Recent content by bmarvs04

Ha! Good catch! that ended up being the problem.. The correct answer was 6200V.. I've been doing problems with microCoulombs so I must've gotten screwed up between the two. Thanks a bunch And by the way, your last line of work is correct.. When I wrote it in my first post I just factored out the Q

Yes, nC is a nanoCoulomb, which is why my calculations were converted to Coulombs (4*10^-5). Also, I think what you're picture is correct.. and yes we are interested in the potential at the surface of the inner sphere. Also, I probably should have said this earlier: my calculations were based...

Homework Statement A conducting sphere 4.5 cm in radius carries 40 nC. It's surrounded by a concentric spherical conducting shell of radius 20 cm carrying -40 nC. Find the potential at the sphere's surface, taking the zero of potential at infinity. Homework Equations Inside Sphere: V = Q...

Thanks a bunch.. It worked perfectly

Homework Statement A 17 meter long vine hangs vertically from a tree on one side of a 10 meter wide gorge, as shown in the figure. Tarzan runs up, hoping to grab the vine, swing over the gorge, and drop vertically off the vine to land on the other side Homework Equations PE = mgh KE =...

So it would be (L-y)/L? I forgot to re-read the problem to see what 'y' was actually measuring. Thanks for everything.. I really appreciate it.

Ok I think I understand. Would it make sense for T = (M + ((y/L)*m))*g then? This makes sense to me because 'y/L' would give you the fraction of rope you are working with, then you could multiply it with 'm' to find the mass of the length you are working with.

I'm sorry, I forgot to define my variables. W is the weight of the system, not the work done. And T is the tension of the rope. Now since the rope is uniform, I know the mass and length have a direct relationship and that the tension will increase when the length (L) is increased. I still...

Homework Statement A block of mass M hangs from a uniform rope of length L and mass m. Find an expression for the tension in the rope as a function of the distance y measured vertically downward from the top of the rope. Homework Equations For the System: W = (m + M)g = T The...

Thanks a bunch! That makes perfect sense to me now. You've been a great help!

Ok thanks, So in order to solve for the vertical velocity, I used the equation Vf^2 = 2(9.8)(1.6) = 5.6m Then do I use trig to solve for horizontal velocity? In other words, Horizontal Velocity = cos(70.4268)*5.6 where 70.4268 is the downward angle of the triangle formed by 1.6m and 4.5m...

Homework Statement "In a chase scene, a movie stuntman is supposed to run right off the flat roof of one city building and land on another roof 1.6 m lower." "If the gap between the buildings is 4.5 m wide, how fast must he run? " Homework Equations Vf^2 = V0^2 + 2a (Xf-X0) This...