physics help please! 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.) Answer in m^2 please. I just can't seem to get this one thank you for your help!
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In this problem, the buoyant force is equal to the weight of the load.
Buoyant force = 1.025 * 10^3 * 9.8 * Volume of displaced sea water
1.18 * 10^6 = 1.045 * 10^4 * V
V = 1.18 * 10^6 ÷ 1.045 * 10^4
The volume of displaced sea water is approximately 112.9 m^3. To determine the cross-sectional area of the ship at water level, divide the volume by the 2 cm.
Area = (1.18 * 10^6 ÷ 1.045 * 10^4) ÷ 0.02
This is approximately 5,650 m^2.
I hope this is helpful for you.