a nice discussion at
Physlink.com .

QUESTION:
What is mass, really? What is inertia, really?
I don't mean to ask for a description, I want to know their source.

ANSWER:
For most of physics, mass is simply an empirical quantity which measures
its inertia, its resistance to being accelerated. Asking what mass is,
"really", is like asking what electric charge is "really", what time is
"really", what length is "really", etc . I know it is like
"begging the question", but you have to start somewhere in science. That
said, science always tries, like you are trying, to push forward. For
example, special relativity lets us have some progress toward
understanding time and length better, somewhat less empirically.
Regarding mass, the recent interest in searching for a
Higgs boson at
accelerators in the US and Europe is because this (so-far unobserved)
particle is predicted to be a quantum of the Higgs field which endows
particles with mass.

QUESTION:
What would happen to an object's velocity if a mass was dropped on it vertically? (the system is frictionless)

ANSWER:
I think you do not really mean the whole system is frictionless. If the
dropped object did not stick to the moving object, the dropped object
would just keep on with the same speed. If the dropped object sticks,
the two proceed at a speed which conserves linear momentum: u= (M /(M+m ))v
where v is the initial speed, u is the final speed,
M is the moving mass, and m is the dropped mass.

QUESTION:
As per Newton's law every body on the earth attracts another body.
this is gravitational force.
why don't we humans experience this force?

ANSWER:
Because gravity is the weakest force in nature. The gravitational force
between two 200 lb men about 3 ft apart would be about one millionth of
a pound.

QUESTION:
Plutonium and Uranium are used for the ease of releasing the locked in energy in matter. My question is does all matter have the same amount of energy, a pound of iron, a pound of calcium, etc. Not the ease of liberating it but its stored energy value?

ANSWER:
The mass energy of a pound of anything is the same as a pound of
anything else. The trick is to find a way to release this energy. When
very heavy nuclei like uranium split in two (fission), a little bit of
mass energy is released; i.e ., the mass of all the fission
products is smaller than the mass of the original uranium atom. But this
does not work for anything. For example, if you take an iron nucleus and
split it in half, you will end up with more mass than you started with,
that is, you must put energy in to make it split. We do not know of any
way to take ordinary matter and convert it into pure energy. The best
known case of totally annihilating mass is when a particle and an
antiparticle encounter each other. But, normal atoms? No way. You might
be interested in my
earlier
discussion of fission and fusion.

QUESTION:
orbit is currently stable and altitude is constant. why does the space station not fall out of the sky and back to earth?

ANSWER:
In fact, the space station is constantly falling, it just never gets
back to earth. I find the easiest way to understand this is to play
around with
Newton's mountain.

QUESTION:
what would happen to a golf ball that was hit in outer-space?

ANSWER:
By "outer-space" I presume you mean no air, no nearby objects (planets,
moons, stars, etc .) In that case, after the ball leaves the club
it will continue with constant speed in a straight line forever.

QUESTION:
How will changing the mass of a object effect the amount of force needed to move the object ?

ANSWER:
There is no answer to this question because any net force will move
something. A tiny force will make an object start moving very slowly if
there are no other forces and a big force will get it moving much
quicker. The thing which varies with mass and force is acceleration.
Newton's second law tells us that a=F/m where a is the
acceleration, F the net force, and m the mass. So, for a given force,
acceleration decreases as mass increases. For a given mass, acceleration
increases as force increases.

QUESTION:
What would happen to you if you reached the speed of light and why can nothing be faster than the speed of light?

ANSWER:
You cannot reach the speed of light. To see why, see the
FAQ page.

QUESTION:
If two objects are traveling at an equal, constant distance on the same path, how much would the one behind have to accelerate in order to catch up with the one ahead of it in a certain amount of time?

ANSWER:
The equation which describes the position x of an object with
constant acceleration a is x=x _{0} +v _{0} t+ Ѕat ^{2}
where t is the time, x _{0} is x at t= 0
and v _{0} is the speed at t =0. If you have two objects
separated by a distance d , going some speed v and, at t =0,
the back one starts accelerating with acceleration a , then if we
choose x= 0 at the position of the accelerating object at t= 0,
the equations describing the two are: x _{behind} =vt +Ѕat ^{2}
and x _{ahead} =d+vt . You are interested in the time
when x _{behind} =x _{ahead} , so Ѕat ^{2} =d ,
so a =2d /t ^{2} .

QUESTION:
is 2012 real the answer bugs me and it wonders me if the wolrds going to end

ANSWER:
I would not lose any sleep over it. It is essentially astrological
nonsense. Read a good article in the
New York Times .

QUESTION:
Why do galaxies appear to be coherent, organised structures (in the case of spiral galaxies at least) when we recieve the light from the stars closest to us hundreds or thousands of years out of sync with the stars furthest away from us? It seems like there should be some distortion or warping of some kind. Is it because the distance travelled by the farthest away stars throught space is negligable in comparison to the time taken for the light to reach us?

ANSWER:
You are right, light from the other side of a galaxy will reach us long
after simultaneously emitted light from the near side will. However, if
the speed of individual stars relative to the center of the galaxy are
small compared to the speed of light, the distance traveled by the more
distant stars will be small compared to the size of the galaxy and it
will not appear to be significantly distorted due to this time
difference.

QUESTION:
Can the amount of spins a wheel makes (EX: 25 spins per second) determine the speed in which a vehicle is going?

ANSWER:
As long as the wheel is rolling without slipping, it advances the
distance of one circumference (2 πR )
per revolution. So, the speed v would be calculated as
circumference divided by the time T to make one revolution T =1/(spins
per second): v= 2 πR /T.

QUESTION:
I am a high school student and would like to know how to simply calculate the difference between how many newtons an astronaut weighs as to how much he would weigh on the moon,

ANSWER:
Since the acceleration due to gravity g on the moon is about 1/6
that on earth and since weight is proportional to g , the moon
weight is about 1/6 the earth weight.

QUESTION:
I was just wondering specifically: If an object accelerating at x leaves the Earth's atmosphere, does it continue to accelerate, or continue to move at a constant rate of the speed (as per newtons first law) it had reached before exiting the atmospher?

ANSWER:
Gravity does not end when the atmosphere ends. If you are a distance of
10 earth radii away, there is still a force of gravity but it is 100
times smaller than the force on earth. So, why does a rocket accelerate
away from earth? Because the force it gets from its engines is greater
than the force of gravity. As long as that engine keeps burning, the
rocket keeps accelerating; and, it accelerates at an accelerating rate
because gravity is getting weaker. Of course, fuel in space is a real
problem, so you just cannot keep accelerating. When the engines shut
off, the rocket will continue with a constant speed providing there are
no other significant forces. Be sure you understand Newton's first law:
it only applies if the total force on something is zero. If not,
Newton's second law applies: the acceleration is equal to the force
divided by the mass.

QUESTION:
Why do we not feel the electrical forces that act on us all the time.

ANSWER:
Do you feel the force of a book you are holding push down on your hand?
Do you feel a hammer hitting your finger? Do you feel the bed holding
you from falling to the floor? All of these are "contact forces" and are
nothing more than macroscopic manifestations of electrical forces. You
might look at an
earlier answer .

QUESTION:
Newton's original second law is as follow:

Lex II: Mutationem motus proportionalem esse vi motrici impressae, et fieri secundum lineam rectam qua vis illa imprimitur.
Translated as:

Law II: The alteration of motion is ever proportional to the motive force impress'd; and is made in the direction of the right line in which that force is impress'd.
The law did not show or imply F=d(mv)/dt. How does this equation come about?

ANSWER:
Let's put this in modern language: the time rate of change
("alteration") of linear momentum ("motion") is proportional to the net
external force (motive force impress'd); the fact that linear momentum
and force are both vectors assures they will be in the same direction
("in the direction of …that force") . We
choose the proportionality constant to be 1. I hope that answers your
question.

QUESTION:
If the charge of electron is “positive” but not "negative", what is the deflective direction by the electric field? What is the deflective direction by the magnetic field?

ANSWER:
I am not sure what this question is asking. If you simply want to know
what the forces would be on a particle with the mass of an electron but
the opposite charge, electric and magnetic forces would both reverse
because they are proportional to the charge. If you were asking what the
world would be like if the electron were a positively charged particle,
it would be exactly the same because the choice of the electron having
negative charge is purely arbitrary.

QUESTION:
I am doing an experiment on centrifugal force (its a fake force I know) and I am putting an object in a cup and showing that it stays in when you spin it around, but what i want to know is the minimum speed needed to keep and object in a spinning cup before it falls out?

ANSWER:
Let's talk about the centripetal force, the force which causes an object
to move in a circle. As I presume you know, F=mv ^{2} /R
where m is the mass, v is the speed, R is the
radius of the circle, and F is the centripetal force. When the
object in the cup is at the top of its circle, there are two forces
acting on it —its own weight mg>mg
and the force down which the cup exerts on it, N . Therefore
F=N+mg . As the cup goes slower and slower, the force N gets
smaller and smaller until it eventually becomes zero. Now all the
centripetal force is provided by the weight, so mg = mv ^{2} /R
or v =√(gR ). If you go
any slower than this, the weight stays the same but the required
centripetal force gets less and the object will not continue on its
circular path but become a projectile.

QUESTION:
I'm an 8th grade student and we are supposed to design an amusement park ride that will operate on the moon. We have to pick a ride that exists on earth and modify it so that it will work on the moon. We have to explain our choice and the modification. We thought we'd pick a roller coaster and make the cars heavier to deal with the lower amount of gravity. Are there any other ways that a roller coaster would operate differently on the moon then it does on earth?

ANSWER:
Think about it: if you make a roller coaster car heavier here on earth,
what is the difference? None, because the acceleration of the car is
independent of its mass (neglecting friction). Remember, Galileo showed
that two dropped objects of different weights hit the ground together.
On the moon, the acceleration due to gravity is smaller, so if all you
do is move the roller coaster to the moon and increase the mass of the
cars, it will go much slower than on earth and increasing the mass of
the cars will have no effect. So, since this is your project, I will
leave it up to you to figure out how you can make the speed at the
bottom of a hill the same as the speed on earth.

QUESTION:
I want to try and calculate the speed of a hockey slap shot when I know the mass of the stick and the speed of the hockey stick I am swinging. Is that possible?
I am curious to see if I get stronger and use a heavier stick and swing the stick faster, how my shot speed would increase over time. It would be a good goal to set for myself that I could track.

ANSWER:
No, you cannot calculate the speed with the information given; you need
to know how elastic the collision is. Also, it is not the usual simple
physics problem of two colliding masses because you are exerting a force
on the stick at the instant of impact which is probably more important
than the mass of the stick. You certainly want the stick to be moving as
fast as possible and you want to be exerting a big force on it. The
tradeoff is that the more massive a stick you use, the more difficult it
is for you to get it going fast. My advice is to try different sticks
until you determine, empirically, which is best for you.

QUESTION:
if you double the frequency in a wave, what happen to the velocity?

ANSWER:
Approximately nothing. Wavelength
λ , frequency
f , and velocity v are related by the simple
equation v= λf. Normally, v is determined by the properties of the medium which
are more or less constant over a relatively wide range of frequencies. A
small effect, called dispersion, is when the velocity depends slightly
on the frequency of the wave; dispersion is why you see a "rainbow" of
colors from cut glass or a diamond.

QUESTION:
In my book about all the study of optics is based on snell's law. It's derivation was not in my book i search in web for its derivation. I got several derivation but in all derivation there is ' dt/dx = 0 ' means light takes the smallest possible path to reach from first to second point. can you tell me why this is assumed in it's derivation.If you have any other derivation for it then you can send me that one.

ANSWER:
It is not the "smallest possible path", rather the path which takes the
least time. This is Fermat's principle which is a result of Huygen's
principle; a qualitative derivation of Fermat's principle may be found
on
Wikepedia . Further discussion and alternative derivations may be
seen
here .

QUESTION:
Do dark matter, the multiverse, and, dark energy exist?
They seem like wild guesses, created more from the need to answer a question than from evidence. Even worse, if I understand the multiverse correctly. It's unprovable. :/ I thought if something was unprovable and/or lacking in evidence, we were supposed to assume it wasn't true. (i.e. the reason I reject the existence of the flying spaghetti monster.) Wouldn't it make more sense to say, "we don't know"?

ANSWER:
I do not think you want to assume it is not true, but certainly not that
it is true either. It is always fun to speculate about what might be,
but real science is in the observing and measuring. So, you and I are
basically on the same page. Read some of my other answers regarding
dark matter and energy .

