A circular swimming pool has a diameter of 20 ft, the sides are 6 ft high, and the depth of the water is 5 ft. How much work (in ft-lb) is required to pump all of the water out over the side

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 = = = 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

good bye

step-by-step explanation:

the answer is d

step-by-step explanation:

y+3=-4(x-2)

y+3=-4x+8

  -3         -3  

y=-4x+5