The density of seawater at a depth where the pressure is 500 atm is [tex]1124kg/m^3[/tex]
Explanation:
The relationship between bulk modulus and pressure is the following:
[tex]B=\rho_0 \frac{\Delta p}{\Delta \rho}[/tex]
where
B is the bulk modulus
[tex]\rho_0[/tex] is the density at surface
[tex]\Delta p[/tex] is the variation of pressure
[tex]\Delta \rho[/tex] is the variation of density
In this problem, we have:
[tex]B=2.3\cdot 10^9 N/m^2[/tex] is the bulk modulus
[tex]\rho_0 =1100 kg/m^3[/tex]
[tex]\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa[/tex] is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)
Therefore, we can find the density of the water where the pressure is 500 atm as follows:
[tex]\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3[/tex]
Learn more about pressure in a fluid:
brainly.com/question/9805263
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