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Elenna [48]
3 years ago
11

Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diamete

r and 6 meters long is carried on the back of a truck and is used to fuel tractors. The axis of the tank is horizontal. The opening on the tractor tank is 5 meters above the top of the tank in the truck. Find the work done in pumping the entire contents of the fuel tank into the tractor.
Physics
1 answer:
almond37 [142]3 years ago
5 0

Answer:

<em>work done in pumping the entire fuel is 1399761 J</em>

Explanation:

weight per volume of the gasoline = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 6 m

The height of the tractor tank above the top of the tank = 5 m

The total volume of the fuel is gotten below

we know that the tank is cylindrical.

<em>we assume that the fuel completely fills the tank.</em>

therefore, the volume of a cylinder =  

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x  x 6 = 42.417 m^3

we then proceed to find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 42.417 = 279952.2 N

therefore,

the work done to pump the fuel through to the 5 m height = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 279952.2 x 5 = <em>1399761 J</em>

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A=Fh
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so
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A fireperson is 50 m from a burning building and directs a stream of water from a fire hose at an angle of 300 above the horizon
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Answer:

We can think the water stream as a solid object that is fired.

The distance between the fireperson and the building is 50m. (i consider that the position of the fireperson is our position = 0)

The angle is 30 above the horizontal. (yo wrote 300, but this has no sense because 300° implies that he is pointing to the ground).

The initial speed of the stream is 40m/s.

First, using the fact that:

x = R*cos(θ)

y = R*sin(θ)

in this case R = 40m/s and θ = 30°

We can use the above relation to find the components of the velocity:

Vx = 40m/s*cos(30°) = 34.64m/s

Vy = 20m/s.

First step:

We want to find the time needed to the stream to hit the buildin.

The horizontal speed is 34.64m/s and the distance to the wall is 50m

So we want that:

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t = 50m/(34.64m/s) = 1.44 seconds.

Now we need to calculate the height of the stream at t = 1.44s

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The velocity equation is:

v(t) = -g*t + 20m/s.

For the position, we integrate again over time:

p(t) = -(1/2)*g*t^2 + 20m/s*t + p0

p0 is the initial height of the stream, this data is not known.

Now, the height at the time t = 1.44s is

p(1.44s) = -5.9m/s^2*(1.44s)^2 + 20m/s*1.44s + po

             = 16.57m + p0

So the height at wich the stream hits the building is 16.57 meters above the initial height of the fire hose.

5 0
3 years ago
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