When a load of 0.97 106 N is placed on a battleship, the ship sinks only 3.0 cm in the water. Estimate the cross-sectional area of the ship at water level. (The density of sea water is 1.025 103 kg/m3.)

Answer 1

rho1*V1 = rho2*V2

where rho1 is the density of the object; V1 is the volume of the ship; rho2 is the density of the water; V2 is the volume of the water

mass of object = 0.97 106/g = 0.97 106/9.8
density of object = 0.97 106/(9.8*0.03*A)
where A is the area

(0.97 106/(9.8*0.03*A))*0.03*A = 1.025 103 * (0.03*A)

(0.97 106/(9.8*0.03*A))= 1.025 103

Solve for A, the area