QUESTION:
Hello, I know that the redshift is a well known example of how we supposedly know that the universe its not only expanding but accelerating, now my question is:
How do scientist know that we are not receiving the light from an object moving faster in the past (like a star 8.000 millions years ago) on an object moving slower in the present (like the earth 8.000 millions years later)?

ANSWER:
You are right, we cannot know what a distant star is doing right now.
Any information we can glean is for how the object was when the light we
are seeing left it. The systematics we see of objects at many different
distances are pretty compelling evidence that nothing drastic is
happening. (Like maybe they just stop or something?)

QUESTION:
Having just read about Brian Naranjo's fusion reaction using Iithium Niobate, I wonder why just accelerating duterium ions and colliding them into a duteride at sufficient energy in a linear accelerator isn't a simple solution to the fusion problem?
What's wrong with this concept?

ANSWER:
It is all a yield problem. Each individual fusion reaction releases a
relatively large amount of energy but it is extremely tiny. You
simply cannot get an intense enough beam with enough deuterons in it to
yield any useable amount of energy. Even if you could, target would get
so hot as to melt away. A related problem is that every encounter of
possibly fusing nuclei does not result in a fusion reaction, that is the
probability of an iteraction works against you too.

QUESTION:
I was surfing today and came across a site which goes on to explain about fusion. My question is if they build this Tokamak reactor the magnetic forces produced in the reactor would be 200,000 time greater than earths magnetic field. My question is will this not affect our planet?
A concerned earthling who would like to remain here a little longer.....

ANSWER:
This is only the field inside the tokamak, it is virtually zero outside.
See the answer to an earlier,
similar question .

QUESTION:
I am an air traffic controller and last night we got into the following discussion which has two sides of beliefs and no side budges. Two planes at the same altitude have to be 5 miles apart. We have a tool that places a 5 mile circle around a plane to give us a visual reference with the location of the plane being the center point. If a plane that has this circle is heading south (180 degrees) and a second plane is on the 5 mile circle to the east of the plane heading west (270 degrees) at the exact same speed. Will these planes ever get closer then 5 miles? One side argues using pythagorean theorem that they will. If each plane moves 2.5 miles they will get to be withinn approx 3.5 miles. The other side argues that the center of the circle continues to move with the plane and therefore you will maintain your 5 miles.

ANSWER:
The easiest way to see this problem is to view the situation from one
plane or the other. Let the plane going south be labelled S and the
plane going west be labelled W. Suppose the speed of each plane is v .
As seen by S, W is going northwest with a speed u=v √2.
The two components of this velocity are u _{x} =-v
and u _{y} =v. I have chosen the y-axis to be north
and the x-axis to be east; the origin, x=y =0, is at plane S. The
coordinates of plane W (relative to plane S) are then x =5-vt ,
y=vt where t is the time since W was due east of S. We can
now calculate the distance R between the two as a function of
t : R =√(x ^{2} +y ^{2} )=√((5-vt )^{2} +(vt )^{2} ).
A little algebra leads to R =5√[1+((2v ^{2} t ^{2} -10vt )/25)].
So, you see, the whole thing boils down to whether the quantity 2v ^{2} t ^{2} -10vt
ever becomes negative, in which case R will be less than 5
miles. Noting that both v (speed) and t (time) are always
positive numbers, when vt is between 0 and 5, R is less
than 5 miles. For example, if v =300 mph, the distance will be
less than 5 miles until time t =(5 mi)/(300 mi/hr)=1/60 hr=1
minute. The closest the two planes will get will be at t =Ѕ min
when R =3.54 mi. Shown to the right is a graph illustrating this
specific numerical example. (Actually, the argument for each of the two
moving 2.5 miles works very well and is correct.)

QUESTION:
If there was two of the same cars but one had 1500 pounds more in one of them which one would stop faster?

ANSWER:
If everything else was the same, the lighter car would come to rest
first.

QUESTION:
Is it possible to produce mass by passing a ray of high intensity through a varying magnetic field of high magnitude?

ANSWER:
By "ray", I presume you mean a ray of light? Or maybe x-rays? Or gamma
rays? Let's just say ray means a beam of electromagnetic radiation of
some sort. The possibility to create mass is determined not by the
intensity of the radiation but by its frequency. The energy E of
an individual photon (the smallest possible quantum of the radiation) is
given by E=hf where h is Planck's constant and f is the
frequency. A photon may be induced to create mass m subject to
the constraint that mc^{2} <hf .
The most common example is
pair production
where an electron-positron pair may be created when a photon of energy
greater than twice the electon rest-mass energy encounters the strong
electric field of a nucleus. Any strong electric or magnetic field could
induce this mass creation and it would not have to be time varying.

QUESTION:
i was doing a expirement and i came to the conclusion that free fall occurs because of the high air resistance is this a true statment ?

ANSWER:
Certainly not. Free fall means that an object accelerates experiencing
only the force of gravity. Air resistance prevents something from free
falling.

QUESTION:
When a shuttle orbits the earth you are falling through space and therfore are weightless. If you are in the middle of deep space and you come to a complete stop in space, moving in no direction and no objects near to exert any real gravity, would you still be weightless and is there any laws you know about that could explain this.

ANSWER:
In orbit you are not weightless; as I have said many times, weight is
the force gravity exerts on something and that does not go away if you
are in free fall. When you are in empty space, however, moving with
constant velocity (or at rest), you are truly weightless.

QUESTION:
the way they explain the fabric of spacetime seems 2 dimensional to me, a star putting a dip in the fabric implies that the star is sitting on the fabric? and a black hole would also implie that the spacetime fabric is flat. i understand the theory of G realitivity how a big star curves spacetime because of gravity, and how the sun keeps earth in orbit and not flying off into deep space, but what keeps it from crashing into the sun?

ANSWER:
This frequently used illustration is not meant to be rigorous. See my
earlier answer
along these lines. Regarding why the earth does not crash into the sun,
it is because the earth is not sitting still but moving. Just like the
moon does not crash into the earth. This can be understood just using
classical physics, general relativity does not need to be invoked. Have
a look at the
Newton's mountain .

QUESTION:
is gravity a force???, in one book i found its just a distortion between two object

ANSWER:
See several earlier answers linked to on the
FAQ page .

QUESTION:
My husband and his friends at work or in a debate about which would fall faster! A thousand pounds of feathers or a Thousand pounds of steel? My opinion was that it depended upon the fact of weather the feathers were in a contained box or container or if they were dumped out of something!

ANSWER:
Your opinion is completely correct —the
faster is determined by how you "package" the feathers or steel. For
example, you could make a huge parachute out of the steel but compress
the feathers into a ball; the feathers would win hands down. The key is
the air drag force on something. If you are interested in a more
quantitative discussion, read on. A falling object has two forces on it,
its own weight down W=mg and air drag F up which can be
pretty well approximated as
F ≈јAv ^{2} ; here m is the mass, g
the acceleration due to gravity (9.8 m/s^{2} ), A the
cross sectional area presented to the air, and v is the speed.
(Incidentally, the equation for F is only correct if you use SI
units: meters, kilograms, seconds, etc .) Notice that the faster
something goes, the greater the air drag; this should make sense to you
if you imagine sticking your hand out the window of a car at low and
high speeds. As the object falls it goes faster and faster so the drag
gets bigger and bigger until finally the drag equals the weight and the
object falls with a constant speed called the terminal velocity v _{t} .
It is easy to solve for this velocity: F=W so mg= јAv _{t} ^{2}
so v _{t} =2√(mg /A ). Since both steel and
feathers have the same m and experience the same g ,
whichever has the smaller A has the larger terminal velocity,
i.e. falls faster. If you were to do the test someplace where there
was no air, like the moon, the two would always tie.

QUESTION:
If a mass is moving near the speed of light and passed close to a second mass traveling at a very low fraction of light speed, would an observer on the slow mass observe the relativistic mass of the object, or would it still appear to have it's rest mass?
My reason for asking is, if you needed to change the orbit of a large slow moving mass (say an asteroid), could you do it by passing a stream of relativistic masses with a small rest mass near the large mass?

ANSWER:
Scroll down just a few questions to see my answer about mass increase in
relativity. You will see that, even for a particle going 99.9% the speed
of light, the mass increase is about a factor of 20. And it takes a huge
amount of energy to accelerate a macroscopic object to that velocity, in
fact the heaviest objects ever accelerated by man to relativistic speeds
are single atoms. Yes, the asteroid would see an increased mass of the
projectile, but your idea is totally impracticable.

QUESTION:
On a cold day--40 degrees--I start my golf round with a golf ball at room temperature of 72 degrees. Will playing with the ball maintain that 72 degrees in the ball or will the ball lose its flexibility eventhough you are hitting the ball every few minutes?

ANSWER:
Somewhat to my surprise, I found that it takes about 6 hours for a golf
ball to come to equilibrium (all the way to the core) with a different
temperature. Also, the increased drag on the ball by the denser cold air
is a more serious cause of shortening the range than is the temperature
of the ball. One reference recommended playing with a couple of balls,
playing one while the other rides in your pocket and switching them
periodically.

QUESTION:
About how fast would you need to go to in order to remain in free fall around the planet? once in stable orbit, what would happen if your speed were to drop?

ANSWER:
Try it for yourself with
Newton's mountain .

QUESTION:
If you are somehow existing at the very center of a star, the gravitational pull in all directions would be equal, hence there would be no gravitational force in any direction. So, why do physics books discuss massive gravitational forces in the center of stars?

ANSWER:
You are right, the gravitational force at the center of a star must be
zero. However, there is an enormous pressure there so you would be
crushed. This pressure is due to the high temperature. The gravity
holding the whole star together keeps this high pressure from blowing
the star apart.

QUESTION:
Einstein states that nothing with mass can accelerate to the speed of light, due to the concept that it's mass will become infinite. Therein begins my question, though easy to ask, I cannot seem to draw an answer from it as my abilities with mathematics is lacking compared to my ability to think about underlying ideas. Why is it that the Large Hadron Collider can accelerate gold atoms to 99% of the speed of light, without a major increase in their mass? On that, does this then mean that A: Their mass only increases exponentially over the next 1%, or B: was Einstein merely wrong?

ANSWER:
Well, Einstein was certainly not wrong! Maybe you just need the
expression to calculate the increase in mass: M /M _{0} =1/ √(1-β ^{2} )
where M is the mass when moving, M _{0} is the mass
when at rest, and β is the fraction of the speed of light. In
your example, β= 0.99 and so
M /M _{0} =7.09; the mass of your gold atom is about
7 times bigger than it would be at rest. I do not know why you said that
they traveled at this speed " … without
a major increase in their mass … ";
a factor of 7 seems pretty major to me —I'd
weigh 1400 lb!
If you had
β= 0.999, then
M /M _{0} =22.4; for
β= 0.9999, then
M /M _{0} =70.7; etc . Notice that if
β= 1,
M /M _{0} = ∞.

QUESTION:
How to estimate the force of the wind of speed 100 kilometer per hour on an adult human being standing facing the wind ?

ANSWER:
100 km/hr is about 28 m/s. A good estimate of the air friction force is
F ≈јAv ^{2} where F is the force in Newtons,
A is the cross sectional area in m^{2} , and v is the
velocity of (or through) the air. If I estimate the cross sectional area
of a person to be about 1 m^{2} , I find
F ≈200 N, about 45 pounds.

QUESTION:
If the sun disappeared how long would we orbit the void left by the sun? is it 8mins? a small part of the gravitational wave must come from the earth, so how can a gravitational wave be traveling at the speed of light from the sun cover a greater distance than another gravitational wave traveling at the speed of light from the earth ? or does a gravitational wave travel the same distance despite the size of the two "attracted" objects? does a wave from the centre of the earth reach the centre of the sun in the same time a wave from the centre of the sun reaches the centre of the earth? I hope that makes some kind of sense, I know what I mean, it's just hard to write it down

ANSWER:
Don't think about gravity waves, just think about the field. The
question is "how fast does a gravitational field propogate?" How long
after mass appears does the field appear at some distance away or, as
you are asking, how long before it disappears if the mass disappears? It
is well known that the electromagnetic field propogates at the speed of
light. It is generally assumed that a gravitational field propogates at
the speed of light also, but it has never been measured (to my
knowledge). See my earlier answer .

QUESTION:
Is dark matter/energy assumed to be pervasive throught the entire universe? That is - does it exist in our solar system, and here on earth? If yes, have physicists had to rethink/recalculate experimental observations (gravity, electromagnetism, etc.) to take into account dark matter/energy? I guess another way to ask it is - has the concept of dark matter/energy changed our fundamental theories on the atomic level, or just on the galactic cluster level?

