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A boat has a porthole window, of area 0.00849 m^2, 6.25 m below the surface. The density of sea water is 1027 kg/m^3. The air inside the boat is at 1 atm. What is the net force on the window?

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Answer:

[tex]325.65\ \text{N}[/tex] outward.

Explanation:

[tex]\rho[/tex] = Density of water = [tex]1027\ \text{kg/m}^3[/tex]

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

h = Height of the water = 6.25 m

A = Area of the window = [tex]0.00849\ \text{m}^2[/tex]

[tex]P_i[/tex] = Internal pressure of the boat = [tex]1\ \text{atm}=101325\ \text{Pa}[/tex]

Pressure by the water on the window

[tex]P=\rho gh\\\Rightarrow P=1027\times 9.81\times 6.25\\\Rightarrow P=62967.9375\ \text{Pa}[/tex]

Net pressure on the window

[tex]P_n=P-P_i=62967.9375-101325=-38357.0625\ \text{Pa}[/tex]

Force is given by

[tex]F=P_nA\\\Rightarrow F=-38357.0625\times 0.00849\\\Rightarrow F=-325.651460625\ \text{N}[/tex]

Net force on the window is [tex]325.65\ \text{N}[/tex] outward.

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