A certain gasoline engine has an efficiency of 35.3%. What would the hot reservoir temperature be for a Carnot engine having tha
t efficiency, if it operates with a cold reservoir temperature of 160°C? (°C)
1 answer:
Answer:
The temperature of hot reservoir is 669.24 K.
Explanation:
It is given that,
Efficiency of gasoline engine, 
Temperature of cold reservoir, 
We need to find the temperature of hot reservoir. The efficiency of Carnot engine is given by :




So, the temperature of hot reservoir is 669.24 K. Hence, this is the required solution.
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Vx = 35.31 [km/h]
Vy = 18.77 [km/h]
Explanation:
In order to solve this problem, we must decompose the velocity component by means of the angle of 28° using the cosine function of the angle.
![v_{x} = 40*cos(28)\\V_{x} = 35.31 [km/h]](https://tex.z-dn.net/?f=v_%7Bx%7D%20%3D%2040%2Acos%2828%29%5C%5CV_%7Bx%7D%20%3D%2035.31%20%5Bkm%2Fh%5D)
In order to find the vertical component, we must use the sine function of the angle.
![V_{y}=40*sin(28)\\V_{y} = 18.77 [km/h]](https://tex.z-dn.net/?f=V_%7By%7D%3D40%2Asin%2828%29%5C%5CV_%7By%7D%20%3D%2018.77%20%5Bkm%2Fh%5D)
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