As shown in the figure, on the horizontal plane, the object block a with a mass of 10kg is tied to one end of a spring which is horizontally stretched, and the other end of the spring is fixed on the trolley. When the trolley is stationary and the spring force on the object block is 5N, the object block is in a static state. If the trolley accelerates to the right along the horizontal ground with an acceleration of 1m / S2 () A. Block a moves relative to the trolley B. the friction force on block a will decrease C. the friction force on block a will remain unchanged D. the elastic force on block a will increase

As shown in the figure, on the horizontal plane, the object block a with a mass of 10kg is tied to one end of a spring which is horizontally stretched, and the other end of the spring is fixed on the trolley. When the trolley is stationary and the spring force on the object block is 5N, the object block is in a static state. If the trolley accelerates to the right along the horizontal ground with an acceleration of 1m / S2 () A. Block a moves relative to the trolley B. the friction force on block a will decrease C. the friction force on block a will remain unchanged D. the elastic force on block a will increase

A. When the object just slides to the left relative to the car, the acceleration is A0 = F + FMM ≥ 5 + 510 = 1m / S2, so when the car's acceleration is a = 1m / S2, the object block a phase

What is 2 microamperes

2μA=2*10^-3mA=2*10^-6A

As shown in the figure, the mass of the object on the trolley plate is m = 8kg. It is pulled by a spring stretched horizontally and is still on the trolley. At this time, the spring force is 6N. Now, the force is applied to the trolley along the horizontal right direction, so that the trolley starts to move from the rest, and the acceleration gradually increases from zero to 1m / S2, The following statement is correct: AC
A. The object and the car always keep relatively static, and the force of the spring on the object never changes
B. The friction force on the object keeps decreasing
C. When the car acceleration (right) is 0.75 m / S2, the object is not affected by friction
D. When the car moves to the right with a uniform acceleration of 1 m / S2, the friction force on the object is 8N
I can't pass the picture, that is, there is a car on the ground. There is a block on the car connected with the car through the spring. Then use f to pull the car and start to move
Question: why does the car and the object always keep static? Is it possible that after the car moves by F, the friction force on the object remains unchanged, and then the object slides backward relative to the car, the spring length is extended, and the object and the car have the same acceleration? Why does the friction force remain unchanged?
Why does the car and the block remain relatively stationary at the beginning, rather than gradually accelerating to a standstill?

Your analysis is wrong
When the car and the object are stationary, the spring should be pulled to the right, the friction between the car and the object is to the left, and the spring has a tension of 6N, indicating that the static friction has 6N, and the maximum static friction FM may be greater than 6N, but never less than 6N
We know that the static friction can vary with the situation, but it can not exceed the maximum static friction FM
Taking the right direction as the positive direction, the friction force between the car and the object can take any value from - 6N to + 6N. The spring tension is proportional to the spring elongation. As long as the object does not move relative to the car, the spring tension will remain unchanged, still 6N. When the static friction force is also right, the maximum combined external force that the object can achieve is 6N + 6N = 12n. Once it exceeds, it cannot guarantee that the object and the car are relatively stationary
According to the conditions in the question, the acceleration provided by the external force to the car will not be greater than 1m / S2, and the maximum combined external force on the object is 8N, which is within the range, so the object and the car remain relatively static
The special case is that the acceleration is 0.75 m / S2, and the external force of the corresponding object is f = ma = 0.75 * 8 = 6N, which is just provided by the spring. At this time, the object is not subject to friction
Question supplement: why does the car and the object remain relatively static at the beginning, instead of the object gradually accelerating to a static state?
The condition that the car and the object remain relatively stationary is that the acceleration is equal
As long as the external force (including spring tension and static friction) can guarantee the acceleration of the car, the object can be stationary relative to the car

The three sides of a triangle differ by 5cm in turn. The perimeter is 120cm. How long is the shortest side? The longest

Let three sides be a-5, a, a + 5
therefore
a-5+a+a+5=120
3a=120
a=40
So the three sides are: 35cm, 40cm, 45cm
So the longest is 45cm
The shortest is 35cm

A wooden box with mass of M is on the horizontal plane, and one end of a light spring is suspended above the wooden box. On the other end, objects a and B with mass of M are hung by thin wires. After cutting the thin wires between a and B, a makes a simple movement. When a reaches the highest point, the pressure of the wooden box on the floor is

Acceleration down, weightlessness,
Because in simple harmonic motion, the acceleration at the lowest point is equal to that at the highest point = mg
Pressure f = mg mg

Primary 5 grade 4 mixed operation 50
Arrange the serial number well, don't mix it up

1.3/7 × 49/9 - 4/3
2.8/9 × 15/36 + 1/27
3.12× 5/6 – 2/9 ×3
4.8× 5/4 + 1/4
5.6÷ 3/8 – 3/8 ÷6
6.4/7 × 5/9 + 3/7 × 5/9
7.5/2 -（ 3/2 + 4/5 ）
8.7/8 + （ 1/8 + 1/9 ）
9.9 × 5/6 + 5/6
10.3/4 × 8/9 - 1/3
11.7 × 5/49 + 3/14
12.6 ×（ 1/2 + 2/3 ）
13.8 × 4/5 + 8 × 11/5
14.31 × 5/6 – 5/6
15.9/7 - （ 2/7 – 10/21 ）
16.5/9 × 18 – 14 × 2/7
17.4/5 × 25/16 + 2/3 × 3/4
18.14 × 8/7 – 5/6 × 12/15
19.17/32 – 3/4 × 9/24
20.3 × 2/9 + 1/3
21.5/7 × 3/25 + 3/7
22.3/14 ×× 2/3 + 1/6
23.1/5 × 2/3 + 5/6
24.9/22 + 1/11 ÷ 1/2
25.5/3 × 11/5 + 4/3
26.45 × 2/3 + 1/3 × 15
27.7/19 + 12/19 × 5/6
28.1/4 + 3/4 ÷ 2/3
29.8/7 × 21/16 + 1/2
30.101 × 1/5 – 1/5 × 21
31.50＋160÷40 （58+370）÷（64-45）
32.120-144÷18+35
33.347+45×2-4160÷52
34（58+37）÷（64-9×5）
35.95÷（64-45）
36.178-145÷5×6+42 420+580-64×21÷28
37.812-700÷（9+31×11） （136+64）×（65-345÷23）
38.85+14×（14+208÷26）
39.（284+16）×（512-8208÷18）
40.120-36×4÷18+35
41.（58+37）÷（64-9×5）
42.(6.8-6.8×0.55)÷8.5
43.0.12× 4.8÷0.12×4.8
44.（3.2×1.5+2.5）÷1.6 （2）3.2×（1.5+2.5）÷1.6
45.6-1.6÷4= 5.38+7.85-5.37=
46.7.2÷0.8-1.2×5= 6-1.19×3-0.43=
47.6.5×（4.8-1.2×4）= 0.68×1.9+0.32×1.9
48.10.15-10.75×0.4-5.7
49.5.8×（3.87-0.13）+4.2×3.74
50.32.52-（6+9.728÷3.2）×2.5
51.[（7.1-5.6）×0.9-1.15] ÷2.5
52.5.4÷[2.6×（3.7-2.9）+0.62]

On College Physics -- moment of inertia,
My foundation is not very good,

1. Moment of inertia: J = ml ^ 2 / 3 = 10 * 36 / 3 = 120 (kg. S ^ 2)
2. Let the angular acceleration be: ε
From the theorem of moment of momentum: J ε = MGL / 2
ε=mgL/(2J)=10*10*6/240=2.5 (rad/s^2)
3. Let the angular acceleration be ε 1 and the angular velocity be ω
From the theorem of moment of momentum: J ε = mglcos θ / 2
ε=mgLcosθ/（2J）=（5√3）/4
By energy conservation: J ω ^ 2 / 2 = mglsin θ / 2
ω^2=mgLsinθ/J=10*10*6*0.5/120=300/120=5/2
ω=√10/2 (rad/s)

The sum of the first n terms of the arithmetic sequence {an} is Sn, and a4-a2 = 8, S10 = 190 (1) the formula for finding the general term (2) P and Q are all positive integers. Is AP multiplied by AQ still the term in the sequence {an}? Please prove

Solution (1) a4-a2 = 2D = 8 -------- d = 4s10 = 10A1 + [10x (10-1)] X4 / 2 = 190 -------- A1 = 1, so the general formula of arithmetic sequence is: an = a1 + (n-1) d = 1 + (n-1) X4 = 4n-3 (n belongs to N +) (2) apaq = (4p-3) (4q-3) = 16pq-12 (P + Q) + 9 = 4 [4pq-3 (P + Q) + 3]

The speed of a car running along a straight road is 36km / h, and the acceleration obtained after braking is 4m / S ^ 21. The speed at the end of 3S after braking
. speed 36km / h = 10m / s,
Notice that the car stops as long as V / a = 10 / 4 = 2.5s
So at the end of three seconds, the speed of the car is zero
The answer is above, why is v / a = 10 / 4 = 2.5s? At the end of 3 seconds, the speed of the car is 0?

Note that this is a driving problem. You must distinguish it from a force pulling an object to make it decelerate. When braking, there is only friction in the horizontal direction. That is to say, the friction makes the car slow down until it stops. After stopping, the friction naturally disappears. At this time, there is no force in the horizontal direction and no longer moves. That is, as long as t ≥ 2.5s, the speed is 0

Xiaoli has 12 more color pictures than Xiaolan, which is just three tenths of Xiaolan's pictures. How many color pictures does Xiaolan have? How many color pictures does Xiaoli have?
To calculate a formula or equation

Xiaolan = 12 △ 3 / 10 = 40 sheets
Xiaoli = 40 + 12 = 52