A sophomore physics question, urgent, a sophomore physics question, please teach, thank you

Answer: Set to n, the capacitor B plate has a charge of Q=Nq, and at the same time, the same amount of -nq is induced on the opposite side of the A board. The different ** charge nq on board A enters the ground through the grounding wire.

The potential difference between the two plates: u=q c=nq c, and the electric field strength during the period: e=u d=nq (cd).

Because the charge of plate B is the same as that of the droplet, the falling droplet is affected by the upward electric field force and has gravity, and when the two are balanced, the droplet will not fall again.

That is: mg=qe, mg=q*nq (cd), and the solution is: n=cdmg q2

I hope my analysis can help you a little!

The upstairs idea is right, but the analysis is wrong.

then the mass of water passing through per second is.

m=10^4

kg has an output energy per second for.

w = mgh + 1 2mv12-1 2mv22 = 10 5 + power. p=w/t=6*10^4w

First of all, let's calculate the output work of a hydropower station in δt time, and it is clear that according to the law of conservation of energy, the mechanical energy of the water can be changed as δe=mgδh+

1) formula. The water flow rate is q=10m3 s. Then equation (1) gives qδtgδh+;

2) formula. Whereas there is a loss of power in hydro turbines and generators, according to known conditions, the generator's can be known.

Output work w= 'qδtgδh+so, output power p=w δt= 'q[gδh+

2. The known power is.

624000W, the output voltage is 240V

Then the current in the original coil circuit is.

i=624000/240=2600a

The energy loss of the transmission line is 5%.

then the power consumption of the transmission line is.

po=624000×5%=31200w

then the transmission line current is.

io=sqrt（31200×3/50）=

It can be obtained according to the relationship between the current in the transformer coil and the number of turns of the coil.

The ratio of turns of the primary secondary coil is.

n1:n2=i2:i1.

n1:n2=：2600

The ratio of the number of turns of the primary and secondary coils of the booster is.

then the output voltage of the booster.

n1:n2=u1：u2

240:n2=：2600

Get u2 = volts.

Because the power consumption of the transmission line is.

po=624000×5%=31200w

Whereas, the transmission line current is.

io=, the transmission line consumes voltage as po=io u

u=volt. Therefore, the input voltage of the step-down voltage must be volt.

then n1:n2=:220

The ratio of the number of turns of the primary and secondary coils of the step-down is .

Upstairs, your answer seems to have a little problem, when gravity is equal to the repulsive force, it is possible to reach the B plate, because there is an initial velocity. It should be considered as work

The work of electric field force w1=uq=n0q*q c

Work on gravity w2=mgd

Prime kinetic energy e=1 2mv2

When W1>=W2+E, the droplets no longer reach the B plate.

n0>=(2gd+v2)mc 2q2---1) satisfies the minimum value of n0 of equation (1).

f=kq1q2/r^2=9*10^9*2*10^-4*4*10^-4/3^2=80n

The magnitude of the Coulomb force with which the two point charges are interacted with is 80N

Mutually exclusive, in opposite directions.

1g of hydrogen is 1mol

The number of hydrogen atoms is.

The energy of complete fusion is.

e is the result of your first question)

This is also a high school question??? I graduated from high school early, but I won't

We can analyze each option based on the known characteristics of the Earth's magnetic field and knowledge of electromagnetics:

a.Magnetic inductance lines are curves that are real in a magnetic field.

Mistake. Inductance lines are virtual curves used to represent the distribution of a magnetic field, they represent the direction and strength of the magnetic field, but they are not real physical entities.

b.The earth is negatively charged if the geomagnetic field is generated by a charge on the earth's surface.

That's right. Although the geomagnetic field is not actually generated by the charge of the earth's surface, if we assume that the geomagnetic field is generated by the charge of the earth's surface, then the earth will be negatively charged. Because in the Northern Hemisphere, the magnetic field lines are directed towards the ground, and the electric field lines generated by the negative charge are directed towards the electric charge, which corresponds to the direction of the magnetic field lines.

c.The positively charged cosmic rays that shoot vertically downward towards the equator are subjected to the Lorentz force to the east of this group.

That's right. According to the right-hand rule, for a positively charged particle, the direction of the Lorentz force experienced in the magnetic field is the cross of the direction of the charge velocity (downward) and the direction of the magnetic field (roughly south to north near the equator), resulting in east.

d.At the equator, a conductor rod with a vertical downward direction is subjected to a force of ampere force to the west.

That's right. According to the ampere force formula and the right-hand rule, when an electric current flows vertically downward along a conductor rod, the ampere force it experiences in the geomagnetic field (roughly from south to north near the equator) is a cross of the current direction (downward) and the magnetic field direction (from south to north), resulting in a westward direction.

Therefore, the correct options are c and d.

What does turning from west to east and east to west have on whether it is positive or negative? The title doesn't want to assume this, so there's no need to dwell on it.

