Recent content by selter01

yes, x is the distance a spring is compressed (from equilibrium)

delta X = Vot + 1/2 at^2. Vo is 0.

hm, then I guess use Gm(p)m(s)/r^2 = m(s)v^2/r you know G, the gravitational constant, you know the mass of the Earth and satelite.. theen, since you don't know v OR r, you can use v = d/t, so Gmp/r = (r/t)^2, solve for r. once you know r, find v, then KE.

they told you time it takes for 1 rev, so if you know the distance [ circumference] and the time, you can find V, so just 1/2 mv^2

you mean it slides down a hill? O.o

they gave you mg and Fn, Fn is your apparent weight. mg is normal weight

ma = -mg + Fn

Once I find out how to solve for one I'll know how to solve for both, so let's just deal with the 5 kg cart. That statement is incorrect, you have to deal with both of them at the same time. So think of them in terms of Energy, 1/2kx^2 = 1/2m1v1^2 + 1/2m2v2^2

I have the acceleration, but nothing else makes sense after that. Care to explain?