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Brilliant_brown [7]
2 years ago
3

A typical compact car experiences a total drag force at 55 m i/h of about 350 N. If this car gets 35 miles per gallon of gasolin

e at this speed, and a liter of gasoline (1 gal = 3.8 L) releases about 3.2 X 107J when burned, what is the car’s efficiency
Physics
1 answer:
densk [106]2 years ago
3 0

Answer:

Efficiency = 16.2124 %

Explanation:

Given:

velocity, v = 55 mi/hr = 24.5872 m/s

Drag force = 350 N

Work done by friction = Force x displacement

now for 1 second,

we get

Power (P) as

P = 350 × 24.5872 = 8605.52 W  

Now,

The consumption of gasoline by car for travelling 35 miles = 3.8 L

Energy released by burning of 3.8 L of gasoline = 3.2 × 10⁷ J

Time taken to travel 35 mile = 35/55 = 0.6363 hour = 0.6363 × 3600 seconds = 2290.90 seconds

Thus,

The power used  = (3.2 x 10⁷ × 3.8)/2290.90 = 53079.57 W

or

Therefore,

The efficiency = [output power/ input power] × 100%

The efficiency = [8605.52 W   /53079.57] x 100%

Efficiency = 16.2124 %

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0.50m/s

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Answer:

t is appropriate to clarify that units such as time and angles the transformation is not in base ten, for example:

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Explanation:

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In the example of velocity, the derivative unit is m / s, which is why it works in the same way that you transform length and time if in the equation it is multiplying it is multiplied and if it is dividing it is divided.

It is appropriate to clarify that units such as time and angles the transformation is not in base ten, for example:

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Answer:

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Explanation:

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r is the distance from the charge

In this problem, we have

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2 years ago
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