In digital circuits, BCD stands for binary-encoded decimal digits. When a number is represented in BCD form, it is divided into a combination of four numbers, each with a range of 0-9. Therefore, BCD numbers can be represented using four-bit binary.

West-to-east and east-to-west refer to the direction in which the current flows. In a circuit, the direction of current flows from the positive (high potential) to the negative (low potential). If the current flows from a high potential to a low potential, then the circuit is forward, and vice versa.

For the Ampere's rule, it is indeed the left-handed rule, which can be used to determine the direction of the current along the conductor as well as the direction of the magnetic field. In the left-handed rule, you can place your thumb, index finger, and middle finger in different positions to determine the direction of the current and the direction of the magnetic field. However, in electrical circuits, we usually use the law of closed circuits to calculate the relationship between current and voltage.

Friend, the following is the analysis and solution process, I hope it will help you!

Solution: Analysis: (The spring in your question should be fixed to the right end!) The whole situation is consistent with the conservation of momentum:

That is, mv0=(m+m)v--- v=1m s··· The maximum elastic potential energy of a spring is when the spring is compressed to the greatest extent! According to the inscription, the iron is bounced back and happens to be at the left end of the board, and the whole process is only the friction between the iron block and the board, because the horizontal plane is smooth!

The process of the iron moving forward and returning to the spring, the friction between the iron block and the wooden board is equal and set to w, and the maximum potential energy of the spring is e

The current situation is impermeable:

1 2 mv0 2=w+e+1 2 (m+m)v 2··· Which disadvantages.

When returning to the process:

w=e+1/2 (m+m)v^2···

Since the maximum potential energy is e=4j!

I only did it once, and I only calculated it once, and I forgot to understand it! Hope it helps!

This problem is to calculate the voltage at the 2nd end of a capacitor. Let the voltage of the upper terminal of the capacitor be U up, and the voltage of the lower terminal of the capacitor is U down; The potential of the negative end of the power supply is 0V, and the positive end is V.

When the rheostat is on the far right:

The potential on the U is the same as that at the negative end of the power supply, which is 0V.

The potential at u is v 2r r = v 2

When the rheostat is on the far left:

The potential on u is v 3r r = v 3

The potential at u is v 2r r = v 2

When the rheostat R is adjusted from right to left between 0 and 2R: the difference becomes smaller when you are up and down is the same, so the capacitor is discharging and the maximum voltage of discharge = v 3

8 resolutions:

Because the wave range difference between source A and O and point B is r=r1—r2=r=45m=, the phenomenon of interference cancellation occurs at point B.

The condition for not hearing sound at any point c in the circumference is:

r = r1—r2 = (2k+1)=5(2k+1) substitute r2=r=45m into the above equation to get: r1=5(2k+1)+ r2So: r1=10k+50 or r1= —10k+40 and 0 < r1 < 90m, so there is:

0 <(10k+50) <90m and 0 <(10k+40) <90m

Obtain :—5 < k < 4

i.e. k, so there are eight places where no sound can be heard until point A is reached.

It's okay to say this question 8 times.

The answer is quite satisfactory.

It is solved by the principle of interference of waves. The difference between the distance between the two horns at the point on the circumference where the sound cannot be heard is an odd multiple of half a wavelength.

Take a good look at the book on waves, the difference between the two and the relationship between wavelength.

You didn't say what else you should want! The conditions are not very sufficient.

He wrote it wrong, he wanted to square it, and you will know that he wrote it wrong by checking it with the unit system:

The unit of BLV R is the current i

Isn't it bil when you multiply BL, is it ampere force F So what he calculates is force, not power!

And the unit of power is f v, so he is still one v short of multiplying it

In **. Hope that helps.

p1=w／t

w=mgh m／t=ρq

p1=mgh／t=ρqgh=1*10^5wp2=p1*η1=5*10^4w

p3=p2*η2=3*10^3w

p3=﹙u2^2﹚／r

u2=﹙p3*r﹚^﹙1／2﹚=300v

The number of turns of a step-up transformer is 4 5

The number of turns of a step-down transformer is 15 11

It can make 470 "220V, 100W" lamps shine normally.

Gravitational potential energy is converted into electrical energy, w=mgh= qgh=1*10 3*2*

Since the output voltage is 240V, the step-down process is performed. The voltage ratio is equal to the turns ratio. Therefore, the turn-to-turn ratio of the transformer is: 12:11

By: w=n*w, n=w w=

First of all, the mercury above the gap will definitely flow out, and the column of mercury below will also rise and flow out. It is easy to know that the atmospheric pressure on the left side is 56cm mercury column, and if it drops h, then there is a universal formula: pv=nrt, and the pressure ratio of mercury is the reciprocal of the height ratio, and the pressure converted to mercury is obtained respectively

40 56=(40+h) (52+h) know h=4 then know how high the closed air column above the left side will become40+4=44cm

No outflow of air column becomes 16cm

Hello, ** not?