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and electric motor lifts an elevator a vertical distance of 18.0m in a time of 30.0s by exerting an upward force of 6.0x10^3 N . what power does the electric motor produce?

Sagot :

Answer:

[tex]\fbox {3.6kW}[/tex]

Explanation:

Formula used :

[tex]\boxed {P = F \times v} \leftarrow \rightarrow \boxed {P = \frac{F \times S}{t}}[/tex]

Given :

⇒ F = 6 × 10³ N

⇒ S = 18 m

⇒ t = 30 s

Solving :

⇒ P = 6 × 10³ × 18 / 30

⇒ P = 2 × 100 × 18

⇒ P = 3600 W

⇒ P = 3.6kW

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