swansont

Gamma. It's the term that tells you the amount of time dilation or length contraction [math]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/math] I didn't see a V, just v for speed

"Degree" also means "amount." Confusing in this context, to be sure.

Not when you get to the point that you study General Relativity. Sitting still in a gravity field is not an inertial frame — freefall is.

Make sense to whom? Nature is under no obligation to be understandable to us. The proton is about 2.8 nuclear magnetons, and the electron is about one Bohr magneton Where did you get that equation?

Not at your beck and call, obviously. Forces get tricky in relativity http://en.wikipedia.org/wiki/Special_relativity#Force If you look at the fourth equation in that section, you'll see that Force and acceleration no longer have a linear dependence.

Depends on how closely you look, and how you define "contact." Sure. If you are moving in a circle, the centripetal acceleration equation must be satisfied.

Charges will move because of an external electric field. The field from that rearrangement is the induced field. Magnetic fields arise from the motion of charges, or from looking at charges in a moving reference frame. Electric and magnetic fields are related through Maxwell's equations.

The quantum of magnetic moment is not derived from classical theory, though. The Bohr magneton is dependent on Planck's constant, which is the quantum of angular momentum.

There were lots of blown opportunities by the Cards. All those penalties, a dropped pick, the inability of their team to tackle Harrison before he was done running 100 yards.

You presume too much. If an object is moving in a circle, there must be a net force toward the center, equal to mv^2/r, thus the normal force can't cancel/balance mg

Quantum means "discrete," as in "not a continuum. There's nothing inherently discrete in either of those equations — any value of Q or M will do.

You have given no references that support this. They have all been for parallel wires, which is not the same thing. Q=0 for wires. I've seen it worked out but can't find the link at the moment. But E^2-B^2 is a Lorentz invariant.

It's by definition and convention. Attractive forces result in negative potential energies when you consider the energy at arbitrarily large distances to be zero. F = -grad(U)

Sione has been autosuspended for has rapid collection of infractions, including trolling, thread hijacking and ad hominem attacks.

A mirror?

Parallel wires and parallel beams are different, which is why I asked the question. There is no electrostatic force in the wires in their rest frame, which is not true of a beam. IIRC, you have to account for the length contraction of the beams, which increases their charge density and the electrostatic repulsion, and cancels the increased magnetic attraction. This effect is also present when you look at the wires. There is no preferred frame.

The work done by the engine is whatever KE you have at the bottom in excess of mgh, which will also be the force exerted by the motorcycle multiplied by the distance.

The T and V you give are not independent — if you know one, you can find the other. They tell you the energy of the ball as it is tossed (assuming you know the mass, as Mokele has pointed out), but without knowing how long it took, you can't calculate the power.

So what exactly is the problem?

Show mathematically that this is true. I have Q1 at one point and Q2 at another. What is the field mid-way between them?

Yes. There is no work being done on the ball after it leaves your hand, which is why the sum of KE + PE, which is the work you originally did, remains constant.

I don't see the relativistic correction.

We assume the world to obey Newton's laws of motion, i.e. we assume we are in an inertial frame. When we aren't, we assume a force that isn't really there, in order to account for the motion.

If B-B was deemed to be not physically meaningful by the test writer, then it would be 1 & 2.

I think that the mainstream view is more like this: Q1. Is there an effect? A. Maybe. There's excess heat claimed, but it's not always clear if the experiment was rigorous enough. Q2. Is the effect replicable? A. Sort of. Sometimes it's replicated and sometimes not. Conclusion: Can't draw one until they get their act together. Q3. Are there by products? A. Maybe. The levels are around background, so it may just be sloppy experiments. Q4. Are the by products those of a chemical or nuclear nature? A. Could be either. Conclusion: Can't draw one until they get their act together. Now the biggy. Q5. Does current theory allow for the observed effect? A. Depends on of there is really an observed effect. Q6. Does current theory tell us where to look for the discrepancies? A. Yes. Conclusion: Can't draw one until they get their act together.