## Homework Statement

In scuba diving, a regulator is used so that the pressure of the air the diver breathes is close to that of the ambient water. A reckless swimmer decides to use a hose sticking out of the surface to breathe underwater while diving in a lake. When the air pressure in the lungs is at a pressure of around 0.130 atm below the ambient pressure, lung injury may occur. Find the depth at which the swimmer would experience such a pressure differential.

p = po +pgd

## The Attempt at a Solution

p = po + pgd
-po/pg = d
-101300 / ((0.13 * 101300) * 9.8) = 0.78

What am I doing wrong??

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Oh wait I see something I completely missed!

p = 1 atm - 0.13 atm = 0.87 atm * 101300 pa = 88131 pa
p - po / pg = d

mgb_phys
Homework Helper
pressure = force / area.
force = m g h
I 1m^3 of sea water has a mass of 1025kg then 1 meter column of water exerts 1025kg / m^2
F = 1025 * 9.8 * h = 10045N * h
Atmosphere = 101.4KPa = 101400 N /m^2

0.13 * 101400 = 10045 h
h = 1.3m

As a quick check, if atmsopheric pressure is 101.4KPa this is about 10Kg/m^2 or 1 atmosphere is 10m of water.