oil is pumped at 31000/liters/min into a tank 25m above the ground.(1lter of oilhas a mass of 0.68kg)

a) what power, in kw, must be delivered by the pump?
b)what is the increase in PE of ht eoil after 10 minutes?
c)Find the increase in PE of the oil after 10minutes if the tank is 5m nabove the ground?

To find the answers to the questions, we can use the formulas for power and potential energy.

a) The power, P, can be calculated using the formula:
P = W/t,
where W is the work done and t is the time taken. In this case, the work done is equal to the change in potential energy (ΔPE) of the oil pumped.
The formula for potential energy is:
PE = mgh,
where m is the mass, g is the acceleration due to gravity, and h is the height.

First, let's find the mass of oil being pumped per minute:
Mass flow rate = 31000 liters/min
1 liter of oil has a mass of 0.68 kg
Mass flow rate = 31000 liters/min * 0.68 kg/liter

Now, let's find the change in potential energy after 10 minutes:
Change in PE = ΔPE = mgh,
where m is the mass of oil pumped in 10 minutes, g is the acceleration due to gravity, and h is the height. We'll assume g is approximately 9.8 m/s².

b) Change in PE = ΔPE = mgh,
where m is the mass of oil pumped in 10 minutes, g is the acceleration due to gravity, and h is the height. However, since the height is given as 25m above the ground, we'll calculate the change in potential energy using this height.

c) Change in PE = ΔPE = mgh,
where m is the mass of oil pumped in 10 minutes, g is the acceleration due to gravity, and h is the height. In this case, the height is given as 5m above the ground, so we'll calculate the change in potential energy using this height.

Let's calculate the answers using these formulas and the given values.