ANSWER:
In my opinion, dark matter does not really exist until it is observed.
As I have said in many earlier
answers , I feel that the observations which lead to the "invention"
of dark matter may, in fact, be evidence that we do not understand
gravity as well as we think we do. This is a minority view, but not
unique (see this article
and references in it). Similarly, dark energy, the apparent
repulsive force which is causing the most distant objects to accelerate
away, can be partly understood as a modification of the current theory
of gravity (general relativity).

QUESTION:
Ignoring relativity and just using Newtonian physics, how much energy would it take to move a 1000 metric ton space ship 10 light years in 10 days? Assume constant acceleration and ignore deceleration. (The ship will just make a "fly by" at some distant star 10 light years away without bothering to try and stop.) I'm sure the amount of energy needed will be enormous, so please try to describe the amount of energy in a way that people can grasp.

ANSWER:
I usually ignore questions with faster-than-light assumptions but (i )
the question stipulates that I calculate the answer classically, without
relativity, and (ii ) the questioner made a donation and I guess I
have my price! Simple kinematics, assuming the space ship starts at
rest, are x = Ѕat ^{2}
and v=at ; using the x and t given by the questioner, I find that
a= 2.5x10^{5} m/s^{2} (which is more than a 20,000
times g —kinda crushing to any passengers) and v =2.2x10^{11} m/s
(nearly 1000 times faster than the speed of light). So, it passes the
distant star with a kinetic energy of T =Ѕmv ^{2} ≈2.4x10^{28}
J. The entire US consumes about 10^{20} J/year, so the
(fictional) spaceship would require the entire energy consumption of the
US for about a quarter of a billion years.

QUESTION:
i've read that capacitors can be used as batteries-does that mean they can be used interchangeably?

ANSWER:
It seems a stretch to me to say a capacitor can be used as a battery. A
capacitor, like a battery, stores energy and that energy can be tapped
by discharging the capacitor. But that is a one-time event —you
get a jolt of energy which quickly is exhausted. A battery, on the other
hand, uses chemistry to give a continuous flow of energy to its load, a
very different situation.

QUESTION:
My sixth grade science students have a question about the spin of Venus. We watched a video that explains that Venus rotates in the opposite direction of Earth. The explanation was that a meteor hit it and caused it to turn over almost 180 degrees. Why didn't Venus continue to spin in the direction it was knocked by the meteor?

ANSWER:
It is probably more accurate to say that some cataclysmic event caused
it. Since most orbits of things in the solar system are approximately in
the plane of the orbits of the planets, it is more likely that the
collision would be on the equatorial plane of Venus than at one of its
poles. Imagine a collision on the equator. If it happened on the side
where the planet was moving in the same direction as the impinging body,
it would speed up the planet's rotation; if it happened on the other
side it slow it down or even reverse it. A polar collision would
seriously alter the direction of the axis of rotation. You are right,
Venus's rotation will forever after retain the memory of the collision
by its new rotation, different from the original; in other words, it
does "… continue
to spin in the direction it was knocked by the meteor … "
but the old rotation is present also — the
two add.

QUESTION:
while watching a kid deliver flyers, he delivered one home then crossed the street to deliver to the home across the street. I wondered whether it would be better to deliver the 20 homes on one side first, then cross the street and deliver all on the other side. Does it take longer by crossing the street every time or should he deliver one side of the street then cross the street and deliver on the other side.
If you can explain how to do the math as well I would be most grateful.

ANSWER:
Well, this is not really physics, is it? Anyhow, it is a nice little
mental exercise. The bottom line is that it depends on how wide the
street is and how closely spaced the houses are. Call the length of the
street L and the width W . The houses are equally spaced
and directly across the street from each other and there are N of
them. For simplicity I will assume the kid delivers to one house,
crosses the street and delivers 2, crosses the street and delivers 2,
etc . He could zig-zag and deliver one at a time before crossing, but
that makes the geometry too complicated for the quick and dirty answer.
So, call x _{1} your way, up one side and back the other:
x _{1} =2L+W . The kid's way is x _{2}
and (I will let you draw a picture and figure this out) x _{2} =L+NW =(2L+W )+((N -1)W -L )=x _{1} +((N -1)W -L ).
So, x _{2} -x _{1} =(N -1)W-L.
For your way to be shortest, x _{2} -x _{1} >0,
so L <(N -1)W, or, W >L /(N -1);
but, L /(N -1)=D where D is the distance
between houses, so your way is shortest if the street is wider than the
distance between houses. That is the solution to your problem for the
simplest street crossing pattern. But, there is probably a more
important consideration —h e
probably did not want to end up where he started if he had more than
just your street to do.

QUESTION:
Why don't we feel the rotation and revolution of the Earth?

ANSWER:
Because the accelerations associated with the rotation and revolution of the
earth are very small compared to the acceleration due to gravity.

QUESTION:
I would like to know when heating a gas why do the molecules collide with the walls more violently, and cause a increased gas pressure?

ANSWER:
Because, as temperature increases, the average speed of the molecules
increases.

QUESTION:
In a nuclear reactor, when a fast neutron strikes a light water moderator what are the byproducts? Is the hydrogen's proton ejected only to undergo beta decay to become a neutron again? What happens to the water molecule?

ANSWER:
For all intents and purposes, when a neutron hits a water molecule
nothing happens to the molecule, the neutron slows down. That is the
purpose of a moderator. Hydrogen is a good moderator because the best
way to slow something down is to collide it with something about the
same mass, hence neutrons hitting protons. A head-on collision results
in the neutron essentially stopping. The disadvantage is that, as the
neutrons thermalize the hydrogen has a relatively large probability of
capturing the neutron, hence removing that neutron from the chain
reaction. Also, free protons do not undergo beta decay, they are stable;
but few if any protons are knocked out of the molecule.

QUESTION:
I have a paintball marker that can fire the same size paintballs, but of multiple weights. Assuming all paintballs have the same cross section and drag. A .2g paintball immediately accelerates to 400 feet per second, then begins decelerating as it passes through the air. What would the formula be to calculate the velocity of the paintball at 100ft? I am attempting to demonstrate that a heavier weight paintball (.3g) retains more momentum at distance than a lighter paintball, despite actually travelling slower and thus may be more likely to break apart and mark. Seeking the formula so I can graph across multiple weights.

ANSWER:
First, to get a full discussion of the approximations used, see my
earlier answer
about air drag for badminton birdies and tennis balls. The approximation
for the air drag used there, and here, is
F ≈јAv ^{2} where A is the cross
sectional area and v is the speed of the object; this formula
requires that A be in m^{2} and v in m/s. I also
assume that in the time of flight, the vertical component of the
velocity never gets as large as the horizontal velocity; in other words,
I approximate that the paintball goes in a straight line. A little
research reveals that the most common paintball size is .68 caliber
which I took to mean .68" in diameter; this gives A =2.3x10^{-4}
m^{2} . While I was looking up the size, I found that the typical
paintball mass was about 3 g, ten times bigger than you quote for your
paintballs, so I will use m =2 g instead or your .2 g. The
pertinent equations are
v=v _{0} /(1+kt ) and x= (v _{0} /k )ln(1+kt )
where k= јAv _{0} /m where x= 100 ft=30.5 m
is the distance traveled, t is the time to go that distance, v
is the speed it has there, and v _{0} =400 ft/s=122 m/s is
the speed it left the gun. When I put in the numbers I get t =0.4
s and v =50 m/s=164 ft/s; this paintball loses more than half its
original speed. In order to do the heavier ball, you need to know what
v _{0} is for it; if it is the same 400 ft/s, then I get
t =0.34 s and v =68 m/s=223 ft/s, definitely faster.
However, I suspect that the initial speed of the heavier ball is slower,
perhaps the two balls have about the same energy. In that case the 3 g
ball would have an initial speed of about 100 m/s=328 ft/s, in which
case I get t =0.42 s and v =56 m/s=184 ft/s, just barely
faster. Finally, your idea about momentum is about right in that the
force to stop the ball in a given time is proportional to its momentum
and the greater force would be more likely to break it (assuming the
cladding is the same). But, momentum is mass times velocity, so the
heavier ball would have to be going with a speed of about (2/3)164=109
ft/s to have the same momentum as the lighter ball. A lot hinges on how
fast the heavier ball leaves the gun.

QUESTION:
My 6 year son asked me if we stand due to gravity then we should be lying down instead of standing up. Can you explain in simple words so that he can understand.

ANSWER:
Lying down or standing up, it doesn't really matter. In both cases the gravity keeps us from floating up to the ceiling. When standing up you have to balance, otherwise you would soon be lying down!

QUESTION:
When an object slides down a ramp with friction, the kinetic energy at the bottom of the ramp is not equal to the potential energy at the top. Why doesn’t this situation violate the law of conservation of energy?

ANSWER:
As the object slides, it gives some of its kinetic energy to the atoms
in the incline and on its own surface, that is, those surfaces get
warmer. Some of the energy goes into the sound made while sliding.
Energy is conserved but the kinetic plus potential energies of the
object are not. The law of conservation energy states that the total
energy of an isolated system does not change; just looking at the
kinetic and potential energy of the object is not including all energy
of the system.

QUESTION:
We think of rock as a solid. If we hit a rock with a hammer, it shatters. So how can solid rocks flow within Earth's mantle to cause the drifting of the continents

ANSWER:
The rock in the earth's core is molten. It is real hot in there.

QUESTION:
I was wondering if energy can be harnessed from a nuclear chain reaction. If the reaction produces so much power, why is it necessary to use steams turbines to generate electricity? Can we not harness the energy from the chain reaction itself?

ANSWER:
The energy released in nuclear fission is in the form of the kinetic
energy of the fragments, essentially heat. You cannot just use heat to
light a light bulb or operate your toaster, you have to convert it to
electricity. Because the fragments are radioactive, you cannot use them
directly but must transfer their energy to something else, usually
water. You could then use that hot water to heat your house, for
example, but it is most efficient to use that hot water to drive
turbines to generate electricity which is easily transported or stored
somehow.

QUESTION:
In nuclear decay say of uranium 238 if I had only two atoms of it. One atom may decay in a few seconds the other might task over 5 billion years to decay. The wave function determines what makes it decay in a certain amount of time, what I want to know is how often per second does the atom 'roll the dice' and see if it's time is up? Why would one see that it's time is up in a few seconds and the other not for 5 billion years, how often is time and chance related? They must be somehow related in a chance per/second relationship or the atom would never decay.

ANSWER:
A quantum system does not "roll the dice"; that is the wrong way to
think of it. The wave function is not something which, for some
particular nucleus, is heading for some particular time to decay, it is
the information which gives the rate at which an ensemble of identical
nuclei will decay. Trying to think of things like half life etc .
does not really make any sense with one or two nuclei. As you suggest,
all you can say about any individual nucleus is that it might or might
not decay at any particular time. If a particular nucleus has a long
half life, then it is likelier that it will be around in a few years
than if it has a short half life.

QUESTION:
Does a photon have a de Broglie wavelength?

ANSWER:
The relationships between energy, momentum, frequency, wavelength, and
speed for a photon were the inspiration for deBroglie's postulation of
the wavelength for a particle,
λ=h /p. For a photon, p=E /c=hf /c=h /λ
which is still λ=h /p. A photon is just a particle with
zero mass. The answer is yes.

QUESTION:
Are electromagnetic fields (or any other kinds of fields) real things or are they a mathematical, theoretical construct intended better to describe the forces in nature?

ANSWER:
Well, it depends on what "real" means to you. Certainly forces are real
and fields are representations of forces. To me, that makes fields real.
I guess you imply that if they are a mathematical construct to describe
forces they are not "real"; I see no reason why not. Maybe it is just a
matter of semantics, not physics.

QUESTION:
It is my understanding that Einstein considered gravity to be a maifestation of acceleration, such as in the elevator thought experiment. This makes intuitive sense but I am hung up on the fact that sometimes we are on the side of the globe that is opposite the direction of accelertion. Why don't we fly off the planet?

ANSWER:
It is not accurate to say that gravity is a "manifestation of
acceleration". Read my earlier
answers relating to the equivalence principle. I do not understand
your question at all; what do you mean by being "opposite the direction
of acceleration"? Accelerations you experience as a result of the
earth's motion are really small compared to the acceleration you feel
because of the earth's gravity.

QUESTION:
When I fill my thermos with lets say water... I hear a sound, the tone goes from low to high, what causes this? (While the thermos is being filled)

ANSWER:
your thermos is acting like a column of air which is closed at one end
and open at another and the length is changing as it fills. It is acting
like an organ pipe of varying length. An organ pipe which is long plays
a low note and one the is short plays a high note. The sound of the
water splashing into the thermos has just about all frequencies but only
those which resonate with the cavity survive very long. It is sort of
like a trombone where you change the note by changing the length of the
air column or one of those slide whistles that has a piston you can pull
back and forth.

QUESTION:
Can an atom retain information about any other atom it was bonded with in the past if even for a very short time after that bond is broken?

ANSWER:
Well, every chemical reaction has a time profile and the reaction
products might not end up in their ground states, so some time will
elapse before you have the reaction products in their ground states.
These times, are very short, like microseconds or nanoseconds or
picoseconds. If you are asking me in a subtle way what I think of
homeopathy, I think it is a crock.

QUESTION:
How did Newton measure the mass of the earth? (Am assuming it was Newton or was it somebody else?)

ANSWER:
Newton's triumph was to be able to explain Kepler's laws which
empirically describe how the solar system works. You might think he
would need the mass of the sun to do that. But, he could not determine
the mass M of the sun because he did not know the value of the
gravitational constant G . What he could determine was their
product, GM . Similarly, he could not find the mass of the earth,
only its product with G . For a planet in a circular orbit, T =2 π √(R ^{3} /(MG ))
where T is the length of the year and R is the radius of
the circle. A way to get the mass of the earth is from the acceleration
due to gravity g=9.8 m/s^{2} which can be shown to be g =M _{earth} G /R ^{2} _{earth} ,
again showing that the product of G with the mass is all you can
determine. You can also get GM for the earth from the period of
the moon (about 28 days). Masses of planets and the sun were only
determined when G was independently measured in the
Cavendish
experiment which directly measured the gravitational attraction
between two known masses. I believe that G is still the most
poorly known of fundamental physical constants. Masses of planets
without moons were notoriously hard to determine before we were able to
shoot spacecraft at them.

QUESTION:
how come when we free fall (base jumping,bungee jumping) we fell weightless??
shouldn't we will be feeling something due to the gravity attraction...??
i

ANSWER:
How do you "feel" gravity? One way is that you feel the floor pushing up
on your feet. But, it is much more than that. All your internal organs
each have weight and the muscles etc . which support them signal
your brain that they are being pulled down. When you are in free fall,
all of you is and so the usual signals to your brain, from the bottoms
of your feet and the muscles holding up your liver all are missing which
your brain interprets as "hey, no gravity".

QUESTION:
i heard a story on npr about magnetic north shiftin or something. i cant remember exactly what was said but i know navigation systems were out of synch b/c of it. ive got a terrible memory. mabey im way off base but it was something to do w/ the poles or something.

ANSWER:
The location of the magnetic north and south poles have always drifted
around. In fact, it is known that many times the polarity of the earth
has actually shifted, that is the current north becomes south and
vice versa . I do not think that much serious navigation is done
anymore using a magnetic compass; if a gps is not used, a gyrocompass is
more accurate than a magnetic compass. Navigation using magnetic
compasses rely on accurate maps of the magnetic field at the earth's
surface and they are periodically updated. Magnetic north and geographic
north have always been different.

QUESTION:
Hasn't anyone considered it to be odd that there is a speed limit to stuff in the universe and that there is light that travels at precisely that speed limit?

ANSWER:
It is only odd if you think that nature should behave according to
preconceived ideas you have. It is a law of physics derived rather
easily from the theory of special relativity and the properties of
electromagnetic waves. Have a look at earlier answers explaining why
there is a
speed limit and why light has
that speed .

QUESTION:
Which are inertial forces? Give me examples in everyday life.

ANSWER:
Inertial forces are more often referred to as fictitious forces. They do
not exist but they appear to in noninertial frames of reference,
accelerating frames. Because Newton's laws are not true in noninertial
frames, adding these fictitious forces is a way of forcing Newton's laws
to be true. A few examples are:

When you accelerate a car, you feel like
there is a force pushing you back into the seat. You are in fact feeling
the force of the back of the seat pushing you forward to accelerate.

Centrifugal force, the force which appears
to push objects out in a rotating vessel, is a fictitious force.

The reason that weather patterns circulate
around low or high pressure regions is that the earth is a rotating and
therefore accelerating frame. The force which causes the air circulation
is called the Coriolis force.

QUESTION:
I don't understand why is nature always seeking a higher entropy state. I once read that nature always seeks lower potential energy states but how is an ice melting, which is an increase in entropy, a lower potential energy state if the melted ice is now at a higher energy state (particles have more kinetic energy).

ANSWER:
The mistake you make is to not consider only isolated systems. Your
example of melting water ignores the source of energy which melts it. If
you put an ice cube in a room of hot air, the energy of the water goes
up and the energy of the room goes down. But, the entropy of the room
plus water increases. In an isolated system, energy remains constant and
entropy increases or stays the same.

QUESTION:
To be honest I dont know much about physics at all and im only in highschool, however I was thinking the big bang theory implies that there was well a huge explosion, and the remains of that explosion was the universe, stars planets etc right?
well in space something will keep moving if there is no alternate force acting upon it, right? so at the moment of the explosion that would have sent the remains at millions and millions of miles an hour, now wouldnt that mean because it was the instance of the creation of the universe that there would be nothing to stop these stars etc and that they would still be travelling at millions of miles an hour because there is no force to stop them?

ANSWER:
But there is a force to "stop them". Every part of the universe
experiences a gravitational force from the rest of the universe trying
to slow it down. Recently, astronomers have discovered that the most
distant parts of the universe appear to be speeding up and attibute this
to an mysterious force which they call dark energy. Whether there is
enough gravitational force in the universe to eventually turn it around
is one of the most important unanswered questions in physics.

QUESTION:
Ok, general relativity states that light bends as it passes through a gravitational field. My question is why would it do this if light is massless? Considering in special relativity that nothing containing mass can reach and equal the speed of "light".

ANSWER:
Certainly, light does not have mass. However, mass is not necessary for
it to be bent by a gravitational field. I have dealt with this questions
several times before; see the FAQ
page.

QUESTION:
Theoretically, how many human hairs would be needed to lift an elephant?

ANSWER:
A typical elephant weighs about 10,000 lb. A healthy human hair can hold
about 3 oz=0.1875 lb. Therefore, N =
10,000 /0.1875 ≈ 53,000.

QUESTION:
If two people are going around a racetrack, one going 5mph and one going 50000mph, my friend is saying that they would age the same in relation to an outside (stationary) observer. I disagree and need supporting evidence. This question was posed because we were discussing whether speed or acceration has an effect on time dialation. I believe that the faster a person (or thing) travels, the slower time goes by. He believes that speed has no affect on time; its the acceleration an object has.

ANSWER:
Well, both acceleration and velocity determine time dilation. The
velocity time dilation is the easiest to deal with, just a simple
formula. Because of the equivalence principle, acceleration is
indistinguishable from gravity and so this effect is usually referred to
gravitational time dilation. You should also be aware that going around
a racetrack, both people have an acceleration (v ^{2} /R ) where v is
the speed and R is the radius of the track. You should also note that
50,000 mph is a very small velocity for these kinds of effects; v
should be comparable to the speed of light, 186,000 miles/second. the
speedier clock in your example will run very slightly slower because of
both its velocity and its acceleration.

QUESTION:
protons in a nucleus are separated by a very small distance.why doesn't the nucleus fly apart because of the strong coulomb repulsion between the protons?

ANSWER:
One can only conclude that there must be another force which, at small
distances, is stronger than the coulomb force and attractive. This force
is called the nuclear force or the strong interaction. Unlike the
coulomb force which is very long range, this force is neglibly small
except at very small distances ( ≈10^{-14}
m). It therefore has no observable effects on atomic structure; atoms
are held together by the coulomb force.

QUESTION:
Please could you solve an 'office' debate we have - Here is the scenario :
If a vehicle (Lets say an HGV) weighs exactly 20 tonnes when empty. The vehicle is then loaded with a pallet that weighs exactly 1 ton.
The vehicle then must weigh 20 tonnes, and the pallet still weighs 1 ton. Therefore the net weight of the vehicle is 21 tonnes.
Heres the question...
The vechile hits a bump in the road (lets discard the speed / inertia etc needed for this part) and the pallet for a split second leaves the bed of the trailer (and is not attached via straps etc to the vehicle) (but the vechile remains grounded), at this point is the net weight of the vehicle reduced to 20 tonnes or (in terms of physics) does the vehicle still actually weigh 21 tonnes?
The argument is that the pallet is no longer touching any part of the vehicle, and at this point if the vechile could be weighed it would weigh 20 tonnes. However the pallet is still actually inside the vehicle....
Could you shed some light on this for us?

ANSWER:
I have an easier way to answer your question since it avoids the truck
having an acceleration in the vertical direction, and therefore the
reading of a scale under the truck will simply be the weight. The
picture is to have the pallet hanging from a rope attached to the roof
of the truck; the truck stands still on a scale. The scale reads 21.
Now, the rope breaks. During the time the pallet is falling the scale
reads 20. When the pallet again is at rest on the floor, the scale will
read 21. (Note that I have ignored any effect of the air in the truck —buoyant
and frictional forces—which is an excellent approximation. If you want
to read a really exhaustive discussion of all the possibilities, see an
earlier answer which
addressed a similar question of sand falling in an hourglass.)

QUESTION:
I was wondering why, during a head-on collision between a moving train and a car that are both travelling at the same speed, the car gets completely crushed but the train is hardly damaged?

ANSWER:
What matters in a collision is momentum, not just speed. Momentum is
mass times velocity. Since the train has much more mass than the car, it
has a much bigger magnitude of momentum. Even though the two feel the
same force, the effect of the force on the car is a huge acceleration
(it changes direction in a very short time) whereas the train has only a
small acceleration (it continues in the same direction with very little
change of speed).

QUESTION:
My daughter wants to know why a stick makes noise when you swing it in the air?

ANSWER:
When you drag a box across the floor, friction takes energy away from
the box. That is why you have to keep pulling, putting energy in, to
keep it moving with a constant speed. Where does that energy go? The
friction is caused by atoms in the box hitting atoms in the floor and
making them move faster. These moving atoms mean that some energy is
converted into thermal energy (the floor gets warmer) and some of the
energy goes into sound energy (you can hear the box sliding). A stick
being swung through the air does the same thing by interacting with the
air molecules —heating the air and
making sound. Sound has a much smaller fraction of the energy than the
heat.

QUESTION:
I have a 16 foot aluminum boat that weights 1000 lbs dry weight.
I have a 12 lb. lead ball that I want to use for an anchor in the river and the bottom is covered with rocks of all sizes. The depth of the river is between 6 feet and 10 feet where I need to anchor. I have seen other people use this same lead ball and it works great. Can you explain to me how a 12 lb. round lead on about a 10 foot long rope can anchor a 16 foot boat weighing 1000 lbs. to a rocky bottom?

ANSWER:
Whether or not the anchor will hold depends on many conditions. If there
were no current, a 1 oz anchor would work. In a modest current, the 12
lb anchor would need to not slide frictionlessly on the bottom and would
likely hold on one of the larger rocks. In a strong current, it is
likely that the 12 lb anchor would not hold unless it got wedged between
a couple of big rocks. The weight of the boat is really not what
matters, it is how hard the current is trying to make the boat move.

QUESTION:
if a ship is lifted from its equilibrium out of still water with a crane, will it initially weigh less until it is clear of the water?

ANSWER:
The weight of the ship does not change because the weight is the
gravitational force the earth exerts on the ship. However, the force
which the crane must exert to lift it is less when it is in the water
because the water pushes up on it (buoyant force).

QUESTION:
Iam a 14 year kid just compleated my gravitational chapter...i hav one doubt i knw it's easy for u to solve please help me...MY question is As per newton law of gravitation every particle of matter attracts every another particle of matter. Then why do positive and positive charge repel each other ?

ANSWER:
Because gravitational and electrical forces are two completely
different, unrelated kinds of forces.

QUESTION:
I am doing an experiment involving a Gaussian gun, I want to determine the kinetic energy of the projectile ball. I know the kinetic energy formula, but I feel it would be difficult to measure the velocity of the ball accurately. A website suggested the below formula:
KE= 1/2 m * (d/t)
M is the mass of the projectile, d is the distance it travels and t is 0.45 seconds. They said you can use t=0.45 seconds as this is the time it takes any object to fall 1 metre (disregarding air friction). Why can you use this? And also is the formula accurate.

ANSWER:
This expression is not even dimensionally correct. Maybe it was meant to be (d /t )^{2} ? Is the gun 1 m above the ground and shooting horizontally? In that case, this would be right if
d /t were squared. The reason it works is because the
motion in the horizontal direction is not accelerated and so the speed
the projectile exits is the horizontal component of the speed over the
whole flight. Vertically, it falls exactly as if it were simply dropped.

