A bunch of physics concepts to fill in the blanks, who would

The acceleration of an object moving in a straight line at a uniform variable speed always remains unchanged throughout the motion, not only the magnitude is not the same, but also the direction is also the same. In the process of learning, we should grasp the characteristics of linear motion with uniform variable speed and use the law to deal with practical problems. In dealing with problems, it is necessary to pay attention to solving multiple solutions to one problem.

When dealing with the linear motion of the particle to do uniform deceleration, it is necessary to consider whether the particle may stop.

5) Pursuit and encounter problems are a common type of kinematic problem, and from the perspective of time and space, encounter refers to reaching the same place at the same time. It can be seen that there must be the following two relationships between the objects that meet: First, there is a certain displacement relationship between the meeting position and the travel time of each object.

If you start from the same place, you will be displaced when you meet.

Spatial conditions. Second, there is also a certain relationship between the motion time of the encountering objects. If the objects are moving at the same time, the movement time is different; If A starts δt earlier than B, then the motion time relation is tA = displacement velocity In order for the objects to meet, the displacement relation and the motion time relation must be satisfied.

The solution method of chasing and encountering the problem is generally to first analyze the motion characteristics of each object to form a clear situation. The objects are then established based on the location of the encounter.

relational equations; Finally, according to the motion characteristics of each object, the relationship between the motion speed is found. Commonly used methods are the formula method and the image method.

d) Acceleration in the direction (this empty will not) be stationary.

5) Displacement (?) Same Equal T B + δt Displacement Time Velocity Velocity.

4) Speed, direction (5) starting position, same, equal, t A t t B δt, while satisfying the relationship between displacement and motion time.

4] Acceleration direction rotation.

5] Distance Equal Same T B Displacement Time Displacement Time.

The object has gone for s with velocity v1 in a straight line and s with velocity v2 in the same direction. His average velocity in these 2s displacements is -

2s (s v1+s v2) = 2v1v2 (v1+v2) If time t is taken by v1 and time t is taken at the same speed in the same direction, the average velocity in time 2t is ——-

v1+v2)t/2t=(v1+v2)/2

Average velocity is defined as the ratio of the displacement of the object's motion to the time it takes for that displacement to occur. This definition should be used in mind.

So the average velocity in a displacement of 2s is v=2v1v2 v1+v2

The average velocity in the 2s displacement is v=1 2(v1+v2).

Flat = distance Total time = 2s (s v1 + s v2) (find time).

Ping = (v1t+v2t) 2t (find the distance).

The time taken for s with v1 is v1, and the time taken for s is v2 with v2, and the total time is (s v1+s v2); Then the total distance is 2s, divide the total distance by the total time, 2s (s v1+s v2), and simplify it to 2v1v2 (v1+v2).

The second question is the same, get (v1+v2) 2.

The first 2s (s v1+s v2) and the second one is (v1+v2) 2

The momentum theorem states that impulse is what makes the momentum of an object change.

Modern physics defines force as the rate of change in the momentum of an object: f = p t (the form of momentum of Newton's second law).

I don't know how to draw, but I'll tell you what I think.

First of all, look at them as a whole, and the bricks weighing 4mg are sandwiched between the two sandwich orange boards, and 1 and 4 are definitely subjected to the friction of 2mg each of the splints in an upward direction.

In the analysis of brick 1, it is now known that the upward friction force subjected to 2mg, the self-weight mg, the brick is at rest indicating that it is balanced. Therefore, brick 1 must also be subjected to a force of magnitude mg but in a downward direction. And this force can only be provided by brick 2.

Hence the first empty fill mg.

Analyzing the force of the second brick, it is said above that brick 2 provides the frictional force of brick 1 of mag mg, according to Newton's third law, brick 2 is subjected to the friction of brick 1 of mag mg, and the direction is upward. Brick 2 has its own weight mg, and the two forces are just balanced, so brick 2 must not be subjected to the friction force of brick 3. The same is true for this question from right to left.

The second blank is filled with 0.

(1) Examine the question well, "the car starts from a standstill somewhere, and at this time, a truck passes by the car at a constant speed of 15m s", so after the car moves, the distance of the truck and the car is the same!

2) For the sake of calculation convenience, the distance is converted into the calculation of the truck, because their movement time and distance are the same, but the movement state is different!!

If you still don't understand, you can ask me!!

Let's start with this problem: when the car starts, the truck passes by the car, then the two objects are moving at the same time and starting point. When a sedan catches up with a truck, both cars end at the same end.

According to the definition of displacement, the displacement of the two cars is the same. Therefore, this 225m is both a sedan and a truck, so the solution is correct.

In addition, there is still ambiguity in the sentence "two moving objects meet, that is, their displacement is equal relative to the same frame of reference". There are two types of encounter problems, one that moves in opposite directions and the other that moves in the same direction. For the former, the starting point is different and the end point is the same, and the displacement may not be the same, so in this case, this sentence is wrong.

For the latter, if the starting and ending points are the same, the displacement is the same, which is true.

Cars and vans travel the same distance. If you look at the title again, it actually has nothing to do with your above sentence.

A: Parallel lines at the end of the arrows over the resultant force - the only solution.

B: through the net force arrow end as a parallel line, truncated f1 length - the only solution C: with the resultant force arrow as the center of the circle, f2 length as the radius as the circle, take the intersection point - no solution, one solution or two solutions.

Range: Greater than minimum resultant force, less than maximum resultant force.

To find the three resultant forces is to combine two by two, and the same is true for more than three.

Actually, it's a matter of drawing a parallelogram, you just have to think about a parallelogram in your heart.

If there are more than three forces, they should be combined in pairs and gradually synthesize.

Whether it is the only solution depends on the number of parallelograms that meet the conditions.

Use the parallelogram rule or the triangle rule.

With the parallelogram rule (1) (2) unique (3) not unique.

The rate of change of velocity is the acceleration of an object. Since the formula a = v t and the formula a = v t is called the rate of change of v, both, like both, are vectors.

But we need to distinguish between three different physical quantities: velocity v, velocity change vt-v0 and velocity rate of change (vt-v0) t.

When the acceleration is positive, the direction of acceleration is the same as the direction of the initial velocity, and the velocity of the object increases, that is, the accelerated motion is done; Conversely, when the acceleration is negative, the direction of acceleration is opposite to the direction of the initial velocity, and the velocity of the object decreases, i.e., it decelerates.

Yes, comrades on the 1st floor are troublesome, don't mislead people.

Yes, the difference between the two velocities is the velocity change value, and the division time is the velocity change rate (the amount of change in velocity per unit of time), i.e., the acceleration.

I don't want to give more points to these questions, and I'll do it.