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riadik2000 [5.3K]
4 years ago
5

A fuel oil tank is an upright cylinder, buried so that its circular top is 14 feet beneath ground level. the tank has a radius o

f 5 feet and is 15 feet high, although the current oil level is only 14 feet deep. calculate the work required to pump all of the oil to the surface. oil weighs 50lb/ft350lb/ft3.
Physics
1 answer:
Ierofanga [76]4 years ago
3 0

Say we have a cylinder that has a height of dx, we see that the cylinder has a volume of: <span>

<span>Vcylinder = πr^2*h = π(5)^2(dx) = 25π dx

Then, the weight of oil in this cylinder is: 

Fcylinder = 50 * Vcylinder = (50)(25π dx) = 1250π dx. 

Then, since the oil x feet from the top of the tank needs to travel x feet to get the top, we have: 

Wcylinder = Force x Distance = (1250π dx)(x) = 1250π x dx. 

<span>Integrating from x1 to x2 ft gives the total work to be: (x1 = distance from top liquid level to ground level; x2 = distance from bottom liquid level to ground level)</span>

<span>W = ∫ 1250π x dx  
<span>W = 1250π ∫ x dx
W = 625π * (x2 – x1)</span></span></span></span>

<span>x2 = 14 ft  + 15 ft = 29 ft</span>

x1 = 14 ft + 1 ft = 15 ft

<span>
W = 625π * (29^2 - 15^2) 
<span>W = 385,000π ft-lbs = 1,209,513.17 ft-lbs</span></span>

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Flura [38]

The given data is incomplete. The complete question is as follows.

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36.  Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)

Explanation:

Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and \mu be the friction coefficient and m be the mass of car.

Hence, the given data is as follows.

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According to Newton's law of second motion the expression for acceleration is as follows.

                      F = ma

                 -\mu N = ma

                 -\mu mg = ma

                      a = -\mu g

Also,    

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              (0)^{2} = u^{2} + 2(-\mu g)s

                  u^{2} = 2(\mu g)s

                            = \sqrt{2(0.36)(9.81 m/s^{2})(84 m)}

                            = 24.36 m/s

Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.

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