1 rare middle school physics problem, 1 physics problem, simple but difficult

21 answers

Anonymous users2024-02-11

For the microscopic structure of atoms, they follow the laws of quantum mechanics.

If we use the theory of classical physics, the resultant force experienced by the electron provides the centripetal force that moves around the nucleus of the atom at high speed.

Quantum mechanics explains: simply speaking, electrons do not revolve around the nucleus, but appear around the nucleus according to probability, electrons do not radiate energy outward when they revolve around the nucleus, and unlike the laws of electromagnetic induction in the macroscopic, electrons will only release photons outward when the energy level transitions.

Classical theories of mechanics and electricity completely fail in the quantum range!!

It is necessary to change the method of analysis, which is more complicated.

If you take physics in college, you will learn o( o

Didn't copy it, hope.
Anonymous users2024-02-10

1 The whole problem can be analyzed in three parts:

1. When the jump is not taken: Xiao Ming maintains the same speed as the train after the train starts, and is relatively stationary.

2. The moment of jumping: When Xiao Ming jumps, at the moment when his feet leave the ground, Xiao Ming and the train still maintain the same speed and move forward.

3. After jumping: When Xiao Ming jumps up, the train is still moving at a very fast speed (assuming that the speed of the train has not changed), after Xiao Ming jumps up, due to inertia, he still maintains the speed of the moment when he leaves the ground, and this speed is the same as the speed of the train. Now it is explained in several cases:

1) Regardless of the resistance of the air, Xiao Ming has the same speed as the train in the air, and will move at the same speed as the train, so he and the train still remain relatively stationary, and when he lands, he will still jump at the place where he starts.

2) Considering the resistance of the air, because the speed of Xiao Ming will change due to the resistance of the air in the air, it is impossible to maintain the same speed as the train, so he will definitely not fall in the same place.
Anonymous users2024-02-09

If the air resistance is not taken into account, due to inertia, Xiao Ming will still be in the original starting position when falling;

But if there is air resistance, it is necessary to calculate how far away Xiao Ming is from the original position when he lands according to the air resistance given.
Anonymous users2024-02-08

If the train travels at a constant speed, it will fall to its original position.

Because even provoked due to inertia he has the same horizontal speed as the car.
Anonymous users2024-02-07

It feels more like a language question, hehe, driving. Xiao Ming jumped down at this time.

I don't think so.
Anonymous users2024-02-06

If you jump high enough, you can think that if he jumps out of the earth, or even directly out of the solar system, he won't be able to fall back
Anonymous users2024-02-05

Yes, Xiao Ming and the train are relatively stationary.
Anonymous users2024-02-04

It shouldn't be in place, you haven't taken into account the resistance of the wind.
Anonymous users2024-02-03

Question 1:

Option A: The maximum displacement to the right is the process of object A from v0 to zero.

The law of conservation of energy: 1 2mv0 2=umgs

Therefore, the maximum displacement s is 1 2v0 2 ug

Option B: The maximum displacement of the trolley to the left is the process of the trolley from v0 to zero.

The law of conservation of energy: 1 2mv0 2=umg ms, so the maximum displacement s is mv0 2 2umg

Option C: Impulse is equal to the frictional force multiplied by the action time.

umg*t=mv

The friction time has been calculated in option d, so the impulse mv=2mmv0 (m+m) option d:

If the frictional force is UMG, then the force gives the object an acceleration of UG and makes the trolley have an acceleration of UMG M. Finally, the friction does not disappear until it is relatively stationary.

That is, the velocity, v, magnitude and direction of the object and the trolley are the same.

Object v1=v0-ug

Trolley V2 = V0-UMG M

When m > m, the velocity direction of the last object is the same as that of the trolley to the right, so -(v0-ug)=v0-umg m

And when m the result is 2mv0 (m+m)ug

Question 2: s1=v0t+1 2*at 2=

s2=v02t+1/2*a(2t)^2=

t=, v0 is the velocity of point a.

It can be calculated as a=f=2mg=2*

f=f-am=u=
Anonymous users2024-02-02

The answer to the multiple-choice question is b

The acceleration of the latter question is.
Anonymous users2024-02-01

2 is a resistor, but 3 is not a resistance but an ammeter.

The following is an analysis of what devices should be in these places you mentioned:

According to the two ammeter parameters given in the question, in order to make their pointer deflection angle as large as possible, 1 is ammeter A1 and 3 is ammeter A2 (3 currents are greater than 1 current).

After determining the position of the two ammeters, considering that the power supply voltage is 3 volts (excluding the internal resistance), and the internal resistance of the ammeter A2 is much smaller than the internal resistance of A1, the value of the resistance at 2 places can be estimated (that is, when the sliding end of the sliding rheostat is at the right end, the A1 ammeter and the 2 resistors are connected in series, and the voltage is added to the voltage of 3 volts after series connection, here is the estimate, so the influence of the A2 ammeter with relatively small internal resistance is ignored).

From 300 microamps* (1 thousand ohms + r) = 3 volts, r = 9 thousand ohms.

It can be seen that the resistor at 2 should be selected as the 9 thousand ohm resistance of term e, i.e. r3.
Anonymous users2024-01-31

Hello subject, this question chooses A, because the ship does not use the hollow method to float on the water, like a large cruise ship, there is basically no space, it is achieved by the ship's own gravity and buoyancy equally, and the buoyancy is large because the ship is large in the water. b, right, according to f buoyancy = density v row g, the submarine changes its own buoyancy by changing its own water inlet and outlet. c, yes, when the density of gas in the balloon is smaller than that of air, it will rise.

d, yes, the air heats and expands, and the air quality is certain at this time, the volume becomes larger, the density becomes smaller, and it is less than the usual air temperature, so the hot air balloon rises.

