Answer:
3.9 × 10^7 J
Explanation:
Given that a tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J
Solution
Since the tank is half full, the height = 2.5m
Pressure = density × gravity × height
Pressure = 900 × 9.8 × 2.5
Pressure = 22050 Pascal
The cross sectional area of the pump will be area of a circle.
A = πr^2
A = π × 15^2
A = 706.858 m^2
Using the formula
Density = mass/volume
Mass = density × volume
Mass = 900 × 706.86 × 2.5
Mass = 1590.435
Energy = mgh
Energy = 1590.435 × 9.8 × 2.5
Energy = 38965657.8 J
Since the work done = energy
Therefore, the work done = 3.9 × 10^7 J
