A physics problem in the third year of high school. Solve the master

1) From the right-hand rule, we know that the current flows from B to A, the plate N is positively charged and Q is negatively charged, and the particles are negatively charged by the upward electric field force and gravity. The induced electromotive force e=blv0, the capacitance voltage u=e 2, the electric field strength e=u d=blv0 2d, and the amount of charge q=2mgd blv0 is obtained from qe=mg

2) AB rod motion velocity v=at, instantaneous induced electromotive force e=blat, electric field force = qblat 2d = mgat v0. In the positive direction, the resultant force of the particle is f=mgat v0-mg, and the acceleration of the particle is =f m=g(at v0-1).

So when tv0 a, the acceleration goes upwards.

AB moves, a current is generated, and there is a voltage at both ends of NQ. U=BLV first judges the direction of the current, and uses the right-hand rule to get that the A end is still positive, and then there is an inter-plate electric field downward. Because particles are subject to gravity, they must produce an upward force in order to be stationary.

Thus, the electrical properties of the particles can be known. is negative. The voltage between the panels is U (R+R) times R=U 2=BLV 2

mg＝qu＝qblv/2

q 2 mg blv

Accelerating the move ab, vi changes, u changes, f==qe =qu d= f slowly increases from 0. So accelerate downwards first and the acceleration slowly decreases. Direction is 0Again. down the deceleration. The acceleration slowly increases.

The speed of the acda ball in the direction of the rod starts and ends at 0Therefore, the initial and final velocity of B is 0, the velocity of B increases first and then decreases, and the mechanical energy of the system is conserved, so AC is correct, when the mechanical energy of A is the smallest, and the velocity of B is the largest, then the rod does not do work on B, and then does negative work on B, so the D term is correct.

a.A keeps decreasing and B increasing.

b.That's right.

c.The decreasing gravitational potential energy of a is equal to the kinetic energy of the increased by a, b dWhen A reaches the bottom, A has a horizontal velocity of 0, B has a horizontal velocity to the right, the force is not 0, and the force of B is to the left.

Since the mechanical energy of the system is conserved, the kinetic energy of ball B increases first and then decreases, so the mechanical energy of ball A decreases first and then increases, and A pairs. Because the kinetic energy of the B ball first increases and then decreases, the light rod first does positive work on the B ball and then does negative work, and B is wrong. C is true, D is false. In the end, the correct answers are A and C.

A is right and B is wrong, for A first do negative work and then do positive work, and for B first do positive work and then do negative work.

C to D to .

According to the analysis of the conservation of mechanical energy of the system, it can be known that:

Before A reaches the bottom end, the light rod has been doing negative work on A and positive work on B, so A has been reducing the mechanical energy and A is wrong;

b correct; When A reaches the bottom, the mechanical energy of the system is conserved, and the velocity of ball B reaches the maximum at this time, which is not 0;

When the A ball reaches the bottom end, the instantaneous velocity is 0, and the instantaneous velocity of B is the largest, but at this time, the instantaneous light rod has no force on the two balls! Since it is a light game, there is no loss of energy, and eventually the two balls A and B will move together to the right. Therefore, it is chosen.

Option A The mechanical energy of the ball does not change due to the absence of friction.

Option B The ball is stationary at the beginning, and then the movement shows that the light club is doing positive work on the ball B. Option BCorrect Option C When the ball A reaches the bottom of the vertical groove, the velocity of the ball B is not zeroOption D When the ball A reaches the bottom, the ball B has a horizontal velocity to the right, while the ball A has no velocity in the horizontal direction, so it can be seen that the mechanical energy of the ball A is the smallest, and the force of the light rod on the ball B is not zero.

Satisfied? Don't forget to give points.

Judging from the picture, the light rod will be stuck at the corners and can't slide down at all?

a,c,d

1.Analyze the physical process first. During the A descent, the B ball moves to the right, accelerates first and then decelerates and finally comes to a standstill.

Because of the overall analysis, the horizontal force of the vertical chute in the horizontal direction of the A and B systems is opposite to the direction of motion to do negative work, which finally cancels out the horizontal kinetic energy converted by the gravitational potential energy. Another angle of analysis, because the last horizontal velocity of ball A is 0, and the distance between b and a remains the same, so the horizontal velocity of the two is the same, and b is also 0, so b will finally be stationary. c correct.

2.There is no gravity on the light club, only the pull (compression) force of the two balls on the light club. Commonly known as the two-force rod, the force on both sides is equal and the direction is opposite.

B accelerates first and then decelerates, so the force of the rod on B first does positive work on B and then does negative work. Therefore, the force of a is first done negative work, and then positive work is done.

