Ask a physics question 20 about references

Whoever is used as a reference is considered to be immobile. You've been confused yourself.

You've summed it up pretty well.,It's a little dizzy around me.,But it feels okay to look at it again.,Which one to use as a reference is to look at it from the perspective of that object.。

If you grasp the core of the problem and be able to use and analyze the problem well, you are very good.

However, for the self-reflection part below, there is still a need to clarify the thinking and be more organized: in fact, the question of the frame of reference is quite easy to understand, but it is sometimes difficult to adapt to our daily behavior and feelings, so you have such an opinion. Here, there are four parts to the research object, right, the train and you, classmates, and the ground.

The purpose of your thinking is that you want to accept from your own way of thinking and psychology that you and the train are stationary relative to the ground, and your classmates are moving relative to the ground. The second frame of reference is involved, which means that you're not simplifying the problem here, it's a little more complicated, but it's more suitable for your own feelings, right?

In fact, the problem is very simple, taking the train as a frame of reference, and the position of the classmate and the train has changed, then it can be said that the classmate is in motion. And if you have to add an unrelated third-party frame of reference, won't it become troublesome, and this understanding of the problem is not conducive to your grasp and understanding of the law, and you will be more troublesome to learn in the future, and it is easy to mess yourself up.

It is added that after you study physics in depth, especially in college, the solution of a problem involves many frames of reference, so we must grasp the essence and apply the laws truly and effectively. In fact, Einstein's theory of relativity was derived from the small problem of frame of reference.

Your spirit of thinking is very good, keep it up, and hope you can become the next Einstein!

This is a matter of reference.

Xiao Ming felt that he was using the ground as a reference as he was advancing, and he was moving forward at this time.

He felt like he was using the train as a reference as he retreated.

Because the speed of the train is much faster than the speed at which he rides.

So when he compares to the train, he feels like he's going backwards.

It was wrong in the first place. It's too much to wrap around. The reference object is immobile under any circumstances. Other pairs.

Taking his classmates as a reference, Xiao Ming is moving. With Xiao Ming as a reference, the classmates are moving.

Option B treats an object as a reference, and the object is to be regarded as stationary. At the same time, the referencing objects are also mutual.

Here's an example

Suppose you are standing still next to the train tracks, and then a train passes by you head-on, because you are stationary, and you are the reference object, the train is moving backwards (the "rear" here refers to behind you).

The references are mutual. Now with the train as a reference, it is regarded as stationary, but in fact, the train is moving, and the distance between you and the train is actually getting farther and farther, since the train is regarded as stationary, it is equivalent to you moving forward (the "front" here refers to your front).

After this "before" and "after" direction is determined, it is unchanged even if the reference object is changed

You're thinking the same way about this question.

Take Xiao Ming as a reference, (change to the same boat that Xiao Ming sits on), take Xiao Ming as stationary, and then you choose a mountain on the shore as the research object, if the mountain you choose is in front of Xiao Ming (the direction that Xiao Ming is facing is the direction of the boat), because the boat is moving, then the distance between Xiao Ming and Qingshan is getting closer and closer. Since Xiao Ming is regarded as stationary, it is equivalent to Qingshan moving in the direction behind Xiao Ming.

If we take the oncoming ship as a reference, the green mountains move forward. As mentioned earlier, changing the reference object does not affect the forward and backward direction that you have already set. At this time, the direction in front of Xiao Ming is still called "front", and the direction behind him is still called "back".

Treat the oncoming car as stationary, and then choose a mountain between Xiao Ming's boat and the oncoming boat as the object of study. The distance between the oncoming ship and the green mountains is getting closer and closer, and since the oncoming ship is regarded as stationary, it is equivalent to the movement of the green mountains forward.

If you draw inferences from other topics, you will soon be able to figure out this kind of problem.

The reference object can be arbitrarily selected, which can be a moving object or a stationary object, and the specific choice of which object is the reference object is determined by the convenience of the research problem;

Because motion and rest are relative, it is impossible to make a certain determination about the state of motion of an object without selecting a reference object in advance, and it is meaningless to say that the object is in motion or at rest

Therefore, option d is correct, and options a, b, and c are incorrect

Therefore, d

a.The trajectories of circles and ellipses must be generated by the action of mental force, and you can look up the knowledge of mental force.

