Recent content by haruspex

If you consider a short element of a very thin ring, tension is the force each such element exerts on its neighbours. The two ends of the element are not quite parallel, so the net force on the element from those tensions is radially inwards. Thus, although the tension is everywhere...

No, ##v_B\sin(\beta)=\omega r_B##, etc. No, it is the rate of increase of the rate of increase of the radius. In a circular orbit it is zero. The component of the acceleration vector in the radial direction is ##\ddot r-r\omega^2##. See e.g...

Coulombs/sec?

No, because the conductive capacity of ##R_3## is shared in some way between those two loops. Given the symmetry, it is shared equally, so in each loop it provides resistance ##2R_3##.

If you mean Kepler III, the radius in that law refers to the semi-major axis. It is a law relating all planets orbiting the same star, not all positions of a given planet. It does not apply to the first part of this question.

Sorry, I should have been clearer. A radial component other than centripetal acceleration, i.e. ## \ddot r##

I don't think you can use that formula in an elliptical orbit. There can be a radial component to the acceleration. What do you know is conserved?

Let me try to be clearer. I am saying that since the velocity only bisects that angle when the range is maximised, a proof that it is bisected must make use of the condition that the range is maximised. The text you posted does not do that, so the 'proof' is incomplete. As to how one proves it...

Clearly that is not true in general (consider firing it straight upwards). It is true if the range is maximised, but I see no attempt to use that it is maximised. So I suspect whoever wrote the proof copied and abbreviated it without understanding it, or is relying on something established...

I cannot see what that has to do with my post #44. The question I raised there was merely one of terminology: having resolved a vector into components in the ##\hat x## and ##\hat y## directions, as ##\vec v=x\hat x+y\hat y## say, what are the "components"? Are they ##x## and ##y## or ##x\hat...

Neat, but it would still have led to an answer that the place to jump in is to the left of A. I would guess @brotherbobby would have been just as puzzled. Either way, the resolution of the puzzlement is that the time of travel is ##\small{t=\dfrac{|x|}{v_2}+\dfrac{\sqrt{d^2+(s-x)^2}}{v_1}}##.

The thing to check is whether your first equation represents reality in all cases. For x<0 it doesn't: it gives a negative time for running on the shore. Consequently, if there is no local minimum with x>0 you get a crazy answer.

##x_1\hat e_1, x_2\hat e_2## are vectors, no allegation required. But please clarify the concern you have.

I repeat, my point is that the usage is not as standard as claimed in post #39. Perfectly respectable sources express conflicting views.

We all, naturally, teach as we were taught (or as we understood at the time, which is not always the same). But as the links I posted illustrate, we were not all taught the same, it seems. This one is interesting: https://www.cuemath.com/geometry/components-of-a-vector/ "The values ##a, b, c...