Mymathlab A diving pool that is 8 m deep and full of water has a viewing window on one of its vertical walls. Find the force on a window that is a%E2%80%8B circle, with a radius of 2m, tangent to the bottom of the pool.

492.6 kN

Explanation:

Since the window as a radius of 2 m, its diameter is thus 4 m. Since the diving pool is 8 m deep, the height of water from the top of the window to the bottom of the pool is 8 - 4 = 4 m. The actual pressure acting on the window is thus

P = ρgh were ρ = density of water = 1000 kg/m³ g = 9.8 m/s² and h = 4 m

P = 1000 kg/m³ × 9.8 m/s² × 4 m = 39200 N/m².

Since P = F/A were F = force and A = area,

F = PA were A = area of window = πr² = π2² = 4π m² = 12.57 m²

F = 39200 N/m² × 12.57 m² = 492601.73 N = 492.6 kN

answer: 31, 5556. 952 seconds

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures