Answer:
[tex]\displaystyle \rho=1,024.74\ kg/m^3[/tex]
Explanation:
Pressure
[tex]P=P_o+\rho gh[/tex]
Where [tex]P_o[/tex] is the atmospheric pressure, [tex]\rho[/tex] is the density of the fluid, and h is the depth
The values are
:
[tex]P_o= 1\ atm= 101,325\ Pa[/tex]
[tex]P=261,000\ Pa[/tex]
[tex]g=9.8\ m/s^2[/tex]
[tex]h=15.9\ m[/tex]
Solving the above equation for [tex]\rho[/tex]
[tex]\displaystyle \rho=\frac{P-P_o}{gh}[/tex]
[tex]\displaystyle \rho=\frac{261,000-101,325}{9.8(15.9)}[/tex]
[tex]\displaystyle \rho=\frac{159,675}{155.82}[/tex]
[tex]\boxed{ \rho=1,024.74\ kg/m^3}[/tex]
