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...
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...
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...