High School Physics, the answer is A, what is wrong with BCD

Figure A is a series circuit, the current of R1 and R2 is the same, the current is not coarse and fine, and it is not used for current regulation;

Figure A is a parallel circuit, R1 and R2 have the same voltage, and are not used for voltage regulation;

Series voltage regulation, coarse adjustment of large resistance, fine adjustment of small resistance.

Parallel current regulation, coarse adjustment with small resistance, and fine adjustment with large resistance.

In the figure, Figure A is the trimming circuit of voltage, and B is the trimming circuit of current, which should be understood! For Figure A, his resistance should be relatively small, you think, divide a resistance into many equal parts, the resistance is small, the resistance value of each equal division is small, so that the change of resistance multiplied by current is small, and the effect of voltage fine-tuning can be achieved! Diagram B.

Equal voltage of current divided by resistance! The larger the resistance, the smaller the current, so to achieve the effect of fine-tuning, the resistance value of each equal division should be larger, so the resistance value of the fine-tuning resistor should be larger than the resistance value of the coarse adjustment resistor!

Series voltage division, parallel shunt, when parallel connection, the coarse adjustment resistance is larger, then the current range of the trimming resistor is too small, the impact is not large, the actual effect is not good, so C is wrong. When the coarse adjustment resistance is large in series, the fine-tuning resistance can achieve the fine-tuning effect.

Series voltage regulation, parallel current regulation.

Figure B is a trimmer circuit suitable for current, and r2 >r1

First of all, this coordinate system is deceptive!

Your calculations start without error, and the ball rises to its highest point in the vertical direction with t = . But the height of the rise at this point h = 1 2 * g * t = m!

That is, in the coordinate diagram, 2 cells represent remorse, and each cell is .

So, based on this finding, let's move on to the calculation:

1 2 * a * t = 3 * = , so a = 15m s

The time taken by the ball to fall must still be , at this time the partial velocity of the ball in the vertical direction vy = 4m s, and the partial velocity in the horizontal direction vx :

vx = a * 2t) =15 * 2 * =12 m/s

Then: v = vx) vy) 4 +12 明森正 = 4 (1 +3 ) 4 *10

So: Spring talk.

v = 4√10 m/s

That is, what you said b and c are all correct answers.

A is the pair, the movement is always not separated, otherwise the internal force of the system needs to do positive work (for example, there is explosives ** between the 12 balls, so that the balls are separated), the mechanical energy of the system is not conserved, and the B item is wrong. On AB, let the angle between the line of each ball and the O point and the horizontal direction be , if there is no pressure between the 23 balls, then the acceleration of the ball along the track direction is a= gcos, by the position of the ball in the figure, the angle of the 2 balls is smaller than 3, so a2=gcos 2>gcos 3=a3, that is, the acceleration of 2 balls is greater than 3 balls, so there must be positive pressure between 23 balls, so 2 must do positive work on 3.

Wrong in item c. On the CD, assuming that there is no pressure between 23, then the acceleration of the ball is GSIN , which is the inclination of the inclination of the inclination of the incline of the CD, and it is found that the acceleration of all the balls is the same, so the assumption that there is no pressure between 23 is correct. So 2 to 3 does not do work.

Item D is wrong. The mechanical energy of the system composed of 4 small balls is conserved, and when it stops after reaching the CD orbit, it should be at the same height as the center of gravity position at AB (i.e., the midpoint of 1234). Since the relationship between angles and angles is unknown, it is impossible to judge.

If the angle is large, the height of CD4 is greater than ab1, and if the angle is small, the height of CD4 is less than ab1. Of course, it is also possible to be equal.

When the four small balls move on the arc, the centrifugal force can be decomposed into horizontal direction and vertical direction, the gravity does not change in the vertical direction, the velocity is the same and there is no interaction, the velocity in the horizontal direction is not equal to produce the interaction force, and the No. 3 ball has a leftward force on the No. 2 ball in the horizontal direction, that is, the No. 3 ball has done work on the No. 2 ball, so item b is wrong.

When the four balls reach the horizontal BC segment, their final velocity will be equal under the interaction force, so it is impossible for them to reach the same height as the initial position of ball 1, hence the D term error.

When the four balls reach the CD segment, because their velocity is already the same in the BC segment, the velocity of the No. 4 ball will first slow down as the altitude increases after entering the CD segment, so as to do work on the No. 3 ball behind it to slow it down, and similarly the No. 3 ball will also do work on the No. 2 ball. So what ball 2 does to ball 3 is not positive, so c is wrong.

The ball is subjected to gravity in the circular orbit, the track support force and the adjacent ball pressure on it, and the angle between the elastic direction of the 2 to 3 balls and the velocity direction of the 3 balls is less than 90 degrees, so the elastic force is positive, and B is wrong.

