For a car with a mass of M = 1000kg, the maximum speed V1 = 12m / s when driving on a straight road at rated power, and the maximum speed V2 = 8m / s when driving on a slope with an increase of 1m for every 20m. Assuming that the friction force on the car is equal under any condition, G is taken as 10m / S2, the following formula is obtained (1) The size of friction force; (2) What is the rated power of automobile engine; (3) The maximum speed a car can reach when going down the slope

For a car with a mass of M = 1000kg, the maximum speed V1 = 12m / s when driving on a straight road at rated power, and the maximum speed V2 = 8m / s when driving on a slope with an increase of 1m for every 20m. Assuming that the friction force on the car is equal under any condition, G is taken as 10m / S2, the following formula is obtained (1) The size of friction force; (2) What is the rated power of automobile engine; (3) The maximum speed a car can reach when going down the slope

1 / 2mv1 ^ 2 = 1 / 2mv2 ^ 2 + MGH + FL, the friction force is: F = [1 / 2m (V1 ^ 2-v2 ^ 2) - MGH] / L = [1 / 2 * 1000 (12 ^ 2-8 ^ 2) - 1000 * 10 * 1] / 20 = 1500n2. When the traction force is: F = F + Mg * H / L = 1500 + 1000 * 10 * 1 / 20 = 2000N, the rated power of the engine is: P = fv2 = 2

As shown in Figure 1, if the straight lines AB, CD and EF intersect at O, ab ⊥ CD, og bisection ∠ AOE, ∠ FOD = 26 °, then the degree of AOG is ()
A 56°
B 58°
C 62°
D 60°

Therefore, AOE = 90 ° + 26 ° = 116 °
Choose B with ∠ AOG = ∠ AOE / 2 = 58 °
Because I didn't see the picture, I drew it by myself. The process is not necessarily right

As shown in the figure, the straight lines AB, CD and EF intersect at point O, and ab ⊥ CD and og bisect ∠ AOE
This is the last question of 7.7 intersection line (2) in the preliminary understanding of figure in Chapter 7 of synchronous exercise of mathematics in grade one of junior high school

Angle DOF = 50 degrees = angle COE angle AOE = 50 + 90 = 140 degrees og bisector angle AOE angle AOG = 140 / 2 = 70 degrees