Recent content by PhysicsThrow

VCC = 9V VEE = -9V VO = 6v rms VI = 2v rms RL = 900Ω Amplifier draws 8mA from each battery. Input sinusoidal current of 1v rms. So the terms relevant to the amplifiers power draw are VCC, VEE and the 8mA current from each battery. I assume I just calculate using P = VI = (9*2)*I...

Homework Statement Calculate the power drawn from the batteries by the amplifier. The batteries are +9v and -9v. The amplifier draws a current of 8mA from each of its power supplies. Homework Equations P = VI The Attempt at a Solution I'm just a little stumped as to what to do for the...

(You're correct about the typo, my bad) Ahh okay, so the first E= equation is the total energy using the lowest point variable H and the second E= equation is the total energy using the highest point variable h. So then it says we equate them by saying k(\sqrt{H^{2}+l^{2}}-l_{0})^{2}-mgH =...

Thank you, thank you, thank you! That makes a lot of sense, thanks. :) Now just to wrap my head around the equating and I'm done! :D

[SIZE="2"]Bump for the previous question above this post? Should I start a new thread as I've now pushed past the original issue? This help then goes on to say that, E=k(\sqrt{H^{2}+l^{2}}-l_{0})^{2}-mgH, H being the distance the particle is lowered from the tops of the poles. and...

Oh gosh now I'm confused. So, the \frac{1}{2}kx^{2} doesn't need the \frac{1}{2} because it's symmetrical, but why does the \frac{1}{2}mv^{2} keep the \frac{1}{2}?

Hmm very interesting. Never encountered that before. EDIT: How come the kinetic energy has a \frac{1}{2}? :o

Yeah sorry, I just worked that out. Now all there is, is... It says how E=\frac{1}{2}mv^2+k(\sqrt{H^{2}+l^{2}}-l_{0})^{2}-mgH which I understand, EXCEPT, they've put k(\sqrt{H^{2}+l^{2}}-l_{0})^{2} instead of \frac{1}{2}k(\sqrt{H^{2}+l^{2}}-l_{0})^{2}. What's the reason for this? Is it because...

EDIT: I worked out the Pythagorean theorem issue before this.

Hmm, still a bit stuck, but I'll give it another hour or so.

Just realized that rearranging the E_{MAX} formula, I believe I get h=\frac{E_{MAX}-\frac{1}{2}kx^{2}-\frac{1}{2}mv^{2}}{mg} which means I'll need to know E_{MAX} first. :S Grr. Yeah good point. I think I'll state in my assumptions that from looking at various clips of this ride in action...

Oh yes of course. Because it's at it's peak, v=0. Okay so x is the extension in length of the rope, I guess I just make a realistic idea up for that? Let's say 10m?

Hmm, Maybe something along the lines of, E_{MAX}=\frac{1}{2}mv^{2}+mgh+\frac{1}{2}kx^{2} [SIZE="2"](E_{MAX}=KE+GPE+EPE) I can then rearrange that to find h. The point at the bottom where KE=0 is easier because v=0, but then that means the height is just 0 as you could guess anyway? But then...

Okay, so I can calculate k with k=\frac{T}{x_{0}} where {x_{0}} is some change in length that I decide. I'm not sure how this would help me find the maximum height and speed though as I've already technically worked out T from T=mg

So, when they're pulled all the way down and clipped to the floor this would occur? EDIT: Actually, no? If they were in equilibrium, F = T meaning mg = T?