You are designing a diving bell to withstand the pressure of seawater at a depth

Question

You are designing a diving bell to withstand the pressure of seawater at a depth of 300 m . A)What is the gauge pressure at this depth? (You can ignore changes in the density of the water with depth.) B)At this depth, what is the net force due to the water outside and the air inside the bell on a circular glass window 31.0 cm in diameter if the pressure inside the diving bell equals the pressure at the surface of the water? (You can ignore the small variation of pressure over the surface of the window.)

Explanation / Answer

the pressure due to depth equation is (rho)*g*h

where:
rho = density of the fluid (seawater, a little more than 1, say 1.003 kg/m^3)
g = acceleration due to gravity (9.8 m/s^2)
h = height of water column. (300m)

Pressure = 1*9.8*300 = 2940

B) given the diameter, we can find the area of the window
A = pi*d^2/4
then Force is pressure / area

A = pi*31^2/4

F = 2940/(pi*0.31^2/4 )

F = 38952.4 N

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