Question

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)​

Answer 1

`\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}`