A gas station stores its gasoline in a tank under the ground. The tank is a cylinder lying horizontally on its side. (In other words, the tank is not standing vertically on one of its flat ends.) If the radius of the cylinder is 1.5 meters, its length is 5 meters, and its top is 5 meters under the ground, find the total amount of work needed to pump the gasoline out of the tank. (The density of gasoline is 673 kilograms per cubic meter; use g=9.8 m/s2.)

Work needed = 1515.15 KJ

Explanation:

The center of mass of a cylinder lying horizontally on its side would lie on the axis of the cylinder at the center of length l.

Depth of center of mass from ground level;Δh = (r + 5) metres

Now, work done to pump the gasoline out of the tank is equal to the gain in potential energy by gasoline on lifting it from center of mass to the ground level.

Thus;

W = ΔU = mgΔh

We know that mass(m) = volume(V) x density(ρ)

So,

W = (ρV)gΔh

Volume(V) = πr²L

Thus;

W = (ρ(πr²L)) * g(r + 5)

We are given;

Density; ρ = 673 kg/m³

Length; L = 5 m

Radius; r = 1.5 m

Acceleration due to gravity;g = 9.8 m/s²

Thus;

W = (673(π•1.5²•5)) * 9.8(1.5 + 5)

W = 1515154.4 J = 1515.15 KJ

d. air pressure decreases as altitude increases.

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer