A skater makes one revolution per second, f = w/(2p) = 1, with her arms and leg stretched (left). By lowering her leg and bringing her arms close to her chest (right) she reduces her moment of inertia from I to I′ = 0.4I. How many revolutions per second, f′ = w¢/(2p), is she making now? Show all your work.

f'=w'/2p=2.49rev/s

Explanation:

The angular momentum L must be equal for both cases.

Hence you have that

L before = L after

I = moment of inertia

w =angular momentum

Hence

f'=w'/2p=2.49rev/s

hope this helps!!

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards