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Mrrafil [7]
3 years ago
15

If 35 J of work was performed in 70 seconds, how much power was used to do this task?

Physics
1 answer:
Irina18 [472]3 years ago
5 0
P = w \times delta \: T
P = Power
W = Work
Delta T = Change of Time

So:
P = 35 Joules
Delta T = 70 secs
W = ?

P = 35 \div 70 \\ P = 0.5 \: watts

* Joule/Sec = Watt

Answer: The power used to do this task was 0.5 Watts.
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<h3>Distance from the center of the meter rule</h3>

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  20 A  (30 - x)↓      x         ↓      20 cm  B 30 cm

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