Answer:
247500 Joule
Step-by-step explanation:
length of tank, l = 4 feet
width of the tank, w = 5 feet
depth of the tank, h = 15 feet
density of liquid, d = 110 pounds per cubit foot
Area of crossection, A = length x width
A = 4 x 5 = 20 ft²
Let the height is dh
Force due to height dy
dF = A x dy x density
dF = 20 x 110 x dy = 2200 dy
Work done in increasing the level by y is
dW = 2200 x y x dy
Total work done is
[tex]W = \int dW = \int_{0}^{15}2200 y dy[/tex]
[tex]W = 1100[y^{2}]_{0}^{15}[/tex]
W = 1100 ( 15 x 15 - 0 )
W = 247500 Joule
Answer:
247500 J
Step-by-step explanation:
We are given that
Length,l=4 feet
Width,b=5 feet
Height,h=15 feet
Weigh,w=110[tex]pounds/ft^3[/tex]
We have to find the work done in pumping all of the liquid out over the top of the tank.
Area of cross-section,A=[tex]l\times b=4\times 5=20ft^2[/tex]
Force increment,[tex]dF=20\times 110dy=2200dy[/tex]
Work increment ,[tex]dW=2200ydy[/tex]
Total work done,W=[tex]\int_{0}^{15}2200ydy[/tex]
[tex]W=2200[\frac{y^2}{2}]^{15}_{0}[/tex]
[tex]W=2200\times \frac{(15)^2}{2}=247500 J[/tex]
