Question

The pressure exerted by a liquid at a depth of 2.5m is 36,750 Pa. What is the density of the liquid, to

the nearest kg/m³?
A pool containing a liquid with a density of 1,000 kg/m³ at the surface of an unknown planet produces a pressure of 8,400 Pa at a depth of 2.4 m. What is the acceleration due to gravity (g)at the surface of the planet?

Answer 1

Answer:
`1500\ kg/m^3,\ 3.5\ m/s^2`

Step-by-step explanation:

Given

The pressure exerted at a depth of
`h=2.5\ m` is
`P=36,750\ Pa`

Also, the gauge pressure is given by

`P=\rho gh`

Putting values

`\Rightarrow 36,750=\rho * 9.8* 2.5\\\\\Rightarrow \rho=(36750)/(24.5)=1500\ kg/m^3`

(b)

The density of the liquid is
`\rho =1000\ kg/m^3`

depth
`h=2.4\ m`

Pressure
`P=8400\ Pa`

We can write

`\Rightarrow P=\rho g'h\\\\\Rightarrow 8400=1000* g'* 2.4\\\\\Rightarrow g'=3.5\ m/s^2`