Solution :
Given
Diameter of the roulette ball = 30 cm
The speed ball spun at the beginning = 150 rpm
The speed of the ball during a period of 5 seconds = 60 rpm
Therefore, change of speed in 5 seconds = 150 - 60
= 90 rpm
Therefore,
90 revolutions in 1 minute
or In 1 minute the ball revolves 90 times
i.e. 1 min = 90 rev
60 sec = 90 rev
1 sec = 90/ 60 rec
5 sec = 
= 75 rev
Therefore, the ball made 75 revolutions during the 5 seconds.
Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

v is the speed of cat, 

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.
'Pressure' is (force) / (area).
The only choice with those units is #1 .
D. 5.0A because this is right and will lead to the right answer okay you got this girl letssssss goooo googoggo Gogol