Studying ball A alone, because the mechanical energy is a non-conservative force to do work, and the force of the vertical chute to a is perpendicular to the direction of velocity, only the work done by the rod to the force of a is considered, so the mechanical energy decreases first and then increases. A is correct, B is false.

3.The smallest mechanical energy of ball A is also the turning point of the light rod from negative work to positive work on A, that is, the turning point from positive work to negative work on ball B, so the force is correct.

The above is my personal opinion, if there is a mistake, please advise. (I have a 3-point difference in physics in the college entrance examination, so I think the accuracy of doing questions is quite high).

The correct answer is acd

The velocity of ball B is zero when ball A reaches the bottom of the vertical groove, because at this point it can be assumed that A moves in a circle with B as the center and AB as the radius (at this moment), so C is correct.

And because the mechanical energy of the system composed of ab is conserved, the final mechanical energy of a is equal to the initial mechanical energy, which is also the maximum. So A is correct, B is wrong.

The mechanical energy of A is the smallest, and the kinetic energy of B is the largest, and this is the time of the demarcation between the positive work and the negative work done by the rod against B, and it is also the demarcation point between the negative work and the positive work done by the rod on A, so D is correct.

First simple, then complex.

For ball B, the kinetic energy has been increasing (with force analysis, it can be found that the rod has always had thrust on ball B), then the mechanical energy of the system has not become a ball, and the mechanical energy of ball A is wrong, B is correct, C is wrong, and the speed of B ball is the maximum.

The state of D is that A reaches the bottom, and the mass of A and B balls should be considered to find the velocity of both, and the conditions are insufficient to determine whether D is true.

So d wrong. Answer B

The initial potential energy of the ball is mgh=

The kinetic energy on the first landing is 1 2mv 2=

The air resistance work is 125-25=100j, and the resistance is 100 25=4nWhen the ball finally stops, the original potential energy is all converted into resistance work, so there is 125=4s, that is, the distance traveled by the ball is a total of s=125 4=meters.

mg*h-f*h=(mv 2) 2 In the whole process, the potential energy is converted into heat energy to obtain mgh=f*s, and s is the distance traveled by the ball.

Considering that in 1s, w=p*1s, q is the flow rate (q*v 2) 2=p, and the horizontal range is 1:2, then the initial velocity is 1:2, so q1:

q2=4：1，v1:v2=1:

2,t1:t2=1;8

The power is the same, that is, the water flow per unit time is the same ((the distance is because the cross-sectional area is small and the flow rate is fast).

If the depth is the same and the base area is 1:2, the volume is 1:2

Then the time is naturally 1:2.

UMG is the maximum static friction, KX is the elastic force of the spring, the car and the object are relatively forbidden, the friction of the box is definitely less than or equal to the resultant force of the external force, the spring is compressed, A, UMG is less than KX, UMG regardless of the left or right, the object is subject to the acceleration of the left, then the car must be to the left, correct.

b, UMG is less than KX, when UMG to the left, then UMG+KX》MC》KX, UMG to the right, then KX》MC》KX-UMG, so the minimum value is correct, but this here can only determine the direction of acceleration, not necessarily the opposite acceleration, but also the right deceleration.

C, UMG is greater than KX, the car can be left or right, correct D, at this time, if UMG is left, UMG+KX》MC》KX, UMG to the right, UMG-KX》MC》0 or KX》MC》0, so the minimum value should be 0

The crux of this question is that the actual friction force is UMG, not UMG.

Because the minimum value of acceleration c of the - mg car is kx-umg m, but the car can accelerate to the left and decelerate to the right, b is wrong.

In the same way, if mg is greater than kx, the maximum value of -mg acceleration is kx+umg m, and the minimum value of acceleration is 0, d is wrong.

Let's look at A and B first. The block and cart are always relatively stationary, so the acceleration is the same. The force on the car is not easy to study, but the blocks are easy to study.

Blocks only collect friction and elasticity. Due to spring-loaded compression, the block must be bounced to the left. As you can tell from the ab option, static friction is less than elastic force, so it doesn't matter if the friction force is left or right.

The combined external forces are all to the left. So choose A and not B.

Look at C and D again. At this point, the frictional force is greater than the elastic force. If the frictional force is left, the resultant force is to the left when superimposed with the elastic force.

Therefore, the acceleration is also to the left. If the frictional force is to the right, the resultant force is superimposed on the elastic force. The acceleration is also to the right.

Therefore c is correct. As for option D, mg is just a formula for the maximum static friction, and the actual static friction is not necessarily him, and the actual static friction may be equal to the elastic force. That is, the acceleration is 0

So d is clearly wrong.

a c d

The reason why option B is wrong is that it is also possible to do deceleration movements.