Question

009 10.0 points

Answer 1

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:

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