Recent content by g-racer

thanks so is 3/2(-1)^2/3 a cubed root squared and therefore =3/2?

Hi there, I am stuck on this problem: the integral of 1/x^(1/3) from -1 to 8. I have broken it up into the integral from -1 to 0 and 0 to 8. I am confused as to how the negative values under a cubed root affect things and whether or not I need to break it up further. I am not sure whether...

ok so 210=kq/r and 400=kq/(r+0.1)^2 setting equal gives 210r=400(r+0.1)^2 which gives the quadratic 400r^2+80r+4=210r which gives 400r^2-130r+4 ...nevermind I had put -40r for some reason! thanks so much

ok thanks so much I had forgot the e in the equation so setting equal for kq I get 210r=400(r+0.1)^2 but then I still get negative square roots...?

I know there should be two but with 1600pi r^2+(320pi-210/k)r +16pi I only get r=-0.1

The electric potential immediately outside a charged conducting sphere is 200 V, and 10.0 cm farther from the center of the sphere the potential is 150 V. Determine (a) the radius of the sphere and (b) the charge on it. The electric potential immediately outside another charged conducting sphere...

I think I might have got it. so wf=wi+αt=wi-2ukgt/R and vf=vi+at=0+ukgt and as vf equals wf.r when they are not slipping we can set the equations equal for wf and solve for t which gives t=wiR/3ukg? thanks

when rolling without slipping v=wr and d=theta.r but I am not sure how to work with this. thanks

ok so angular acceleration = -2ukg/R and linear acceleration = ukmg/m=ukg ? I am not sure where to go from here for part c

A wheel spinning clockwise on its axis at with angular velocity ω0 drops to the horizontal ground. It initially has no center-of-mass velocity. The coefficient of kinetic friction between the ground and the barrel is µ. The radius of the wheel is R, and it is a solid disc of mass M. Express your...

ok i think I get it so when he has jumped his angular momentum will be mvr=36x1.5x2. and this plus the angular momentum of the merry go round will equal the initial angular momentum of the merry go round and fred before he jumps? for an answer I got 0.3157rad/s so the merry go round slows down?

sorry I still don't get it. so is the angular momentum of just the merry go round conserved or the merry go round and fred running?

Ok thanks. So his horizontal velocity will be his angular velocity times the radius? but this would only be when he is is the air then he runs at 1.5m/s. I don't understand how him running will affect the angular velocity of the merry go round, or does it not affect it?

Fred, mass 36 kg, stands at the center of a merry-go-round that has mass 50.0 kg and radius 2.0m and that rotates one full revolution every 4.5 seconds. Treat the merry go round as a solid disc, and Fred as a point object. As it turns, he walks to the edge of the merry-go-round, then jumps off...