Sophomore Physics 50 points! Physics questions in the second year of high school, high scores for

To do physics problems, we must grasp the essence of physics.

For example, this question.

The key lies in the understanding of the conservation of energy, mainly in the conserved field.

Both the gravitational field and the electric field are conservative fields.

I started the analysis process:

Determine whether the charge carried by the particle is positive or negative: the particle is subjected to only two forces, gravity: vertical downward; The electric field force should be vertically upward, otherwise the particle must accelerate vertically downward, and it is impossible for the velocity to be zero at point b (zero velocity does not mean that the force is zero!).

Therefore, according to the electric field lines pointing vertically downward, it can be determined that the particle is negatively charged.

Determine the potential difference between points A and B. Starting with the conservation of energy, e=ep+ek. There are points A and B ek=0, so the A to B process is to convert the gravitational potential energy (mgh>0) into electric potential energy (qu>0). So, answer B is correct.

There are several ways to judge the strength of the electric field, and I will list them accordingly

1.If there is an electric field distribution map, the conclusion can be drawn directly from the distribution density of the electric field lines.

2.From the analysis of the electric potential difference and the charged properties of the particles, by the analysis, q<0, by the analysis qu>0, so it can be known that the UAB <0, so it can be judged that the electric field is convergent, that is, the electric field line can be accelerated to be generated by a negative charge.

3.Start with the analysis of the movement process and force. Think about the process, it must be the downward acceleration first, that is, gravity qe; Then decelerate to zero speed, that is, mg In summary, you should know the solution.

The most important thing is that I should be analyzed in this way, and the high school physics topics can be simplified. Actually, it's not that difficult.

Good luck with the analytical approach.

c.The B-point field is stronger than the A-point field.

d.A is stronger than B field.

The answer is a, c because this system takes gravity into account, and gravity is always constant. The gravitational force is greater than the electric field force at the beginning of point A to point B, and the direction of the electric field force is upward, and the electric field force gradually increases as the acceleration movement from A starts from point B, so the velocity to point B is zero.

1.The acceleration due to gravity is downward, and the acceleration generated by the electric field can only be neutralized upward, so it must be negatively charged.

The total acceleration at time is downward, and the total acceleration at time B is upward, so the electric field at A is weak and the electric field at B is strong.

So d is wrong.

The velocity is the same at all times, so the work done by gravity is equal to the negative work done by the electric field force, w=mgh=u*q so b is correct.

The wrong one is d

AB is clearly right.

aIf there is a positive charge, both the gravitational force and the electric field force are downward, and there will be no rest.

b is obtained by the conservation of mechanical energy.

From point A to point B, the electric field force becomes larger, and E becomes largerThe B-point field is stronger than the A-point field.

So the answer is d

It's L1 over there, huh? Can you have a picture? I set the length of the cutting edge l, and the length of the uncut is x, and the specific data is substituted for yourself.

1. When the photo frame falls freely from a height of H, the acceleration is a uniform acceleration motion with g, when the photo frame has not fully entered the frame due to the gradual increase of ampere force, so the photo frame does an acceleration motion with a gradual decrease in acceleration, when the acceleration is equal to 0, the ampere force is equal to gravity, and then the uniform motion is done. When all the frames are entered, the magnetic flux does not change, and no induced current is generated, so the acceleration is a uniform acceleration motion of g.

2. Let the velocity when entering the magnetic field be v0

According to the kinetic energy theorem, there is: mv0 2=mgh

Get v0 = 2g

The photo frame moves in the magnetic field, and the ampere force gradually increases, so the photo frame does an accelerated motion with the acceleration gradually decreasing, and when the acceleration is equal to 0, the ampere force is equal to gravity.

Yes: mg=b l v r. Leave the data to yourself!

After that, it moves at a constant speed, and when all the frames enter, the magnetic flux does not change, and no induced current is generated, and then the acceleration is g.

