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4 years ago

9

It takes 100,832 J of work to lift an elevator 18.3 meters. If this is done in 21.0 seconds, what is the average power of the el

evator during the process?

1 answer:

Aneli [31]4 years ago

6 0

Answer:

P=4801.5

Explanation:

Given :

work done = W = 100,832 J

time = 21.0 sec

Find:

P = ?

Formula:

P = W/t

Solution:

P = W/t

P = 100,832/21.0

= 4801.52 J/s or Watts

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