An underground tank full of water has the following shape
1) hemisphere of radius 5 m at the bottom
2) a cylinder of radius 5 m and height 10 m in the middle
3) a circular cone with base radius 5 and height 4 m at the top
The top of the tan is 2 m below the ground surface and is connected to the surface by a spout. Find the work required to empty the tank by pumping all of the water out of the tank up to the surface
Density of water = 1000kg/m^3
Gravity = 10m/s^2
I am doing it so I find the work done of all three (three different parts) then adding the all 3 work done to get the final answer. I really just need help solving the cone I do not even know where to begin! Someone please help me solve this!
I can help you with finding the volume of the different shapes
volume of hemisphere = (1/2)(4/3)π(5^3) = (50/3)π
volume of cyliner = π(5^2)(10) = 250π
volume of cone = (1/3)π(5^2(4) = (100/3)π
The rest of the question deals with concepts of physics, of which I have little knowledge.
To find the work required to empty the tank, we need to calculate the work done for each part separately and then add them up.
Let's start with the hemisphere at the bottom of the tank:
1) Hemisphere:
The volume of a hemisphere is given by V = (2/3)πr^3, where r is the radius. In this case, r = 5 m. So the volume of the hemisphere is V = (2/3)π(5^3) = 250π m^3.
To calculate the work done in pumping out the water from the hemisphere, we need to consider the weight of the water. The weight of the water is given by W = mg, where m is the mass of the water and g is the acceleration due to gravity.
The mass of water is equal to the density times the volume: m = ρV, where ρ is the density of water (1000 kg/m^3). So, m = 1000 * 250π = 250000π kg.
Now, the work done in pumping out the water from the hemisphere is given by W = mgh, where h is the height difference between the center of mass of the water and the surface. Since the hemisphere is at the bottom, h = 0. Therefore, the work done in pumping out the water from the hemisphere is zero.
Next, let's move on to the cylinder in the middle of the tank:
2) Cylinder:
The volume of a cylinder is given by V = πr^2h, where r is the radius and h is the height. In this case, r = 5 m and h = 10 m. So the volume of the cylinder is V = π(5^2)(10) = 250π m^3.
Similar to the hemisphere, the mass of water in the cylinder is m = ρV = 1000 * 250π = 250000π kg.
To calculate the work done, we need to find the height difference, h. The center of mass of the water in the cylinder is at half the height of the cylinder, which is h/2 = 10/2 = 5 m below the ground surface.
The work done in pumping out the water from the cylinder is given by W = mgh = 250000π * 10 * 5 = 12500000π J.
Finally, let's consider the top part of the tank, which is a cone:
3) Cone:
The volume of a cone is given by V = (1/3)πr^2h, where r is the radius and h is the height. In this case, r = 5 m and h = 4 m. So the volume of the cone is V = (1/3)π(5^2)(4) = 100π m^3.
Similar to the previous parts, the mass of water in the cone is m = ρV = 1000 * 100π = 100000π kg.
To calculate the work done, we need to find the height difference, h. The center of mass of the water in the cone is at one-third of the height of the cone, which is h/3 = 4/3 m below the ground surface.
The work done in pumping out the water from the cone is given by W = mgh = 100000π * 10 * (4/3) = 4000000π/3 J.
Now, to find the total work done, we add up the work done for each part:
Total work done = Work done for the hemisphere + Work done for the cylinder + Work done for the cone
Total work done = 0 + 12500000π + 4000000π/3
Therefore, the total work required to empty the tank by pumping all the water out of the tank up to the surface is approximately 12500000π + 4000000π/3 J.