Recent content by ShEeRMiLiTaNt

Hello Everyone, I need to solve this problem: How many seconds does it take to charge a 7.7 x 10-7 farad capacitor to 70V if it is initially uncharged, then connected to a 190V battery through a 11000 ohm resistor. What I have so far: q = CV q = (7.7 x 10-7f)(70V) q = 5.39 x 10-5C q...

ok so the position of the bead initially is simply r(0) = A1+A2 because e0= 1 then v(0) = 2\pif(A1+A2) is v(0) = 0 so that i can find one of the A's?

r(t) but is that right? thanks again

I am not sure why this is...why is b so important that we had to relate it to \omega just to show that is the position function Any ways...ummm good question would these be something like this? 2\pift(A1-A2)?

Thank you very very much! so \omega= 2\pif? Do i plug that back into r(t)?

b2(A1ebt+A2e-bt)= \omega2r b2(A1ebt+A2e-bt)= \omega2(A1ebt+A2e-bt) b2= \omega2 this is where i am stuck how do i prove 0 = 0?

I don't get how those equal zero, none of the values cancel out, wouldn't i need another equation that relates some of the values in order to be able to cancel them out?

ok so i took the double derivative and it equals r"(t) = A1b2ebt+A2b2e-bt whats next boss?

Right you are i am still unaware of differential equations, i don't get what you're trying to get me to do here you want me to plug in r(t)=A1ebt +A2e–bt into d^2r/dt^2 = \omega^2r? Do i have to take the double derivative of r(t) first?

ok so \omega = sqrt(a/r) d^2r/dt^2 = a^2/r ?

let u = tangential velocity let v = radial velocity = \omega r dv/dt = \omega dr/dt at = du/dt = d/dt(\omegar) = \omega dr/dt ar = d^2r/dt^2 = dv/dt= (\omega^2)r so \omega dr/dt = \omega^2 r dr/dt = \omega r or dr/r = \omega dt ln r = \omega dt + c' r = e^(\omegat + e^c') = C...

F = ma = mw^2r where w = 2pif ?

Homework Statement A small bead of mass m is free to slide along a long, thin rod without any friction. The rod rotates in a horizontal plane about a vertical axis passing through its end at a constant rate of f revolutions per second. Show that the displacement of the bead as a function of...