By how much does the H20 pressure exceed atmospheric pressure at 3 meters below the surface of a pool with fresh H20.
I don't know what equation to use.
You have to figure the weight of a column of water of some height h, and horizontal area A.
Weight= mg= density*volume*g
=density*hA*g
then
Pressure= WEight/A= density*h*g
There is a rule of thumb you ought to memorize: 10 meters of water is about equal to one atmosphere of pressure.
To calculate the pressure at a certain depth in water, you can use the equation:
Pressure = density * gravity * depth
First, you need to know the density of water, which is approximately 1000 kg/m^3.
Next, you need to find the depth in meters. In this case, it is 3 meters.
The acceleration due to gravity is approximately 9.8 m/s^2.
Using these values, you can substitute them into the equation to find the pressure:
Pressure = 1000 kg/m^3 * 9.8 m/s^2 * 3 m
= 29400 pascals (Pa)
To convert pascals to atmospheres, you can use the conversion factor:
1 atmosphere = 101325 pascals
So, the pressure at 3 meters below the surface of the pool with fresh water exceeds atmospheric pressure by:
29400 Pa / 101325 Pa = 0.289 atmospheres
Therefore, the H20 pressure exceeds atmospheric pressure by approximately 0.289 atmospheres.