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For a huge luxury liner to move with constant velocity, its engines need to supply a forward thrust of 7.45 105 N. What is the magnitude of the resistive force exerted by the water on the cruise ship

User Latinos
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1 Answer

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Answer:

7.45 10^5N.

Step-by-step explanation:

according to newtons second law of motion;


\sum F_x = ma_x\\F_(app) - F_r = ma_x

Fapp is the applied force

Fr is the resistive force

m is the mass of the luxury

a is the acceleration

Since the huge luxury liner move with constant velocity, then acceleration is zero i.e a = 0. The equation becomes;


F_(app) - F_f = m(0)\\F_(app) - F_f =0\\F_(app) = F_f

This shows that the applied force will be equal to the resistive force if the velocity is constant.

Given Fr = 7.45 10^5 N therefore the resistive force will also be 7.45 10^5N.

Hence the magnitude of the resistive force exerted by the water on the cruise ship is 7.45 10^5N.

User GrouchyPanda
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