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Kelsie Sawayn

Feb 20, 2025

A crane lifts the 18,000 kg steel hull ship out of the water.

A crane lifts the 18,000 kg steel hull ship out of the water.Density of steel is 7.8 x 10^3 kg/m^3 Density of water is 1000 kg/m^3A) Water displaced with the ship fully submerged in it? No idea how to do this partB) Tension on cable, ship fully submerged? B= Den (water) * V (steel) G (10) = 1000 (18000/7800) 10 = 23 077 NC) Tension on cable, ship out of water?tension = mg (g=10) 180,000N thanks
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Hermina Ferry

Feb 20, 2025

A)Volume of water displaced = Volume of submerged shipVw = (18,000 kg) / (7.8e^3 kg/m^3)Vw = 2.3077 m^3Weight of water displaced = (density of water)(volume of water)(acceleration due to gravity)Ww = (1000 kg/m^3)(2.3077 m^3)(9.81 m/s^2)Ww = 2.2638e4 NB)By Archimedes’ principle, the buoyant force on the ship equals the amount of fluid displaced.(Tension of cable) + (Weight of displaced water) = Weight of shipT + 2.2638e4 N = (18,000 kg)(9.81 m/s^2)T = 1.539e5 NC)Your approach to this one is correct.... Show More
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