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?

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:

(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:

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

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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

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .