Question

The pressure at the top of a liquid

Answer 1

Answer:

$\displaystyle \rho=1,024.74\ kg/m^3$

Explanation:

Pressure

$P=P_o+\rho gh$

Where $P_o$ is the atmospheric pressure, $\rho$ is the density of the fluid, and h is the depth

The values are :

$P_o= 1\ atm= 101,325\ Pa$

$P=261,000\ Pa$

$g=9.8\ m/s^2$

$h=15.9\ m$

Solving the above equation for $\rho$

$\displaystyle \rho=\frac{P-P_o}{gh}$

$\displaystyle \rho=\frac{261,000-101,325}{9.8(15.9)}$

$\displaystyle \rho=\frac{159,675}{155.82}$

$\boxed{ \rho=1,024.74\ kg/m^3}$