Recent content by Brodo17

A boy pulls a sled of mass 5kg with a rope that makes an angle of 60 degrees w/ respect to the horizonal surface of the frozen pond. The boy pulls on the rope with a force of 10N and the sled moves with constant velocity. What is the coefficient of friction between the sled and the ice...

What do you mean I forgot about the 40N force? Where would I put it in my equation? Actually, If I go (5kg)(2) = mewFN + 40N -30N = mewFN -30 / (-9.8 x 5) = .61 Thanks

A 5kg block is placed on top of a 15kg block that rests on a frictionless table. The surface between the top and bottom blocks is roughened so that there is no slipping between the blocks. A 40N horizontal force is applied to the top block. What is the minimum coefficient of static friction...

Never mind :) I was feeling daring and imputed my answer, it was right Thanks for all your help!

Alright... so I just found the FB on the crate, as well as the FB on the balloon. Now I divided that by 9.8 and got a mass. (since mg = pvg and pvg is the sum of the buoyant forces) Does that sound right?

ok so I calculated the mass of the water which would have been displaced... where can I go from there though? Where does the buoyant force that the water is exerting on the crate come into play?

In the problem I am working a shipping crate is underneath the water, the dimensions of the crate are given. They tell you that when a balloon is inflated to a radius of 1.3m the crate starts to rise. What is the mass of the crate? FB = Wfluid P= F/AI calculated the buoyant force acting on...

ok so I pick a random point and calculate the force each charge is exerting at that point? I really don't understand where to go from there though.

I don't understand! Could someone please solve the problem and show me the solution.

Two carges of + 1.5 x 10 ^-6 C and + 3.0 X 10^-6 C are .20m apart. Where is the electric field between them equal to zero? Do I use the equation Kq2/r^2 ? I also have Kq1q2 / r^2 I am pretty sure the problem can be solved using those two equations, and most likely the first equation...

A 10,000kg space shuttle moving east at 3000km/h wishes to change its course by 10°. It does so by ejecting an object at a speed of 5000km/h (South). Calculate the mass of the ejected object. Equations Momentum before = momentum after momentum is mass x velocity I am honesty pretty...

PS. How do I make it say that the Thread is solved?

Alright, thanks for the help everyone. I should be ok now.

Im just wondering what exactly that equation is and when it can be used. I've never seen that before.

The answer in the back of the book says 3.56 x 10^3 N...