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Analysis: a, according to the relationship between fluid pressure and flow, the car will be affected by an upward lift when driving, so its pressure on the ground will be less than the gravity of the car itself. B, traction and resistance are a pair of balancing forces in the horizontal direction, and the work done to overcome resistance is equal to the work done by traction. Need to use the formula W=Pt to calculate. C, the work done by traction is calculated as b, according to the formula f = w/s.

D can be calculated to understand the concept of heat engine efficiency: the ratio of energy used to do useful work to energy released by complete combustion of fuel.

Solution: A, the car is streamlined. When the car is running, the air above the car has high velocity and low pressure, while the air below has low velocity and strong pressure, resulting in upward pressure difference. So there will be upward lift, which makes the pressure of the car on the ground less than the gravity of the car itself. Total vehicle weight g = mg = 1.5× 103kg. W = pt = 23×103 w×1× 3600s = 8.28×107j, so b is wrong. C, the traction of this car is: F=

W/s = 8.28×107j/72×103m =1.15×103n, so c is correct.

D. The heat released by complete combustion of gasoline is q = MQ = 6kg× 4.6x107j/kg = 2.76x108j; η = w/q = 8.28×107j/2.76×108j = 30%, so d is correct.