Recent content by Hollumber

Yes, I don't think the bead would move, but that would be a very ideal case. Realistically, it would be shifted one way or another, and then its height determined by the angular velocity of the ring. Think of a ball at the top of a pyramid. It's in "dynamic equilibrium," where it will...

Yes, you are definitely on the right track. Right now, it looks like you have calculated the circuit current (current through the battery). If this was a series circuit, then everything would also have the same current, but because there are some parallel components, you must calculate how much...

I don't think you have to worry about torque here. Just find the normal force upward as it depends on how high the step is and set it equal to the weight.

If you want the "potential energy for one whole cycle", then it means you want, essentially, a value in units of Joule-seconds. Then divide this by T to get the average. To get units of J*s from an equation giving J, an integral over time can be performed. (Hint: Should the integral go from 0...

I can't think of one either.

It gives you the ratio between their effectiveness per molecule. How to change this ratio considering the effectiveness per mass instead of molecule?

if cos(x-pi/2) = sin(x), what does cos(x+pi/2) equal?

Mm, I wouldn't trust that. Must be some sort of misprint or mistake...unless my 6th grade teacher was wrong all this time :o

Yes Heisen, your thinking is correct. Final KE is (integrated V + initial V)^2 *m/2, since the integrated v only accounts for speed gained due to force and doesn't consider initial v (so we add it in afterwards).

After researching this so much, I finally realized where I was going wrong. (NP/NS)^2 = RP/RS is used to calculate the turn ratio of the transformer to get max power to the load resistor (RS). If this ratio doesn't hold true, then Req = RS(NP/NS)^2, where Req is the equivalent resistance that...

I think that's around the value I got as well (don't remember, but it's reasonable). Did you use (-)1.3 for delta y? If not, then your velocity is probably a little low.

Yes, adding the velocity due to the force to initial velocity gives final velocity. Yes, this can be converted into KE. Work is equal to the difference of KE's.

Hmm, I think you can assume all the energy from the spring transfers into the object such that the spring stops decompressing at its equilibrium length. So delta s is just the distance it is compressed, and delta y is how high the block is from where it started.

You should show the steps you took to arrive at your answer.

I'm not saying what you did is wrong, but to make sure you are consistent with your signs. What values did you use for delta s and delta y?