The gravitational force of the falling ball is constant downward, the acceleration equation is listed in the xy direction, and then the displacement equation is obtained by integrating, and the trajectory equation is obtained by removing the parameters and merging, which is a parabolic equation.

Parabola, there is no air resistance, which is equivalent to a flat throwing motion, so it is a parabola;

ABC is right.,The reason for the first, second, and third speed of the universe.,Think about it for yourself.。。。

Your poor physics and electricity are in junior high school.

In the second year of high school, electricity is re-learned. (That is, the content in junior high school should be relearned in high school (in fact, the original knowledge should be learned first, and then deepened.) For example, if you learn Ohm's law in junior high school, you will learn it again in high school, but in the later stage, you have to extend Ohm's law from part of the circuit to all circuits.

Therefore, you should learn at this time and then take out the previous content and review it. As long as you are serious, you will definitely learn well.

Short circuit and open circuit of the circuit are two different situations. Their impact on the circuit depends on the specific situation.

If the short circuit is between the positive and negative poles of the power supply, then the entire circuit is dead. Similarly, if the circuit is broken at the positive or negative pole of the power supply, the entire circuit will also run out of power. Therefore, from the external situation, there is no electricity, but the cause is different. So analyze it carefully.

Physics in high school is a difficult subject. The emphasis is on understanding, and the problem of the circuit is a part of the easier to learn, and it is related to the reality of life. Think about it more, the idea is the most important.

A short circuit is to directly regard this electrical appliance as a wire, and for an open circuit, it is to see it as two wires that are not connected. EMF and road voltage are different concepts. You can look up the related books.

We know that light is a transverse wave, and in the process of propagation, there will be crests and troughs, and in the area where two columns of coherent waves are superimposed on each other, bright streaks will appear where the crests, crests or troughs meet, and dark streaks will appear where the crests and troughs meet. When the single-slit S moves slightly upwards from the symmetry axis position of the double-slit S1 and S2, if it just moves to the odd times of the half-wavelength difference between the optical path of the single-slit S1 and S2, the light waves generated at the double-slit S1 and S2 are still coherent light, but their phase difference is (2N+1), and the vibration direction of the two wave sources is just opposite, and because the optical paths of S1 and S2 to the P point on the optical screen are equal, the number of waves contained in the waves is equal, so dark fringes will appear at the P point. This contradicts the idea of "** bright lines", so answer B in the above question is incorrect.

How to determine the location of the first bright lines? According to the above analysis, we can see that whether the P spot on the light screen appears dark or bright will be affected by the position of the single slit S. However, in double-slit interference, there is no doubt that the "** point" must be a "bright line".

Therefore, when judging the position of the "** point", it should be: "From the measurement of the single slit S, the point containing the same number of waves in the optical path of the double slit S1 and S2 to the optical screen is the "** point", because the point must be a bright line, and has nothing to do with the vibration at S1 and S2. Therefore, the correct answer to the above question should be D.

Related to the optical path difference, ** the fringe optical path difference is 0That is, ss1 + s1p = ss2 + s2p. In addition, the conditions for interference to occur are that the vibration frequency is the same, there is a stable phase difference, and there are components with the same vibration direction, and the conditions are met, and interference can be generated. d

To understand it is very simple, the interference of light requires the optical path difference to be equal to a specific value, but the default is 0, you can use a stroke, starting from s, all the way to p, passing through the two beams of light of s1 and s2, the difference of the distance is equal to 0When S moves up, the optical path through S1 becomes shorter, so P moves downward.

Attached: Optical path, (in the same medium) refers to the distance traveled by light.

When the single-slit S moves slightly upwards from the ** symmetry axis position of the double-slit S1 and S2, it can still produce bright lines at the light screen.

But comparatively speaking, s can be regarded as the light source of S1 and S2.

S moves up slightly, S1, S2 remain unchanged, then the image formed on the light screen will move slightly down, so the answer is D.

It's true that I chose B, but now I forgot about this knowledge, and I often did this kind of question in my junior year of high school, so it's best to discuss it with my classmates, or read the book, which should be in the book.

Because light travels in a straight line, it will cause displacement on the light screen, so I don't need to talk about interference.

The interference of light is only related to the frequency of light, which is combined with the linear propagation of light, so answer d