Physics masters, please enter the good additional 200 and do what you say

Newton's first law is the "Newton's first law" when any object remains at rest or in a constant linear motion until other objects exert force on it to change this state. It should have been learned

The reason why it is said that a car (object) also needs force in order to maintain a uniform linear motion is because the car will be subjected to the friction force of the road surface when it moves, and the force required for the car to maintain its motion is to overcome this friction. Because it is very difficult to have an object that can move without friction (the resistance of air is also a frictional force) in real life, it is really difficult to find a counterexample ...... you wantHowever, in space, this is a common example because of the vacuum state where there is no air. For example, when an astronaut walks out of a space capsule, there must be a rope to connect him to the spacecraft.

If there is no rope outside the capsule, once the astronaut starts moving, he will never stop, and will float forever in space. This is what Newton's first law says, that he will always maintain a uniform motion.

Your question is a bit vague......But it can be said that it takes force to change the state of motion, and it does not take force to maintain the state of motion. That is, the change of an object from a state of rest to a state of motion or from a state of motion to a state of rest requires a force to be applied to it; No force is required to simply remain stationary or in motion. Force is needed to make a stationary object move, and the object does not need force when it is already moving.

Haha, now that we're going to get to the point of arguing, we're going to start studying languages

It's just a discrepancy in language, and I'm sure you already know this issue thoroughly, which is why you can bring up so many questions here. You don't need anyone to help you answer it anymore, I wish you success in your studies and happy growth

Haha, just learning physics, you have to be a little imaginative, don't drill the horns! I also think that there is no example of movement on a ball without leaving the action of force!

Think about it, what is that sentence: Give me a fulcrum, and I can lift the earth! It's all theoretical, it's impossible to achieve, but it's theoretical! Theory is only used to guide practice, and the sublimation of practice is theory!

Don't fight theory with theory, we will always read and return to practice!

Reason 2; It takes force to make a stationary object move, >> definitely does.

But you can't use this as a reason to exercise the need for strength.

To go from rest to motion is to change the state of motion of an object, and motion is to maintain this state of motion, and you yourself are a mistake, what do you call someone else?

It's like if you ask somebody if you ask someone if you need force for a stationary object, and you say you need to do it because it takes force to make a moving object rest, and that reason has nothing to do with the need for a stationary object to ask someone to refute you, how to refute your argument.

9. If you don't know that a stationary object needs force, you can ask me this question, write it in your question).

Ignoring the most critical stuff.

Movement and stillness are mutual.

Let's just say that the cup on the table doesn't give him force, and it's right to say that he's stationary, and it's right to say that he's moving. Relative to the table is still, relative to the sun is moving. To say whether an object is at rest or in motion, one must say what is the reference.

What your teacher is talking about is that it takes force to change the speed of motion of an object. Whether it is to make a moving object come to rest or to make a stationary object move is to change the speed of motion, a force is required.

Motion requires the initiation of force but does not need the support of force, for example, to move at a uniform speed, the external force is zero.

The teacher is talking about the ideal state, but in reality there is no movement without force. But you have to study physics in an ideal state. You'll know why when you're older.

A force is a physical quantity that changes the state of physical motion, not a physical quantity that makes a physical motion.

It's useless for dashagua to tell you, you don't understand what others say.

At the very least, you're asking the wrong question.

1. The car maintains a uniform linear motion under the action of traction and friction given to it by the ground. (If you ignore other relatively small forces such as air resistance.) ）

If there is no traction, it will slow down to a standstill, and that's because it is subject to friction.

If all external forces disappear, the car moving in a straight line at a uniform speed will maintain a straight motion at a uniform speed.

2. Force is required to make a stationary object move, but it only means that the force changes the motion state of the object, and it cannot be said that the object motion requires force.

In the absence of an external force, the moving object remains in motion.

I love liberal arts. Came to the wrong house.

1. The velocity of the movement to point b is 0Then the ball is on its angular bisector by the electric field force and the gravitational force when the resultant force and the pulling force are in a straight line. So there is tan30=mg qe mg=qetan30 i.e. qe= 3mg

At point B, the tensile force t = mgcos30 + qecos60 = mgcos30+ 3mgcos60 = 3mg

Second, there is no contradiction. o Point intensity is zero. The other says that starting from point O, it becomes stronger first and then weaker. It's about starting from 0 to get stronger.

It says that it becomes dense first and then sparse, and the field strength of zero can be regarded as the most sparse!

It is known that the muzzle velocity is v0, and the velocity at t after a period of time is v1 and the velocity v2 at the same time t.

Analysis: According to the characteristics of horizontal throwing motion, V1 is orthogonally decomposed into horizontal component V1X and vertical component V1Y

then v1x v0 , v1y root number (v1 2 v0 2) and v1y g* t

In the same way, v2 is also orthogonally decomposed into the horizontal component v2x, and the vertical component v2y gives v2x v0 , v2y g*(2t).

Visible, v2y 2 * v1y

Then the v2 root number (v2x 2 v2y 2) is the v2 root number [ v0 2 ( 2 * v1y) 2 ] root { v0 2 [2 * root number (v1 2 v0 2)] 2 }

Root number (4 * v1 2 3 * v0 2).