QUESTION:
Take any equation, e.g. F[N])=m[kg]a[(m/s)/s/]. (SI units)
If the mass is to be measured in, say, grams, the new form of the equation needs to be:
F = (m/1000)a
Why is this? - why can't the [kg], in the first equation, simply be substituted 1kg=1000g.
Hence: F=1000m(g)a
But that's clearly wrong.
It seems counterintuitive.

ANSWER:
I always teach my students to multiply by 1 such that the units come out
the way you need them to come out. So, let's do an example with F=ma
where m =2 g and a =5 cm/s/hr. Then F =(2
g )x[1
kg/1000
g ]x(5
cm /s/ hr )x[1
m/100
cm ]x[1
hr /3600
s]=2.78x10^{-8} kg-m/s/s= 2.78x10^{-8} N. Note that all
the ratios in brackets [ratio]=1 and I choose the way to write the
ratios so that the unwanted units (g, cm, and hr) cancel out.

QUESTION:
I have a question about Blackbody Radiation; I looked it up on Wikipedia but didn't quite understand it. What exactly is a Blackbody, and how can it theoretically emit infinite energy, and how is this related to Blackbody Radiation?

ANSWER:
A black body (BB) is a perfect absorber of radiation. Of course, there
is no such thing, but many things are excellent approximations. One
example which is very nearly a BB is a tiny hole in a cavity: radiation
which enters the hole is highly unlikely to find its way back out the
hole. If you have a BB, it will also radiate energy which is called BB
radiation. So, if you want to study BB radiation, look at the light
coming out of the tiny hole in a hollow metal ball. It turns out that
the spectrum has a continuous distributions of energies and is brightest
at a wavelength which gets smaller as the object gets hotter; for
example, red radiation is not "as hot" as blue radiation. BB radiation
played a very important role in physics. At the end of the 19^{th}
century the nature of BB radiation was well known and physicists were
trying to explain it. It turns out that classical physics predicted a
spectrum which contained an infinite amount of energy, obviously not
possible (that must be what you are referring to). This was referred to
as the "ultraviolet catastrophe" because the spectrum just got bigger
without bound for small wavelengths. The problem was finally solved by
Max Planck by proposing the first ever application of quantum physics.

QUESTION:
Can you explain the theory behind the Schrodinger equation to me? How exactly do we compute atomic orbitals, when electrons are too small and too elusive to be seen directly?

ANSWER:
The theory is contained in an
earlier answer . Note
that this is about as nontechnical as I can make it, but still requires
a pretty good grasp of elementary physics and elementary calculus.
Regarding your second question, computing orbitals is done by finding
the solutions to the Schr ц dinger
equation for the atom and just because you cannot see something does not
mean you can't understand it by other means (like properties of atoms,
atomic spectra, etc .).

QUESTION:
when i was in junior high, a classmate (very smart girl) asked our science teacher why electrons don't just fall into the atomic nucleus; after all, they're negatively charged and should be "captured" by the positively charged protons. that's when we were given a lesson in the existence of orbitals and the teacher said that nature/quantum mechanical laws require that electrons "orbit" the nucleus. but are there circumstances wherein electrons do "fall" into a nucleus? and are neutrons actually a composite of a proton, electron and neutrino?

ANSWER:
First ask yourself "why don't the planets just fall into the sun? After
all, they're attracted by the gravitational force and should be
'captured'". But they don't because they are in stable orbits where that
force of attraction causes them to move in (nearly) circular orbits
instead of falling. Although there are differences which arise because
of quantum mechanics and because charged particles radiate energy when
they accelerate, the idea for orbital stability is the same as for
gravity. Oh, and a neutron is not simply a composite of its decay
products, it is a particle which happens to decay.

QUESTION:
If you stopped an Earth satellite dead in its tracks, it would simply crash into Earth. Why, then, don't the communications satellites that "hover motionless" above the same spot on Earth crash into Earth?

ANSWER:
Because they only appear to hover motionless. Don't forget, the earth is
spinning around once every 24 hours. The period of a communication
satellite is 24 hours, it is not standing still.

QUESTION:
Why does the amount of reflection increase and the amount of refraction decrease as the angle of incidence increases? This is not homework and I have asked my teacher and she is not sure either can you please answer this for me?

ANSWER:
It is actually not a simple problem easily explained. Also, it depends on the polarization of the light and
your statement that that reflection increases as the angle of incidence
increases is not always true. For example, plane polarized light with
magnetic fields parallel to the reflecting surface will have zero
reflection at an angle called Brewster's angle. This is the idea behind
polarized sunglasses. You can read about this in the Wikepedia article
on Brewster's
angle .
For a more general discussion, see the article on
Fresnel's
equations . Yet another violation of your basic assumption is
total
internal reflection where, if the index of refraction of the first
medium is greater than the second; for angles greater than the critical
angle there is 100% reflection. This is the idea behind fiber optical
cables.

QUESTION:
How to find the mass of a piece of lead without using a scale or density?

ANSWER:
Exert a known force on it and measure its acceleration.

QUESTION:
When we say the half life of certain subtsnace is say (for example Carbon 14) is 5730 years, we mean that there we can expect on average 50% of the atoms to decay? If so , why do some decay and some dont? If we think there is no reason, that we cannot predict which one will decay and which one wont, does this give us grounds to doubt the notion of causality?

ANSWER:
Decay is a statistical process and, yes, halflife is what you describe.
The fact that something happens statistically does mean there is no
reason that it happens. The reason that a particular decay occurs is
that the wave functions of the initial and final states (which describe
the natures of the states) are able in nature to be connected by a
physical process. It also has nothing to do with causality since there
is a cause which does not come after the effect.

QUESTION:
When I turn on a flash light, the light speeds off at 186,000 mps. When I turn off the flashlight, where does the light go?

ANSWER:
It keeps going away until it encounters something which either reflects
or absorbs it. Of course, no more is starting at the flashlight so it is
far away from you pretty quickly.

QUESTION:
Betelgeuse is 640 lightyears from earth. So if I were on Betelgeuse today looking at the earth, would I be looking at earth 640 years in the past? Meaning If my telescope was strong enough could I see the dinosaurs for example?

ANSWER:
The answer to your first question is yes. The answer to your second
question is: do you really think there were dinosaurs roaming the earth
640 years ago?

QUESTION:
I have learned that if an object shall escape from a gravity field the total mechanical energy (potential+kinetic) must be zero or more. In one way, mathematically, I can accept this. But sience a gravity field has an "infinite long extent" I dont understand why an object cant return. Even though the gravity-force will become very small, it will still act. And even though it will take long time, why wont this force manage to return the object?

ANSWER:
Well, the problem you are fussing about is an idealization. It is correct
only if "infinitely far away" actually is realizable (the universe is
finite) and if the earth is the only thing in the universe (it isn't).
Nevertheless, if the idealization were correct and if we could give
something precisely zero total energy at the earth's surface, the object
would certainly come perfectly to rest after an inifinte amount of time.
But, of course, in a finite and changing universe, there is no such
thing as infinite time either. So stop fussing and just give it a
smidgin of energy above zero and you can be assured that it will not
come back. Whoops, that is not true either because it could be deflected
by some other object and come back. The lesson here, I think, is that
you can talk about things as they are approximately in physics, and
still learn a great deal about how nature works. Incidentally, we do not
really have good data proving that gravity is the simple 1/r ^{2}
force we think it is at values of r larger than about the size of
the solar system.

QUESTION:
My question is about mass and gravitational time dilation. It is accepted that a clock on our planet's surface runs more slowly than a clock on a mountain. So how much bigger (e.g. in spherical volume, or diameter) would the earth need to be for both clocks to run at half the speed?

ANSWER:
I do not know what you mean by both clocks running at half the speed. This gets a little technical, but if you are a distance
R from the center of a nonrotating spherical mass M where
R is greater than the radius of the sphere, and if you use the
Schwartzchild metric, then t=t _{∞} √[1-(2GM /(Rc ^{2} ))]
where t is elapsed time on the your clock and
t _{∞ } is the elapsed
time on a clock at R =∞. Here G is the universal
gravitational constant and c is the speed of light. So, a clock
at the earth's surface would run only about 10^{-7} % slower than
an infinitely distant clock. So there is certainly no radius where the
time would be 50% slower. You would have to both increase mass and
decrease size to have a noticably slower clock. If 2GM /(Rc ^{2} )≈0.75,
t ≈0.5t _{∞} .
These kinds of numbers only happen near black holes, I believe.

QUESTION:
The size(range of electrons) of 1s orbital of Helium would be equal to that of heavier elements(let's say Radon).Yes or no.Why??

ANSWER:
The size of the 1s orbital gets smaller as the atomic number (Z )
gets greater. The reason is that the 1s is the innermost orbital and has
a radius which is inversely proportional to Z . The electron sees
the much larger positive charge of the nucleus and little effect from
the outer electrons. The size of the whole atom, however, is fairly
constant over the whole periodic table because the inner electrons
screen much of the nuclear charge and so the last electron sees a
positive charge of about +e , like hydrogen.

QUESTION:
If a plasma rocket were able to maintain an acceleration that mimicked 1
G to humans on board, how soon would the rocket reach the closest star?

ANSWER:
This is a somewhat complicated question because the rocket can never
exceed the speed of light so, although the force on it is constant, the
acceleration (to an outside observer) is not. I have actually worked
this problem out in an earlier answer which you can have a look at. In
that answer I give the position as a function of time to be x =(mc ^{2} /F )(√[1+(Ft /(mc ))^{2} ]-1).
In your case, F=mg . Solving for t , I find t= √[(x /c )^{2} +(2x /g )].
I chose to calculate t for 4 light years, approximately the
distance to the closest star, and found t ≈5 years. This is the
time elapsed on earth for the trip. For a passenger on the rocket the
time will be much shorter because the velocity for much of the trip is
apparently near the speed of light (it takes only about 5 years to go 4
light years).

QUESTION:
Why do bubbles of C02 escape the bottle of soda water when the cap is undone?

ANSWER:
The reason soda is "fizzy" is that carbon dioxide is dissolved in it.
When the cap is on the pressure inside the bottle is a little above
atmospheric pressure which allows there to be more CO_{2} to be
dissolved than at atmospheric pressure. When the cap is removed the
pressure lowers and CO_{2} is released.

QUESTION:
I don't understand how rockets work in space. I understand forward motion to be a result of force exerted backwards. Considering that space is supposed to be a vacuum, how is it possible that the energy exerted by the rockets creates the backward pressure to propel it forward?

ANSWER:
You are not describing how rockets work, in space or otherwise. Imagine
you are in empty space and throw a ball. You exert a force on the ball
which causes it to move away in the direction of the force you exerted.
But, Newton's third law says if you exert a force on the ball, the ball
exerts an equal and opposite force on you; so you recoil and move away
in the opposite direction from the ball. That is how a rocket works,
like throwing out many tiny balls.

QUESTION:
When two "large" objects, i.e. a hammer and a chisel, meet, which of the four fundamental forces prevent the intermingling?

ANSWER:
The electromagnetic force. See an
earlier answer .

QUESTION:
Assuming that your molecules and atoms do not burn up due to velocity at which you are going, at what velocity would you have to be going, in order to pass through a solid object unharmed?

ANSWER:
This never happens. See an
earlier answer .

QUESTION:
I've been reading "elementary modern physics" (circa 1960) by Weidner & Sells with respect to "fission & fusion nuclear reactions".
The text says that in these highly exothermic reactions of very heavy nuclides, the fractional conversion of mass to energy is greater for a fusion reaction (0.66%) than for a fission reaction(0.09%). Both are certainly less than 1%. Wow....This is incredible!
This seems consistent with some mass-energy work I've done in older classical texts. Frankley, I originally thought I must be wrong due to the low conversion efficiency.
Most working masses I've seen are usually 10-20 kilograms. With such a low % conversion efficiency, where does all the remaining nuclear material go? The initial reaction occurs in a micro second. It seems the material gets blown out radially before the chain reaction proceeds?
Does the remaining material become the so-called "radioactive fallout"?

ANSWER:
Do you mean that you think energy conversions on the order of 1% are small? Compare
that to chemistry (coal and gasoline), where we get most of our energy,
and where the mass conversions are more like 10^{-7} %! Regarding where the mass not converted into energy goes, it is mostly the fission or fusion products.
When uranium fissions, it disintegrates into two nuclei (usually) and
several neutrons. If you add up all that mass, you find less than the
uranium nucleus had. When hydrogen fuses, it becomes helium and the
helium nucleus has less mass than the hydrogen which fused to make it.
In the case of fission, the reaction products are indeed unstable and
their radioactivity (often of long half life) is the source of the
"radioactive waste" problem associated with spent fuel rods at nuclear
reactors. You might be interested in an earlier answer on
fission and fusion .

QUESTION:
If you climbed a rope to the top of the bleachers, instead of walking up the steps to reach the top, you would do more work.

ANSWER:
Is that a statement or a question? In each case, the amount of work you
expended to raise yourself, that is to increase your potential energy,
is the same. However, this is not just a physics problem, it is also a
biology problem. I recently read that when you climb stairs, only about
20% of the work you do is lifting yourself, the rest of the energy is
presumably against frictional forces in your joints, etc . This fraction would likely be different for
climbing a rope, but I have no idea which would be larger.

QUESTION:
Why can't you observe Brownian motion in easily visible objects such bits of paper floating in water?

ANSWER:
Think of a bb striking a bowling ball —the
bowling ball does not noticeably recoil. But, if the bb strikes a small
marble, you might observe the recoil of the marble. The atoms move with
very large speed but they have very small masses, so to see a recoil,
the thing you are looking at has to be very small.

QUESTION:
Is there a magnet powerful enough that when suspended from a string will actually pull itself north? Or is there a way to make such a magnet?
Really hope you can answer this for me

ANSWER:
There are two things to consider: First, the earth's magnetic field is very
weak. But, more important, there is no net force on a magnetic dipole in
a uniform magnetic field, only a net torque. Because the earth is so big
compared to any magnet you are likely to have, the field is nearly
uniform over the volume occupied by the magnet. So, no matter how stong
the magnet is, it will experience a torque which tries to align it with
the field (which is how a compass works) but it will experience no
net force.

QUESTION:
What a nice idea. What is the diameter of a sphere occupied by 1 pound of CO2 (at sea level, if specifying an altitude is required)? Similarly, what are the dimensions of a cube so occupied? I am preparing material for elementary school-age children and senior-age adults. I would like to demonstrate spatially what this amount of CO2 might look like. Then, I would like to give examples of activities that generate a pound. Also, I am trying to translate these dimensions into recognizable every day objects. Web searches have lead to few results and have been of questionable validity. So far, one claims is that a pound is about the size of a dishwasher. Another, a Pilates (exercise) ball which is about 2.5 feet across. These two do not seem likely to be equivalent. I would prefer to have the measurements if possible and to generate examples from that.

ANSWER:
The density of CO_{2} at 20^{0} C and 1 atmosphere
pressure is 0.115 lb/ft^{3} . Therefore the volume of 1 lb of the
gas is 1/0.115=8.7 ft^{3} . A cube of side L has a volume
of L ^{3} , so L =^{3} √8.7=2.06
ft. A sphere of radius R has a volume of 4πR ^{3} /3, so
R = ^{3} √(3x8.7/(4π ))=1.28
ft. So the Pilates ball is about right and a dishwasher is about right
too (about 2 feet on a side?).

QUESTION:
When comparing the volume of a gas released to the volume of the liquid
in which it came, why is the gas a larger volume than the liquid?

ANSWER:
In a liquid (or solid) the forces between nearby atoms are able to hold
the atoms in a close-packed configuration. As the material heats up, the
atoms in a liquid start moving around faster and eventually move fast
enough to pop out at the surface. They are now moving around fast enough
that the force of attraction to other atoms in the gas is not big enough
to bind them together.

QUESTION:
I am sure this is a pretty silly questions, but I am very curious about it. I wanted to know if it takes more effort to jump from a moving vehicle to another moving vehicle or while simply standing still.
Mainly, I am thinking that there would be factors like wind resistance involved while jumping in motion.

ANSWER:
You are right, air drag will play a role, particularly at higher speeds.
The force which the wind exerts on you as you jump from one car to the
other is approximately proportional to the square of the speed. For
example, the force on you at 40 mph will be 4 times larger than at 20
mph. At lower speeds you would be able to jump more or less as you would
if the cars were at rest. As the speed increased you would have to
"lead" the jump, that is, jump toward a point ahead of where you wanted
to land because the wind will blow you back.

QUESTION:
I was reading my physics textbook to prepare for a test and this sentence, I can't put into my own words. "All electromagnetic radiation, including x-rays, can be visualized as two perpendicular sine waves that travel at the speed of light". What does it mean?

ANSWER:
See an
earlier answer .

QUESTION:
I'm reading a book on Galaxy formation and it says "This baryonic matter
would radiate away energy in ways that dark matter could not.
Consequently, the baryonic matter settles into the centre of the dark
matter halo, where it forms the visible part of the galaxy". My question
is - why would matter that radiates energy more efficiently (I don't
know if this assumption is true based on the text) settle in the center?

ANSWER:
As noted on my site, I avoid detailed astrophysics questions. Also, I am
a total skeptic on dark matter —I do
not deny its existence, I just am waiting for somebody to actually
directly observe it as opposed to inferring it. But, I think I can
answer your question from my perspective. The normal matter in the
universe is normally electrically charged, and accellerated charges
radiate electromagnetic radiation. Hence, kinetic energy of charged
particles tends, because of collisions, to lose kinetic energy. As
particles slow down, their collective gravitational attraction will tend
to bring them together. The dark matter, on the other hand, would remain
"hot" and not tend to "settle" to the center.

QUESTION:
if all the forces are part of equal and opposite pairs, then why don't they all cancel and forces always sum to zero?

ANSWER:
It all depends on what you are looking at. If you look at an isolated
system, then all the forces do add up to zero. But, if you are looking
at an object which is acted upon by other things, the forces do not add up
to zero. Drop a ball and the net force on it is its weight (neglecting
air drag), not zero. The reaction force (the force which the ball exerts
on the earth, equal and opposite the ball's weight) is not a force on
the ball . However, all the forces between the atoms which compose
the ball do add up to zero and you do not need to worry about them.

QUESTION:
would an object traveling in mid air with a weighted core take any more or less force to initiate a rotation on an axis than an object with evenly distrubuted weight of the same size, shape, and mass?
I am currently in a debate with someone who argues that an object with a weighted core would take less force to initiate a rotation than that of the same object whos weight was evenly distributed.
the object in question is a wakeskate which is like a skateboard for water (similar to a wakeboard but the board is not attatched to your feet) and the arguement is whether a board with a weighted core would take less effort to do a trick once the board was airborn.

ANSWER:
What determines how easily something is to get spinning about a
particular axis is the moment of inertia. The closer the mass is to the
axis of rotation, the smaller the moment of inertia and so the easier it
is to get rotating. One example is a hoop with all its mass at the
circumference (like a bike wheel, approximately) and a disk of the same
mass. The disk is easier to get spinning because the mass is, on
average, closer to the axis. But, to compare the two they must have the same
mass, so you would not help things by adding a core of lead to the center
of your board; you would have to "move" some of the mass closer to the
axis of rotation. It also depends on where your intended axis of
rotation is. Suppose your board is a rectangle with a long side and a
short side and much of the mass were along the center of the long
direction. This would make it easier to spin along the long axis but not
along the short axis.

QUESTION:
You have a long rope. You tide it into a tree and go walking around the world. When you come back to the tree from the other side. You push the rope tight and untide it from the tree and and tide the ropes together. Is this possible, or will it float.

ANSWER:
Sorry, this could not possibly work with a real rope. For one thing, the
rope would sag under its own weight before you got to the horizon from
the tree. But, just suppose you had this rope in a circle a few meters
above the ground and all the way around the world. Look at a small piece
of the rope and see if it is in equilibrium. There is its own weight,
straight down, and the tension in the rope pulling left and down and
right and down. No forces up, so how could that little piece of the rope
be in equilibrium?

QUESTION:
My book says (when talking about a piece of copper with an excess of charge): "Once static equilibrium is established with all of the excess charge on the surface, no further movement of charge occurs. Similarly, excess positive charge also moves to the surface of a conductor. In general, at equilibrium under electrostatic conditions, any excess charge resides on the surface of a conductor"
Ok, I understand that the negative charge can lie on the surface (electrons moves readily in copper), but how can positive charge lie on the surface of the conductor if protons are fixed to the nucleus?

ANSWER:
Excess positive charge usually means missing electrons. When a
positively charged conductor comes to equilibrium, all the atoms missing
electrons are on the surface.

QUESTION:
I have a question about the double slit experiment. What is it about observing the light that effects whether it's a particle or a wave? Is the observing instrument in some way interacting with the light? Why would just observing it make any difference? Is that just the way things work at the subatomic level?

ANSWER:
It is a mistake to ask whether light is a particle or a wave. It is both
simultaneously. If you design an experiment to prove it is a wave, you
will succeed. If you design an experiment to prove it is a particle, you
will succeed.

QUESTION:
what is the acceleration of a snowflake?

ANSWER:
If there is no wind, a snowflake would have acquired its terminal
velocity almost immediately, that is, its acceleration is zero.

QUESTION:
If a photon is traveling towards a body of mass already at the speed of light, what would be the effect of the gravitation on the photon? Would it accelerate?

ANSWER:
The photon gains energy as it "falls". But, it does not gain speed. Its
frequency increases, its wavelength decreases. This is called a
gravitational blueshift.

QUESTION:
if you have an indestructible box and fill it with enough water such that if the water were frozen its volume would exceed the interior space of the box, what would happen? would the water freeze until the space was filled, leaving some liquid? would the water freeze completely with the ice being more dense?

ANSWER:
Well, site ground rules forbid "indestructible" boxes since there is no such
thing. Still, I will answer your question because you could make a very
strong box. The problem with your question is that it assumes that "ice"
is just one thing, the solid form of water. In fact, there are many
different kinds (phases) of ice depending on the temperature and
pressure. In your case, as you tried to freeze the water, its tendency
to expand would greatly increase the pressure. You can see a
P-T phase diagram
for water on Wikepedia. As you see, there is no simple answer to your
question.

QUESTION:
I am puzzled. In your FAQ's about length contraction you state 'I often make a big deal about relativity being about how things are, not how things appear'. If the length contraction of a moving object is about what 'is' - not about how it 'appears'- then what can we say about an accelerating object's increase in mass? Is that an 'appearance' or an 'is'? If the latter, are we saying that the object's mass increases but it occupies less space?
I believe Einstein favored the approach of 'what can be measured' rather than 'how things are'.

ANSWER:
When I talk about "how things are" I mean that there is an operational
definition which allows me to make a measurement, so I am certainly in
agreement with Einstein (whew!). Length, the distance between two
points, is defined to be the difference between two position measurments
made at the same time. A time interval is the difference between two
time measurements made at the same location in space. Mass does not play
the same fundamental role in relativity that it does in classical
physics. In classical physics it is a measure of how much resistance
(inertia) something has to acceleration for a given force (Newton's
second law, F=ma ). It is not, as I have
previously discussed , necessary to say mass increases with speed.
Since acceleration is not an important quantity in relativity, you can
get rid of it completely and
redefine linear momentum . Then all you ever talk about is rest mass,
the inertia something has when at rest. I don't get your remark about
"occupies less space". You can certainly interpret the redefinition of
momentum as a redefinition of mass. There is no question that an object
always resists acceleration, it is just that it has more inertia as it
goes faster, not a constant amount like in classical physics.

QUESTION:
I'm trying to understand Maxwell's but I can't!
There is a lot of maths and I'm still 15 which means I'm not ready for that kind of level!
But how can I be a physicist without understanding Maxwell's ? Could you please explain it to me sir?

ANSWER:
Since you cannot do the mathematics, I can only give you a brief
qualitative explanation of what Maxwell's four equations say. Maxwell's
equations tell you everything there is to know about electricity and
magnetism. They also contain the theory of special relativity; they
predict the speed of light and demonstrate that light is nothing more
than varying electric and magnetic fields. You need to know what an
electric field and a magnetic field are: electric fields, when present,
cause an electric charge to feel a force and magnetic fields cause a
moving electric field to feel a force. Here are Maxwell's equations:

Electric charges cause electric fields; a
point charge causes a field which falls off like 1/r ^{2} ,
r being the distance from the charge.

Electric currents (moving charges) cause
magnetic fields.

Changing electric fields cause magnetic
fields.

Changing magnetic fields cause electric
fields.

That is about what you can do without math.

QUESTION:
I understand that elements differ from one another by the number of electrons in the inner and outer orbits. I also understand that electrons can be acquired or lost when an element comes into contact certain others. Why then cannot one element be transmuted into another by adding or subtracting electrons?

ANSWER:
Your understanding is flawed. Different elements have different numbers
of electrons, but they also have different numbers of protons in their
nuclei. For example, if you take an electron away from a helium atom you
get a positive helium ion (a helium atom missing one electron), not
hydrogen. The thing you have to do to cause transmutation is to change
the nucleus. This is why alchemists in the Middle Ages were unable to
transform lead into gold.

