The maximum depth d_max that a diver can snorkel is set by the density of the water and the fact that human lungs can function against a maximum pressure difference (between inside and outside the chest cavity) of 0.050 atm. What is the difference in d_max for fresh water and the water of the Dead Sea (the saltiest natural water in the world, with a density of 1.5 × 10³kg/m³)?

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boffeemadrid boffeemadrid · Answer 1 · 2019-12-27T07:15:42+01:00

Answer:

0.17214 m

Explanation:

P = Pressure difference between inside and outside the chest cavity = 0.05 atm

[tex]\rho[/tex] = Density of fresh water = 1000 kg/m³

[tex]\rho_d[/tex] = Density of water in the Dead Sea= 1500 kg/m³

h = Height

In fresh water

[tex]P=\rho gh_1\\\Rightarrow h_1=\dfrac{P}{\rho g}\\\Rightarrow h_1=\dfrac{0.05\times 101325}{1000\times 9.81}\\\Rightarrow h_1=0.51643\ m[/tex]

In dead sea

[tex]P=\rho_d gh_2\\\Rightarrow h_2=\dfrac{P}{\rho_d g}\\\Rightarrow h_2=\dfrac{0.05\times 101325}{1500\times 9.81}\\\Rightarrow h_2=0.34429\ m[/tex]

The difference in height is 0.51643-0.34429 = 0.17214 m

Vespertilio Vespertilio · Answer 2 · 2019-12-28T01:57:48+01:00

Answer:

17.18 m

Explanation:

density of fresh water = 1000 kg /m^3

density of dead sea = 1500 kg/m^3

Pressure, P = 5 atm = 5 x 1.01 x 10^5 = 5.05 x 10^5 Pa

Let h be the depth in fresh water and h' be the depth in sea water

P = h x density of fresh water x g

5.05 x 10^5 = h x 1000 x 9.8

h = 51.53 m

Now for dead sea

P = h' x density of dead sea x g

5.05 x 10^5 = h' x 1500 x 9.8

h' = 34.35 m

So, the difference in height

d = h - h'

d = 51.53 - 34.35

h' = 17.18 m