A remote ranch has a cylindrical water storage tank. It has a vertical central axis, a diameter of 16 ft, the sides are 5 ft hig
h, and it is full to the brim. The depth of this water is 4 ft. How much work (in ft-lb) would be required to pump all of the water over the upper rim? (Use the fact that water weighs 62.5 lb/ft3. Round your answer to the nearest integer.)
1 answer:
Answer and Step-by-step explanation:
Given: diameter = 16ft height = 5 ft. Depth = 4ft
Work =?
As we know that work = force x distance
W= distance x density x volume
W= ʃ04 (5-y) 62.5x ∏ (d/2)2dy (area = ∏r2 , r= d/2)
= 62.5 ʃ04 (5-y) ∏ (4y) dy
=62.5∏ ʃ04 (5-y) 4ydy
=62.5∏ ʃ04 20y-4y2dy
=62.5∏ |20y2/2 – 4y3/3|04
=62.5∏ |74.66|
=14659ft.lbs (approximately)
You might be interested in
Answer:
Expected Profit = $220,700
Step-by-step explanation:
(SEE ATTACHMENT BELOW)
Given
Wholesale = $20
Retail = $30
Total profit made = Profit made from wholesale + profit made from retails
We get total profit made from wholesale by getting the number of units sold
Number of units sold = 2000 * 0.1 + 5000 * 0.3 + 10000 * 0.4 + 20000 * 0.2
Units sold = 200 + 1500 + 4000 + 4000
Units sold = 9700
Wholesale Profit = Units * Wholesale price
Wholesale Profit = 9700 * $20
Wholesale Profit = $194,000
Similarly, We get total profit made from retail by getting the number of units sold
Number of units sold = 600 * 0.4 + 1000 * 0.5 + 1500 * 0.1
Units sold = 240 + 500 + 150
Units sold = 890
Retail Profit = Units * Retail price
Retail Profit = 890 * $30
Retail Profit = $26,700
Total Profit = Wholesale Profit + Retail Profit
Total Profit = $194,000 + $26,700
Total Profit = $220,700
