Answer:
130.826 kilojoules
Step-by-step explanation:
The Work required to pump water = pgV
Where, p = density of water = 1000 kg/m³
g = acceleration due to gravity = 9.81ms-¹
V = volume of water
Since, radius r = 5 ft and height h = 9ft
Volume of water in the cylindrical tank = (2/3)πr²h
V = (2/3)π *5²*9 = 471.24 ft³ = 471.24 * 0.0283m³ = 13.336m³
Work required = 1000*9.81*13.336
W = 130.826 KJ
Therefore, the work required to pump water 2/3 of the volume of the cylindrical tank = 130.826 kilojoules.
