How can I review physics well in my third year of high school!!! 20 points

I am a physics teacher, and I used to be the third prize in the Anhui Physics Competition. Learning physics was easy and fun.

1.First of all, don't be afraid, if you can learn physics well, that's weird. Also, don't be afraid of difficulties, and have the bite to dare to challenge.

2.Books are the key, and you must read them thoroughly. My physics teacher said, "You have to read the book badly, and look at the sleepy ones when you go to bed at noon."

3.If you understand the book, you must consolidate. Do a set of questions, time-limited, even if it is just one point, you have to finish it yourself at the specified time. One set is enough, I won't figure it out, and then I will do the whole set again to see the effect. And that's not all!

4.Change the subject! Try to change the question yourself so that you can understand the idea and method of doing the question, as well as the traps that the person who wrote the question may have to set. This is the highest level of learning physics as a student.

The main thing is to face it bravely, fail again and again, and get up again and again. Life is not endless, and the struggle is endless. Believe in yourself!

Since the ball is stationary at point a, it can be seen that the electric field force is horizontally to the right, negatively charged, and eq=mgtan53=4mg3

When there is a force of gravity equal at the pull force in C, the velocity at point C is 0

According to the kinetic energy theorem mgl-eql=-mvb 2 2, vb=(root number 6gl) 3 can be solved

Move to point D, the velocity of VD at point D is 53 degrees with the initial velocity, then the resultant force of the electric field force and gravity is constant force f = root number (mg) 2 + (eq) 2 = 5mg 3, f is at right angles to vd, then start from point d to do a flat throwing motion, x=2lcos53=vdt

2lsin53=ft^2/2m

The joint solution yields vd = (root number 3GL)2

You are the "Invisible Dream", I am the subject of the Invisible You! Can't help you......

Choose CD oblique split to move to the left, the heavy ball must swing to the left with the oblique split, and slide upwards on the inclined plane, the friction direction of the oblique split on the heavy ball is downward along the inclined surface, and there is a component to do positive work in the positive direction of the horizontal displacement of the heavy ball and negative work in the negative direction of vertical displacement, but in the end, the center of gravity of the heavy ball rises, and the gravitational potential energy increases, so the friction between mm does positive work on m, and ab is wrong.

Since the oblique split moves at a uniform speed under the action of F, the ground is smooth, and only the force of the heavy ball M on the oblique split is balanced with F, which is equal to the work done by F, and C is paired.

Because there is a relative sliding between the small ball and the oblique split, there is a pair of interacting elastic force and a pair of interacting friction force in the process respectively, because the displacement of the horizontal direction of the heavy ball (the heavy ball suspension line finally has a swing angle to the left) is less than the horizontal displacement of the oblique split, the vertical component of the heavy ball friction force does negative work, and there is no displacement in the vertical direction of the oblique split, and the vertical component of the two frictional force components is equal in size, so the friction force does different work on the heavy ball and the oblique split, and the sum of the work done by the support force and the frictional force is equal, so m, The absolute value of the work done by the elastic force between m and the work done on m is not equal, d pair.

It is the support force that does the work on m.

Friction does work on m and no work on m.

So A is right and B is wrong.

The elastic force between F and mm is unknown, and it is impossible to judge that the absolute value is not equal, and C is wrong.

d is the relationship between the action and reaction forces, and the absolute value is equal.

Option: a

After the object is put up, it moves upwards relative to the conveyor belt, so the friction should be downward along the slope. The initial speed of the object is 0, and the acceleration motion is envied. The frictional force f=un=, and the component of gravity in the inclined plane direction is, so the resultant force in the inclined plane direction is, and the acceleration a=8.

The object needs time to accelerate to the same speed as the conveyor belt, at which point the object displaces s=2 2 2a=, leaving a distance of 3m. The object then continues to accelerate faster than the conveyor belt, so that the friction in turn rises along the inclined plane. The resultant force in the inclined direction is, and the acceleration is.

With s=v0t+, the column equation: 3=2t+, t=1s is solved. So the total time.

The object first accelerates the linear motion f = ma f = mgsin30 + f f = umgcon30 to get a = ?(Do the math yourself) After accelerating a period of Sun Quarrel time, when it reaches v=2m s motion, then the displacement of V=AT1 S1=1 2ati is determined by the maximum static friction (the maximum static friction is approximately equal to the sliding friction, which is actually slightly greater) f=umgcon30 is greater than the force that makes the object slide down, mgsin30, so the object can move at a uniform speed s2=vt2 s2=l-s1 t, total=t1+t2=?Or do the math yourself)

The title is wrong! Without the mass of the object, it is useless to just give friction...I would like to check the next question.

