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1) The object is at rest because the velocity on the second graph is equal to zero.

2) t 5, when the object is in a constant velocity, the frictional force is equal to the tensile force is equal to 2

3) The object is not in a constant velocity state, is not affected by the equilibrium force, is affected by the non-equilibrium force, is in a uniform acceleration state, the acceleration is equal to 2, and the frictional force on the object is the sliding friction force is equal to the tensile force when the uniform velocity is equal to 2

1: Resting state.

2：2n3:2n

Detailed Analysis:

1: According to the second diagram in Figure B, at 1s, v 0 then it can be seen that the object is not moving and is at rest.

2:5s between 4s and 6s f = 2n, v = 4m s From the second figure of Figure B, it can be seen that the velocity of the object remains constant in this period of time, and the force is balanced, and the analysis of the horizontal force is only f and friction, then the friction is equal to f, and the friction is 2n

3: The vertical direction of the analysis object is only gravity and support force, from the second figure of Figure B, it can be seen that in the process of 2s 8s, the speed is not zero, indicating that the object is in motion in the process, then the friction in the process belongs to the sliding friction force, the sliding friction force is only related to the sliding friction coefficient and the pressure on the object, in this topic, these two factors have remained unchanged, so the friction force during the 2s-8s period is the same, from the second question, the friction force is 2n

Analysis: The focus of this question is to list the area of the triangle ABC and eliminate the algebraic term area "B".

1. Let the area between A and B be C;

2. Semicircular area = A + C = 200 (calculation process omitted), C = 200 - A;

3. Known: A-B=68, then A=B+68;Therefore, C = 200 -B-684, find the area of the triangle, and eliminate "B".

S triangle abc = C + B = (200 -B -68) + B = 200 -685, so: bc = 2 s triangle abc ab = 2 (200 -68) 40

Solution: In ABC: BAC=90° because AB2+AC2=BC2; In a triangular ADC, because 16 2 + 12 2 = 20 2

So dca=90°

So ab cd (the inner wrong angles are equal, and the two lines are parallel).

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Proof: DE AC, DF AB

The quadrilateral aedf is a parallelogram.

de//ac

Bucket ADE= CAD

AD bisects the BAC

bad=∠cad

ade=∠bad

ae=de quadrilateral, aedf is rhomboid, and parallelogram with equal adjacent sides is rhomboid