Respuesta :
Answer:
Step-by-step explanation:
Given that tank has the shape of inverted circular cone
Height = 10 feet and radius = 4 feet
After water pumped out height = 5 ft
Hence volume of water pumped out = [tex]\frac{1}{3} \pi R^2 h-\frac{1}{3} \pi r^2 h'[/tex]
Here we have r/h is constant always
Hence [tex]\frac{r}{h'} =\frac{R}{h} \\r=Rh'/h = 4(5)/10 = 2[/tex]
Substitute to get volume of water pumped out = [tex]\frac{1}{3} \pi (10*16-4*5)=\frac{140\pi}{3}[/tex]
Mass of water = density x volume = [tex]62.4(\frac{140\pi}{3})\\= 2912\pi lbs[/tex]
Work done = force x displacement
= mass x accen x displacement
Here acceleration = gravity = 32.2 ft/sec^2
Displaement = height reduced = 5 ft
W=2912(32.2)(5)\pi =468832\pi foot -pound
The tank as a cone.
As per the question, the tank is given a shape of an inverted circular cone has a point to the bottom with an height of radius of 4 feet. The tank is full of water the pipe can be cued to pump out the water from the top and until which the tank ill have a level of 5 feet from the bottom.
Thus the answer is W equals to 468832 foot-pound
- As per the given information the tank consists of the inverted circular cone the Height of cone is equal to 10 feet and radius = 4 feet After water pumped out height = 5 ft. Thus the volume of water pumped out
- Here we have r/h is constant always
- Hence the Substitute to get volume of water pumped out equals to the Mass of water = density x volume = Work done = force x displacement = mass x accents x displacement. Here acceleration = gravity = 32.2 ft/sec^2. Displacement = height reduced = 5 ft.
- Hence the W equals to 2912(32.2) that is 468832\pi foot-pound.
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