When a 25000-kg fighter airplane lands on the deck of the aircraft carrier, the carrier sinks 0.23cm deeper into the water.

a) Determine the cross-sectional area of the carrier. Assume the density of seawater is 1025 kg/m3.

A = ???

10604 square meters

Explanation:

0.23 cm = 0.0023m

Assume the carrier has a shape of a rectangular box. When the carrier sinks 0.0023m deeper into water, the extra volume submerged into water is the same as the extra water volume being replaced. This extra volume would add an additional buoyant force to counter balance the extra weight created by the 25000 kg fighter.

With g being constant, the mass of the extra water displaced is the same as the mass of the fighter airplane . And mass of the water displaced is its volume V times its density

We assume the carrier has a shape of a rectangular box, so this extra displaced volume is the extra depth d = 0.0023 m times cross-section area A

c. magnitude and direction