The car was traveling at a constant speed of 25 meters per second. The driver suddenly found that there was an emergency ahead. After 0.5 seconds, he began to brake and taxied for 60 seconds after 4.5 seconds The average speed of the car between the time the driver found the situation and the complete stop

The car was traveling at a constant speed of 25 meters per second. The driver suddenly found that there was an emergency ahead. After 0.5 seconds, he began to brake and taxied for 60 seconds after 4.5 seconds The average speed of the car between the time the driver found the situation and the complete stop

Initial speed = 25 * 60 * 60 / 1000 = 90km / h 0.5s driving distance = 90 * 0.5 / 3600 = 0.0125km 4.5s driving distance = 60 / 1000 = 0.06km total distance in 5S = 0.0125 + 0.06 = 0.0725km average speed = 0.0725/5 * 3600 = 52.2km/h
Conclusion: during the period from discovery to complete stop, the average speed of the car is 52.2km/h

A car is traveling at a constant speed of 30m / s. when the driver suddenly finds an emergency ahead, he starts braking after 0.6s, taxis for 52m after 4.4S, and the car stops. The average speed from finding the situation to stopping is______ m/s．

The driving distance of 0.6s vehicle is s = VT = 30m / s × 0.6s=18m，
The average speed from discovery to stop is v = s total
Ttotal = 18m + 52m
0.6s+4.4s=70m
5s=14m/s，
So the answer is: 14

(answer by two methods) a vehicle with a mass of M = 1000kg and a speed of V0 = 10m / s. after turning off the engine, it taxied on the horizontal ground for a distance of L = 5M and stopped, Try to find the resistance of the car. Two methods

① Method 1: kinetic energy theorem
-W resistance = 0-1 / 2mv0 ²
W resistance = f resistance L
The solution shows that f resistance = 10000n
② Method 2: Newton's second law
F resistance = ma
V0 ²= 2aL
The solution shows that f resistance = 10000n

A 1000kg vehicle is driven at 10m / s. now let it slow down evenly within a distance of 12.5m and stop to find the required resistance

Let the required resistance be f, then
Deceleration acceleration is a = f / M = f / 1000,
So there
V ²/ 2A = s, i.e
ten ²/ (2*f/1000)=12.5,
The solution is f = 4000N,
That is, the required resistance is 4000N

Find a math blank Use 2, 3, 4, 5, 6, 7 and 8 to form two digits. Take out any two digits. They are multiple relations: (). Among the two digits, only the number of common factor 1: () Do you want to fill in the number in brackets or fill in every case? It's too much to fill in

To tell you the truth, the title is a little ambiguous. Use 2, 3, 4, 5, 6, 7 and 8 to form two digits, and take out two digits arbitrarily. They are multiple relations: if you group the 7 * 6 = 42 digits first, and then select any two digits (15 cases: 23 / 46 24 / 48 24 / 72 26 / 52 26 / 78 27 / 54 28 / 56 28 / 84 32 / 64 34 / 68 36 / 72 37 / 74 38 / 76 42 / 84 43 / 86) If it is understood that two double digits are formed at the same time, the six groups 24 / 48, 24 / 72, 26 / 52, 28 / 84, 37 / 74, 42 / 84 should be removed, and there are nine groups left, but the title seems to say the former;
In the second question, if 42 double digits are first grouped according to the former, it will be too many. Take the smallest number 23 as an example. Except for 23 and 46, the other 40 double digits meet the conditions. In this way, there are eight prime numbers 23, 37, 43, 47, 53, 67, 73 and 83
23 corresponds to 40; 37 deduct 23 / 37 / 74 to 39; 43 deduct 23 / 37 / 43 / 86 to 38; 47 deduct 23 / 37 / 43 / 47 to 38; 53 deduct 23 / 37 / 43 / 47 / 53 to 37; 67 deduct the first 5 and its own 36; 73 is 35; 83 = 34, 40 + 39 + 38 + 38 + 37 + 36 + 35 + 34 = 297 in total; Add 24 / 25, 24 / 35... There are too many, and there are no rules. You can only look at them in double digits
What grade are you in? I guess the teacher asked you to fill in with examples

If the radius of the inscribed circle of the triangle is R and the lengths of the three sides are a, B and C respectively, the area of the triangle is s = 1 2R (a + B + C). According to the analogy, if the radius of the inscribed sphere of the tetrahedron is R and the areas of the four surfaces are S1, S2, S3 and S4 respectively, the volume of the tetrahedron is () A. V=1 6R（S1+S2+S3+S4） B. V=1 4R（S1+S2+S3+S4） C. V=1 3R（S1+S2+S3+S4） D. V=1 2R（S1+S2+S3+S4）

Let the center of the inscribed ball of the tetrahedron be o,
Then the distance from the ball center O to the four faces is r,
So the volume of the tetrahedron is equal to o as the vertex,
The sum of the volumes of four triangular pyramids with four faces as the bottom
I.e. v = 1
3R（S1+S2+S3+S4）．
Therefore: C