Recent content by xXmarkXx

no...q1=25uC = 2.5*10^-6 right? same with q2.

If i am not mistaken, shouldn't q1=2.5E-6 and q2=2.0E-6? That is how i got .009. I'm still kind of confused. What is F31?? The force created from q3 right?

I found the r (the distance between q2 and q1) to be root 5. Then i made the equation ((9*10^9)(2.0*10^-6)(2.5*10^-6))/5 which = .009 .009*2=.018N since q2=q3.

Take q1=25uC at (0,1), q2=20uC. at (2,0), and q3=? at (2,2). IF the force on q1 points in the -x direction, (a) what is q3 and (b) what is the magnitude of the force on q1? F=k(q1,q2)/r^2 coulombs law I know since q1 points in the -x direction that q3=q2. So (a)=20uC. I'm not sure...

Again, i don't know where to start on this one. I know that mgh=mgl(1-cos(theta)) but i don't know how to apply this in the problem. I'm not sure where to go.

So nobody wants to help a man out? ic...thanks guys!

Homework Statement Show that the potential energy of a simple pendulum is proportional to the square of the angular displacement in the small amplitude limit. Homework Equations U=mgh sin^2wt + cos^2wt=1 The Attempt at a Solution I can't figure out where to start.

and then i got 120pie = w0 how bout that? correct o mundo? Thanks hage and werg for the help and being patient with me. I'm a slow learner.

my fault, i checked it over again and now i got -3pie = angular acceleration. Would you guys at least tell me if i am on the right track or not?

and now i used pie=a and solved for the angular velocity, which i had gotten before to be equal to 40*pie...agreed??

right, i end up with the angular acceleration to be just pie.. vi=-40a, then i plugged that into the other equation and solved for a. which is pie. Is that what you guys got??

where did the 1400*2pie come from??

hage, where did you get 1400*2pie...is it not 1200 revolutions? in 40 sec? so wouldn't it be 1200*2pie?

anybody still willing to help me out?

ok, i ended up at 40pie=w0 i just labeled it incorrectly. But you still disagree?? if so, could you point me in the correct direction.