Answer:
The pressure is [tex]P = 1.31*10^{5} \ Pa[/tex]
Explanation:
From the question we are told that
The depth of the swimming pool is [tex]d = 3.00 \ m[/tex]
The density of water is [tex]\rho = 1.00*10^{3} \ kg /m^3[/tex]
Generally the total pressure exerted on the bottom of the swimming pool is mathematically represented as
[tex]P = P_o + \rho * g * h[/tex]
Here [tex]P_o[/tex] is the atmospheric pressure with value
[tex]P_o = 101325 \ Pa[/tex]
So
[tex]P = 101325 + [1000 * 9.8 * 3][/tex]
=> [tex]P = 130725 \ Pa[/tex]
=> [tex]P = 1.31*10^{5} \ Pa[/tex]
