How is high school physics so incomprehensible? Problems with High School Physics?

1. Learn mathematics well, mathematics is the language of physics.

2. Mechanics grasps balance, and electricity grasps equations

After the understanding of the various mechanical concepts, the equilibrium equations of the forces (stationary, uniform linear line, floating, levitation, levers, block pulleys, thermal equilibrium、、、 are often used

Electricity is to use the laws of physics to list the electrical equations after clarifying the direction of the current, and solve them. (There only needs to be one change in the circuit: the resistance changes, the switch changes, the temperature changes, and so on.) The individual physical quantities in the circuit will be redistributed).

3. Pay attention to the method of problem solving: experience repeatedly. All kinds of problems are the superposition of simple problems, and finding the relationship between each simple problem is the key to solving the problem.

4. Pay attention to several conservation laws: the law of conservation of energy, the law of conservation of momentum, the conservation of mechanical energy, the conservation of charge, and the conservation of mass、、、 according to their equations, which are simple and clear.

Of course not, this is still a mastery method.

I have to say that high school physics is based on high school mathematics to solve, that is, to use elementary mathematics tools to solve, relatively speaking, the calculation is relatively simple.

If you want to be proficient in physics knowledge, as far as the content of high school is concerned, it is very important to be very familiar with the representation symbols of each physical quantity, there are not many things in high school physics, and you can sort it out after combing, such as acceleration, force, work, velocity, mass... At the same time, you also need to know the units of these quantities, because sometimes the units can be calculated to get the units of the corresponding quantities, for example, the units of mass and acceleration can be calculated to get the units of force.

After mastering the representation and units of quantities, it is necessary to know the physical meaning of the mathematical quantities that are specifically represented by quantities, such as the speed of the quantity that reflects the speed of motion.

After that, it is the understanding and memorization of the formula, using the physical quantities written down before, repeatedly deriving, finding the connection between the formulas, and deducing and simplifying at will, these are all a process of accumulation and proficiency, you are only a freshman in high school, so you have time to be more solid.

After that, it is not difficult to analyze the known and problems of the question stem, contact the lower quantity, find the physical meaning, and basically solve the problem.

If you want to master it all, you have to keep getting in touch with new types, and learn some college physics and Olympiad physics appropriately.

According to the textbook catalog, summarize the content and key points by yourself, and then compare them with the textbook, and you will know your mastery. It takes patience to do it repeatedly.

This, our teacher said that you need to persevere, maybe you can't understand, but one day you will suddenly open up, learn to summarize the question type, don't do it blindly, and strive to do everything you have done, and then eat the formula thoroughly, to keep changing everything, will actually be those things, but the process of hard work is very hard, I hope it will help you, I am a girl, it's really hard to learn physics, how to do it, it's really not open, so I changed to literature, in fact, I like science very much.

Knowledge points (concepts, formulas, laws, theorems, etc.) should be kept in mind and made to be accurate and not confused.

To figure out the type of questions, each question type has a clear solution idea, and the solution of the same question type is mostly similar.

Find out the test point, listen carefully to the teacher in class, which common test, in what form, etc., take good notes, common test knowledge and question types, etc., and mark them well, accumulate what content is often tested in each exam, and you can judge a rough idea of what is often tested in each exam, and you can do a good job of review and preparation before the exam.

Pay attention to the application of mathematical knowledge, including trigonometric functions, solving systems of equations, plane geometry, etc.

Remember to be careless, Zhang Guan Li Dai, and be sure to answer as required.

Physics requires comprehension and a lot of problems to understand what has been learned.

The fixed pulley can change the direction of the action: when the fixed pulley under the ceiling is hung on both sides, the gravity on both sides acts on each other through the rope, and when the mass of the two sides is not the same (ma>mb), the two forces are not unbalanced, so there is a relationship of "force = mass acceleration" between the force of the two objects moving together as a "whole" and the hidden change of the overall mass and acceleration, where "mag-mbg" is the "combined external force" when the whole accelerates the motion.

mag is the weight of weight a:=ma gmbg is the weight of weight b:=mb g because the two weights a and b are connected by a rope around the fixed pulley.

Both of them move together and have the same acceleration.

Without considering the friction of the pulley, consider these two punching weights as a whole. At this point, the resultant force they experience is equal to:

f = ga - gb = mag - mbg = ma - mb)g

So let's dress up: f = ma+mb) *a

i.e.: f = ma+mb) *a = m-mb)g

Mag-MBG refers to the difference in gravity between MA and MB (G is the acceleration due to gravity) and acts on a system with two weights as a whole.

It's envy that I didn't quarrel with my brother Cha Liang like this:

AB and C acceleration are the same, AB and C before the impact of the velocity is 4-VC and VC respectively, after the impact the corresponding velocity becomes VC and 4-VC, the final velocity becomes V, C increases from rest to VC, and then decreases from 4-VC to V, the total time is equal to T=VC A+(4-VC-V) A=2S, it can be understood in this way, and it can be written in the form of the question at once, it can be regarded as such a summary, I hope it will help you.

Because whether it is an object or an inclined plane, when they are separated, it is when the external force of their hall is revoked, so it is no longer accelerated, that is, the acceleration is the last moment. So the speed is at its maximum at this time.

