swansont

Feel free to do the calculations necessary to back this up.

I think you'll find that there is, somewhere, and angle between the magnetic field and the current flow. The field of the magnet has to bend around on itself. Or, to look at it another way, the current flow creates a magnetic field that can interact with the field of the permanent magnet. ... pause to Google ... This is similar to Faraday's first demonstration of motor action. more here or click here http://sci-toys.com/scitoys/scitoys/electro/homopolar/homopolar.html and scroll down to "homopolar motor" another thing to Google would be "Maxwell's motor"

i_a already addressed this. The atmospheric effects you propose do not account for what is observed.

Work is the energy transfer by a force acting through a displacement. No force means no work. (No displacement means no work as well) One body may travel to another place at constant velocity, which means no work is done in doing so. There may be forces that cancel, meaning there is work done by each of these forces, but the sum will be zero, as in the example Bignose gave.

Klaynos asked you to cite evidence, not wikipedia.

No, the twins won't be the same age, The twin that underwent an acceleration changed from one inertial frame to the other. That breaks the symmetry. Since acceleration is not relative you can tell, between the two, whose clock is correct. This is explained in some detail at many places on the 'net.

No work is done. [math]W = \Delta KE =\int F.ds[/math]

I would probably not say that without knowing what a "hammer throw ball" is. How you describe motion depends on what coordinate system you choose. You would probably not look at the moon rotation of you were looking at the earth-moon system, since talking about the orbit and phase-lock adequately describes that, and already includes the behavior of the moon. But when you ask the question of whether the moon rotates on its own axis, you are looking solely at the moon, and the answer to that is "yes."

Gravity propagates at the speed of light. The relevant term in GR is the energy density, not the mass, so annihilation won't change the gravity anyway, AFAIK. Thanks for playing. And what the heck is spectral relativity?

The container must be able to be compressed without permanently deforming, so it can return to the original shape. Like a plastic soda bottle.

There's another effect you can use, which makes an "artesian diver." If the object that is to sink/float has an air bubble in it, its bouyancy will depend on the surrounding pressure, which affects the size of the bubble. http://www.newton.dep.anl.gov/askasci/gen99/gen99599.htm

It required the recognition that light was an electromagnetic wave, and then that it applied to mechanical systems. Einstein's paper was "On the Electrodynamics of Moving Bodies"

... which is known as Archimedes principle, in case you want to investigate further. The bouyancy force is equal to the weight of the displaced fluid.

No, you just need to apply F = ma properly.

Please, if you expect people to respond, have the courtesy to not use text-message shorthand. ———— I'm afraid that experience demonstrates that this is an inherently confusing topic. The earth is not an inertial frame of reference. There are conditions under which this has a material effect on results, meaning you have to pay attention to these effects and how big they are. You can choose a frame and calibrate to that frame, but this is by convenience. There is nothing in the physics that makes one frame preferred over another. Again, you have chosen a frame. The physics will still work in any other frame. There is no test that will prove that any one inertial frame is actually at rest.

No force means no work. But as m4rc has discussed, this does not hold during a collision, since there will be a force. Klaynos has mentioned the proper equation, and this ties in to Newton's third law; during the collision, each body acts on the other with equal magnitude forces, but opposite in direction. Momentum will be conserved, and this is the relevant quantity to calculate. Under some circumstances, kinetic energy will be conserved as well, but this does not hold universally.

Astroman has not posted again, and this looks like a homework problem, so I am loath to work it for him. But there are two tensions to consider, and if the mass is accelerating, they will be unequal. But I don't think there's enough information given to actually solve the problem.

By definition, the moon rotates on its axis, once per revolution. The car in your example is, too. Just because one might not realize this does not make it untrue.

The statement that "space is curved" is a shorthand for saying that the coordinate system best used to describe it is non-Euclidian. e.g. the shortest distance between two points is no longer a straight line in Cartesian coordinates. It's not a comment about whether space is infinite in extent or not — that's a separate issue.

If you assume this, I think you end up with a physically impossible situation, i.e. there will be some point where you can't make the forces cancel, assuming a nonelastic rope.

The force P is not the only vertical force. But I agree that the system is not in equilibrium.

Yes, the net force is zero. You can have forces present, but they must sum (as vectors) to zero.

Nature is under no mandate to behave as you want it to. But in the frame of the clock, the distance is unchanged, i.e. the length of time measured by the clock is dependent on your frame of reference. (And it doesn't matter what kind of clock it is). So there is no calibration you can do to make multiple observers (in different frames) agree on what the clock is reading.

Each object feels a force GMm/r^2 From F = ma, they will respectively have accelerations of GM/r^2 (for mass m) and Gm/r^2 (for mass M). From there on it's just a little calculus, integrating and applying the boundary conditions (once for velocity and twice for position). If they are point masses they should meet up at their center of mass, e.g. if m = M then they will meet at r0/2

The fact that both the large (100x) and small masses will accelerate does not support the conclusion that G isn't constant. It just means you have to apply the physics correctly.