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When the submarine's density is equal to the density of the surrounding seawater, the submarine will maintain depth.

If a 103,200 kg submarine takes on 2,100 kg of water to maintain depth at 1,000 feet, where the density of seawater is approximately 1,033 kg/m\(^3\), what is the total displacement (volume) of the submarine in m\(^3\)?

Report your answer to four significant figures without a written unit.

Answer :

Answer:

[tex]2.023 m^3[/tex] is the total displacement (volume) of the submarine.

Explanation:

Mass of water carried by submarine at 1000 ft depth = m = 2100 kg

The density of seawater at 1000 ft depth = d = [tex]1033 kg/m^3[/tex]

Volume of the water displaced = V= ?

Total displacement of the submarine = Volume of the water displaced = V

[tex]Density=\frac{Mass}{Volume}[/tex]

[tex]V=\frac{m}{d}=\frac{2100 kg}{1033 kg/m^3}=2.023 m^3[/tex]

[tex]2.023 m^3[/tex] is the total displacement (volume) of the submarine.

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