A gas station stores its gasoline in a tank underground. The tank is a cylinder lying horizontally on its side. The radius is 5 ft, the length is 15 ft, and the top of the tank is 10 feet under the ground. Assume the tank is full and all of the gasoline will be pumped to the surface of the ground. The density of gasoline is 42 lb/ ft3. Consider a slice of gasoline that is delta y ft thick and located y ft above the center of the cylinder. Use delta or the CalcPad for delta . Leave pi in your answer. Volume of slice : ft3 Displacement of slice : ft Find the endpoints of the integral needed to find the exact work required to pump all the gasoline to the surface of the ground. Lower endpoint = Upper endpoint =

The exact work required to pump all the gasoline to the surface of the ground is π × 5.094 × 10⁶ j

Step-by-step explanation:

Here, we note that volume of a slice is given by

Length × Width × Height

Length of slice = Length of cylinder = 15 ft

Since the slice height is Δy ft thick and located y ft above the center of the cylinder, then

Width of slice = 2 × √(r² - y²)

Where:

r = Radius of the cylinder =5 ft

∴ Width of slice = 2 × √(25 - y²)

∴Volume of slice = 15 ×2 × √(25 - y²)×Δy

Mass of slice then = 42 × 15 ×2 × √(25 - y²)×Δy = 1260 × √(25 - y²)×Δy

The force required to lift the slice is the weight of the slice, which is given by

32.2 × 1260 × √(25 - y²)×Δy = 40572 × √(25 - y²)×Δy N (Newtons)

The work done by the force is the product of the force and the distsnce through which the force acts.

Work done = 40752×(10-y) × √(25 - y²)×Δy

Therefore total work done is given by

= 5094000·π J = π × 5.094 × 10⁶ j

The exact work required to pump all the gasoline to the surface of the ground = π × 5.094 × 10⁶ j

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00