QUESTION:
Neutrons are unstable and decay into other particles after a short time if not bound in the nucleus of an atom. Given this, why is it that there are any neutrons in the universe today since the universe was so hot and energetic for its first 300,000 years that atoms did not form. How did neutrons survive for those first 300,000 years without decaying?

ANSWER:
For the reason you state, it is fair that to say that the number of free
neutrons in the universe at any given time is very small compared to
everything else. The early universe was mostly just protons, electrons,
and photons. So H atoms form, then stars form, and fusion begins in
stars. The various fusion reactions create He (containing neutrons) and
all the heavier elements up to Fe. Essentially what happens is that
protons get converted into neutrons plus positrons and the neutrons stay
in nuclei where they do not decay.

QUESTION:
Assuming I had a nulcear reactor to make electricity And some form of electromechanical propulsion that was totally self contained and could move me at a constant acceleration of 1G. Why would I not be able to go faster than the speed of light. Having crunched the numbers a constant 1g force would take less than a year to reach that speed. Is it because of spacetime dilation or what?

ANSWER:
First of all, acceleration does not play the important role in
relativity that it does in Newtonian mechanics, F=ma is no longer
a valid equation. The simple fact is that, if I watch you in your
amazing ship go by, it is impossible for me to measure a constant
acceleration for you as you approach the speed of light. If you are
within 5 m/s of the speed of light and you think your acceleration is 10
m/s^{2} , I cannot possibly see that because your speed after 1 s
would be 5 m/s above the speed of light. I have discussed this in great
detail in
earlier answers .

QUESTION:
Why does some light bounce off the surface of water and some light pass right through? For example, when we look into a swimming pool, we can see the bottom, but the light is also reflecting off the surface.

ANSWER:
That is just the nature of waves. When they encounter a discontinuity of media, they can both reflect and refract. There are special cases where one gets 100% reflection (which is what allows fiber optic cables to work) or where one gets 0% reflection (polarized light at special angles).

QUESTION:
If the moon were to gain mass, say from attracting the passing dust, and eventually became more massive than the Earth, it's gravitational force would be stronger than that of Earth. Would this result in the Earth begining to orbit the moon as opposed to the other way around? Also would we then refer to the moon as a planet and the Earth as a moon or is this based on the size of the body and not on their orbits?

ANSWER:
The moon does not orbit around the earth. Rather, the two orbit about
their center of mass which is a point inside the earth for their current
masses (but not at the center of the earth). For example, if the moon
were to become as massive as the earth, the two would orbit around a
point halfway between them. I do not know much about astronomy naming
practice, but if one were not much more massive than the other, it would
not be clear which was the planet and which the satellite. They would
probably be called something like binary planets much as bound stars are
called binary stars.

QUESTION:
Is the volume of a liquid in a cup of water quantized? Can you add an arbitrarily small amount to a cup of water or is there some smallest amount that you can add or subtract?

ANSWER:
Well, you can't add a half molecule of water, can you?

QUESTION:
Is it possible to have a negative Mass Defect? When I calculate the total atomic mass for Neon (10P + 10N + 10E) the result is 20.16478 amu, but on the periodic table it lists the observed atomic mass to be 20.180. This would result in a negative Mass Defect of -0.01522. Where am I going wrong? Should I not use the mass of an electron since it is moving near the speed of light and thus has a varying mass?

ANSWER:
I would guess that you are using the atomic mass for the "proton"; that
is, what is usually tabulated is the mass of H and so by adding the
electrons you are double counting.

QUESTION:
what is the origin of spin of objects in nature, from electron to galaxy? where from the energy comes to continue spin?

ANSWER:
To get something moving takes energy, to keep it moving does not. To get
something spinning takes energy, to keep it spinning does not. Imagine a
huge cloud of gas in space. The odds are that it will have a net
rotation, although very slow, about its certer of mass. As it contracts
to form a star, it spins faster and so all stars will be observed to be
spinning.

QUESTION:
Is there any formula, or rule, that can accurately predict an isotope's chance for stability if given the amounts of protons and neutrons? My teacher gave us a strange "rule" to follow, that I have have doubts about. She said that if the isotope's atomic number is less than 20 (Z<20), then="" neutrons="" and="" protons="" must="" be="" at="" a="" 1:1="" ratio.="" if="" z="">20, then the ratio fo N/P can be slightly above 1 for best chance of stability.
This seems to work "ok" for isotopes with an "even-even" configuration of neutrons and protons, but it's when she applies the same rule to "odd-even" configurations that the problems become apparent. Look at Lithium on the periodic table as one of many examples of where this rule doesn't work. Is this a real rule or just something she made up?

ANSWER:
There is no simple rule by which you can tell if a nucleus will be stable.
However, your teacher did not just "make this up". Often in science it
is useful to have "rules of thumb" which give a good qualitative
description of something. One of the most striking features of stable
nuclei is the systematics of N/Z. We talk about a "line of stability",
the trajectory of stable nuclei which roughly run through the middle of
nearby stable nuclei. If you look at light nuclei, you see that this
line runs along N=Z. However, above A ≈40
the lines tends to curve toward N>Z and it gets more and more unbalanced
as A increases; uranium has about 60% more neutrons than protons.
Observing systematics
can be very helpful in understanding nature and providing a starting
point for formulating theories.

QUESTION:
How much centripetal force would a hydrogen atom exert if it were traveling at 99.999% the speed of light around the LHC?

ANSWER:
The hydrogen atom (actually, just the proton) has a centripetal force
exerted on it, not by it. Anyhow, my function is not to do arithmetic,
it is to do physics. You need simply to calculate F=mv ^{2} /R
where v is the speed (0.99999x3x10^{8} m/s), m
is the relativistic mass m _{0} / √(1-0.99999^{2} ),
R is the radius of the LHC in m,
and m _{0} is the proton rest mass in kg.

QUESTION:
How would you calculate the amount of force necessary to pull a suction cup from a wall?

ANSWER:
Assuming all you had to do is overcome the vacuum behind it, it would be
the area of the suction cup times atmospheric pressure.

QUESTION:
imagine an astronaut takes a bicycle wheel with her into deep space holding both sides of the axis she starts the wheel spinning then she lets it go and flies back to earth . assuming no one and nothing touched the wheel if you looked through a telescope at the bicycle wheel ten years from now would it still be spinning

ANSWER:
If it is really empty space, then once it started spinning it would
never stop.

QUESTION:
Is it possible to establish an electric field in a box, inside which there is a vacuum?

ANSWER:
Sure. An electric field can exist in a vacuum.

QUESTION:
Could you please explain what spin and magnetic moment are when referring to elementary particles?

ANSWER:
Spin normally refers to intrinsic angular momentum; that is the property
of a spinning mass which quantifies the speed of spinning and the mass
and shape of the spinning object. Magnetic moment is a measure of how
strong a dipole magnetic field the particle has; elemantary particles
usually look like tiny bar magnets and magnetic moment tells how strong
that magnet is.

QUESTION:
What would travel furthest? A whifle ball or a ping pong ball? Why?

ANSWER:
I presume you mean if they are launched with identical velocities. It is
a problem where air friction has a major impact. And, it is very
difficult to do and depends on how large the initial velocity is. To get
an idea of how difficult it can be, see an
earlier answer
where I compared tennis balls and badminton birdies.

QUESTION:
If you push a rock up a hill and raise its gravitational potential energy but not its kinetic energy why does this not violate the work energy theorem?

ANSWER:
"The work energy theorem" means different things to different people. If
it is that the work done equals the change in kinetic energy, then if
there is no friction, the work done is zero because you do positive work
and gravity does an equal amount of negative work. If it is the work
done equals change in total energy, then if there is no friction, the
positive work you do is used to increase the potential energy.

QUESTION:
Alright - hopefully you can help me clear up my issue with this concept. I made up this question for some students and I understand how the math works, but why it does is beyond me - hope you can help!
A man who jumps straight up with an initial velocity of 3.70 m/s (enough a person with a 2.36 m reach to jump and touch the rim on a 10 ft hoop: 3.05m high) will travel up the 3.05 m. Easy math, easy understanding. What is hard to understand is how if that man was to run horizontally with that same speed (3.70 m/s) how he would also be able to reach that same height? This is PE=KE stuff and also easy, just hard to believe. Do you have a way you could better explain this? It just seems even in a perfect world with no wasted energy that some of the energy would not be enough compared to just jumping straight up. Instead of all of the velocity being concentrated in one direction (up), its broken into 2 components going up and out. I know energy does not have vectors, but it still baffles me - thanks!

ANSWER:
The problem, I think, is that PE=KE simply does not get it except for
the simplest problems. The way to view it is (PE+KE)_{before} =(PE+KE)_{after} .
So, choosing the ground as PE=0, the first problem has PE_{before} =0
and PE_{after} =mgh =9.8mh ,
KE_{before} = Ѕm (3.7)^{2} =6.85m
and KE_{after} =0, so 6.85m +0=0+9.8mh or h =0.70
m; add to that your 2.36 m reach and you get 3.06 m. In your second
(more puzzling to you) problem, you still need to get to h =0.70
m, but you start with a horizontal component of your velocity v _{x} =3.7
m/s. But to go up, you must give yourself a vertical component of your
velocity of v _{y} . When you get to the top you do not
have zero kinetic energy, but the kinetic energy due to your horizontal
velocity (which never changes). So the way to set it up is
KE_{before} = Ѕm [(3.7)^{2} +v _{y} ^{2} ],
KE_{after} =Ѕm (3.7)^{2} , PE_{before} =0, PE_{after} =(9.8)(0.70)m .
If you put this all together, you will find that you need a (suprise!)
vertical component of velocity of v_{y} =3.7 m/s.

FOLLOWUP QUESTION:
Alright I left a crucial part out of my last question about the vertical jumping velocity and the horizontal running velocity. The velocity that the man jumps vertically with 3.7 m/s allows him to reach up to the 3.05 m rim. The second man runs horizontally - grabs a rope (pendulum) - and this should take him up to the same height given his same reach. The first problem you can do like you said with the potential and kinetic energies or constant acceleration formulas. The second I can do as well. I'm ok on the math, i just don't know how that could be? The first man is directing all of his energy in one direction while the second man is grabbing a rope which would take him both vertically AND horizontally. How can the same velocity get that man to travel the same vertical height but in the second case - a further distance?

ANSWER:
The method is still the same. But now he starts with a velocity of 3.7
m/s and ends (at the top of the swing) with zero velocity. So, applying
(PE+KE)_{before} =(PE+KE)_{after} , (0+ Ѕm (3.7)^{2} )=(9.8mh+ 0).
Note that this looks just like your first problem—it starts out with all
kinetic and ends up with all potential energy. The shape of the path is
irrelevant, i.e. , if it starts at the ground with some speed and ends up
somewhere else at rest, then it makes no difference how it got there.
Don't worry about the "further distance" because no energy is lost or
gained by the horizontal motion.

QUESTION:
if a convex lens having focal length "f" is cut into four equal parts , then what is the focal length of one part? whats the case with a concave lens and also with concave and convex mirrors?

ANSWER:
For thin spherical lenses, the focal length is determined by the radius
of the sphere(s) and the index of refraction of the lens. Every piece of
the lens has these same properties. Mirrors are the same but, of course,
have no index of refraction to worry about.

QUESTION:
Which factor remains constant during refraction? If it is frequency then how can the wavelength change but frequency remain the same?

ANSWER:
It is frequency which remains constant. Frequency f , wavelength
λ , and velocity v of a wave are related by fλ=v and
so, if v decreases, λ decreases.

QUESTION:
How is the speed/trajectory of the ISS calculated/adjusted? For example...as NASA and the other international partners add parts/modules to the station, it's mass increases; at what mass would the station need to be sped up or elevated in altitude/orbit to negate the added mass an avoid a decaying orbit?

ANSWER:
I am sure that the space station has thrusters used to perform minor
maneuvers and attitude adjustments. However the problem you anticpate,
the orbit changing due to changes in mass, is not an issue. The orbit,
its radius, shape, period, etc ., are independent of mass of the
satellite; a pea next to the ISS will orbit right along with it.

QUESTION:
It has been many years since I took math and physics. Can you please show me the total solution of this formula. I would have to go back to my college physics class to solve:
RCF = 1.12r [(RPM/1000)squared] where r is 83mm and RCF is 2500g.
I no longer know how to handle the [(RPM/1000)squared.] I know there are calculators to do this but I want to see the soluton.

ANSWER:
I have never seen this particular equation before (not surprising, since
the units are unphysics-like) but, after a little searching I found that
RCF must be relative centrifugal force . So, this must be
something centrifuge guys use. As I have often noted, centrifugal force
is a fictitious force, no such thing. It is a force invented to make
Newton's laws work for an accelerating system like a centrifuge; if you
are in a centrifuge, you feel like there is a force pushing you outward
but there isn't really. But, for this system it is pretty easy because
the centrifugal force is equal and opposite the real force, the
centripetal force , which is holding you in circular motion. The
magnitude of the centripetal force (and therefore of the centrifugal
force) is given by a very well-known equation, F =mr ω ^{2}
where m is the mass in kg, r the radius of the circle
in meters, and
ω
is the angular velocity in s^{-1} . I believe that RCF is F /mg ,
the times greater than the weight of the object that the centripetal
force is, a dimensionless quantity, where g =9.8 m/s^{2}
is the acceleration due to gravity. So, we have RCF=(r ω ^{2} /g )
but only if all the units are expressed as given above. So, it is now
just the tedium of converting: r in mm to r in m requires
us to multiply by 10^{-3} and
ω expressed in rev/min/1000 requires that we multiply by
[(1 min/60 s)x(2π radians/1 rev)x1000]^{2}
and so, RCF=(r ω ^{2} /9.8)x10^{-3} x[(1
min/60 s)x(2π radians/1 rev)x1000]^{2} =1.119
r ω ^{2
} where r is expressed in mm and
ω is expressed in rev/(1000 min).
Since this works out so beautifully, I am confused by your factor of
g on your value of RCF; RCF should be a dimensionless quantity and
so I might assume that 2500g=2500x9.8=24,500. Or maybe the g is
stuck on to indicate that the force would be 2500 times greater than the
weight on earth; that would be more likely because that is what RCF is.
So, let's take the second assumption, that RCF=2500=1.119x83x ω ^{2} = 92.9and
so
ω=√( 2500/92.9)=5.19
rev/1000 min=519 RPM. Rereading your question, I think you were just
interested in the last couple of sentences, but I had to be sure I knew
where this came from before I tried to get a number from it.

QUESTION:
I'd like to ask, thinking molecularly, what happens when a ''bubble'' of air is in the water. I'm not so curious about the journey of the air bubble (it'll quickly travel upwards) but I'm wondering what's happening exactly in moleular terms to the air bubble. Specfically, is the ''air bubble'' a group of many molecules moving at high speed and being pushed by the slower moving water molecules (liquid water)?

ANSWER:
In any gas the average speed of the molecules is determined by the
temperature. The water has nothing to do with speed of the gas molecules
except as a determiner of temperature. The volume of the bubble will get
larger as it rises because the pressure in the water decreases. But, the
process is probably what we call adiabatic, that is, no significant
amount of heat flows in or out of the bubble; the result is that as the
bubble rises its temperature decreases and so the average speed of the
molecules decreases.

QUESTION:
If I aim a laser pointer at a 45 degree angle through a cup of water, the beam is refracted downward as predicted by Snell's law. I understand the mathematics of what happens. The reason the laser light is supposedly refracted when it hits the water is that the light travels more slowly in the water than air. Can you explain why the slowing down of the light beam causes it to refract.

ANSWER:
Think about the light as a series of plane wave fronts approaching the
surface of the water. As each wave front hits the surface, part of it
enters at a s slower speed so the wavefronts get closer together and
gets bent as shown in the accompanying figure.

QUESTION:
is that true that Albert Einstein didn't believe in black holes? why?

ANSWER:
Einstein died in 1955. At that time the astronomical landscape was much
more meager than today. Since the astrophysics of supernovae, neutron
stars, etc. had not been developed yet. So, I guess black holes just
sounded a little far out.

QUESTION:
this may be a silly but i was wondering, if one second has infinite time intervals how does time move through that second?

ANSWER:
This sort of sounds like Zeno's paradox: an arrow must first go half the
distance, then half the remaining distance, etc ., so it never
gets there. The answer is that an infinite sum may have a finite value.
If you think about the second being divided into an infinite number of
parts of equal duration, each interval is infinetesmal (essentially
zero) and the infinite sum of zeros adds up to one second.

QUESTION:
Does a magnetic force change a system's momentum? So far all I've found is that it doesn't change the magnitude but it changes direction and I don't really get why.

ANSWER:
The magnetic force on a moving charge is always perpendicular to the
velocity vector. Therefore, the direction but not the magnitude of the
momentum changes.

QUESTION:
How can I calculate the resitivity of seawater?
I have a tank of seawater that I am applying a current to via dipoles. This is to stimulate fish to feed by creating an electric field that resembles prey. To calculate their sensitivity to the fields I need resistivity to be input into an equation. This is what I can't figure out.

ANSWER:
I found a number for the conductivity of seawater which is about
σ = 5
Ω^{-1} /m. The resistivity is the reciprocal of this, ρ=σ ^{-1} .
So, the resistivity is about ρ= 0.2 Ωm. Keep in mind that this
depends on both temperature and salinity, but should be close enough for
your application. You can find a table of conductivity for various
salinities and termperatures at
http://www.kayelaby.npl.co.uk/general_physics/2_7/2_7_9.html (1 S= 1
Ω^{-1} ).

QUESTION:
If a 4000 pound object moving at 40 miles per hour hits a stationary object weighing 100 pounds, how far will the stationary object move?

ANSWER:
There is no way to determine this from the information given. The most important missing piece of information is how much energy is lost in the collision? Anything from perfectly inelastic (the two stick together) to elastic (they bounce off like billiard balls). Also, does the stationary object become airborne? And, if so, what angle does it depart at?
But, I can do some rough calculations imagining this is a car hitting a
person.

The car is sufficiently
heavier than the person so that we can approximate its mass as infinite.
In this case, the range of possible recoil velocities of the person is
40 mph (perfectly inelastic) to 80 mph (perfectly elastic). I would
guess that the collision would be more like perfectly inelastic, so
let's suppose the recoil is at 50 mph.

Suppose the recoil angle is 40^{0} .
The equation for the range of a projectile is R=v ^{2} sin(2

θ )/g
where v is the speed (in m/s) and g is acceleration
due to gravity 9.8 m/s^{2} ; I have neglected air friction. This
gives about 50 m for the distance.

You can play around with various possible parameters using a
handy calculator on the web.

QUESTION:
does a body weigh less in air or vaccum when measured by a spring balance

ANSWER:
Less in air because of the upward buoyant force (the same force which
lifts a helium balloon).

QUESTION:
Me and my father were arguing over a physics related question and it's been bugging me to get this question answered! Anyways the question was: If you were to drop a ball from a tall building but with no air resistance what would the acceleration be?

ANSWER:
The acceleration due to gravity is 32 ft/s/s. This means that as each
second ticks by, the speed increases by 32 ft/s. Expressed in mph, as
each second ticks by, the speed increases by 21.8 mph. For example, 5
seconds after you drop it it has a speed of 109 mph.

QUESTION:
Assume that it would be possible to create an environment with no
gravitational forces and frictionless. Two objects traveling at the same
velocity (say 100knots in the same direction). However, they are
seperated by about 1000 yards on parallel paths. If it be possible to
fire a gun with the absence of oxygen in space, and you shot a gun from
the object on the right to the object on the left at an exact 90 degree
angle from the trajectory of that object, would the bullet still tavel
at that exact 90 degree angle and miss the other object? Or, would the
bullet still hold that same forward speed and stay with the two craft
and hit its target?

ANSWER:
As seen by you and the other object, the bullet would go directly from
you to the other object (which appears at rest relative to you). As seen
by someone you are passing, the bullet would have a velocity which is
larger than you see because your forward velocity is imparted to the
bullet; he would still see the bullet hit the other object.

QUESTION:
I have an experiment that I am doing, but I do not know the formulas for what I need to find out. Hopefully you will help me out. I am dealing with a vertically oriented circle. I can get the radius of the circle and the amount of weight that will be at the end of that radius. Starting at the top of the circle at a 0 speed, I need to know how to figure how fast the weight will drop to the bottom of the circle, how may pounds of force will be generated at the bottom of the circle, how many pounds of momentum will be generated and how much force is needed to get the weight back to the top of the circle. Any help will be greatly appreciated.

ANSWER:
Well, your units are not what physicists like to work in (we don't like
English units). A fundamental is that a weight (W ) is a force
exerted on the object and it is related to the mass (m ) by the
equation W=mg .where g= 32 ft/s^{2} is the
acceleration due to gravity; so, for example, a 32 lb weight would have
a mass of 1 lb-s^{2} /ft (see why we don't like English units?).
The easiest way to find the speed at the bottom is from energy
conservation —the potential energy
at the top WD equals the kinetic energy at the bottom Ѕmv ^{2} =ЅWv ^{2} /g
where D is the diameter of the circle and v is the speed
at the bottom. Solving, you see that the weight does not matter, v =√(2gD )=8√D ;
the answer will be in ft/s if D is in ft. The question " how
may pounds of force will be generated at the bottom of the circle" has
no meaning in physics. The mass at the bottom of the circle has two
forces on it, its own weight W down and an upward force from the
circle mv ^{2} /(D /2)=(2W /g )v ^{2} /D .
Putting in what v is and doing all the arithemetic, I find the
force the circle exerts up to be 4W . Therefore the net force on
the moving mass is 3W up and the force on the ring itself is 4W
down. Momentum is not measured in pounds because it is mv . The
momentum at the bottom is therefore Wv /g = јW √D
(note, though, that the units are lb-seconds). It takes no force for the
mass to go back to the top, it will do it on its own (again, assuming no
friction). Be sure to note that everything here is right only if all
times are seconds, all forces are pounds, and all lengths are feet.

QUESTION:
My question is, the hadron collider is it causing all the bird and fish deaths that have been happening? Also could it be causing all the floods and earth quakes that have been getting worse in the past few years. I do not really understand the hadron collider at all. But it seams not to far fetched that is could in fact be throwing off the magnetic field that the birds need and doing something to the oxygen levels in our water . But again that is why I am asking you.

ANSWER:
The magnets are designed so that the intense magnetic field is where the
hadrons are and if you go maybe 100 meters away from the magnet its
field would be much much smaller than the earth's magnetic field.
Believe me, birds dying in Arkansas has nothing to do with the collider
in Switzerland. If you need more evidence that the two are not
connected, the LHC was turned off on December 6 and the blackbird die
off was January 6.

QUESTION:
how do i figure out the energy required to move a 166 kg mass vertically 28 meters. If I can convert this to nutritional calories also instead of newtons or joules. This would really help me out.

ANSWER:
The work done is weight times height, mgh =(166 kg)(9.8 m/s^{2} )(28
m)=45,550 J=45.6 kJ. Now, the food calorie is 4.2 kJ, so about 11
calories are burned doing this work.

QUESTION:
I am trying to do a micro-Compressed Air Energy Storage Project and and trying to see if a set of air tanks can power a pneumatic motor for 8 hours.
The tanks I will be using are 5000psi and total 405 fluid liters.
The motor will operate at 90psi and use 160-212cfm. (depending on which model I pick)
How do I calculate the rate of discharge over a period of time? I can't find a formula.

ANSWER:
I am not sure I get the picture. First, the pressures you quote are
probably gauge pressures, that is, pressure above atmospheric pressure
which is 14.7 psi. Hence the two pressures are 5014.7 psi and 104.7 psi.
I presume cfm is cubic feet per minute, so let's convert the tank volume
to cubic feet: 405 liters=14.3 ft^{3 } . Finally, assuming that
the tank and motor are at the same temperature, we can apply Boyle's law
which states that the product of the pressure and the volume are the
same, P_{tank} V_{tank} =P_{motor} V_{motor} .
Doing the arithmetic, the volume of air available to the motor at its
pressure is V_{motor} =5014.7x14.3/104.7=685 ft^{3} . So, if I
understand the problem correctly, the smaller motor could run
685/160=4.3 minutes and the larger one 685/212=3.2 minutes. That is a
long way from 8 hours!

QUESTION:
I read that the empty space within a coffee cup contains enough zero point energy to boil all the earths oceans. What is zero point energy and how would one harness it if the technology were available.

ANSWER:
Zero-point energy refers to the minimum energy a quantum mechanical
system may have, not the energy in "empty space". You are perhaps
thinking of vacuum polarization. But, back to zero point energy. A
simple system is a particle in a box; the particle may not be at rest
(have zero kinetic energy) because that would violate the uncertainty
principle which states that the position and momentum (essentially
speed) may not be known with arbitrary accuracy. So, the reason you cannot
tap this energy of the particle is because of the laws of nature; it has
nothing to do with technology. However, there is a technology which can
tap zero-point energy —it is called
chemistry. For example, an O_{2} molecule has a certain
zero-point energy and two H_{2} molecules have a certain value
for the sum of their zero-point energies; when combined, we get two
water molecules, O_{2} +2H_{2} —>2H_{2} O+energy.
You may think of the energy as coming from the lower zero-point energy
of the water molecules.

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