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## Main Question or Discussion Point

Brief Summary of Question:

One submarine starts with a neutral buoyancy 100m below the surface when it displaces 8.9m^3 of water from its ballast tanks. If the mass of the submarine is 69 metric tons what is the initial, vertical acceleration of the submarine immediately after displacing the water?

[ρ(water)=1,000 kgm-3, 1 atm=105 N/m2, g=9.8ms-2]

Related Equations: F(buoyancy) = pgV, where p=density of fluid, g=acc., V = displacement of the water due to sub.

Work done:

Used the buoyancy formula to calculate the displacement of the water by the sub. = 69m^3. (Not sure how the 100m below the surface played into this equation, as well it would seem we need the difference in contained water to calculate the acceleration upwards after ejection)

Guessing we need to use this in tandem with the 8.9m^3 that gets ejected from its tanks.

One submarine starts with a neutral buoyancy 100m below the surface when it displaces 8.9m^3 of water from its ballast tanks. If the mass of the submarine is 69 metric tons what is the initial, vertical acceleration of the submarine immediately after displacing the water?

[ρ(water)=1,000 kgm-3, 1 atm=105 N/m2, g=9.8ms-2]

Related Equations: F(buoyancy) = pgV, where p=density of fluid, g=acc., V = displacement of the water due to sub.

F = ma

Neutral buoyancy --> F(buoyancy) = Weight

Neutral buoyancy --> F(buoyancy) = Weight

Work done:

Used the buoyancy formula to calculate the displacement of the water by the sub. = 69m^3. (Not sure how the 100m below the surface played into this equation, as well it would seem we need the difference in contained water to calculate the acceleration upwards after ejection)

Guessing we need to use this in tandem with the 8.9m^3 that gets ejected from its tanks.