Consider the following depiction of a tank with radius r = 9 meters with a spout of height h = 4.5 meters.
A tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g and 3.14 for π. Round your answer to the nearest integer.)
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To find the work required to pump the oil out of the spout, we need to calculate the gravitational potential energy of the oil in the tank.
The formula for gravitational potential energy is given by:
PE = m * g * h
where:
PE is the gravitational potential energy
m is the mass of the object
g is the acceleration due to gravity
h is the height
To find the mass of the oil in the tank, we use the formula:
m = ρ * V
where:
m is the mass
ρ is the density
V is the volume
In this case, the tank is halfway filled, which means the volume of oil is half the volume of the tank. The volume of a cylinder is given by:
V = π * r^2 * h
Substituting the given values, we get:
V = 3.14 * (9^2) * 4.5
Next, we can find the mass of the oil:
m = 900 * V
Now we have the mass of the oil and we can substitute it into the formula for gravitational potential energy:
PE = m * g * h
Substituting the values of m, g, and h, we get:
PE = (900 * V) * 9.8 * 4.5
Now we can find the work required to pump the oil out of the spout. The work is equal to the change in potential energy, which is the difference between the potential energy when the tank is half full and when it is empty. Since the tank is half full, the initial potential energy is:
PE_initial = 0.5 * ((900 * V) * 9.8 * 4.5)
When the tank is empty, the potential energy is zero. So the final potential energy is:
PE_final = 0
The work required is therefore the difference between the initial potential energy and the final potential energy:
W = PE_initial - PE_final
W = 0.5 * ((900 * V) * 9.8 * 4.5) - 0
Now we can substitute the value of V and calculate the work:
V = 3.14 * (9^2) * 4.5
W = 0.5 * ((900 * V) * 9.8 * 4.5)
Using the given values and rounding the answer to the nearest integer, the work required to pump the oil out of the spout is obtained.