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2 years ago

11

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? Explain your answer.

1 answer:

34kurt2 years ago

6 0

Answer:

4.80 kW

Explanation:

Power = work / time

P = 100,832 J / 21.0 s

P ≈ 4800 W

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Answer: 3 m.

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3 years ago

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Elena L [17]

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From the question we are told that

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$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

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