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**[SOLVED] Fluid Mechanics**

**1. Homework Statement**

You are designing a diving bell to withstand the pressure of seawater at a depth of 230.

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

got 2.32x10^6 Pa - masteringphysics said this is correct

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 34.0cm 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.)

**2. Homework Equations**

rho = M/V

p = p0 + (rho)gh

p = F/A

F1/A1 = F2/A2

gauge pressure --> p-p0

**3. The Attempt at a Solution**

F = pA = (2.32x10^6).340m = 7.9x10^5 N - incorrect

F = pA = (2.32x10^6)(.340*230) = 1.81x10^8 N - incorrect

F = pA = (2.32x10^6)(.340^2) = 2.7x10^5 N - incorrect

I think I'm having a hard time visualizing what the question is asking, any help is greaty appreciated.