Answer:
19467649.76 lb-ft^2/s^2
Step-by-step explanation:
diameter of the pool d = 20 ft
radius = d/2 = 20/2 = 10 ft
height of pool side h = 6 ft
depth of water d = 5 ft
the force on the bottom of the pool due to the water in the pool is
F = pgdA
where p = density of water = 62.4 lb/ft^3
g = acceleration due to gravity = 32.17 ft/s^2
Area A = [tex]\pi r^{2}[/tex] = [tex]3.142 * 10^{2}[/tex] = 314.2 ft^2
Force on pool bottom = 64.2 x 32.17 x 5 x 314.2 = 3244608.29 lb-ft/s^2
work done = force times the height the water will be pumped
work = F x h = 3244608.29 x 6 = 19467649.76 lb-ft^2/s^2
The work (in ft-lb) is required to pump all of the water out over the side is :
Formula:
F = pgdA
p = density of water = 62.4 lb/ft^3
g = acceleration due to gravity = 32.17 ft/s^2
Area A = [tex]\pi r2\\[/tex] = 314.2 ft^2
The work (in ft-lb) is required to pump all of the water out over the side is 19467649.76 lb-ft^2/s^2.
Learn more :
https://brainly.com/question/13102487referrer=searchResults
