submarine floats on the ocean with 2/3 of its volume submerged when its on the s

Question

submarine floats on the ocean with 2/3 of its volume submerged when its on the surface.  the total volume of the sub id 88,000 ft^3. a) what is the weight of the submarine?  b) what is the minimum weight must be taken into the sub in order  to sybmerge?

weight = 3,807,232 lbs

Explanation / Answer

a). for the static equilibrium of submarine; buoyant force acting in upward direction = force of gravity acting in downward direction => p*(V submerged)*g = weight of submarine => Weight of submarine = 1000*[(2/3)*(88000*0.0283 m^3)]*9.8 => W = 16270613.33 N => W = 16270613.33* 0.224808942443 lbs => W = 3657779.376 lbs b). for complete submergence; p*V*g = weight => weight = 1000*(88000*0.0283 m^3)*9.8 => weight = 24405920 N => weight = 24405920*0.224808942443 lbs => Weight = 5486669.065 lbs extra weight required= 5486669.065 - 3657779.376 = 1828889.689 lbs

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