3. In the process of the photo frame falling from free fall to entering the magnetic field, the gravitational potential energy of the photo frame is converted into the kinetic energy and heat of the photo frame. When all the frames are entered, the height of the frames falling is x+h

According to the conservation of energy, there is: mg(x+h)=mv 2+q

So q=mg(x+h)-mv 2

(1) Because v increases (ampere force increases), a decreases, and finally the velocity is constant....Therefore, the wireframe first does the acceleration motion with reduced acceleration, and finally the constant velocity....

2) Because of the constant velocity. So f amp = bil = (b 2 * l 2 * v ) r ( with i = e r ...e=blv (HLV). Since there is no picture. I don't know if you call L1 or L2, so you won't substitute the data.

F amp = (b 2 * l 2 * v ) r = mg. It can be solved that v (3) first determines how much negative work is done by the ampere force for the electric heat generated during the process of entering the magnetic field. Then use the kinetic energy theorem

1 2 mV 2-0 = mgh + w amp.

q = -w amp = mgh-1 2 mv 2

I hope to help you, if you don't understand, please continue to ask

(1) When A and B move together, the sliding friction of the ice face it will be evenly decelerated and accelerated.

a=μ2g= v2 2s =1m/s2

The kinetic friction factor between the plank and the ice surface is obtained 2=

2) The small object A is subjected to the sliding friction of the wooden board on the long wooden board, and does a uniform deceleration motion and acceleration.

a1=μ1g= m/s2

When the small object A slides on the plank, the plank B is affected by the sliding friction of the small object A and the sliding friction of the ice surface, and makes a uniform acceleration motion, and there is 1mg- 2(2m)g=mA2

The solution is accelerated as a2 = m s2

Let the initial velocity of the small object when it hits the plank be v10, and the velocities of a and b after time t are the same as v

From the motion of the long plank, v=a2t, the gliding time t= v a2 = is solved

The initial velocity of a small block against the plank v10=v+a1t= m s

The distance that a small block a slides on a long plank b is s=s1-s2=v10t- 1 2 a1t2- 1 2 a2t2= m

3) The greater the initial velocity of the small block A, the greater the distance it will slide on the long plank B, and when the sliding distance reaches the rightmost end of the plank B, the velocity of both is equal (set to v'), in this case the initial velocity of A is the maximum initial velocity to ensure that it does not slide off the plank, set to v0

There is v0t- 1 2 a1t2- 1 2 a2t2=l

v0-v′=a1t

v′=a2t

In order to ensure that the small object does not slide off the right end of the plank, the initial velocity of the small object rushing onto the long plank is not greater than the maximum initial velocity v0=

2(a1+a2)l =

Answer: (1) The dynamic friction factor between the wooden board and the ice surface is.

2) The distance at which the small block glides relative to the long plank is.

3) In order to ensure that the small object does not slip off the right end of the plank, the initial velocity of the small object rushing onto the long plank may be v0 3m s

in the speed selector.

eq=qvb1

v=e b1 to obtain v=2*10 5m s

Under the action of the magnetic field B2.

qvb2=mv squared r

r=（mv）/（qb2）

The ratio of proton to deuterium nucleus radius is equal to the ratio of mass divided by charge charge.

So it's equal to 1:1

<1> Particles move in a uniform linear motion in the velocity selector, qvb=qe, v=e b1=

2>qvb2=mv 2 r r=mv qbr proton = m*2*10 5 q* the specific charge of the proton is 1 So r proton = deuterium nucleus specific charge is, the reciprocal is 2, so r deuterium nucleus is 5*10 4d = 2 * r deuterium nucleus -2 * proton = 10 5-5*10 4=5*10 4 That's it, be patient, thank you.

By the nature of the mass spectrometer:

The particles injected into the B1 magnetic field receive the Lorentz force and the electric field force.

qvb1=eq

At the same time, we get v=e b1= m s

The mass of the proton is 1 U, the charge charge is a q c The mass of the deuterium nucleus is 2 U (approximate as a linear superposition of a proton and a neutron), and the charge is a q

So, by. Both enter the deflected magnetic field B2 with the same velocity, qvb2=m*v2 r

De: For protons, r quality = m quality v qb2 =

For deuterium nucleus, r deuterium = m deuterium v qb2 = 5 * 10 -3m so the distance difference d = 2 * (r deuterium - r mass) = 5 * 10 -3m This student must study physics hard, otherwise he will not be able to mix it when he goes to college.

According to the inscription, it can be seen that in the process of moving a small block from point m to point n, the friction force (rough horizontal plane) and the point charge q exert on the block (electric field force) in the horizontal direction opposite to the direction of motion, and the friction force is opposite to the electric field force.

Since the block is moved from rest by the force of the electric field, the direction of the electric field force is the same as the direction of motion, and the direction of friction is opposite.

The tangent direction of each point of the electric field line is consistent with the direction of the field strength at that point. The electric field line can only describe the direction of the electric field and qualitatively describe the strength of the electric field, which determines the direction of the electric field force and the direction of acceleration (Newton's second law) of the charged particles at each point on the electric field line, and the electric field line is not necessarily the trajectory of the charged particles in the electric field.

LZ: Not very complete.

Small Blocks with Positive or Negative Bands. is the key to judgment.

One. It's with the right ... is to the left. Positive: The direction of motion is the same as the direction of the electric field.

Two. Negative means to the right. Negative: The direction of motion is opposite to the direction of the electric field.

The direction of the field strength can be determined by looking at the electric field lines, but the force can also be determined according to the motion state of the object: in this problem, "no initial velocity release", moving to the left under the action of the electric field force, the force is of course to the left.

Since the block has no initial velocity, it is only subjected to electric field forces in the horizontal direction.

It moves to the left, so it can only be subjected to the electric field force to the left.

The direction of the electric field force is not necessarily determined by the direction of the electric field lines.

Sometimes it also depends on the electrical properties of the charge

The concept of a geostationary satellite is that the period of an artificial satellite around the Earth is equal to the period of the Earth's rotation, and it will be stationary above the Earth. From this it is certain that Satellite B must be on Building A.

Gravitational force provides the centripetal force, g m earth m satellite r 2 = m satellite r = (g m earth r 3) (1 2).

Angular velocity of satellite B: b = (g m Earth (NR) 3) (1 2)Angular velocity of satellite C: C = (G m Earth (9 16 x NR) 3) (1 2).

4096 729 x g m earth (nr) 3) (1 2) = 64 27 (g m earth (nr) 3) (1 2) = 64 27 b

It is known that satellite b is a synchronous satellite, so b = 1 24 (lap hours) c = 64 27 b

64/27 x1/24

8 81 (lap hours).

Satellite B turned 1 after 48 hours of 24 hours x 48 hours = 2 turns of building A under Satellite B.

Satellite B turned 8 81 times after 48 hours Hour x 48 hours = Circle, so the answer is D.

You think of it this way: How long did the puppy run? Start from A and B and run until A and B meet.

The puppy's speed is stable and has been 6km h

Therefore, it is good to only ask for the time from departure to meeting between A and B.

The time from departure to meeting = of A and B

So the distance the puppy runs = 6

Child, it's just a formula. s vt forget it!!

You solve this problem in the wrong order, you solve the complex first, and then look at the simple one. So the conclusion is wrong. All questions should be as simple as possible to complex.

You start with the case of C, but the case of C is more complicated, and there are two objects on it. You should start by looking at the two objects A and B above. Looking at a first, there is no relative movement trend between a and c, so the friction f1 between a and c is 0.

Looking at B again, B is subject to a tensile force of magnitude f to the right, and because B is in equilibrium by force, B must be balanced by the frictional force of magnitude F to the left and the tensile force of magnitude f to the right. So the frictional force between b and c f2=f. Finally, if we look at C, C is subjected to a tensile force of magnitude F to the left, and also to a frictional force of magnitude F to the right (friction between B and C), and the force has been balanced.

So there is no friction between c and the ground, i.e., f3=0.

That is, f1 = 0, f2 = f, f3 = 0