Respuesta :
Answer:
Therefore work required to pump out of all liquid is 81p ft-lb.
Step-by-step explanation:
Given that,
A rectangle tank dimension is 3 feet long, 6 feet wide and 3 feet tall.
The density of the liquid in the rectangular tank is p.
The volume of water [tex]\triangle y[/tex] of strip is
=long ×wide ×[tex]\triangle y[/tex]
=(3 ×6 ×[tex]\triangle y[/tex]) ft³
Force is = (3 ×6 ×[tex]\triangle y[/tex])×p pound
Total work done
[tex]W=\int_0^3 3\times 6 \times p\times \triangle y(3- y)[/tex]
[tex]=\int_0^318p (3- y)dy[/tex]
[tex]=18p\int_0^3(3-y)dy[/tex]
[tex]=18p[3y-\frac{y^2}{2}]_0^3[/tex]
[tex]=18p[(3.3-\frac{3^2}{2})-(3.0-\frac{0^2}{2})][/tex]
[tex]=18p[9-\frac{9}{2}][/tex]
[tex]=18p\times \frac92[/tex]
=81p ft-lb
Therefore work required to pump out of all liquid is 81p ft-lb.
The overall done work is "81p ft-lb", and further calculations can be defined as follows:
Work calculation:
Rectangular tank Dimension:
length =[tex]3 \ feet[/tex]
width = [tex]6 \ feet[/tex]
height= [tex]3\ feet[/tex]
Calculating the volume of water by using [tex]\Delta y[/tex] the strip:
=length * wide *[tex]\Delta y[/tex]
[tex]=(3 \times 6 \times \Delta y) \ ft^3\\\\[/tex]
Force [tex]F= (3\times 6 \times \Delta y ) \times p \ pound\\\\[/tex]
Calculating the total work:
[tex]W = \int^3_{0} 3 \times 6 \times p \times \Delta y (3 - y)\\\\[/tex]
[tex]=\int^{3}_{0} 18p(3 -y )dy\\\\ =18p \int^{3}_{0} (3 - y)dy \\\\ = 18p [3y - \frac{y^2}{2}]^{3}_{0}\\\\ =18p [ 9-\frac{9}{2}] \\\\= 18 p\times \frac{9}{2}\\\\= 81 \ ft- lb\\\\[/tex]
As a result, the amount of work necessary to pump out all of the liquid is 81p ft-lb.
Find out more about the work here:
brainly.com/question/62183
