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 hemisphere I do not even know where to begin! Someone please help me solve this!
To find the work required to empty the tank, you need to calculate the work done for each individual part of the tank, and then add them together.
Let's start by calculating the work required to empty the hemisphere at the bottom of the tank.
The work done to empty a hemisphere can be calculated using the formula:
Work = Force x Distance
The force required to lift the water is equal to its weight, which can be calculated using the formula:
Weight = Density x Volume x Gravity
In this case, the density of water is given as 1000 kg/m^3, and the volume of the hemisphere can be calculated using the formula for the volume of a hemisphere:
Volume = (2/3) * π * Radius^3
So, the weight of the water in the hemisphere is:
Weight = Density x Volume x Gravity
= 1000 kg/m^3 x [(2/3) * π * (5m)^3] x 10 m/s^2
Now, we need to calculate the distance over which the water needs to be lifted. In this case, the height of the hemisphere is equal to its radius, so the distance is 5m.
Now we can calculate the work done to empty the hemisphere by multiplying the weight and the distance:
Work_Hemisphere = Weight x Distance
= (1000 kg/m^3 x [(2/3) * π * (5m)^3] x 10 m/s^2) x 5m
= (1000 kg/m^3 x (2/3) x π x 125m^3) x 10 m/s^2 x 5m
You can now simplify and calculate this for the numerical answer. Repeat these steps for the cylinder and the cone parts of the tank, and finally, add up the work done for all three parts to get the total work required to empty the tank.
To find the work required to empty the tank, we need to consider the three different parts separately: the hemisphere, the cylinder, and the cone. Let's start with the hemisphere.
The work done in pumping water out of the hemisphere can be calculated by considering small incremental volumes of water being pumped out and then integrating them.
First, we need to find the incremental work done in pumping an incremental volume of water out of the hemisphere.
The differential volume element of the hemisphere can be expressed as:
dV = πr²dh,
where r is the radius of the hemisphere, and dh is a small height element along the axis of the hemisphere.
To calculate the incremental work done, we need to consider a small volume of water dV at a height h above the surface of the water. The gravitational force acting on this small volume can be expressed as:
dF = ρgdV,
where ρ is the density of water, g is the acceleration due to gravity.
The incremental work done can be calculated as:
dW = dF * h,
Integrating both sides, we get:
W = ∫(dF * h),
where the limits of integration are from 0 to the height of the hemisphere, which is equal to the radius of the hemisphere (since the spout is at the top of the tank).
Now, let's substitute the values into the integral:
W = ∫(ρg * πr² * h) dh,
W = ρgπr² * ∫(h * dh),
W = ρgπr² * [h²/2],
W = (1/2) ρgπr² * h².
Now, we can substitute the given values:
W = (1/2) * 1000 * 10 * π * 5² * 5²,
W = 12,500π J.
So, the work required to pump out the hemisphere is 12,500π J.
Now, you can proceed with finding the work required for the cylinder and the cone using similar calculations, and then add up the three results to get the final answer.