The pressure at the top of a liquid is 1.00 atm. At a depth of 15.9 m, the fluid pressure is 261000 Pa. What is the density of the fluid? (Unit = kg/m^3)

Question

asked Sep 9, 2021 217k views

1 Answer

Oimitro · Answer 1 · 2021-09-13T08:42:18+0000

Answer:

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

Step-by-step 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
$\rho$

$\displaystyle \rho=(P-P_o)/(gh)$

$\displaystyle \rho=(261,000-101,325)/(9.8(15.9))$

$\displaystyle \rho=(159,675)/(155.82)$

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