Recent content by Tom2

Homework Statement To integrate a function (the function itself is not important) over the region Q. Q is bounded by the sphere x²+y²+z²=2 (ρ=sqrt2) and the cylinder x²+y²=1 (ρ=cscφ). To avoid any confusion, for the coordinates (ρ,φ,θ), θ is essentially the same θ from polar coordinates in 2...

Homework Statement Let V be the 2-dim subspace of R^3 spanned by V1 = (1,1,1) and V2 = (-2,0,1). Find the orthogonal compliment Vperpendicular. 2. Relevant equation X + Y + Z = 0 -5X + Y + 4Z = 0 The Attempt at a Solution Firstly, I orthogonalize the basis for V and get the vectors (1,1,1)...

Homework Statement You are 60kg and you are holding a 10kg weight as you stand at rest on a frictionless surface. you can throw the weight 5m on ground where there is friction to keep you from moving backwards. How far will the 10kg weight be from you after you throw it on the frictionless...

Thank you

Oh ok so what you are saying is that the speed of the watermelon as it moves away from you on stable ground will be the same as the speed of the watermelon as it moves away from you when u are on a platform on the frictionless surface. So in both cases the watermelon ends up 5m away from "you"...

What do you mean by the same relative speed? what are you referring to as being the same?

I've been trying to figure out what you said for a while now and I think I get a little of what your saying but unfortunately I am unable to fully understand what you mean. The watermelon is thrown with some velocity Vo and takes a time t to be displaced 5m. Like you said Vo and t don't really...

What effect will throwing the watermelon cause. I know that since the watermelon will move forward the {you + platform} system will move backwards, and that the net momentum must still be 0, but nothing about any velocity is known so I don't know how to start this off.

Ok here it goes.. You are back at rest at the left end of the 3.00m board (also at rest) on the frictionless surface but you now have a large 4kg watermelon in hand. If you throw the watermelon to the right, how far will the watermelon land relative to the right end of the platform if you throw...

Thanks a lot! If you don't mind there is another (final) part of this problem I didn't put up because I thaught that i would have more trouble with this than I actually did. May I ask about the final part of the problem or should I put up a new thread

By sum do you just mean addition? So that would be 1/2(50kg)(2.00m/s)^2 + 1/2(100kg)(-1.00m/s)^2

posted before I finished .. but if at t = 1s you will be at the other end of it and it will have moved -1m in that time

Kinetic Energy Equation Is 1/2MV^2 M will be just the total mass of the system which is 150kg. Not sure about velocity though.And I see that I did make a mistake solving for displacement i can't think of an equation off the top of my head for it but thanks to how the problem is structured I...

Homework Statement You stand at rest on one end of a 3.00m long platform (also at rest). The platform is on a smooth friction-less surface. If you walk from one end of the platform to the other at the constant rate of 2.00m/s, what is the kinetic energy of the system (you and the platform)...