Question

When the submarine's density is equal to the density of the surrounding seawater, the submarine will maintain depth. If a 103200 kg submarine takes on 2100 kg of water to maintain depth at 1000 feet, where the density of seawater is approximately 1033 kg/m3, what is the total displacement (volume) of the submarine in m3 (Report your answer to 4 sig figs without a written unit)?

Answer 1

Answer:

$2.023 m^3$ 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 = $1033 kg/m^3$

Volume of the water displaced = V= ?

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

$Density=\frac{Mass}{Volume}$

$V=\frac{m}{d}=\frac{2100 kg}{1033 kg/m^3}=2.023 m^3$

$2.023 m^3$ is the total displacement (volume) of the submarine.