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Aliun [14]
2 years ago
7

A 2650-lb car is traveling at sea level at a constant speed. its engine is running at 4500 rev/min and is producing 175 ft-lb of

torque. it has a drivetrain efficiency of 90%, a drive axle slippage of 2%, 15- inch–radius wheels, and an overall gear reduction ratio of 3 to 1. if the car's frontal area is 21.5 ft 2 , what is its drag coefficient?

Physics
2 answers:
7nadin3 [17]2 years ago
8 0
The drag coefficient is given by   
C =( (F - fW) * 2 ) / ( A*V*V*r)  
 V = speed of the vehicle = 192.42 
 f = cofficient of rolling resistance = 0.0231 
 F = traffic effort = 378 
 r = density of air = 0.002378 
 A = frontal area of the vehicle = 21.5 
W = weight of the vehicle = 2650 
 by substituting all the values in the above equation  
 drag coefficient = C = 0.167
Zepler [3.9K]2 years ago
6 0
Check the attached file for the answer.

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Answer:

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Explanation:

given,

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using Snell's law

\dfrac{sin\ i}{sin\ r} = \dfrac{n_2}{n_1}

\dfrac{sin\ 62^0}{sin\ r} = \dfrac{1.58}{1.70}

1.7 ×sin 62 ^0 = 1.58× sin r

sinr = \dfrac{1.7\times sin 62^0}{1.58}

sin r = 0.95

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angle of refraction =r = 71.8⁰

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