When the velocity is V, the vertical velocity is sqrt(v 2-v0 2). Therefore the flight time t=vy g=sqrt(v 2-v0 2) g

After such a long time, the vertical velocity is g*2t=2sqrt(v 2-v0 2).

At this point, the velocity v'=sqrt(v0^2+(2sqrt(v^2-v0^2))^2)=sqrt(4v^2-3v0^2)

If you learn the kinetic energy theorem, if you learn it, mgh1=1 2mv 2-1 2mv0 2, according to the law of kinematics, the second vertical displacement is 3h1, so mg3h1=1 2mv2 2-1 2mv 2, the two formulas come out by comparison.

Well, I'll teach you.

The concept of the arm is to extend the line of action of this force (that is, the direction of the stretch of the spring) and pass the distance of the perpendicular line made by the fulcrum.

And when you do that, you'll see that there's going to be a right-angled triangle, and the length of the arm is a right-angled side, and the length of the lever is four squares long.

So here, in the same triangle, the right-angled edge is smaller than the hypotenuse, so the force is greater than 1n.

Pulling down diagonally, the force arm becomes smaller and the force becomes larger.

Originally, if it was pulled vertically, it would be 2*2 4n, but now it is pulled diagonally, and the force used is greater, and the drawing is very simple, you can extend the straight line in the direction of the force, and then make a line perpendicular to the straight line from the fulcrum, that is, the distance from the fulcrum to the straight line multiplied by the force, and it can be solved.

You think about the seesaw a little bit, if I'm as heavy as you, then you and I are both at the same distance on either side of the seesaw, just right to balance, and after a few days I'm getting thinner, and you're still in the same position on the seesaw, and I'm going to have to move out a little bit to be able to balance with you. The same goes for the rod scale, the scale is on the rod, and the number is larger when you move it outward.

<>My circuit, and yours, is completely equivalent, you analyze it yourself, after the current comes out of the positive electrode, will it take the gray resistance?

Isn't it just down from the slide (arrow) and then go green?

Your technique is correct, but your analysis is wrong. From the positive pole to the arrow, there are two, one directly down, and the other is a gray resistor, and the preference is that it should be the arrow, not the gray resistor.

That is true. You have to look at it this way, if you are in the middle of this diagram, you should look at it, the red line part has resistance, the green line part also has resistance, according to the principle of laziness, I can form a loop in the green line, I will not go to the red line.

The purpose of taking AB as the research object is to study the tension of the rope between AB. If there is no electric field, there are no charged balls. Then it's a matter of gravity, and acceleration or something is g.

And now it is nothing more than an extra Coulomb force QE on the b-ball, in the direction of downward (the direction of the negative charge force is opposite to the direction of the field strength). The next thing is who is acting on this electric field force, the direct effect is B, then B's velocity acceleration will be greater than G, which is A's acceleration only by gravity is G, then in this case, B will pull A away. Like, you are running in front and a person is walking in the back, and you build a rope in the middle of the two, and you must be pulling him, and the rope is continuous, and your two speeds must be the same.

The rope must have a pulling force. And there is a person in front of you who is also walking, and there is a rope between him and you. This rope must not be straightened.

Surely you are not pulled by him, and the tension of the rope will be 0

If you have to choose ABC as the research subject, you can also do it.

The external force of the ABC system is as follows: F = 10mg + EQF = MAAA + MBAB + MCAC

Since aa=ab, ac=g is substituted.

F = 5mA + 3mA + 2mg = 10mg + EQA = G + EQ 8m

In the analysis, AB was selected as the research object, considering that AB acceleration is the same, and it is more convenient to use the whole method to calculate.

AB acceleration is greater than g, and of course there is no tension between BC.

Imagine if the B ball is not charged, cut the OA, the three balls are all in free fall, the acceleration is G, there is no tension between the ropes, the B ball is electrified, the acceleration is greater than G, and the speed is greater than the C ball at any time, how can it be possible to rely on the C ball to pull it.

If you change the C ball with a negative point, and the AB ball is not charged, it is different, and there must be tension between BC.

The first question, which one to choose as the research object, is based on the experience accumulated in the problem, ignoring the rope between ABC, the electric field force and gravity of the ball B, and the direction is the same. Then, at the moment when the OA rope is cut, the acceleration of ball B is g+x, and the acceleration of ball A is g, then, ball A is pulled by ball B to run, then the rope of the AB section is tightened, and the ball AB does not move with each other. Then AB can be seen as a whole.

The acceleration of the ball c is also g, and the BC will gradually get closer to each other, and there is a mutual motion, so it cannot be regarded as a whole. Then what about the addition of ropes, AB is not moving with each other, then the acceleration of AB should be the same, so when calculating the acceleration of A, AB should be calculated together.

The second question, simple consideration, B acceleration is greater than C, then the next second, B's speed will be greater than C, the distance between B and C will also decrease, the rope is not tightened, of course, there is no tension.

Why AB as the research object? If you take ABC as a whole to study, how can you analyze the forces between ABs (within the system)? Therefore, to require the acceleration of A, AB must be taken as the object of study.

Like you said, OA is affected by three balls, and you confuse the whole with the isolation. If you take ABC, you can only find the force of the OA line

Why a>g??The BC line really has no pull here. This is due to the inertia of the (C-ball), and the C-ball has not yet started to fall at that moment