Can you send me the answers to the unit quality assessment of Senior 1 Physics? 20

Teacher's book? There are very few books on the Internet, and generally only teachers have such books, and they cannot be published on the Internet because of copyright issues. No, you can type it out and we'll discuss it.

1. Select B for the first question

The same is to walk am, that is, to turn once.

If you push flat. Apply work. mgμa=

If rolled. Its center of gravity changes from the original A 2 to the root number of two a, and the change value is two minus one a

Therefore, people do work mg* half of the root of two minus one a

Calculate the flat push and do less work.

2. Select B for the second question

Because the carriage moves at a uniform acceleration.

So the analysis of human forces:

In the horizontal direction, the person is subjected to the thrust of the carriage to the person (opposite to the size of the person to the carriage, that is, opposite to the direction of the movement of the car) and the friction of the car to the person (the person and the car are relatively stationary, so the direction is forward. Consistent with the direction of vehicle speed).

Because both people and vehicles have positive acceleration, the resultant force received by people is in the same direction as the speed of the vehicle, that is, the frictional force is greater than the thrust, that is, the force of the team and the people are combined to move forward, according to Newton's third law. So the force of the person on the car is forced, and the resultant force is backwards. So.

People do negative work on the car. I'm exhausted.

3 can't. Because it's true that oil droplets can rise, but he can't drip. If it can't drip, there will be no way to design it later.

I'm tired ... Don't write it.

So many questions!! I don't have that time!!

1.Let's set the gravitational acceleration to g first. If it is a flat push, then it is necessary to overcome the frictional force to do the work, because it is a slow advance, so the forward process can be regarded as a constant speed, then the force is equal to the frictional force.

When turning to the past, because there is no relative sliding with the ground, there is no need to overcome friction to do work, and the work done by lifting the object is converted into gravitational potential energy, and the center of gravity is the center of the square, so the conversion amount of gravitational potential energy is (root number 2-1) 2*mga. When the object reaches its highest point, it rolls by its own gravity, so there is no need for human work. In this way, it should be a flat push with less work, choose b

2.In addition to the thrust of the hand, the force of the person on the carriage also has the friction between the foot and the ground, and the two are equal. Therefore, people and carriages are moving at a constant speed, so people do not do work on the carriages. Choose C

(1) Correct, velocity is a vector, both size and direction, the size indicates the speed of the movement, and the direction indicates the direction of movement.

2) False, the average speed is the bit removed by time, and the average speed is the distance divided by time.

3) Error, for example, if a person starts from a certain place and returns to a certain place, then his displacement is zero, and the average velocity is also zero, and obviously, in this process, the velocity is not zero.

4) Correct, δt infinitely tends to zero, which is instantaneous, and the average velocity at this time is instantaneous velocity. (Note that it is not the displacement that tends to zero, but the time).

1. Correct.

2. It is the ratio of displacement and practice, wrong.

3.The average velocity is 0, and the velocity is not necessarily equal to 0

4. Correct. The average velocity for a short period of time is the instantaneous velocity.

1.It should be the velocity that indicates the speed of the movement of the particles. Therefore 1 is false.

2.Average speed = distance divided by time. Therefore 2 is correct.

Average velocity = displacement divided by time.

3.Mistake. When the starting velocity = -2 and the end velocity is equal to 2, the average velocity = 0, but the velocity is not equal to 0

4.That's right. Because the time is short enough, it is negligible.

Hello, see that your question is about to be squeezed out of the list of questions by new questions, and if the question is not expired, points will be deducted and bounty points will be forfeited!So I'll give you a few pieces of advice:

First, you can choose to ask questions in the correct classification of Suipai or go to the professional ** forum related to your question, so that there will be more people who know the answer to your question, and there will be more people.

Second, you can get to know more knowledgeable netizens, keep in touch with netizens who have answered your questions often, and ask these friends directly when you encounter questions, and they will be more sincere and enthusiastic to find answers for you.

Third, what should be done by oneself must be done by oneself, and some things must be solved by one's own ingenuity, and it is impossible for others to do it for you!Only what you do is truly your own, others can only provide you with guidance and advice, and ultimately rely on yourself.

You don't have my answer, but please be sure of my advice!

Although my answer may not solve your guess Zhou Chang problem, it will definitely make it better for you to use it

Follow-up problems are generally divided into two categories:1Whether A can catch up with B.

You can suppose that you can catch up, the time it takes is t, and then the displacement of A is equal to the displacement of B plus the distance between A and B to list the equation, calculate delta, (that is, the square of b minus 4ac), if it is greater than or equal to 0, you can catch up. (When there are two solutions, one is generally rounded off, depending on the specific situation).

2.If you can't catch up, ask you when they are closest or farthest apart, as the second floor said, when they have the same speed, if they can't catch up (that is, when the speed is equal, the displacement of A is still less than the displacement of B plus the distance between A and B), then you will never be able to catch up. In fact, you can still set a time t for this kind of problem, and it is sometimes convenient to solve physics problems mathematically.

Of course, it's good to draw math images, but doing this all the time will reduce your ability to analyze physics, so it's recommended to only use it for exams. Hope it helps you a little.

To find the critical condition, it is generally the displacement relationship when the speed of the two vehicles is equal.

The catch-up problems involving acceleration in the first year of high school are roughly as follows:

1. A chases at a constant speed and car B accelerates uniformly from a standstill.

At this time, the critical point at which A can catch up with B is: when the speed of car B is equal to car A, it happens to be caught up by A, that is, if A does not catch up with B at this time, then it will not be able to catch up in the future.

2. A evenly slows down to chase car B that is driving at a constant speed.

At this time, the critical point at which A can catch up with B is: B happens to be caught up by A when the speed is equal to A's car, that is, if A does not catch up with B at this time, then it will not be able to catch up in the future.

3. Uniform deceleration and even acceleration, same as above.

This problem is too general, that is, to calculate the distance difference between two objects, sometimes it is easier to solve it using an image.

Catch up with the problem, another.

Very simple thinking.

Suppose A and B are separated by s, and a number of conditions are given, and A is asked if he can catch up with B.

In fact, it is to ask about the relationship between their displacement difference and the size of S, if the question is whether A can catch up with B, then let the displacement of A be S1, and the displacement of B is S2, if S1-S2 is greater than or equal to the initial distance S, it is to catch up. If it's less than, you can't catch up. In other words, they are like a difference so far that they use the speed difference to cover these distances, and they just ask if there is enough time.

If one deceleration object is chasing another object. If the pursued physics is at a constant velocity, then the pursuer must catch up before the speed is reduced to the pursued. Otherwise, the rest will only get farther and farther away.

Then the time should be calculated according to the time before slowing down to the speed of the pursuer.

This type of problem can also be converted into a geometry problem, which calculates the area of the figure according to the "velocity time" axis. The area of this graph is the displacement. As shown in the figure below, regardless of whether the initial velocity is zero or not, and whether it is a uniform acceleration or a uniform deceleration, the area of the graph composed of velocity v and time t and the coordinate axis is the displacement.

Remember, physics is something you think, not something you can do by listening to lectures and doing problems.

It's even simpler if both objects are moving at a uniform speed or if the pursued person is stationary. s (v1-v2) = t, i.e. the distance divided by the velocity difference is equal to the time it takes to catch up.