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4 years ago

13

A supersonic aircraft consumes 5320 imperial gallons of kerosene per hour of flight and flies an average of 14 hours per day. it

takes roughly seven tons of crude oil to produce one ton of kerosene. the density of kerosene is 0 965 g/cm . how many planes would it take to consume the entire annual world production of 4 02 10 metric tons of crude oil?

1 answer:

jolli1 [7]4 years ago

4 0

5320*14= 74480 gallons. <span>
First let us convert gallons to cm^3 and we get 2.819374858x10^8 cm^3.
We then multiply with density and then times 1000 to get it in terms of tons.
Then multiply the number by 7 to get how many tons of crude oil a single plane uses in 14 hrs.
Finally divide this by the production of crude oil.

<span>I got 4737 airplanes.</span></span>

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b) power output at t=4.0s
The velocity at t=4.0 s is
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$v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s$

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What is 63,235 inches in km?<br><br> Round your answer to 2 decimal places

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Answer:

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Explanation:

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3 years ago

You are a member of a geological team in Central Africa. Your team comes upon a wide river that is flowing east. You must determ

Dovator [93]

Answer:

(a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

Explanation:

Given that,

Constant speed = 6.00 m/s

Time = 20.1 sec

Speed = 9.00 m/s

Time = 11.2 sec

We need to write a equation for to travel due north across the river,

Using equation for north

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Put the value in the equation

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We need to write a equation for to travel due south across the river,

Using equation for south

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Put the value in the equation

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(a). We need to calculate the wide of the river

Using equation (I) and (II)

$45=\dfrac{w^2}{125.44}-\dfrac{w^2}{404.01}$

$45=w^2(0.00549)$

$w^2=\dfrac{45}{0.00549}$

$w=\sqrt{\dfrac{45}{0.00549}}$

$w=90.5$

We need to calculate the current speed

Using equation (I)

$36-c^2=\dfrac{(90.5)^2}{(20.1)^2}$

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We need to calculate the distance

Using formula of distance

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$d=59.4\ m$

Hence, (a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

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