Junior high school physics teachers or seniors, please enter.

I hope you'd better find a teacher to give you a make-up class at home, see the simple place yourself, let the teacher help you summarize the difficulties, give you a thorough explanation, so that your efficiency will be much higher, will save a lot of time, your biggest problem now is that there is no time, don't just rely on yourself to do the question or read the book, that will definitely not work, after the knowledge points are understood, quickly do the question, it is best to be basic according to the chapter, and then do mock questions. That's the fastest. Be sure to stick to it!

Complete the basic questions and understand the knowledge points in the book in 15 days, use the 10 days to do mock questions, write down the questions or knowledge points that you will not know every day in a book, and read them every day, as long as you persist, you will definitely get good results.

Although the high school entrance examination is very simple, it is also very difficult for your junior high school children to study so much in such a short time, and there are so many subjects, there are indeed a lot of things to understand in junior high school, and there is also the practical application of knowledge.

Brother, I'm a freshman in high school, if I'm not mistaken.

The key points are: several methods of density measurement.

Leverage issues. Gas pressure.

Electrical design steps measure u,i

There are also experimental steps in the book, familiar with the small knowledge points, memorize it.

Physics is very simple.

Just do more basic questions and adjust your mentality.

Do a few sets of must-haves for the high school entrance examination and find out how you feel.

Junior high school physics can be divided into the following sections: Force and Motion, Density, Pressure, and Buoyancy, Simple Mechanics, Mechanical Work, Sound and Light, Heat and Internal Energy, Circuits, Magnets, Atoms, and Galaxies. Grasp the key points of each part, there will be no problem in the high school entrance examination, believe in yourself.

i believe i can！

Grasp the textbook, understand the basic knowledge, and master all the best experiments in the textbook.

I generally scored more than 95 points in physics, and the secret is to understand more, I understand all the formulas, and I have read more wrong questions; Be sure to understand! The high school entrance examination is actually very simple, you don't need to be too nervous, first lay a solid foundation.

Find an outline to review the bibliography, first know the general knowledge, slowly study the details, make up a little bit is a little bit, all do this, you should be able to too.

Summary. Hello, what's the problem?

Physics teacher in junior high school.

Hello, what's the problem?

23 questions for teachers.

Okay, send the answer right away.

Hello, the first question: from the leverage balance condition: m0 * ob = 5 * oam0 = 3kgm * oa = m0 * ocm = 40kg

Hello, OA, OB, OC is the length of the question.

Hello, m=12kgm*oa=m0*ocm=12kg. Is that all there is to it?

Hello, the first question: the lever balance condition: m0 * ob = 5 * oam0 = 3kg two together is the first question.

Hello, m=12kgm*oa=m0*ocm=12kg teacher, you continue.

Good. It's gone. Real * 6 = Real =

The answer to the second question.

Ask about custom messages].

Hello, I am a physics teacher, and the process of solving the problem is as follows:

The voltmeter measures the voltage of R1, and when the sliding rheostat slider P is at the midpoint, the voltage at both ends of R1 is 4V; When the sliding rheostat slider p is at the maximum, the voltage of R1 will decrease, and the change of 2V is indicated in the title, so it is 4V-2V=2V, the power is 4W, and the square of R1=U P=1 ohm; Let the maximum resistance of the sliding rheostat be 2R, the sliding vane p of the sliding rheostat will increase by 2V from the midpoint position, and the resistance of the sliding rheostat will increase from R to 2R, so the voltage at both ends of the sliding rheostat is 4V, so the distribution of the power supply voltage is 2V at both ends of R1, and the sliding rheostat is 4V, so the maximum value of the sliding rheostat is 2 ohms, and the power supply voltage is 6V

Continue to ask questions about the unclear, and hope to give points, thank you.

To measure the resistance, the component must be disconnected from the circuit, so the first blank fill: S2 small bulb resistance is generally a few ohms to tens of ohms, for example, 8 ohms, so the selector switch is 1, and the reading is 8 ohms.

The multimeter is used as a voltmeter, which is connected in parallel with the small beads, so the voltage of the small beads is measured.

To measure the small bead current, the ammeter must be connected in series in the circuit, so S1 is closed and S2 is disconnected

S1 is disconnected because the ohmmeter cannot measure live.

1 ohmmeter median value is 15 The scale around the median value is relatively uniform, the measurement is more accurate, and the resistance value of the small bead is about 10 ohms.

8 ohms reading is fine, and we chose a magnification of 1

l Parallel to the light.

S1 S2 Let the multimeter be connected in series with the bulb.

Of course there are ampere. Because when exercising, an induced electromotive force will be generated, and if there is an electromotive force, there will be an electric current, and if there is a current, there will be ampere force. In fact, there can be a formula f=b 2l 2*v r, which is deduced according to the voltage and ampere force conditions of electromagnetic induction, and can be used casually in a uniform magnetic field.

The negative work done by the ampere force is the ** energy consumed by the resistor r.

Compulsory 2 in the first year of high school

h and t of an object in free fall are obtained from h=1 2 gt 2.

The acceleration due to gravity on the surface of the Moon g=2h (t 2).

Solution: There is a question to know, h, t is known. According to the formula h=

It can be seen that g=2*h t 2.

h and t of an object in free fall are obtained from h=1 2 gt 2.

The acceleration due to gravity on the surface of the Moon g=2h (t 2).

The spacecraft orbits the lunar cycle t

by mg=m42t2r

The radius of the moon can be obtained r=gt 2 4 2

1800n reason: the worker tightens the rope to provide 300n tension, and the sand and box pair is 300n 3 = 900n

And because the average density of sand and boxes is 2 times that of the river, the total gravity of the sand and the box is 2 times of the buoyancy, the sand box rises at a constant speed, and the sum of the pulling force of the movable pulley on the sandbox and the buoyancy of the water on the sandbox is equal to the total gravity of the sandbox, that is, the pulling force is equal to the buoyancy and is equal to half of the total gravity.

So, the total gravity is equal to 2 times the pulling force.