Answer:
The depth will be equal to 6141.96 m
Explanation:
pressure on the submarine [tex]P_{sea}[/tex] = 62 MPa = 62 x 10^6 Pa
we also know that [tex]P_{sea}[/tex] = ρgh
where
ρ is the density of sea water = 1029 kg/m^3
g is acceleration due to gravity = 9.81 m/s^2
h is the depth below the water that this pressure acts
substituting values, we have
[tex]P_{sea}[/tex] = 1029 x 9.81 x h = 10094.49h
The gauge pressure within the submarine [tex]P_{g}[/tex] = 101 kPa = 101000 Pa
this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.
Equating the pressure [tex]P_{sea}[/tex], we have
62 x 10^6 = 10094.49h
depth h = 6141.96 m
