What is the total force on the bottom of a 2.0-m-diameter by 2.0-m-deep round wading pool due to the weight of the air and the weight of the water? (Note the pressure contribution from the atmosphere is 1.0 × 105 N/m2, the density of water is 1 000 kg/m3, and g = 9.8 m/s2.)Select one:a.​2.3 × 105 Nb.5.3 × 106 Nc.3.8 × 105 Nd.15 × 106 Ne.3.8 × 106 N

To solve this problem, we need to calculate the force due to the weight of the air and the weight of the water separately, and then add them together.

1. Calculate the force due to the weight of the water:

First, we need to find the volume of the water in the pool. Since the pool is a cylinder, we can use the formula for the volume of a cylinder, which is V = πr²h. The radius r is half the diameter, so r = 1.0 m. The height h is the depth of the pool, which is 2.0 m. So, the volume V = π(1.0 m)²(2.0 m) = 2π m³.

The mass of the water is the volume times the density, so m = Vρ = 2π m³ * 1000 kg/m³ = 2000π kg.

The weight of the water is the mass times the acceleration due to gravity, so W = mg = 2000π kg * 9.8 m/s² = 19600π N.

2. Calculate the force due to the weight of the air:

The pressure contribution from the atmosphere is given as 1.0 × 10⁵ N/m². This is the force per unit area, so the total force is this pressure times the area of the bottom of the pool.

The area A of the bottom of the pool is πr² = π(1.0 m)² = π m².

So, the force due to the weight of the air is F = PA = 1.0 × 10⁵ N/m² * π m² = 100000π N.

3. Add the forces together:

The total force on the bottom of the pool is the sum of these two forces, so F_total = W + F = 19600π N + 100000π N = 119600π N.

This is approximately 3.76 × 10⁵ N, so the closest answer is c. 3.8 × 10⁵ N.