Question

If a scuba diver descends too quickly into the sea, the internal pressure on each eardrum remains at atmospheric pressure while the external pressure increases due to the increased water depth. At sufficient depths, the difference between the external and internal pressures can rupture aneardrum. Eardrums can rupture when the pressure difference is as little as 35 kPa. What is the depth at which this pressure difference could occur? The density of seawater is 1025 kg/m3.

Answer 1

Answer:

3.48076 m

Explanation:

$\Delta P$ = Pressure difference = 35 kPa

$\rho$ = Density of water = 1025 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Depth of water

Pressure difference is given by

$\Delta P=\rho gh\\\Rightarrow h=\frac{\Delta P}{\rho g}\\\Rightarrow h=\frac{35000}{1025\times 9.81}\\\Rightarrow h=3.48076\ m$

The depth at which this pressure difference occurs is 3.48076 m

Answer 2

Answer:

No answer just a comment....

Explanation:

I am 12 and I am Scuba certified and I can go all the way down to 60 Ft. It's amazing!