A uniform cylindrical grinding wheel of mass 50.0 kg and diameter 1.0 m is turned on by an electric motor. The friction in the bearings is negligible. What torque must be applied to the wheel to bring it from rest to 120 rev/min in 20 revolutions?

The torque must be applied to the wheel is 15.7 N-m.  

Explanation:

Given that,

Mass of the wheel of cylinder, m = 50 kg

Diameter of the wheel, d = 1 m

Radius, r = 0.5 m      

Initial speed off the wheel is 0

Final angular speed of the wheel, 120 rev/min = 12.56 rad/s

Angular displacement,

The torque is given by :

So, the torque must be applied to the wheel is 15.7 N-m.

before we can answer, we have to see the choices that are printed there with the question.   that's the only way we can point you to the correct one.

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.