The electric motor of an elevator(lift) uses 630kJ of electric energy when raising the elevator and passengers, of total mass 12
1 answer:
Answer:
Efficiency = 0.575 = 57.5%
Explanation:
First, we will calculate the output energy of the elevator, which is equal to the potential energy acquired by the load.
where,
O = Output = ?
mg = W = Weight = 12500 N
h = height = 29 m
Therefore,
The input is given as:
I = 630 KJ
Thus the efficiency will be:
<u>Efficiency = 0.575 = 57.5%</u>
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B)
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