A 5.0-m-diameter merry-go-round is turning with a 3.7 s period. Part A What is the speed of a child on the rim

Speed of a child on the rim is 4.25 m/s.

Explanation:

Given;

diameter of the merry-go-round, d = 5.0 m

period of the motion, t = 3.7 s

one complete rotation of the merry-go-round = πd

one complete rotation of the merry-go-round = π(5) = 15.71 m

Speed is given as distance / time

speed of a child on the rim = 15.71 / 3.7

speed of a child on the rim = 4.25 m/s

Therefore, speed of a child on the rim is 4.25 m/s.

im guessing the west of station b

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