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Bad White [126]
3 years ago
5

This graph provides data on the amount of water needed to grow four different crops, each used to produce ethanol, a fuel. For e

ach crop, the data shows how many liters of water are needed to produce one liter of ethanol fuel. Which of the following would require using the most water? *
Producing 100 L of ethanol from sugar cane
Producing 200 L of ethanol from sugar beet
Producing 300 L of ethanol from potatoes
Producing 400 L of ethanol from corn

Physics
1 answer:
Degger [83]3 years ago
3 0

Answer:

Producing 300 L of ethanol from potatoes

Explanation:

From the diagram, one liter of ethanol production from sugar cane requires 2000 liters of water. Hence, in order to produce 100 L of ethanol from sugar cane, 2000 x 100 = 200,000 L of water.

1000 liters of water is needed to produce 1 liter of ethanol from sugar beet. Hence, 200 x 1000 = 200,000 L of water will be needed to produce 200 liters of ethanol.

1000 liters of water is also required to produce 1 liter of ethanol from potatoes, hence, 300 x 1000 = 300,000 L of water would be required to produce 300 L of ethanol from the same material

About 500 liters of water is required to produce 1 liter of ethanol from corn, hence, 400 x 500 = 200,000 L of water would be needed to produce 400 L of ethanol from corn.

<u>In conclusion, producing 300 L of ethanol from potatoes would require using the most water among all the options.</u>

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One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing
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Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

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(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

Along X- axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}

v_{1}=u_{1}+u_{2}

Put the value into the formula

v_{1}=2.2i-0.80i

v_{1}=1.4i\ m/s

Along Y-axis

0=m_{1}v_{1}+m_{2}v_{2}

m_{1}v_{1}=-m_{2}v_{2}

v_{1}=-v_{2}

Put the value into the formula

v_{1}=-1.36j\ m/s

Then the final speed of the first ball

v_{1}=\sqrt{(1.4)^2+(1.36)^2}

v_{1}=1.95\ m/s

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

\tan\theta=\dfrac{v_{2}}{v_{1}}

\tan\theta=\dfrac{-1.36}{1.4}

\theta=\tan^{-1}\dfrac{-1.36}{1.4}

\theta=-44.16^{\circ}

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

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Answer:

No.

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The "guide to Engineering and land surveying" for professional engineers and land surveyors by the California board reviews that an unlicensed person cannot be a sole owner of an engineering business, unless there is partnership with a licensed engineer.

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4 years ago
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The minimum stopping distance of a car moving at 20.5 mi/h is 11.6 m. Under the same conditions (so that the maximum braking for
pshichka [43]

Answer:

d = 69 .57 meter

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Force = ( workdone /distance ) = ( 41.99 m / 11.6 )   =  3.619 m N

Second case

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d = ?

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Since the braking force remains the same .

3.619 m = ( 251.7768 m / d )

d = 69 .57 meter

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Answer:

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Explanation:

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