Asked by sabeen
A small domestic elevator has a mass of 454 kg and can ascend at a rate of 0.180 m/s. What is the average power that must be supplied for the elevator to move at this rate ? answer in watts.
Answers
Answered by
Henry
Fe = 454kg * 9.8N/kg = 4449.2N.
Pe Fd/t = FV = 4449.2 * 0.180m/s = 801W
Pe Fd/t = FV = 4449.2 * 0.180m/s = 801W
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