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tamaranim1 [39]
3 years ago
9

009 10.0 points

Physics
1 answer:
Ilya [14]3 years ago
3 0

The density is 1106 kg/m^3

Explanation:

The bulk modulus of a liquid is defined as

B=\rho_0 \frac{\Delta p}{\Delta \rho}

where

\rho_0 is the density at the reference level

\Delta p is the change in pressure

\Delta \rho is the change in density

The equation can be re-arranged as

\Delta \rho = \frac{\rho_0 \Delta p}{B}

In this problem, we have:

\rho_0 = 1100 kg/m^3 (density at the surface)

B=2.3\cdot 10^9 N/m^2 (bulk modulus of seawater)

\Delta p = 126 - 1 = 125 atm = (125\cdot 1.01\cdot 10^5) Pa (pressure difference between the point where p = 126 atm and the pressure at sea level, which is 1 atm)

Therefore,

\Delta \rho = \frac{(1100)(125\cdot 1.01\cdot 10^5)}{2.3\cdot 10^9}=6.0 kg/m^3

So, the density at the depth where the pressure is 126 atm is

\rho = \rho_0 +\Delta \rho = 1100+6=1106 kg/m^3

Learn more about pressure in fluids:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

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