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.
Answer:
W=1705.2 J
Explanation:
Given that
mass ,m= 60 kg
Acceleration due to gravity ,g= 9.8 m/s²
Height ,h= 2.9 m
As we know that work done by a force given as
W = F . d
F=force
d=Displacement
W=work done by force
Now by putting the values
F= m g (Acting downward )
d= h (Upward)
W= m g h ( work done against the force)
W= 60 x 9.8 x 2.9 J
W=1705.2 J
Therefore the answer will be 1705.2 J.
Explanation:
Efficiency is defined as the ratio between the useful output over the total amount consumed. 
The fan does 500W of useful work while wasting 300 W. The total power consumption is 800 W (500 + 300).
If one of two interacting charges is doubled, the force between the charges will double.
Explanation:
The force between two charges is given by Coulomb's law
K=constant= 9 x 10⁹ N m²/C²
q1= charge on first particle
q2= charge on second particle
r= distance between the two charges
Now if the first charge is doubled,
we get 
F'= 2 F
Thus the force gets doubled.
