Physics masters, come to a sophomore physics problem

1）c=q/u ; e=u/d 。E=q cd=ne Cl is obtained.

2) Let's say there are up to n. Then the velocity of the nth electron is exactly 0 when it reaches the b plate

From v1*v1-v0*v0=2as, v0*v0=2*(ee m)*l is solved to obtain n=m*v0*v0*c 2e*e+1.

3) The first electron moves at a constant velocity with velocity v0.

The last electron moves in a uniform deceleration motion and the velocity v0 drops to v0.

So t2-t1=l v0.

Solution: 1. When n electrons arrive on board B, let the voltage at both ends be U, and there is C=Q U to get U=Ne C

and the electric field strength e=u l. So e=ne lc

2. If n electrons collide with the B plate, the speed of the Nth electron when it reaches the B board is just 0

According to the kinetic energy theorem, there are: 0-mv 2=-neu and c=ne u, n=v e (cm 2)3, the time of the first motion t1=v l

The time of the nth movement t2 = v a

There are 2 problems, e=u l=v l (m 2c) electric field force f=

a=f/m=ev/lm×√(m/2c)

Exercise time t2 = v a=ve l 1 (2cm) t=t2-t1=ev l 1 (2cm)-v l

(1) Let the velocity of the particle pass through the q point as v, and the angle between its direction and the positive direction of the x-axis is 60°, and the particle does a flat-like motion with a uniform electric field, and the solution of V=VO 60°=2VO is obtained by having V=Cos60° at the Q point

2) Let the magnitude of the electric field strength of the uniform electric field be E, and the particles will move like a flat throw.

x-axis displacement x=vot, y-direction displacement y=at 2 2.a)

From the meaning of the title, y=h-xtan30°b)

At the q point there is vy v=sin60°

So vy=at=3vo, y=vyt2= 3vot2.c)

Synapsis (a)(b)(c) gives 3vot 2=h+vot tan30° solution t=2 3 h vo....d)

a=f/m=qe/m at=√3vo e=mvo^2/qh

3) The time it takes for a particle to move from point q to point m'Charged particles move in a uniform circular motion in a uniform magnetic field, from the inscription, VQ is perpendicular Oa, vm is perpendicular to O, so the center of the circle is at point O, let the radius be r, the period is t, then it can be seen from the title r=x cos30°=3h

qvb=mv^2/r b=2mvo/(3qh) t=2∏m/qb

t'=t-t/6=5t/6 t'=5∏ h/(2vo)

1.Let the voltages at the two points of a and b be u, which is obtained by the kinetic energy theorem: slag limb type u*(-q) = (1 2)*m(2v) 2-(1 2)*mv 2

Therefore u=-3*mv 2 (2*q).

2.The electric field intensity e=-3*mv 2 (2*qd), and the direction of the electric field cong line is horizontally to the left.

This is a question of deep filial piety.

You can remember it like this.

When looking at an object in the water from the air, the depth becomes smaller, and the actual depth is divided by the refractive index, and the depth of the bird and fish is 3m

The height of the object in the air is sensitive or in the water and the height becomes larger, the actual height is multiplied by the refractive index, and the height of the fish and the bird is 4m and the second vacancy is 8m

Both voids are 7 because they are directly below and the light is straight.

Rate of change of magnetic flux = change of magnetic field b * area s time t, b, s, t are the same, so 1:1

Therefore, the electromotive force ratio is also 1:1

Energy is conserved, so the power ratio is 1:1

It is correct to let the bs distance be x,a, because, b,s displacement is the same, then x=n (n is an integer), then as=2x=2n, which is also an integer multiple of the phase difference wavelength, should always be the same displacement.

b is not true, because bs is the opposite, indicating x=(n+1 2), then as (2n 1) should always be the same.

c is right, because cs is the same, indicating x=n ; Then dc=2x=2n; , as in the case of A.

d is also true, because cs is the opposite, indicating x=(n+1 2); then dc=2x=(2n+1); It is also an integer multiple of the wavelength, so it is always the same.

The right one is ACD

Select ACD

A pair, b and s are always the same, then the distance is an integer multiple of the wavelength, so a and s are also integer multiples of the wavelength, so a and s are always the same.

b is false, because if the wavelength is as, then a and b are always opposite, i.e., a is inverse to s, then b is the same as s.

c Pair. If the displacements of S and C are always opposite, the distance is an odd multiple of half a wavelength, and Cd is an even multiple of half a wavelength, that is, a multiple of wavelength, and D and C are always the same.

C is the opposite of D, so C is wrong, and the same principle is true.

Pick B. The ground floor was well explained.

It's a classic physics experiment, don't you know?

The answer is b parsing:

When the round hole is large, the light travels along a straight line, and a bright circle is obtained on the screen, and the round hole is reduced, and the bright circle is also reduced;

Then get an inverted image of a small bulb on the screen, this phenomenon is called small hole imaging, and the light is still propagating in a straight line;

When the hole is reduced to a certain extent, the light is diffracted and some light and dark rings are obtained on the screen, and the hole is further reduced until it is closed, the number of rings increases, the area expands, and the brightness of the ring decreases until it is completely dark

Only when the wavelength of light is much smaller than that of the obstacle, the light can be regarded as straight, and when the size of the obstacle can be compared with the wavelength of the light or even smaller than the wavelength of the light, the diffraction phenomenon of light is very obvious With the change of physical conditions, the physical phenomena that occur can also change accordingly, so when analyzing physical phenomena, we should pay attention to the process from quantitative change to qualitative change

The answer is B

b Diffraction of light, go back and check the relevant knowledge points in the book, you will be clear.

I choose CBecause the total resistance of the user decreases when the output voltage remains unchanged, p=u 2 is smaller and p is larger.

As a result, the current on the transmission line increases.

Loss on the transmission line p=i 2*r

It also increases.

c The user is a parallel circuit, the more lights are turned on, the smaller the resistance, and the greater the power.

Because the power of the primary coil = the power of the secondary coil, the power of the primary coil increases, the voltage remains the same, and the current increases.

According to Power = Avon R, the power on the wire increases.

A wrong transmission power is determined by the load, b is also related to the current of the load, c see upstairs d there is a loss in the power transmission process u -- the amount d option is misprinted! It should be "The output voltage of the step-up transformer is equal to."

The voltage will only be related to the number of turns, so the voltage on the side of the step-down will not change because of the user's use, that is, when the number of electrical appliances used by the user changes, it will only change the working current, that is, the more electrical appliances, the greater the resistance value, the smaller the current, according to the transformer current relationship, the good point of the transmission line can be calculated, "I2R is a good wattage."