Question

Certain underwater vehicles are capable of exploring some of the deepest regions of the ocean. consider the density of seawater to be 1025 kg/m3. if such a vehicle were submerged to a depth of 8513 m below the surface, what force would be exerted on the vehicle's observation window (radius = 9.3 cm)?

Answer 1

The pressure exerted by the water at a depth of h=8513 m below the surface is given by Stevin's law:

`p=p_a + \rho g h`
where

`p_a =1.013\cdot 10^5 Pa` is the atmospheric pressure

`\rho=1025 kg/m^3` is the water density

`g=9.81 m/s^2` is the gravitational acceleration

`h=8513 m` is the depth

If we plug the numbers into the formula, we find the pressure exerted on the window of the vehicle:

`p=1.013 \cdot 10^5 Pa + (1025 kg/m^3)(9.81 m/s^2)(8513 m)=8.57 \cdot 10^7 Pa`

The area of the window is

`A=\pi r^2 = \pi (0.093 m)^2 = 0.027 m^2`

And so the force exerted on the window is

`F=pA=(8.57 \cdot 10^7 Pa)(0.027 m^2)=2.31 \cdot 10^6 N`