High school physics judgment questions, high school physics judgment questions

1. Wrong [can only say that the distance is equal to the size of the displacement, not that the distance is displacement] 2. Wrong [The elastic force of the rod is not necessarily along the rod].

3. False [The force of the horse-drawn carriage and the force of the carriage and the horse are equal in magnitude] 4. False [The static friction force has nothing to do with pressure].

1.False (the distance is s=vt, where v is the average velocity, but when the initial velocity and the final velocity are very different, the displacement is x=v0t+1 2at).

2.Wrong (the way the light bar is subjected to elastic force is a pull or compression) 3Wrong (friction between the ground and the car).

4.Wrong (maximum static friction, pressure can be infinite) hope.

1.Yes, 2Wrong, the direction of the rope is definitely along the rope, the direction of the rod is not necessarily along the rod, 3No, they are a pair of action and reaction forces, they are equal.

4.Wrong, there is no static friction without a relative sliding tendency

1.Correct, 2, wrong, the elastic direction of the light rod can be along the rod, but not along the rod.

3.Wrong. The reaction force is always equal in magnitude and opposite in the direction due to the greater force of the carriage than the resistance.

4. False, static friction has nothing to do with pressure.

1. Yes. 2. The elastic force of the wrong rod is not necessarily along the rod.

3. The force of the horse-drawn carriage and the force of the carriage and the horse-drawn carriage are the interaction forces, and the magnitude is equal.

4. False, static friction has nothing to do with pressure.

Categories: Education Science >> Entrance Examination >> College Entrance Examination.

Problem Description: 1) It only makes sense to determine whether an object is stationary or in motion after selecting a frame of reference ( ).

2) When the object moves in a straight line in a certain direction, the magnitude and distance of the displacement must be the same ( ).

3) The formula Average velocity = initial velocity + final velocity 2 is applicable to all variable speed linear motion ( ).

4) The mass of object A is greater than the mass of object B, then A and B start to do free fall motion from the same height at the same time, and A falls to the ground before B ( ).

5) The object accelerates in a straight line from stationary, and when the acceleration decreases to zero, the speed of the moving object also becomes zero ( ).

Please make a judgment on these five questions and explain why, thank you)

Analysis: 1)...yes

The relativity of motion, the usual "rest" is relative to the reference frame of the earth, and the choice of the solar frame of reference is congratulated by the movement family.

2)..no

Linear acceleration motion is different. Oh, modified: the distance is cumulative, scalar, and the displacement is the sagittal potato, and only the distance between the start and end is calculated.

3)..no

This formula works when the velocity varies evenly, not the other time.

4)..no

The acceleration is the same, the initial velocity is the same, and the ground is the same at the same time.

5)..no

The acceleration decreases to 0, which is equivalent to a decrease in the velocity "increase" to 0, and the velocity will no longer continue to increase. But the speed "accumulation" result is not zero

Question 1, Question 1: The direction of the final velocity is in the same direction as the letter search f2, and the magnitude is 6m s

Explanation: The acceleration of the starting object under the action of f1 is, and under the action of f2, the acceleration is -3m s, and v = repentance s

The second question: in the first 4 seconds, because they are all uniform acceleration motion, the end velocity is 6m s, which can be regarded as the average speed of 3m s, in the same way, the initial velocity in the last 4 seconds is 6m s, the end velocity is -6m s, and the average velocity is 0m s, then the total displacement can be calculated x=3*4-0=12, so the distance from the starting point is 12

0s:0m/s; ;2s:3m/s; 3s：；4s:6m/s; 5s:3m/s; 6s:0m/s; 7s:-3m/s; 8s:-6m/s

Question 2: First analyze a small ball, you decompose the gravity into a reverse component f1 of the wire tension, remember that the angle between f1 and gravity is 1, one is perpendicular to that component f2, then, f1 balances the wire tension, and the acceleration of the ball f2 is a=mg*sin 1 m=g*sin 1In fact, it is not difficult to find that 1 is the angle between the wire and the vertical direction, then you can connect the ball with the center of the circle to make the diameter in the vertical direction, then at this time, the center angle is 2 times 1, and the actual displacement to the low end is x=r*sin(2 1 2)=r*sin 1, do this, and then use x=1 2 at 2, you can find that 1 can be about, that is, the actual time has nothing to do with the angle, so no matter falling from **, the time is the same.

When is the object separated from the plank?

When the object descends by gravity, the tension of the spring and the support force of the plank, when the acceleration of the object falling (less than) is equal to the acceleration of the plank, and the support force of the plank is reduced to zero, the two are separated.

In this case, mg-kx=ma, where x is the type variable.

x=(mg-ma) k

And kl=mg, i.e., k=mg l

x=（g-a）l/g

The separation of the previous work is a uniform acceleration, x=1 2at 2

t = 2 (g-a) l ga under the root number

Solution: The pressure is equal to the component of gravity perpendicular to the inclined plane, so the pressure n=mgcos30°=10

The magnitude of the static friction force on the object is equal to the component of gravity along the inclined plane, so the static friction force is f=mgsin30°=10

The direction is diagonally upwards along the inclined plane.