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A tugboat pulls a ship with a constant net horizontal force of 5.00 × 103 N and causes the ship to move through a harbor. How much work is done on the ship if it moves a distance of 3.00 km?

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opudodennis

Answer:

15 MW

Explanation:

Work done is a product of force and distance moved.

W=fd where f represnt force while d is distance

Substituting f with 5 kN and 3 000 for d then

W= 5*3000=15000 kW which is equivalent to 15 MW

whitneytr12

The work done on the ship when a tugboat pulls it with a constant net horizontal force of 5.00x10³ N over a distance of 3.00 km is 15.0x10⁶ J.  

We can find the work done on the ship as follows:

[tex] W = F*d [/tex] (1)

Where:

F: is the net horizontal force = 5.00x10³ N

d: is the displacement = 3.00 km = 3000 m

By entering the above values into equation (1), we have:

[tex] W = F*d = 5.00 \cdot 10^{3} N*3000 m = 15.0 \cdot 10^{6} J [/tex]  

Therefore, the work done on the ship is 15.0x10⁶ J.

Find more here:

  • https://brainly.com/question/8617144?referrer=searchResults
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I hope it helps you!

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