The velocity of the four balls is the same after reaching the plane, and the velocity acceleration of the subsequent movement is the same except for the position, so there is no elastic force between the balls after they are on the CD slope, and no work is done, and c is wrong.

Considering the four spheres as a system, assuming that d is correct, then the center of mass of the system should be equal according to the conservation of mechanical energy of the system, but the centroid of the four balls at the initial moment is not at the height of the center of mass of the sphere, so the assumption is not true, and d is wrong.

The acceleration of 1 ball in a circular orbit is the largest, and decreases in order of 2, 3, and 4 (you can see it by analyzing the force). So the 4 balls are sticking all the time. The acceleration of 2 balls is greater than 3 balls, and it must have been done at the beginning, B is wrong.

C is even more obvious, the velocity of the four balls at each point on the slope is the same, and there is no ball-to-ball work. d, because A, so, they can be regarded as a whole, the whole, just look at the center of gravity, the center must be the same height, then the height of the 4 balls and the original 1 ball, it is obviously different, a slope, an arc, can only say that the height of the center of gravity remains the same.

bThere's nothing to say... c, the speed of the 4 balls is equal before the uphill climb, and the 4 balls in the ascent phase are equivalent to not interfering with each other until the 4th ball rises to the top.

D is also not true, because when the four balls are out of point B, they are equivalent to a completely inelastic collision, and the energy is lost, so they cannot rise to the original height.

B does positive work, C does not do work, and D position is higher than the original.

For personal answers, you refer to the following.

Let the angle between the A rope and the vertical direction be

When the car is at a constant speed, there is.

t1sinθ=t2

When the car accelerates, the ball accelerates with the car, the acceleration is the same as the car, set to A, and the direction is to the right.

There is t1sin -t2=ma, and the A direction is to the right.

As can be seen from the equation, as long as the calculated acceleration a is positive. That is, t1sin t2, as long as this condition is met.

A: T1 increases, T2 decreases, yes.

b: T1 increases, T2 does not change, yes.

c: T1 does not change, T2 decreases, yes.

d: T1 increases, T2 increases, as long as the value of T2 is less than T1Sin. So the possible situation abcd

The ball accelerates evenly to the right, and a force to the right is applied to the ball, (f=ma).

This force is provided by the component in the horizontal direction of increasing t1, and the other forces or components remain constant.

Therefore, answer (b) is correct.

Previous considerations were too incomplete.

I think B should be chosen

Lou mainly thinks it's right to say it.

This is a dynamic analysis problem, and it is easy to understand if you take into account the included angle when you analyze the force.

You must be clear that when the fine-tuning refers to the change of the sliding rheostat, the whole circuit does not change much, while the coarse adjustment refers to the fast and large change.

Figure A is suitable for the voltage trimming circuit, then the coarse adjustment resistance should be larger, r1>r2 a is correct.

It is known from a that b is incorrect.

Figure B is suitable for current trimming, so d is incorrect When you want to fine-tune, the resistance of R2 should be large, so that when it changes, the current change rate can be small.

So c is incorrect.

A is a series circuit, B is a parallel circuit, in a series circuit, you adjust the size of one of the resistance values, the voltage at both ends will change, and the voltage at two points of the parallel circuit remains unchanged, so A is voltage regulation B is current regulation, according to v = IR, the greater the resistance change value, the greater the voltage change value, so the coarse adjustment resistance in A should be large, and the smaller the resistance in B according to I=V r, the greater the current change value, so in the current regulation, the coarse adjustment resistance should be small.

Series voltage division, parallel shunt, so A is voltage regulation, B is current regulation.

Coarse adjustment refers to the large change in the data, and fine-tuning refers to the small change in the tuning.

Figure A R1>R2, so that the voltage of R1 can be adjusted to change greatly.

Figure B regulates the current of R1 with a large change, so the resistance is small.

Figure A is a voltage divider circuit with a fine-tuning voltage. Figure B is a shunt circuit, and the two resistors should be equal.

5.Correct answer c

The capacitor is connected to the power supply, the inter-plate voltage U remains unchanged, the charged oil droplets are negatively charged, mg=qu d moves up on the plate d increases, the inter-plate field strength e=u d decreases d wrong, the electric field force decreases the downward movement of the oil droplets, a wrong p point potential p=el decreases b wrong, c= s 4 kd decreases q=cu decreases c correct.

No matter what meter A and B are, after the switch is closed, the current can pass through R1 alone, so R1 and R2 cannot be connected in parallel. For multiple-choice questions, B and C can be excluded directly.

The detailed analysis is that the B answer is short-circuited, and the C answer is parallel, but the multiple-choice question should seek the most concise way to answer the question.

If A is an ammeter and B is a voltmeter, then R2 is shorted (because the ammeter resistance is negligible) and D is wrong.

B, A and B are both ammeters, the circuit is short-circuited, C, A and B are both voltmeters R2 is not connected to the circuit D, A is the ammeter, B is the voltmeter, R2 is short-circuited.