Recent content by nabaa

this means you can equate 1/2kq1q2/r^2 = 1/2mv^2 multiply your answer by two because there are two charges. you cannot multiply by two initially or it will mess up the algebra, multiply your answer.

I believe the equation for an Electric fields goes as follows: kq1q2/r^2 not kq1q2/r

please, anyone who understands work in an electric field I know W = -PE

can anyone help?

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thank you so much! mine too!

differentiating once would yield -wAsin(wt) twice would yield -w^2Acos(wt) so w^2 must = 3g/2L w = sqrt(3g/2L) function = Acos(sqrt(3g/2L)t), yes?

cosine because at angle = 0 cosine is 1 A in front, okayAcost/(3g/2L)?? ??

to be honest I don't understand what's wrong with them? isn't the double derivative of 3g/2Lcost = −3g/2Lcost

ohhh so it would end up being (3g/2L)(cost), and the negatives would still cancel (integrate) out

is that right?

C = 3g/2L so the equation is d^2angle(t)/dt^2 = −(3g/2L)(sin(t)) and angle(t) =(3g/2L)(sin(t))

oh.. sin(X) or cos(x)

well usually in this instance i'd use integration but i don't know how to do so with a double derivative, or are you saying we don't need integration?

oops x should be angle, right? meaning: MgLangle/2 = (ML^2/3)(d^2angle/dt^2) (MgL/2)/(ML^2/3) = (d^2angle/dt^2)/angle 3g/2L = (d^2angle/dt^2)/angle how do i isolate angle, then?