Physics, asked by rakesh8933, 4 months ago

a) 34.3 kW
A elevator is designed to lift a load of 1000 kg through 6 floors of a building, averaging 3.5 m per
floor in 6 seconds. The power of the engine of the elevator, neglecting other losses is given by
b) 27.5 kW
c) 20.5 kW
d) 15.2 kW [ ]

Answers

Answered by ushmon
0

Answer:

b) 27.5 kW will be the answer please mark me as brainliest

Answered by Anonymous
7

Given the elevator is going up a height of 6×3.5m in 6 seconds

increase in potential energy of the system PE will be

△PE=mgh

now, power is defined as energy used or spent per unit time

P=△PE/△t

P= \frac{1000 \times 9.8 \times 6 \times 3.5}{6}

P = 3.43 x 10^{4} W

Similar questions