When a ball falls from a height, where does it end up?

2.What is the acceleration of a god-destroying object after it is thrown?

3.What is the difference between force and potential?

4.What is the Momentum Theorem?

5.What is the light mu Yu Xiao Xing Xun Han?

6.What is Electric Field?

7.What is Electric Field Force?

8.What is Polarization?

9.What is Magnetic Field?

10.What is Magnetic Induction?

Another question post of yours on the same topic, Uncle has passed, please move to check it out.

This question involves the conversion of kinetic and potential energy in physics, as well as the basic principles of inclined plane mechanics.

First, we need to understand the dynamics of the movement of blocks downwards along the inclined plane. As a block slides down an inclined plane, it is subjected to two forces: gravity g and the inclined force n supporting it.

The magnitude of these two forces is mg (m is the mass of the block, g is the acceleration due to gravity) and nsin (the angle between the block and the inclined plane).

As the block slides down the slope, its kinetic energy e k = 1 2 * m * v 2, where v is the speed of the block. Since the support force nsin is always perpendicular to the inclined plane, the component of the block on the inclined plane v s = vcos. So, the acceleration of the block sliding down the slope a = g*sin.

Now let's analyze what happens when the block reaches its maximum velocity downwards along the slope. When a block's velocity reaches its maximum, its kinetic energy is also maximum. According to the kinetic energy formula e k = 1 2 * m * v 2, we can get the maximum velocity v max = sqrt (2 * e k m).

At this point, the acceleration of the block A = GSIN also reaches the maximum value, i.e. A Max = GSIN.

Next, let's analyze the situation where the ramp velocity reaches the maximum value. When a block's velocity reaches its maximum, its kinetic energy is also maximum. According to the kinetic energy formula e k = 1 2 * m * v 2, we can get the maximum velocity of the block on the inclined plane v s max = sqrt(2 * e k mg)).

At this point, the acceleration of the block on the inclined plane a s = nsin m = gsin m also reaches the maximum value, i.e. a s max = g*sin m.

Finally, let's analyze the separation of the slope from the block. When a block's velocity reaches its maximum, its kinetic energy is also maximum. According to the kinetic energy formula e k = 1 2 * m * v 2, we can carefully obtain the maximum velocity v s max = sqrt(2 * e k mg)).

At this point, the acceleration of the block on the inclined plane a s = nsin m = gsin m also reaches the maximum value, i.e. a s max = g*sin m.

In summary, when the block reaches its maximum downward velocity along the inclined plane, the inclined plane velocity also reaches its maximum, and it is when the inclined plane separates from the block. This is because in this process, the combined force of the kinetic energy and support force of the block reaches the maximum, so that the speed of the block's downward movement along the inclined plane and the speed of the inclined plane reach the maximum value.

First of all, if the charge of A-T is less than 4q, the slope of the V-T diagram at the end will not be close to 0, but will increase positively.

Secondly, I don't think the explanation of the answer to this question is easy to understand, and I think it would be much better to consider it from the perspective of the force of the negative charge. From the V-T diagram, the negative charge from C-D is a deceleration motion with reduced acceleration, and the acceleration and velocity are 0 at point D, and from the acceleration point we can know that the negative charge is a state of force equilibrium at point D, and at this time, the positive charge charge at point A can be calculated as +4q by using the Coulomb's law equation

The answer to this question involves the law of conservation of kinetic energy and the law of conservation of momentum. When a block moves down an inclined plane, it is supported by gravity and the inclined beam, which is perpendicular to the inclined plane and gravity is downward along the inclined plane. As a result, the kinetic energy of a block moving on an inclined plane decreases due to the force of the inclined plane on the block, while the velocity of the block increases.

When a block moves down the inclined plane, its kinetic energy increases, and when the block leaves the inclined plane, its kinetic energy reaches its maximum value, i.e., the law of conservation of kinetic energy holds true.

At the same time, the friction between the block and the inclined plane will reduce the speed of the block as it moves downward along the inclined plane, and the amount of friction is related to the contact force between the block and the inclined plane. When the velocity of a block reaches a certain value, the bevel will have little support for the block, causing little friction, and the speed of the block will reach its maximum value. When the block and the inclined plane are separated, the friction disappears, the law of conservation of momentum holds, the momentum of the block remains the same, and the velocity remains the same.

Therefore, when the block reaches its maximum downward velocity along the slope, the speed of the slope is also maximum, and this is when the block and the slope are separated.

If the roughness of the inclined plane is insufficient ( due to the smooth ground, the direction of the pressure (elastic force) of the block on the inclined body and the horizontal component of the frictional force are to the right, and the inclined body is accelerated to the right, if the roughness of the inclined plane is insufficient, the pressure is always applied to the inclined body in the process of sliding up and then sliding down, and the horizontal component of the friction to the left is not as large as the horizontal component to the right, and the resultant force received by the inclined body has been pushed to the right to accelerate the motion until the block slides to the bottom of the inclined plane and the two are separated. After separation, the bevel is no longer under pressure, no longer accelerates, and the velocity reaches its maximum.

If the slope is smooth enough, the block will come to rest relative to the slope after sliding to the highest point, and the two will maintain a uniform motion to the right at the same speed, and the speed will naturally reach the maximum sail handicap.