As shown in the figure, object a weighs 100N, and object B weighs 20n. The maximum static friction between a and the horizontal table is 30n. When the whole system is at rest, how much static friction does a suffer? If we increase the gravity of B gradually and keep the system still, what is the maximum gravity of B?

As shown in the figure, object a weighs 100N, and object B weighs 20n. The maximum static friction between a and the horizontal table is 30n. When the whole system is at rest, how much static friction does a suffer? If we increase the gravity of B gradually and keep the system still, what is the maximum gravity of B?

Taking node o as the research object, this paper establishes a rectangular coordinate system: TA = tcos45 ° on X-axis & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ① & nbsp; & nbsp; & nbsp; on Y-axis: TB = GB = tsin45 ° ② & nbsp; & nbsp; on x-axis, it obtains TA = gbtan45 ° on X-axis & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; and substituting it into its value TA = 20 & nbsp; N taking a as the research object, the static friction f = TA ′ = TA = 20 & nbsp; N. When the gravity of B is gradually increased, the system should be in equilibrium. When a reaches the maximum static friction, the gravity of object B reaches the maximum. From the above expression, we can see that: GBM = famtan45 ° = 30n A: the static friction of a is 20n, and the maximum gravity of object B is 30n

On the operation of integral
The best operation of integral is square difference

(a + B-C) (A-C-B) = (A-C) square - b square = a square - 2Ac + C square - b square

A car is moving in a straight line at a speed of 10 meters per second on a straight road. It suddenly stops and the acceleration is 5 meters per second
Find (1) the displacement of the movement within 5 seconds after the start of the car brake? (note the time of the car brake)
(2) What is the displacement of the car in one second after the start of braking?
(3) The displacement of the car in one second before it stops moving? (solved by reverse thinking method)

（1）：v0=10,vt=0,a=-5,
Braking time:
t=(vt-v0)/a=（0-10）/-5=2s
The movement displacement within 5 seconds is as follows:
S=v0*t+1/2*a*t^2=10*2+1/2*（-5）*2*2=10m
（2）：v0=10,a=-5,t=1
S=v0*t+1/2*a*t^2=10*1+1/2*（-5）*1*1=7.5m
(3) This question can be inversely considered as the displacement of the vehicle after accelerating for one second at a speed of 5 meters per second square from the stopped state, then: V0 = 0, a = 5, t = 1
S=v0*t+1/2*a*t^2=0*1+1/2*5*1*1=2.5m
In addition, this question can also be solved according to questions (1) and (2) above
It can be seen from (1): the car can stop braking in 2 seconds, and the braking displacement is 10 meters
It can be seen from (2): the displacement within 1 second after braking is 7.5 meters
The results are as follows:
The displacement within 1s before the vehicle stops moving (i.e. from 1s after the vehicle brakes to stop) is as follows:
10-7.5 =2.5m
The answer is: car For The rice is enough

If M > 1, am-1 + am + 1-am2 = 0, s2m-1 = 38, then M is equal to ()
A. 38B. 20C. 10D. 9

According to the property of arithmetic sequence, we can get: am-1 + am + 1 = 2am, then am-1 + am + 1-am2 = am (2-AM) = 0, the solution is: am = 0 or am = 2, if am is equal to 0, it is obvious that s2m-1 = (2m − 1) (a1 + A2M − 1) 2 = (2m-1) am = 38 does not hold, so am = 2, s2m-1 = (2m-1) am = 4M-2 = 38, the solution is m = 10

When a car brakes at an initial speed of 20 m / S2, the braking process is regarded as a uniform deceleration linear motion. If the acceleration is 5 m / S2 when braking, calculate the speed and speed of the car after braking for five seconds
displacement

In this kind of problem, we should pay attention to a problem, that is, when the speed of the car is reduced to zero, it will stop. So when using the formula, we can't directly replace the data, we should combine with the actual situation analysis. In this problem, time t should replace 4S, so VT = 0, x = 20m / s * 4s-1 / 2 [5m / S ^ 2 * (4S) ^ 2] = 40m

The round track on the playground is 400 meters long. Xiao Ming and Xiao Gang start from the same point and walk in opposite directions. After 2 minutes, they meet each other. If they walk in the same direction, they chase each other after 20 minutes
How many meters per minute do Xiao Gang and Xiao Ming walk?

2 runners per minute
400 △ 2 = 200 (m)
Xiao Gang runs more every minute than Xiao Ming
400 △ 20 = 20 (m)
Xiao Gang runs every minute
(200 + 20) △ 2 = 110 (m)
Xiao Ming runs every minute
200-110 = 90 (m)

I want to ask a math problem of grade one
Observe the law of the following column: 2,5,10,17,26 Then the number 2006 is____ .
Just one question! Help!

N ^ 2 + 1 when n = 2006 n ^ 2 + 1 = 2006 * 2006 + 1 = 4024037

On the number axis, it is known that a is two unit lengths to the left of B, C and D are on both sides of the origin, and the distance to the origin is equal. Find the value of | A-B | + 2 (c + D)

C and D are on both sides of the origin, and the distance to the origin is equal, so C = - D, then c + D = 0

Make a hat with calico. The top is cylindrical and the brim is circular. The radius and height of the brim are 0.9dm and the brim width is 1.1dm
How many square meters of cloth does it take to make such a hat

1. Don't calculate scrap (actual consumption)
（1.1+0.9）^2*3.14+0.9*3.14*0.9=15.1034d㎡=16d㎡
2. Including waste
（1.1+0.9）*（1.1+0.9）*4+0.9*3.14*0.9=18.5434d㎡ =19d㎡

Addition and subtraction of integral
There are two kinds of rice experimental fields in a farm, the first is 23 mu, the average yield per mu is a kg, the second is 37 mu, the average yield per mu is B kg, so the average yield per mu of these two kinds of rice experimental fields is () kg

There are two kinds of rice experimental fields in a farm, the first is 23 mu, the average yield per mu is a kg, the second is 37 mu, the average yield per mu is B kg, so the average yield per mu of these two kinds of rice experimental fields is () kg
（23a+37b）/（23+37）=（23a+37b）/60