Can anyone help me solve this physics problem, help solve a physics problem, thank you

Well, I'll give you a detailed process.

Since the ball is positively charged, the electric field force eq is downward, because the final uniform motion must be the force balance! And the other force must be the friction of the rod against the ball, as for how this friction ff is generated? Because there is also a magnetic field in the I region, since the ball is subjected to an eq greater than the Lorentz force at first (so it accelerates at first), and the Lorentz force = qvb.

Therefore, as the velocity increases, the Lorentz force also increases, and the left hand knows that the Lorentz force is horizontal to the left, so there will be pressure on the rod, and then the rod will produce friction FF on the ball.

In simple words, an increase in velocity leads to an increase in the Lorentz force, an increase in the Lorentz force leads to an increase in pressure, and an increase in pressure leads to an increase in friction.

Khan Okay, don't say much. Cut to the chase. I use u instead of the coefficient of friction

1) By eq=ff, i.e. e0q=uqvb0, so what, hehe. v=e0 divided by ub0.

2) From the formula MV square ratio r is equal to qvb, we know that b=mv divided by qr. From the vertical in and vertical out of the trajectory in the fourth quadrant, we know that the center of the circle in the fourth quadrant that does uniform circular motion is the origin o, so r=d. Substitute the v result of the first question to know, and calculate it yourself.

Aha. 3) As for the third question, let me think about it again. You look at the one above.

Well, if you want to go back to quadrant i, then it is possible for the ball to be tangent to it as it moves to the diagonal line of the fourth quadrant, so change the radius r of the ball's motion in the fourth quadrant. From the MV square ratio r=qvb to know that r=mv is better than qb, so b is bigger, r is smaller, it's ok!

Questions (1) and (2) have been solved by Xiao Ming in the eighth grade, and (3) is not a difficult problem, of course, as far as your current high school knowledge is concerned, you can only change the size of b to calculate a definite value, so that the trajectory of the ball is tangent to OP, and naturally enters the ith quadrant vertically. The rest is a simple geometry problem, trust you to have no problem.

1, b 2, decrease.

First, we must first understand the parallel shunt and series voltage divider; And in a simple parallel circuit, when the resistance increases, the total resistance decreases (why think about it yourself).

First of all, assuming that R1 is broken, and L1 and R1 are regarded as a whole resistance, the resistance value becomes larger (the resistance is infinite). The latter L2 and so on can be regarded as another overall resistance, and its resistance value remains unchanged. Since the two resistors are connected in series, the total resistance becomes larger and the total current becomes smaller (i.e., the current through L1 becomes smaller).

In this case, it can be judged that A is not right. It is correct to know b with an exclusivity rate. (This is a concise way to do the problem, it is better to think about b).

2. When R2 is broken, the overall resistance of R2 and R3 becomes larger, and the sum of the resistance values of R2 and R4 becomes larger. Then the overall resistance of L2 and the previous R3 and R4 becomes larger. If the supply voltage does not change, the voltage division of the large resistor becomes larger, and the total current becomes smaller.

Since the voltage at both ends of L2 increases, the current passing through L2 decreases, and the current plus the current passing through R3 is the total current, so it is easy to know that the current passing through R3 becomes smaller.

The reason why the total resistance decreases when the resistance of a simple parallel circuit increases:

Because of 1 r (1 r1) + (1 r2) + ....

Assuming that R2 becomes larger and the rest of the resistance remains the same, then 1 R2 becomes smaller, 1 R becomes smaller, and R becomes larger.

1. R1 is broken.

2. When R2 is broken, the current of R3 increases.

1.B should be chosen. Note:

R1 is connected in parallel to L1, and L2 is connected in parallel to the whole above it. If R1 is disconnected, the total resistance of the two resistors in parallel with L1 becomes larger (two resistors in parallel, the total resistance is less than any of the resistances), and the total resistance in the circuit becomes larger and the total current decreases. When R1 is disconnected, the total current flows through L1, so L1 will be dimmed.

Therefore, with the method of elimination, b. should be chosen

The R3 current decreases when the circuit is opened. Note: If the total resistance of the same circuit increases, the total current decreases.

At this time, the total resistance value of R1 and L1 in parallel is unchanged, and the terminal voltage decreases. As the voltage at L2 increases, the current flowing through it increases, while the total current decreases, and the current flowing through R3, R4 and the ammeter decreases.

