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 220m

A) What is the gauge pressure at this depth? (You can ignore changes in the density of the water with depth.)

I find for A is P=2.22*10^6Pa

B) find 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 35.0

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

a)

gauge pressure pg = rho*g*h

where,

rho is density = 1030 kg/m^3

h is depth=220 m

pg=1030*9.8*220

=2.22068*10^6 Pascal......is answer

b)

net pressure due to water at the depth is Pw = Po + rho*g*h

and pressure due to air inside the bell is PB = Po

here,

Po is presuure at the surface of water

now,

net force due to the water outside and the air inside the circular glass window is

Fnet = (Pw -PB)*A

= (Po + rho*g*h - Po)*A

= (rho*g*h)*A

where

A is area of the crossection of the circular glass

A=pi*r^2

and radius= 35 cm/2=0.175 m

Fnet= 2.22068*10^6*3.14*0.175^2

=2.13546*10^5 N ...is answer

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