Learning physics is very process-oriented, a process of cognition, understanding, and application.

1.Cognition: Use the things or phenomena around you or even some examples narrated by the teacher to help yourself fully understand it and become interested in it.

2.Comprehension: Use understanding to memorize formulas, theorems, experiments, etc. It can be understood and memorized in clever ways such as visual thinking. For example, what is a vacuum can be understood in this way: a vacuum is really empty, there is nothing.

3.Application: One is to cope with the exam, and the other is to explain some physical phenomena around you.

So, when studying, first of all, don't be afraid, because your previous experience of not learning well may suggest something to you, which may lead to a vicious cycle for you. Try to tell yourself, "I can do it!! In fact, psychological suggestion is very useful!

However, in order to increase your confidence, it is best to do a good job of pre-study and be aware of it.

Secondly, you should follow the teacher's train of thought in class, take some notes appropriately, and write down some knowledge points that are not clearly stated in the book or even omitted and are easy to make mistakes. Take time out of class to practice more, don't prevaricate for any reason, thus giving up the best time to practice, which can only lead to tragedy in the end.

Last but not least, you must make sure to make a timely summary. For example, the paper of the last exam has been sent down, and although it has been carefully corrected, you still have to think about why it was wrong. How is the correct answer calculated?

Will it be wrong if I get it again next time? Wait a minute.

I think that through these study methods, I will definitely be able to learn physics well.

First of all, we should cultivate an interest in learning physics, and secondly, we should listen carefully in class, ask questions if we don't understand, review after class, complete homework carefully, and pay attention to the combination of work and rest.

Some concepts and formulas of physics must be firmly remembered, pay attention to the connection with the textbook when doing the questions, and the thinking must learn to change, sometimes the two questions are different, but the knowledge points examined are the same, so we must remember the concepts and formulas in the textbook.

If you are a downstream student, you can memorize only one formula and one concept every day, which will help you improve your physics score a lot. If you are a midstream student, then you have to review what you have learned in class, and you must memorize everything the teacher said; If you're an upstreamer student, you'll need to find a learning method that works for you, and it's best to review what you learned yesterday every night.

Finally, it should be noted that physics is a science subject, not only to master the knowledge in the textbook, but also to look at the example questions from some review materials or tutorial books, and pay attention to the knowledge points summarized above.

I don't know if it helps you?

Think more, in fact, the knowledge of physics is very neat, close the book can string together the knowledge you have learned, so that it means that you have almost mastered the knowledge in the book, familiar books, natural things when applying, physics is not like mathematics, mathematics does more problems, physics also has to do problems, but it is not that the more questions you do, the better, the key is to do a general word on the basis of your understanding. Hope it helps.

40 ohms. It can be thought of according to the principle of multi-meter ohms.

Let the full bias current when the two meter pens are docked in the ohmic gear is i, and the electromotive force of the internal power supply of the meter is u, then the internal resistance of the meter (in r) = u i (1).

When measuring a standard resistance of 20 ohms, the pointer is exactly in the middle of the dial, indicating that the current is i 2, and u = i 2*(r in + r measurement) (2).

Syntagonism (1) and (2) know that r = 20 ohms.

Measure an unknown resistance with the same gear, and the pointer points at 1 3 of the full scale to indicate that the current is i 3 at this time, and u = i 3 * (r in + r measurement) (3).

Synoptic (1), (2), (3) knows r = 40 ohms.

Since g is the gravitational acceleration of the ground, and the satellite is at the center of the earth 2r at this time, the centripetal acceleration g here'=/g

**Substitution: mg = gmm r 2

For satellites there are: MV 2 (2R)=GMM (2R) 2 Simultaneous and V=Root (

ps.And g'=mv^2/(2r)

The gravitational force between the two balls is centripetal force, and then v can be solved.

Because gravity is not a centripetal force at this time, it cannot be calculated as such.

I'm guessing your A board is under (1) to get speed v

2) If it can fly out horizontally from the plate, it means that the period of flying out is t, 2t, 3t......Then the plate length is ntv0(n=1,2,3......）

3) When flying in, then by 1 4u0q = to get v, and then by the plate distance =

You should give the diagram out, and I'll give you the answer! Don't know where the n-point is**?

1.1/2qu0/md(t/2)(t/2)=d/2---2)

There are no schematic diagrams of metal plates!