Recent content by ippo90

You have a circular capacitor (radius = R). Find the magnetic field from the central axis, for r > R? I have solved the first one for r<R by using ampere's law: /B dr = my(I + d/dt(flux of electric field)) the answer is hard to write in a post like this, but I will do if necessary...

just ignore my last post. I thought about what you said in you post, and it came out=) Thank you for your kind help. one thing I not sure of is that when I integrated the torque i got something like hFt=Iw where h is height, t is time and w is the angul. velocity. here I chose the...

Thank you for the reply. The problem is I don't know where to start. Can you help me?

Exams tomorow! please help Homework Statement A billiard ball is hit with a horizontal force F. The force hits the ball with the height h over the ball's center. the initial velocity is v0 and the the velocity afterward is (9/7)v0. Show that the height h is (4/5)R, where R is the radii of the...

Yay! I did it, thanks to you! =) Thank you for having such patience with me. You're awesome ^^

Now I know that MV=mv(bx) where v(bx) is the velocity of the box from the ground. v(b) can be expressed like this: v(bx)=v(rel)cos(a)+V and v(by)=-v(rel)sin(a) we also know that: mgh=1/2mv(bx)2+1/2mv(by)2+1/2MV2 but the algebra is so hard. But the analysis is correct?

That was a wonderful explanation, now everything makes sense! earlier I thought that v(box) was pointing the same direction as the slope, and v(rel) was pointing with ''the angle + a little more''. now I will try and see if I arrive at the correct answer=)

I really want to understand this, but it's really hard to visualize. This is a stupid request but is it OK if you upload a pic where you draw in the vector for v(rel) and v for the box? MS paint is good for me =)

I somehow can't visualize the directions of v(box) and v(rel). I tried to use the eq. from you post and got Mv2=mvbx and mgh=1/2mvbx2+1/2Mv22 this should be correct

ok.. I've totally lost it. what do I need to know to begin with?

even for the momentum you don't have to include both masses? is it correct that v(box) and v(rel) have the same direction? hence v(box)x=v(box)cos(a) and v(rel)x=v(rel)cos(a) ??

ok I now understand that v(box)=v(rel)+v(slope). if I now write the law for conservation of momentum, it will be: (m+M)v(slope)=m(v(rel)+v(slope))/cos(a)

to be honest, I'm totally lost now. why shouldn't I use (M+m)? the box is still on the slope. I don't know how to use the velocity relative to the slope ='(

the answer is supposed to be v2'2=2m2ghcos2(a)/[(M+m)(M+msin2(a))] did you get this? my attempt: mgh=1/2mv2+1/2(M+m)v2'2 where v is the velocity for the box when it gotten down (not in the x direction) velocity in the x direction is: v=v1'/cos(a) I plugged it into the eq...

I don't think I got that. What velocity do I find when I use this method? velocity of box relative to the slope? if so how do I find the velocity I need? I'm so stuck =P