Physics in the first year of high school will not be TAT

Analysis: The key to answering this question is to see whether the relative position of the object under study and the selected reference object has changed, and if so, the object is in motionIf there is no change, the object is relatively stationary

Answer: Solution: Synchronous communication satellite refers to the geosynchronous satellite, the direction of rotation and the angle of rotation of the satellite and the earth are the same, and there is no change in the relative position between the earth and the satellite, so with the earth as the reference system, the satellite is stationary between the satellite and the sun, the position is constantly changing, that is, if the sun is used as the reference system, the satellite is in motion, so a is correct, bcd is wrong;So choose A

Definition: An artificial satellite orbiting the Earth with the same period as the Earth's rotation period, choose A

There is no quick fix to improving comprehension. Can give you a little advice:

After you have learned a knowledge point, you need to do the corresponding exercises (moderate is good), and then you can think about "whether the problem can be solved by the previous method, is it simpler or more complex", what kind of problem can be dealt with with with this knowledge point, and what are the prerequisites for using this knowledge point?

I believe that going through the above process will be helpful to you in understanding and summarizing skills.

The problem is not which discipline or what kind of problem, all the so-called problems are the same.

Remember the saying "there is no qualitative leap without the accumulation of quantity", if you do not have a qualitative change, it can only indicate a problem, your amount has not been accumulated in place, so don't say "already worked hard", you still have space!

If you have a clear understanding of the process of motion, and understand which forces do positive work and which forces do negative work, you can use the kinetic energy theorem.

If you don't know the process of motion, you can just think about it, and you can use the conservation theorem like the conservation of mechanical energy, the conservation of energy, and so on.

In fact, when you write out the conservation theorem, put the kinetic energy aside from the power amplifier equation and the kinetic energy on the side, you will find that this equation is the kinetic energy theorem.

So there's no need to dwell on the issue of energy.

The two questions are almost the same, and the solution ideas are differentHowever, for the same question, there are more than two ways to solve the problem. Ay...

I'm so good at physics, I didn't know that there was a formula w=ek, hehe.

It's just the wrong way to be attached. What you know is not in line with what it should be. To be clear, you have misunderstood something ...

Ideas for solving the problem?In front of everyone, it's like a decision.,It's like a lot of tricks to understand the problem.。。。 I've never looked at any solution ideas.

No matter what the example question is, I always only look at one thing, and that is the answer... As for how each one understands, do not limit yourself to the minds of others ... Think from whichever angle you think you can figure out...

Approach all issues in your own way so that you don't get confused in other people's minds. Of course, there is one point, this own way is not limited to what kind of problem solving ideas. I've always been first-class for multiple-choice questions, fill-in-the-blank questions, true/false questions, and the speed of doing questions...

It's fast and I haven't even finished reading the question.,The answer is known.,The fastest ten seconds less than a question.。。。

However, doing calculation problems, although it is not that I can't, just because I can't write the process. All processes are crowded. Use your own thinking to sort out ideas that others can understand. At least I'm not going to lag behind here, and the accuracy rate is quite high...

I have a characteristic of doing math problems and physics problems, the answer is the first to squeeze the process... Choose a fill-in-the-blank question, and the answer will be over in seconds or tens of seconds... It's all the idea of not knowing how to write the process.。。。

It's all about understanding with your own potential, not confined to other people's thinking. I'm better than anyone else... At least at that time, he was a top master of physics in school...

1) Let the velocity of the small slider be V when it can reach the Q point, then QE+MG=MV R (the circular motion can reach the highest point condition).

Solution: mv = qe+mg=

In the process of the slider moving from the beginning to the q point, the kinetic energy theorem has:

mg+qe)x-(mg+qe)·2r=½mv²-½mvo² (x=

vo²=49

Solution: vo=7m s

2) Let the velocity of the slider when it reaches point P be v1, then the process from the beginning of motion to point P, the kinetic energy theorem:

mg+qe)x-(mg+qe)r=½mv1²-½mvo²mv1²=

At point p, the circumferential motion is obtained: the supporting force of the orbit on the slider f=mv1 r = by Newton's third law: the pressure is.

The first question is the question of conservation of energy: l is the starting length of the small slider from n, vq is the velocity of the small slider at q point, and vp is the velocity of the small slider at p point.

l（eq+mg)+2r(mg+eq)+1/2mvq²=1/2mv0²..

mvq²/r=eq+mg ..

Simultaneous can be found v0

The second question: 1 2mvp -(mg+eq)r=1 2mvq .

fp=mvp²/r ..

Simultaneous can be used to obtain FP

The meaning of the question shows that the charged body must be negatively charged, and the object moves in a straight line with uniform deceleration, f=eq eq=tan53°mg e=tan53° mg q tan37°=1 tan53° e=mgcos37° qsin37°

Select AD

1. Acceleration due to electric field force: a1=eq m=e acceleration caused by Lorenz force: a2=bqv m=2b acceleration: a=a2-a1=2b-e

According to the electric field and magnetic field change diagram, it can be seen that

2 t 4, a=2m s, direction+y;

At 4 t 6, a=-2m s, direction-y;

Other time periods a=0

The ball flies to point m, the time it takes: t = (2+4 ) 2=1+2 (seconds) of which there is a +y acceleration of 2m s in 1 second

Longitudinal displacement: that is, when the ball flies out of the MN line, it will be offset by 1 meter in the +y direction.

The ordinate of point m is 6 meters, so the ordinate of point Q is 5 meters.

2. If it is shot from point n, because the ordinate of point n is 0 meters, the ordinate of point q is -1 meter.

(1) The amount of momentum change of a small object mv-0=mv

2) According to the conservation of momentum, the amount of momentum change of people = the amount of momentum change of small objects = mv (3) Let the speed of people be v, from question (2): mv = mv, so v=mv m I hope my answer can satisfy you, if you have any questions, please continue to ask, please adopt if you are satisfied, your adoption is the motivation for me to answer and know!

I can't see it clearly, but it seems to be a momentum problem, you can try to solve it with the law of conservation of momentum, this problem is very basic, read the book carefully.

Let the distance of the armed police soldier accelerate and descend is h1, the distance of deceleration and descent is (h-h1), and the final velocity of the acceleration phase is equal to the initial velocity of the deceleration phase.

is vmax, which is obtained from the relation between the topic and the displacement velocity of the uniform variable speed motion:

Free fall process: vmax2=2gh1

Uniform deceleration motion process: V2-Vmax2=-2A(H-H1) is solved by two formulas: H1=(V 2+2Ah) 2(G+A) = mThe maximum speed of the armed police soldier is:

Deceleration time: t2=(vmax v) a = the minimum time for sliding: t=t1+t2= s+ s= sThe minimum time for the armed police soldier to descend is that the distance to accelerate the descent is.

It's that simple, no, give me I'll tell you.

From the decomposition of motion, it can be seen that the motion can be divided into horizontal motion and vertical motion.

So the horizontal velocity is v=10 3m s

When landing, the vertical velocity is v=30m s

So the time of the fall is 3s

The height of the throwing point is 45m

Grasp the direction of this problem to solve!

As follows**! <>

Word hopes you are satisfied!