Yes, they are the geodesics, but then you can find the arc length of any arbitrary curve by using the argument above.
You're right about it only working on Lie groups, of course.
Why not? The coordinate charts don't have to be isometric to R^n with the standard inner product.
It seems like you can generalize the argument given by the OP for standard Euclidean space. Obviously you need a metric (but Euclidean space has one too). Then cover your curve in finitely many...
But how do you define a "curve" that doesn't have a parametrization? For instance, if you say a "curve" is a connected compact 1-dimensional submanifold with boundary, it's easy to show by a gluing argument that this "curve" is just a piecewise regular curve in the ordinary sense, and you can...
Wan et al's "Value Sets over Finite Fields" give this result as their Corollary 2.4; their paper may be worth a look.
If there's a simple counting argument, I don't see it.
That's not quite true. The impulse is independent of the duration, the average force is not. To get the average force you can divide the impulse by the duration.
Hint: Show that for \deg f \neq 1, q, \frac{q}{\deg f} has a remainder. Then do \deg f = 1, q as special cases.
EDIT: Also note that the inequality, as stated, isn't true: consider f = x^{q+1}-x, which has only one value, 0, yet 1 \not\geq 1+\frac{q-1}{q+1}. You need a floor around your fraction.
Yeah, that would give you the impulse - the force integrated over the entire collision. To compute the force at any given time you need the duration of collision and a lot more information - colliding objects tend to compress, then expand, so the force isn't constant.
Permanent magnets are advertised as having some scalar strength, say, 1.5 Teslas, depending on the composition of the magnet but not its volume. I'm confused about what this means. Shouldn't the magnitude of the magnetic field vary depending on where you measure it? And if I epoxy together two...