Recent content by physics10189

I think my equation in the y direction is wrong any thoughts?

Homework Statement On an air hockey table, two pucks collide with each other. Puck A have .05 kgand inital velocity of 5 m/s in the x direction. It collides with puck B which is initially at rest and has a mass of .1 kg. The collision is not a head on. AFter the collision, puck A flies...

Ok thanks then.

Well it doesn't matter if C is something or L is something...I just want to know what is L with the given values, more of plug and chug situation. I just want to know if I made my conversion right with the given equation.

Homework Statement Well, I am not sure about my conversion and I need people to make sure I am doing it right. I have a period, T, of 17.29 ms. To find the frquency it is 1/T= 57.836 s^-1. So the inductive is L=((1/(2*pi*frequency))^2)/C which C is 110*10^-6 and I got L=68.84 mh...

Well I have no clue what you are talking about the first reply could you explain it better? For the second reply I believe this could he a 30 60 triangle but its sides are 1:2:sqrt(3) could you tell me why it is sqrt (2) instead?

Well how do you know the distance is .65 sqrt(2) m?

So what does Interval mean?? The north pole question is another question I would like to ask and cramster got that new r value for the solution

If this is true about 360/60 then what about this question What is the electric potential at the north pole? So 360/90= 4 however this is not true in cramster I use 90 degree because that is the angle from the origin to north but cramster do not use that they use the same degree 60...

V=(9*10^9)(4*10^-6)/(.65) however i looked at cramster and it said V=6(9*10^9)(4*10^-6)/(.65) why is that? is it because charges can be at 10 degree difference so 60/10 = 6 charges?

Homework Statement A sphere with radius 65 cm has its center at the origin. Equal charges of 4 μC are placed at 60◦ intervals along the equator of the sphere. What is the electric potential at the origin? Homework Equations V=KQ/R The Attempt at a Solution I just plug the...

ok i got the answer correct but i am not understand when you mean by both charges are pushing to the right. This i what i see as. q1=5c----->E=0<------q2=-6c to me this diagram makes sense because q1 is positive and q2 is negative and the other one that doesn't make sense to me is that...

Well the first charge is pushing to the right and the second charge is pushing to the left thus making first plus and second negative because of their displacement and the position of the eletric field at x=0 but if we do forget about the +/- then this cannot be solve because the E3=-E1-E2...

Homework Statement a point charge of +5c is located at x=-2cm, and a second point charge of -6c is located at x=1cm where should a thid charge of +4c be placed so that electric field at x=0cm is zero? Homework Equations E(total)=E1+E2+E3 E=k*q/(r^2) The Attempt at a Solution...

Tension=(velocity^2)*((mass)/(length)) which i have neither of the length of the string nor the mass of it.