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

23932242.5 Pa

Step-by-step explanation:

`P_a` = Atmospheric pressure =
`1.013* 10^5\ Pa`

`P_w` = Pressure of seawater

`\rho` = Density of sea water =
`1.025* 10^3\ kg/m^3`

h = Depth of shipwreck =
`2.37* 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* 10^5+1.025* 10^3* 9.81* 2.37* 10^3\\\Rightarrow P_(ab)=23932242.5\ Pa`

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