The density is [tex]1106 kg/m^3[/tex]
Explanation:
The bulk modulus of a liquid is defined as
[tex]B=\rho_0 \frac{\Delta p}{\Delta \rho}[/tex]
where
[tex]\rho_0[/tex] is the density at the reference level
[tex]\Delta p[/tex] is the change in pressure
[tex]\Delta \rho[/tex] is the change in density
The equation can be re-arranged as
[tex]\Delta \rho = \frac{\rho_0 \Delta p}{B}[/tex]
In this problem, we have:
[tex]\rho_0 = 1100 kg/m^3[/tex] (density at the surface)
[tex]B=2.3\cdot 10^9 N/m^2[/tex] (bulk modulus of seawater)
[tex]\Delta p = 126 - 1 = 125 atm = (125\cdot 1.01\cdot 10^5) Pa[/tex] (pressure difference between the point where p = 126 atm and the pressure at sea level, which is 1 atm)
Therefore,
[tex]\Delta \rho = \frac{(1100)(125\cdot 1.01\cdot 10^5)}{2.3\cdot 10^9}=6.0 kg/m^3[/tex]
So, the density at the depth where the pressure is 126 atm is
[tex]\rho = \rho_0 +\Delta \rho = 1100+6=1106 kg/m^3[/tex]
Learn more about pressure in fluids:
brainly.com/question/4868239
brainly.com/question/2438000
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