Asked by Lauren
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.)
Answers
Answered by
Jennifer
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
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
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