The main purpose of this question is to let you understand the change of resistance value when the resistors are connected in series and parallel.

This question takes the common washing machine in life as the proposition point, and mainly examines the timing method, rated power, electric work and energy and conversion. The washing time and dehydration time of the washing machine are both the working hours of the washing machine, although the rated voltage is the same when washing clothes, the rated power and the electric power during dehydration are not the same, therefore, it is necessary to solve the work done by the current under different conditions separately. The electrical energy consumed by washing clothes should be the sum of the work done by the electric currents of washing and dehydration.

Because both the washing motor and the dehydrating motor of the washing machine have resistance, the mechanical energy w2 converted when the current is done can be obtained by subtracting the energy consumed by the current through the conductor resistance to generate heat by subtracting the total work.

Solution: W wash = UIT1 = 220V

w off = ui2t2 = 220v

W1 = W wash + W off = 132000J + 52800J = 184800J

w wash = i21r1t1 = (

w off = i22r2t2 = (

W2 = W1 - (W wash + W off) = 1848000J - (6000J + 3456J) = 175344J

Essence: This question focuses on the common washing machines in life, and guides students to pay attention to the objects in life and the physical problems contained in them. This question mainly examines the energy conversion of electrical appliances, and it is necessary to distinguish that the total energy is composed of the energy consumed by the washing and dehydration of the washing machine.

Misunderstanding analysis: This problem is easy to consider only the work done by the current when the washing machine is washed, as well as the energy conversion, so that the total energy calculated is missing the work done by the current when the washing machine is dehydrated.

1.The maximum acceleration that can be obtained by the object in the car is, when the acceleration of the car is, it does not exceed this limit, so the acceleration of the object is also.

Force=2From the first question, it can be seen that the brother posture exceeds the limit of the field dust, so the object will slide, and the friction force to the car is , and the combined external force is f=

3.Let the time t be required, the acceleration of the object is 3, and the acceleration of the trolley is the )*of the object)*t2=1 of the trolley, and the solution t=......

If you look at the title first, it says that the stars were in the center of the universe at the beginning of the universe, and now they are at the edge of the universe, which shows that the speed of the stars is the same as the spread of the universe. So the lifetime of the universe is equal to the time of planetary motion, i.e., t=r v=10 26 (years, i.e., about 17.6 billion years.

I have the impression that the universe expands and contracts at the speed of light"Large**"It is the beginning, and the end of the universe should be the time when it shrinks to the point where it began.

If so, it can be counted as being present from"Large**"It has been 10 to 26 so much time since the beginning.

So much time at the speed of light will run 10 26 so many distances (the units here should be unified, I omitted them).

At this time, there is no doubt that it is already in the contraction phase, because in the expansion phase, the sun cannot catch up with the speed of light and is at the edge of the universe. That is to say, at this time the universe has passed the maximum radius, shrunk to the current level, the following is simple, it is too troublesome to use numbers, set 10 26=a

Now from the big ** is the time that has passed A.

The speed of light travels a distance during this time.

So the maximum distance that the universe can expand is (a - a) 2 + a

This distance divided by the speed of light is half the lifetime of the universe, and multiplied by 2 is the lifetime of the universe.

2*/c

1 v=s t=800 10=80m s can't use p fv because the forehand is digging and is the external force, and we only know the traction force and don't know the friction force, so we can't use it.

2 Hehui scatters the external force of 0, resistance = traction = the fourth power) n

1.What is the maximum speed?

v=s/t=800/10=80m/s

2.What is the resistance experienced when the train is traveling at maximum speed?

Driving at a constant speed on the track, the resultant force of the smile is zero. Resistance = traction = four touches to the power) n,

10*1000 1=10 4, that is, you need to lift your feet 10 4 times after running a 10km distance.

A does work on shoes: w1=m1gh*10 4=

B does work on shoes: w2=m2gh*10 4=

w1-w2=

Every step per shoe: Shoe A requires more gravitational potential energy than shoe B.

Total number of leg lifts for 10km: (10*1000) 1=1*10 4 times.

So the extra work is: 10 4* (two shoes so multiplied by 2).

Hehe, I passed the class test in the first year of high school.

Set the speed v of the soldier and the speed of the team u

Use the team as a frame of reference.

1000 (v-u) +1000 (v+u) = 15*60 solution. Solve it out v=