Asked by luma
When a load of 1(10)^6 N is placed on a ship, the ship only sinks 2.5 cm into the ocean (sea water). Estimate the cross-sectional area of the ship at water level.
Answers
Answered by
Damon
How amusing - you happened upon a Naval Architect.
increased volume of ship in the water = Area * .025 meters
mass of water displaced = density of seawater * that volume
density of seawater is about 1029 kg/m^3
so
mass of water displaced = 1029 * .025 * A
Buoyancy force due to that water = mass * 9.81 m/s^2 gravity
so
additional buoyant force up
= 9.81*1029 * .025 * A
so Archimedes says
1,000,000 = 9.81 * 1029 * .025 * A
increased volume of ship in the water = Area * .025 meters
mass of water displaced = density of seawater * that volume
density of seawater is about 1029 kg/m^3
so
mass of water displaced = 1029 * .025 * A
Buoyancy force due to that water = mass * 9.81 m/s^2 gravity
so
additional buoyant force up
= 9.81*1029 * .025 * A
so Archimedes says
1,000,000 = 9.81 * 1029 * .025 * A
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.