An electric motor consumes 10.8 kJ of electrical energy in 1.00 min . Part A If one-third of this energy goes into heat and other forms of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 2500 rpm

0.458 N/m

Explanation:

Power = Energy/time

From the question,

Note: two-third of the energy into the motor,

Therefore,

P = 2/3(E)/t Equation 1

Where P = power of the motor, E = Electrical Energy, t = time

Given: E = 10.8 kJ = 10800 J, t = 1.00 min = 1/60 s

Substitute into equation 1

P = 10800(1/60)(2/3)

P = 120 W.

But,

The power of a rotating motor is given as,

P = Tω Equation 2

Where T = Torque of the engine, ω = angular velocity of the engine

Make T the subject of the equation

T = P/ω Equation 3

Given: P = 120 W, ω = 2500 rpm = (2500×0.1047) rad/s = 261.75 rad/s

Substitute into equation  3

T = 120/261.75

T = 0.458 N/m

Then the torque develop in the engine = 0.458 N/m

the maxium height is 34m/s

the answer is c. the lower legs are levers, and the knees are fulcrums. the ankles hold the loads.