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The power of a car running at a constant speed is 120 kW, the speed is 60 km / h, and the car runs for 2 hours. What the engine does is the traction of the truck?
W=Pt=120000W×2×3600s=8.64×10^8J
F=P/v=120000/(60/3.6)N=7200N
The efficiency of a car engine is 30%. When the car runs at a constant speed for 30 minutes, it uses 6kg gasoline. Ask the output power of the car engine during this period
Calorific value of gasoline: 4.6x10 ^ 7j/kg
Total heat generated in 30 minutes w = 6 * 46000000 = 276000000
Energy used w = w * 30% = 82800000
Power P = w / T = 82800000 / (30 * 60) = 46000
(Physics of the second grade of junior high school) a trolley car running at a constant speed with a speed of 30 km / h and a power of 60 kW
(1) The distance of trams within 2 minutes;
(2) The work done by the tram in 2 minutes;
(3) The traction of a tram
1．S＝vt＝30Km／h＊（2／60）h＝1Km
2．W=Pt＝60000w＊2＊60s＝7200000J
3．F＝P／V=60000w／（25／3）m／s＝7200N
A car has a power of 30 kilowatts. It travels 20 kilometers at a speed of 10 km / s on a horizontal road. Q 1. How much work does a car do during this period
How much work does the car do during this period
What's the traction of the car?
Known: P = 30kW = 30000w; v = 10km / S = 10000m / S; s = 20km = 20000m;
Find: w =? F =?
S=vt,t=S/v=20/10=2（s）
P=W/t,W=Pt=30000×2=60000（J）
P=Fv,F=P/v=30000/10000=3(N)
A: slightly
Tips: v = 10km / S = 36000km / h, 36000km / h car! There is no such fast car, please check the data you provided
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