Ask for a detailed analysis of the physics topic in the first year of high school, thank you

Question 1: In general, whether an object can be regarded as a particle or not, the key is whether the object we want to study can ignore its size and shape, the boat in the option is obviously not, because the boat is not a uniform object, so c is correct.

Question 4: Speed limit generally refers to the average rate, that is, the average speed. That is to say, within this limited speed range, it is possible, but it cannot be greater than this limited speed.

Question 7: The graph in item d is not a displacement-time image, but only a coordinate graph, so D is not selected.

Question 10: Based on (v-v0) t=a, it can be concluded that the final velocity of an object must indeed be greater than the initial velocity by 1m s in any given second. And the problem that you assume that the initial velocity is negative is related to the selection of the positive direction, assuming that the positive direction I choose is to the right, then since the velocity is negative, the velocity direction of the object is to the left, and because the object is moving in a straight line with uniform acceleration, the direction of acceleration is to the right, so the object is accelerating to the right, that is, it is accelerating in the opposite direction, which means that the magic velocity is not smaller, but larger.

If you really don't understand, you can think of the number line, the left side is negative, the right side is positive, the number line goes to the right, the number will get bigger and bigger, for example, (-3) will be greater than (-4).

Hope it helps!

1. The ship is swaying from side to side because of the moment action, and the center of mass of the ship does not move;

4. It is not an instantaneous speed, but a rate... This kind of question is boring to say);

7. Child, this is the image of the trajectory... Not a displacement-time image... Displacement-time is a function of the relationship...

10. Child, the title is very clear, it is a uniform acceleration movement, and the rate is increased. . .

1. When an object is regarded as a particle, only its mass is considered, not the shape, and the boat swaying with the wind must consider the state, so it cannot be regarded as a particle.

2. It is not an instantaneous velocity, but a velocity, and the velocity is not considered as a direction, but only as a speed.

3. It is impossible for a displacement-time image to have two intersections on the same line at the same time.

4. For example, 2>1, -1>-2, uniform acceleration linear motion, are added.

10."The object moves in a straight line with uniform acceleration" What is uniform acceleration? That is, the acceleration and the initial velocity are in the same direction! Relative uniform acceleration is uniform deceleration, and acceleration is opposite to initial velocity.

Look back and take a good look and you'll understand. Hope.

。。。My mistake is to ask for a black hosta.

1 Because the object is at rest and only receives two forces, one gravitational and the other is the pulling force of the rope, these two forces constitute a balance of two forces, so it must be collinear and the center of gravity will be below the rope.

2 In this question, you can think of a spring as a spring scale, the upper end of the spring is the hook of the spring scale, and the reading of the spring scale is equal to the tension of the hook end (this is because the spring studied in high school is lightweight). In this problem, although the lower ends of the four springs are different, the upper ends are all pulled by the same force, so the readings should be the same, and the spring scale readings are proportional to the elongation (this is because of Hooke's law), so the elongation of the four springs is the same.

The elastic force in the light spring is equal everywhere, so the elastic force in these four springs is equal to f, i.e., f=kx, and k is the same, so x is the same, i.e., l 1 = l 2 = l 3 = l 4, independent of the other forces acting on the block and the state of motion of the block.

a force is the cause of changing the state of motion of the object, for this problem, since f1 f2 is always true, the resultant force always has an acceleration to the right of the object, so its velocity always increases until it is maximum.

f1>f2 so f1-f2>0 ma = f1-f2>0 so a 0, the velocity gradually increases. But f1 is decreasing, so a is also decreasing, and when a is 0, the velocity increases to its maximum.

The acceleration has been to the right, and I know that it is reduced to zero, so I have been accelerating to the maximum!!

Because f1 is greater than f2, the object moves to the right, and when f1 is equal to f2, the object is not affected by the force and moves at a uniform speed, so the velocity reaches its maximum.

- - Let's put it this way: The object starts moving to the right because f-junction=f1-f2 is to the right.

After that, f1 gradually decreases, but it is still greater than f2, so the direction of velocity is still to the right, but the resultant force decreases, so the acceleration also decreases, and finally f1 = f2 when the acceleration decreases to zero, and the velocity is at its maximum.

The answer is c, the free fall motion is a uniform accelerating linear motion with an initial velocity of 0 and an acceleration of g.

According to the formula v 2-v0 2 = 2ah, the initial velocity is 0 and the acceleration g is g and the height is h substituted into it, which can be obtained.

v 2 = 2 gh, when the altitude is 1 4h, let the velocity at that time be v'. The same formula is still available.

v'^2=2g 1/4h.It is not difficult to see V from these two formulas'=1/2v。This is the foundation to do more questions What should I do if I don't understand this, I am a college student studying in Guangzhou, and I was good in high school in physics, so ask me if you have any questions.

The square of c v is equal to 2gh in the same way; The square of the velocity is equal to 2g multiplied by a quarter of h so that the velocity is solved as v 2

This question can be understood in this way: since the speed of light is much greater than the speed of sound, the time it takes for light to travel is negligible. The distance of lightning is observed at the distance of the person who opens the skin, and the distance at which the sound travels, such as a tomb.

As for the solution, the slag side relatives v = 330m s = 1 3km s. So s=v*t=

As for d, the answer is obvious, if the speed of sound changes, then the result will definitely be different.

You can understand it this way, take t = 1s, use 1 3 =, s 330m, this uses the sound 1s propagation is about 1 3km to calculate, where the search for practical t 3 is to use t * (1 3) hail training, and the speed of light is much greater than the speed of sound, you can ignore the influence of its source finger, except for b other answers are not correct.

The speed of light is too great, much greater than the speed of sound, and it can be ignored.

Acceleration a, v=54, v=0 (final velocity) s1=125, abbreviated s2=225v 2-v 2=2*a*s1

a=54*54/(2*125)

Friction f=ma, m=100*1000kg, traction ff-f) = ma2, (potato talk a2 is the acceleration when the hand is speed) a2 = v 2 2*225, the formula is the same as the one that finds a.