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Use the orthogonal decomposition method to analyze the force, and then use Niu Er:

1) u=0, the object slides along the inclined plane: gsin a2)u 0, the object goes up along the inclined plane: -gsin a3)u≠0, the object slides along the inclined plane:

g(sin a- u cos a)4) u≠0, the object goes up along the inclined plane: -g(sin a+ u cos a) see your "question supplement", reply as follows:

Take the most difficult fourth problem as an example, the object is subjected to gravity, support force and friction, a total of three forces, establish a coordinate system on the force diagram, take the x-axis along the inclined plane direction, and the y-axis along the vertical inclined plane, decompose the gravity force to the coordinate axis, and the resultant force of the component force (mgsin a) and the frictional force along the inclined plane direction is the cause of the acceleration, according to Niu Er, -mgsin a-f=ma, where f is the frictional force, f=un=umgcosa (n and mgcosa are equal in size, The direction is opposite, i.e., the resultant force on the y-axis is zero), substitution, and the solution is: a=-g(sin a+ u cos a). The negative sign in the equation indicates that it is in the opposite direction to the initial velocity.

In other cases, the method is the same, but there is less friction.

1) The object is supported by gravity and inclined planes.

The gravitational force produces acceleration along the components of the inclined plane.

mgsina=ma is solved to obtain a=gsina direction along the slope downward.

2) The object is supported by gravity and inclined planes.

The gravitational force produces acceleration along the components of the inclined plane.

mgsina=ma is solved to obtain a=gsina direction along the slope downward.

3) The object is subjected to gravity, the supporting force of the inclined plane, and the sliding friction force upward along the inclined surface.

Mgsina- mgcosa=ma in the inclined direction is solved in the direction of a=gsina- gcosa and downward in the direction of the inclined plane.

4) The object is subjected to gravity, the supporting force of the inclined plane, and the sliding friction force that goes down along the inclined surface.

Mgsina- mgcosa=ma in the inclined direction is solved in the direction of a=gsina- gcosa and downward in the direction of the inclined plane.

Let the acceleration of the object be a, the velocity to point A is v0, and the time taken to pass through the AB and BC segments is t, then we have:

l1=v0t+at2/2 ……l1+l2=2v0t+2at2………Together:

l2－l1=at2………3l1－l2=2v0t………Let the distance between o and a be l, then there is:

l=v02/2a ……Together:

l=(3l1－l2)2/8(l2－l1)

From the problem, the velocity of b is (l1+l2) 2t, the acceleration is (l2-l1) t 2, the velocity at the intermediate moment between a and b is l1 t, which is obtained by va 2=2ax, xoa = (3l1-l2) 2 8(l2-l1).

First of all, it should be clear that when the tablecloth disc is dragged to separate from the tablecloth, the acceleration of the disc is u1g, and the acceleration after separation is -u2g, and the disc is then slid on the table for a period of time (uniform deceleration) and then stopped after the two are separated.

The reason why there is a restriction on A is because the restriction A does not give the same amount of time to accelerate the disk. Separation time t1, the difference between the distance traveled between the two is l 2, (let the table length l), l 2 = 1 2a(t1)2-1 2(u1g)(t1)2 find t1 (which contains l, which can be eliminated later).

When the disc separates from the tablecloth at time t1, the velocity of the disc is u1gt1 and the distance of gliding is s=

s1 of squared= and then do a uniform deceleration, obtained by the formula v2 (meaning of square) = 2as, s2 = [u1gt1] 2 (meaning of square) (2u2g).

Then the column condition is the condition that does not fall, i.e. s1+s2 <

In this way, the condition of a can be found ... The computer input is too strenuous, and the camera has wood by your side... Take a look at it for yourself...

The square table has a side length of 2L, and the acceleration of the disc on the tablecloth is U1g (mA'= u1mg), the acceleration of the disc on the table is u2g (the same goes for it).

The disc takes t when it is detached from the tablecloth

at^2/2-u1gt^2/2=l

The distance at which the disc slides on the cloth s1 = u1gt 2 2 The distance at which the disc slides on the table s2 = v 2 2u2g=(u1gt) 2 2u2gs1+s2 l

From the above equations, a (2u1u2+u1 2)g u2 can be obtained

Car B is the first to pass the next mark.

Suppose the initial velocity of the three cars is V, and then the speed of A remains the same, B accelerates first and then decelerates, while C is the opposite of B. And the speed at which they pass through the second signpost is the same, and the velocity here v A = v B = v C. It can be seen that B has already started to slow down before reaching the next marker, otherwise he will be faster than A when he arrives, i.e. B's minimum speed is the same as A's.

In the same way, C has already started to accelerate when he reaches the next road mark, otherwise it will be less than A's speed. That is, the maximum velocity of C is the same as that of A. So the average speed of the whole journey to the next marker is the maximum, so B is the first to reach the next marker.

I feel that the conditions given by the question are too few.

It should be car B, you can draw an image and compare it.

Because the area of the image of velocity and time is the displacement.

Because the displacement of the next signpost is the same, and the velocity is the same, you can see by drawing an image that it takes the shortest time to walk the same area.

Draw a V-T plot and compare definite integrals.

C first passes the next mark, A B C starts at the same speed, first say B, B first accelerates and passes through the next mark at the same time as deceleration and A (constant speed), indicating that B has begun to slow down in order to pass at the same time as A, so B begins to become slower than A after passing at the same time. Speaking of C, C is decelerating first and accelerating, and it also passes a certain road sign at the same time as A, so when passing, A has accelerated, and the speed is greater than A when passing. So after passing at the same time, A's velocity is the maximum, and it's still accelerating, so A's passes first.

Mathematical analytic application of geometric knowledge to physics. Comparing the above two diagrams, mathematically they are the same, they are both primary function images, and the corresponding linear equation form is of course the same, so if you take the right graph as the left graph, and replace the corresponding x with s, and y with f, it is easy to understand f=-s+18

Option b is correct.

Analysis: Because the suspension line of ball C is vertical, it means that the force of ball A and B on ball C is equal and the direction is opposite, and it is known that ball A is negatively charged. And because the distance between AB is equal to the distance between BC, the power of A ball must be 4Q. ......b right.

In the case of knowing that A is negatively charged and B is positively charged, according to the characteristics that the suspension line of ball A is deviated to the left, it can be determined that ball C must be negatively charged, and the charge of ball C must be greater than 4Q (Rab=RBC). ...ACD Wrong.

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