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A 45 kg object is lifted vertically at a constant speed to a height of 9.0 m by a 7.5x10² w electric motor. If this motor is 25% efficient in converting electric energy to mechanical energy, how long does the motor take to lift the object?​

Sagot :

onyebuchinnaji

Answer:

the time taken for the motor to lift the object is 21.17 s.

Explanation:

Given;

mass of the object, m = 45 kg

height through which the object was lifted, h = 9.0 m

electrical power used by the motor, P = 750 W

Efficiency of the electrical motor, n = 25% = 0.25

The electrical energy used by the motor in lifting the object is calculated as;

E = P x t

where;

t is the time taken for the motor to lift the object

E = 750 x t

E = 750t

The electrical energy converted by the motor to mechanical energy is calculated as;

P.E = 0.25(750t)

P.E = 187.5t

Recall, P.E = mgh

mgh = 187.5t

45 x 9.8 x 9 = 187.5t

3969 = 187.5t

t = 3969/187.5

t = 21.17 s

Therefore, the time taken for the motor to lift the object is 21.17 s.

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