Question

A spherical submersible 1.84 m in radius, armed with multiple cameras, descends under water in a region of the Atlantic Ocean known for shipwrecks and finds its first shipwreck at a depth of 2.37 ✕ 103 m. Seawater has density 1.025 ✕ 103 kg/m3, and the air pressure at the ocean's surface is 1.013 ✕ 105 Pa. What is the absolute pressure at the depth of the shipwreck?

Answer 1

Answer:

23932242.5 Pa

Explanation:

$P_a$ = Atmospheric pressure = $1.013\times 10^5\ Pa$

$P_w$ = Pressure of seawater

$\rho$ = Density of sea water = $1.025\times 10^3\ kg/m^3$

h = Depth of shipwreck = $2.37\times 10^3\ m$

g = Acceleration due to gravity = 9.81 m/s²

The absolute pressure is given by

$P_{ab}=P_a+P_w\\\Rightarrow P_{ab}=1.013\times 10^5+1.025\times 10^3\times 9.81\times 2.37\times 10^3\\\Rightarrow P_{ab}=23932242.5\ Pa$

The absolute pressure at the depth of the shipwreck is 23932242.5 Pa