There is an object with a mass of 0.2kg on the horizontal turntable. The distance between the object and the rotating shaft is r = 0.1M. The object and the rotating shaft are connected by a string, and the string is just straight What I want to ask is, why does the maximum static friction force provide centripetal force when there is no tension in the rope, and why must it be the "maximum" static friction force?

There is an object with a mass of 0.2kg on the horizontal turntable. The distance between the object and the rotating shaft is r = 0.1M. The object and the rotating shaft are connected by a string, and the string is just straight What I want to ask is, why does the maximum static friction force provide centripetal force when there is no tension in the rope, and why must it be the "maximum" static friction force?

When there is no tension in the rope, if the angular velocity of the turntable is small, the required centripetal force is provided by the static friction, and the static friction is less than the maximum static friction
With the increase of angular velocity, the static friction also increases. When the angular velocity reaches a certain value, the static friction reaches the maximum
If there is no rope, when the angular velocity is greater than ω 0, the object will slide relative to the turntable!
Now because of the existence of the rope, when the angular velocity is greater than ω 0, the rope will produce tension
When the angular velocity is less than or equal to ω 0, the rope has no tension; when the angular velocity is greater than ω 0, the rope has tension

In the sequence {an}, A1 = 1, when n > = 2, Sn = n2an (n's Square * an), find the general term an. A1 = 1, not = 1 / 2

When they are ≥ 2, Sn = n (n-1) sn-sn-sn-1 (a (n-1) sn-sn-1 (a (n-1) sn-sn-1 (a (n-1) sn-sn-1 (a (n-1) sn-sn-1 (a (n-1) sn-sn-sn-1 = an = n \#178; and (n-1) (n-1) (n-1) n \\\\35\\\\\\\\\\\\\\\\\\\\\\\\\\\(n-1) a (n-1) an / a (n-1) = (...)

There are two cars a and B on the straight road. Car a starts to drive from a standstill at the acceleration of a = 0.5m/s ^ 2, and car B starts to drive at the acceleration of VO = 5m at XO = 200m in front of car a/
When does a catch up with B? How fast does a catch up with B? How far is a from the starting point? 2. When is the maximum distance between a and B? What is the distance?

1, s a = v0t + 1 / 2at ^ 2, V0 = 0. Let a catch up with B after time t, S B + 200 = s a, 5T + 200 = 0.5 * T ^ 2, t = 40s,
At this time, V A = 0.5 * 40 = 20m / s, s a = 400m, so after 40 seconds, a catches up with B. at this time, the speed is 20m / s, and a is 400m away from the starting point
2. When the speed of Party A and Party B is the same, the displacement is the largest, V A = 0.5 * t = 5m / s, t = 10s, the distance L = s B + 200-s a = 5 * 10 + 200-0.25 * 100 = 125m, so the distance of Party A and Party B is the largest, 125m at 10s

Xiao Ming and Xiao Hua practice running on a 400 meter circular track. They start from the same point at the same time and walk in opposite directions. Xiao Ming runs 4.5 meters per second and Xiao Hua runs 4.5 meters per second
5.5 meters. After how many seconds, the two met for the second time?

When they met for the second time, they ran two laps
Meeting Time
=（400*2）/（4.5+5.5)
=800/10
=80 (seconds)

4 test paper
4. A, who has applied the problem to solve the equation, is a newspaper
Good people help··········
The answer to volume a and volume B is newspaper
Good people help··········

One day in Changshu, the temperature is 5 ℃ in the morning and rises 3 ℃ at noon. In the afternoon, due to the cold air going south, it drops 9 ℃ at night, so the temperature at night is ℃. 2

Using line AB with length of 2008 units on the number axis, it can cover () integer points
The answer is:
If the starting point is an integer point, there are 2008 + 1 = 2009
If the starting point is not on an integer point, there are 2008
Why do you cover more than one point at an integer point?

If you regard 2008 as 2, you can figure it out by yourself

The soldiers are determined to break the enemy's blockhouse

The soldiers had to make up their minds to break the enemy's blockhouse!

（1）2（4x-0.5） （2）-3（1-6／1x） （3）-x+（2x-2）-（3x+5）
（4）3a²+a²-（2a²-2a）+（3a-a²） （5）（5a+4c+7b）+（5c-3b-6a）
（6）（8xy-x²+y²）-（x²-y²+8xy） （7）（2x²-2／1+3x）-4（x-x²+2／1）
（8)3x²-[7x-（4x-3）-2x²]

（1）2（4x-0.5）=8x-1（2）-3（1-6／1x）=-3+18/1x（3）-x+（2x-2）-（3x+5）=-x+2x-2-3x-15=-2x-13（4）3a²+a²-（2a²-2a）+（3a-a²）=4a²-2a²+2a+3a-a²=a²+5a （5）（5a+...

Choose any 30 natural numbers from the 900 three digit numbers of 100-999, at least how many of them are equal to each other?

The sum of the numbers is between 1 and 27
So at least the sum of three logarithms is equal

The speed of freight cars is 60 km / h, and that of passenger cars is 80 km / h,
On the way, the bus broke down and continued to drive for 1.5 hours. As a result, the bus and the truck met at the midpoint of the two places. How many kilometers are there between a and B?

1.5-hour freight train: 60 × 1.5 = 90 (km)
Passenger cars travel more than freight cars per hour: 80-60 = 20 (km)
The passenger car runs for 4.5 hours, 20 kilometers more than the freight car
Therefore, it takes 4.5 hours for the bus to reach the midpoint, so it takes 4.5 × 2 = 9 (hours) to complete the whole journey
Distance between a and B: 80 × 9 = 720 (km)