Question

N in-ground swimming pool has the dimensions shown in the drawing. It is filled with water to a uniform depth of 3.00 m. The density of water = 1.00 × 103 kg/m3. What is the total pressure exerted on the bottom of the swimming pool? 1.97 × 105 Pa 2.49 × 105 Pa 2.94 × 104 Pa 1.80 × 105 Pa 1.31 × 105 Pa

Answer 1

Answer:

The pressure is  $P = 1.31*10^{5} \ Pa$

Explanation:

From the question we are told that

    The depth of the swimming pool is  $d = 3.00 \ m$

     The density of water is  $\rho = 1.00*10^{3} \ kg /m^3$

Generally the  total pressure exerted on the bottom of the swimming pool is mathematically represented as

          $P = P_o + \rho * g * h$

Here $P_o$ is the atmospheric pressure with value

        $P_o = 101325 \ Pa$

So

        $P = 101325 + [1000 * 9.8 * 3]$

=>     $P = 130725 \ Pa$

=>    $P = 1.31*10^{5} \ Pa$