The Work required to pump water out of the spout would be equal to 464,062.5 π
Given data :
3 ft, 4 ft, 5 ft A right triangular prism tank with a spout is given.
weight of water = 62.5 Ib/ft³
inner radius ( r ) = 3 x 2 =6 ft
Outer radius ( R ) = 3 x 4 = 12 ft
Height ( h ) = 3 x 5 = 15 ft
To determine the work required to pump water out of the spout
y = Vertical distance
dv = cross-section volume
dw = weight
The radius of the tank
Radius = ( r - R ) y/h + R
Radius = ( 6 - 12 ) y/15 + 12
So, Radius = 12 - 0.4y
dv ( cross section volume )
[tex]\pi r^2\\\\\pi (12 - 0.4y)^2dy\\[/tex]
Weight of cross-section ( dw ) = 62.5 ( 12.5 - 0.4y )² π dy
The work done to pump water out of the spout will be
Work done = ∫¹⁵₀ 62.5 ( 12.5 - 0.4y )² π dy
∴ Work done ( W ) = 464,062.5
Hence we can conclude that the Work required to pump water out of the spout would be 464,062.5 π
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