When a load of 1.18 ✕ 10^6 N is placed on a battleship, the ship sinks only 2.0 cm in the water.?
When a load of 1.18 ✕ 10^6 N is placed on a battleship, the ship sinks only 2.0 cm in the water. Estimate the cross-sectional area of the ship at water level. (The density of sea water is 1.025 ✕ 10^3 kg/m3.)
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2cm. = 0.02 metre.
The load is 1,180,000 N.
Density of sea water = 1,025 kg/m^3.
(1,180,000/g) = 120,408.2kg.
Sea water displaced = (120,408.2/1,025) = 117.47 m^3.
Cross sectional area of ship = (117.47/0.02) = 5,873.5 m^2.
Will change if you do not use 9.8 for g.