Recent content by pietastesgood

Aha! I understand now, thanks!

Homework Statement Part f. Calculated current = 0.5 A. Voltage = 30 V. Resistance = 40 Ω Homework Equations P=IV P=I^2*R P=V^2/R The Attempt at a Solution I've been curious about this for a while. Sometimes when I am asked to find the power dissipated through a resistor, and...

Oh, right. cos(wt)=cos(t/√(LC)), so w=1/√(LC) There we go!

Well, taking the second derivative of q0*cos[t/√(LC)] yields -(q0/(LC))*cos(t/√(LC)) Setting that equal to -q/LC, simplifying, I get q0cos(t/√(LC))=q again. Pretty sure I didn't reach the right conclusion there.

Sorry, but I'm not sure what else I can do with plugging in q = q0*cos[t/√(LC)] into the second order differential equation, other than replacing -Q/LC with -q0*cos[t/√(LC)]/LC.

d^2q/dt^2 = -q0*cos[t/√(LC)]/LC I'm only in Calculus II, so I don't really follow how to solve the second order differential equation.

I got some of it after looking up a video on Youtube. ε=Q/C for a capacitor Q/C - L(dI/dt) = 0 Q/C - L(-d^{2}q/dt^{2}) = 0 d^{2}q/dt^{2} = -Q/LC However, I don't quite follow how the angular frequency ω=1/\sqrt{LC} from that second ODE.

Aha! That makes sense. Thank you! Now, just the oscillator problem.

Oops, I'll make sure to spread the problems apart more next time. I understand number 18 now. E=-dV/dr, so the strength of the electric field is the change in voltage/change in distance. So since the equipotential lines for 20 and 30 V are .04 m apart, it would be a change of 10 V/a change of...

Great, thank you for the clarifications! Also, I'm not sure if I'm allowed to link another help post, but I was having a bit of trouble on the 2008 AP test here https://www.physicsforums.com/showthread.php?p=4378951 If you could take a look, that would be awesome!

Homework Statement Answer is B. Answer is E. Answer is E. Homework Equations E=F/q E=V/d Right hand rule for inductors The Attempt at a Solution Quite confused for these problems. For number 18, I'm quite baffled as to where the 0.04 meters even comes from. What I had...

Okay, that makes sense! Thank you! I should probably brush up on angular momentum, haha!

Wait, for what kind of orbits is v^2=Gm/r valid? Also, I tried the energy approach but it doesn't seem to follow what you said (I probably did it wrong). According to v^2=Gm/r, the velocity is inversely proportional to r^.5. But with the energy approach, where KE=-E(total), which equals GMm/2r...

Homework Statement Answer is D. Answer is E. Homework Equations 14. v=√(Gm/r) 17. p=mv τ=rxF L=Iω The Attempt at a Solution Number 14 made perfect sense until I saw the answer, which is D. If the velocity of the spacecraft increases, according to v=√(Gm/r), the radius...

Alrighty, I think I got it. Thanks for all the help and being so patient!