Recent content by Ertosthnes

Hey JyN - I just finished my bachelor's in math and found that my experience was very similar to yours for many of my classes. The professor would stand in front of the board and regurgitate the material in the book for the period of the class, everyone would take notes and then class would let...

Start with something that makes sense, like the fact that you have more momentum with a smaller wavelength. So basically, p=h/\lambda, where h is some constant. Since \hbar=h/2\pi, p=h/\lambda=2\pi\hbar/\lambda=\hbar k, where k=2\pi/\lambda, ie, k is just a vector in the direction of...

Can you elaborate?

100 on both, I found that test very easy. On the other hand, I pretty much never multitask.

Okay, in inertial reference frames, two particles with the same charge will always repel each other. Even if they were moving in parallel at high speeds, and thus producing magnetic fields, special relativity would come in and balance the forces from the electric and magnetic fields so that...

I was reading and came across this statement: If t > 2n^2 is an integer, then t! > (n^2)^(t-n^2) I'm not sure why it is true. I don't know what equations are relevant. My feeling is that you don't need anything more than algebra, but perhaps it would also follow from the gamma function...

Well in this case, I don't know how to interpret the answer. I only used Mathematica in this case because I had a suspicion that I didn't know how to solve that type of differential equation, but I was curious as to what the answer might look like. If the problem had involved an equation that...

Oh, I see. I guess the acceleration is non-constant because the force of the weight does not remain perpendicular to the bar? I don't think I know how to solve differential equations like dd(theta)-(rmg/I)cos(theta) = 0. I entered it into mathematica and got this...

If that's all right, then to determine the difference in speeds after a given time, I think that using the equation: \theta = \omega t + (1/2) \alpha t^2 would do the trick, after substitution into w2 - w1 = {rm \sqrt{2gh}sin \theta}/I . Since the initial w is zero, and \alpha =...

Zombie Feynman says, "Ideas are tested by experiment." However, I think it probably depends on how hot the suit is. Unless the suit was extremely hot, it would probably not have much impact on the height that you are able to dive into water without incurring injury. This is because water has...

Sorry about the notation. Okay, for the first case with a weight hanging from the bar: \DeltaL = Rx\DeltaP Iw1 = rmvsin\theta =>w1 = rmvsin\theta/I For the second case with the weight that fell from a height h: \DeltaL = Rx\DeltaP \DeltaL = rm(v+\sqrt{2gh}-0)sin\theta Iw2 =...

It's F = (mv^2)/r, directed radially inward.

Here are the formulas you will need to solve the problem: v = rw w = 2pi/T By the way, the "velocity of a centripetal force" does not have much meaning. Force is a vector; it doesn't have velocity. You might as well have asked what the mass of the force was, or the radius of the force. I...

The phase velocity isn't important. If you have angular velocity (w) then you can always find T using T = 2pi/w.

Regarding my previous post, I'm thinking that the answer to my last question is that w2 = w1 + (r*mv)/I, where v = sqrt(2gh). That would mean that w(f) = (w1 + r*mv/I) + alpha*t, which would answer the question for a given period of time. The given rotation part seems messier. Basically...