Answer:
rpm
Explanation:
Given that rotational kinetic energy = 
Mass of the fly wheel (m) = 19.7 kg
Radius of the fly wheel (r) = 0.351 m
Moment of inertia (I) = 
Rotational K.E is illustrated as 





Since 1 rpm = 



The power equals energy divided by the time
P=E/t
P=1200/60
P=20W
Hope you get it!
Answer:
The magnitude of the torque is 263.5 N.
Explanation:
Given that,
Applied force = 31 N
Distance from the axis = 8.5 m
She applies her force perpendicularly to a line drawn from the axis of rotation
So, The angle is 90°
We need to calculate the torque
Using formula of torque

Where, F = force
d = distance
Put the value into the formula


Hence, The magnitude of the torque is 263.5 N.