Recent content by vpea

Thanks for that! I tried looking on the internet for the moment of Inertia of the Earth and was getting all sorts of weird numbers no wonder it wasn't working.

I found the moment of Inertia (mr^2) to be = 2.43*10^38. Then found the angular velocity to be (2pi)/(24*60*60). Using the equation I=m*r^2, I got 1.77*10^34

I'm stuck on the second part of a problem and can't seem to get the right answer: Calculate the magnitude of the angular momentum of the Earth due to its rotation around an axis through the north and south poles. Treat the Earth as a uniform sphere of radius 6.38*10^6 that makes one...

A boat of mass 250kg is coasting, with its engine in neutral, through the water at speed 3.00m/s when it starts to rain. The rain is falling vertically, and it accumulates in the boat at the rate of 10.0kg/hr. Assume that the boat is subject to a drag force due to water resistance. The drag is...

Excellent. Thanks for clearing up my confusion :)

I tried it using that formula but I'm pretty sure I stuffed it up somewhere. So I got: 1/2*m_1*v^2 + 1/2*m_2*v^2 = -m_1*g*h - m_2*g*h Now I took the height of m_1 (the larger mass) to now be -s (since it moves down) and the height of m_2 (the smaller mass) to be s (since it moves up). I'm not...

I've been trying to use the hints you gave out but I'm just not getting anywhere. I don't know how to go about dealing with the conservation of energy in the SYSTEM. And the 2nd part of the question is to find the mass of the lighter block and I'm pretty sure I can't do that without first...