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Elis [28]
3 years ago
8

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 othe

r 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
Physics
1 answer:
lina2011 [118]3 years ago
8 0

Answer:

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

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