To sum up, this question chooses A.
Anonymous users2024-01-30

5.(3) b (4) analysis idea: the position of the plane mirror will not affect the law of its imaging, the purpose of our experiment is to ** the characteristics between the object and the image when the plane mirror imaging, so we should record the relevant data of the object and the image several times instead of the plane mirror.

10.(1) Omitted.

2) The image formed by the luminous point c will move away from the lens; The image of the CD gradually gets larger.

3) The velocity of the image is greater than the velocity of the object.
Anonymous users2024-01-29

1. a. The magnified image distance The object distance is magnified.

2. A, the image becomes larger, the image distance becomes larger, the object distance becomes smaller, the image distance becomes larger, and the image becomes larger.

3. a. The retina is equivalent to a light screen.

4. C. Mirage.

5. (3) Which of the following do you think should be taken in order to confirm the reliability of the above imaging features: (

b Keeping the position of the glass plate unchanged, change the position of the A candle several times and perform the same operation as in (2) above.

4) Why should this action be taken? (I mainly didn't write this question, please give me an analysis and give me the answer).

Increase the number of experiments, avoid chances, and make the conclusions universal.

6. B. "Fish" is a virtual image formed by the refraction of light, and "cloud" is a virtual image formed by the reflection of light.

7. B. In the projector, it becomes a magnified and inverted real image.

8. C. The size of the image remains unchanged, and the distance from the image to the mirror becomes smaller.

9. C, upright magnification.

10. (2) According to the analysis of the optical path diagram drawn in Figure C, when the CD moves to the left, the image formed by the luminous point C will move to the left; The image of the CD becomes smaller.

3) When CD moves between F< U< 2F, the speed at which the image it forms is greater than the speed at which the object CD moves.
Anonymous users2024-01-28

1,a, the distance of the magnified image object, such as the larger.

2, a, when the image becomes larger, the distance of the image becomes larger, the distance of the object becomes smaller, the distance of the image becomes larger, and the image becomes larger.

3, a 4, retina, ?Mirage.

5 (3) To confirm the reliability of the imaging characteristics, you think the following actions should be taken equivalent to the light screen.

b.Keeping the glass plate in the same position, repeatedly changing the position of a candle carried out by the same operation above (2).

4) Why should you take the action? (Not the question I wrote, I solved the solution and gave me the answer).

Increase the number of experiments to avoid accidental, generalized conclusions.

6, b, "fish" is a virtual image formed by the refraction of light, and a virtual "cloud" is a reflection of light.

7, b, zoom in to the projector upside down the real image.

8, c, the size of the image is the same, such as the mirror distance becomes smaller.

9,c, upright magnification.

10, (2) according to the drawing in the figure. Analysis of the optical path diagram of the propoxy group when the CD is moved to the left and the image formed by the luminescent point C moves to the left; The disc image becomes smaller.

3) When cd from FWhen moving between

Anonymous users2024-01-27

1 Rape is a measuring tool that can hold a certain weight of something, the density of oil is smaller than that of soy sauce, and the volume of soy sauce of the same quality is smaller than the volume of oil, so the volume of oil measured with soy sauce grape will become smaller, so it is a loss. 2. First measure the mass of the cup and then measure the weight of the cup filled with water, then you can find the volume of the cup, and then measure the weight of the filled oil to find the density. Because it's written on a mobile phone, you can't be too detailed, so you can summarize the steps yourself.

3 Benefits: Abnormal expansion leads to the freezing of the lake surface in winter, not all of it, so that the fish can survive and do not affect the aquaculture industry

Anonymous users2024-01-26

1.You've ever seen a bamboo tube for holding water, a stainless steel tool with a vertical handle at the lower end that resembles a bamboo tube that can hold water. The density of oil is smaller than that of water, so it is smaller than soy sauce, and the volume of soy sauce of the same mass is smaller than the volume of oil, so the volume of grape oil of soy sauce will become smaller, so it is a loss.

2.Weigh out the mass of the empty bottle m1

Fill the empty bottle with water and weigh the mass m2

Pour out the water, fill it with soy sauce and weigh out the mass m3

Calculate(m2-m1) (m3-m1)= water v soy sauce v, i.e. soy sauce = (m3-m1) water (m2-m1).

3.The benefits of the paradoxical expansion of water: The large number of glaciers floating on the Arctic water stores large quantities of fresh water for human survival.

Disadvantages: An airtight container filled with water can break when it freezes, e.g. a beer bottle cannot be frozen.

Anonymous users2024-01-25

Forehead... Unable to insert** Let the masses of the two objects be m and m respectively take a small section of rope dl in contact with the rod, the tension of the left and right ropes is t1 and t2, and the angle of dl to the center of the circle is d. Then DL is trapped by friction and makes the filial piety friction force f=U(T1+T2)*D 2.

There is f=t1-t2. Because DL tends to 0, T1-T2 DT, T1+T2 2T. The slipperable belt obtains u*2t*d 2=dt, and the shift item obtains u*d =dt t, and the Wang manuscript integral, the left side is from 0 to , and the right side is from mg to mg, so there is u = in (mg mg), that is, m m = e u, which is the critical condition for its sliding.

Anonymous users2024-01-24

The difference in gravity between the two objects < = the maximum static friction that can be generated at this time.

Anonymous users2024-01-23

Suppose that when it comes to sliding, the back side is heavier than the left, the left side is heavier mg, and the right side is heavier mg. According to n=(t+t+dt), the world code is vertically *d, f=u*n=dt. dt d = t*u, t(0) = mg.

t=m*g*e (u*).

So if you don't slide, m*g*e (u)> m*g to get m m or press dt t=d *u upstairs

ln(mg mg) = u

The basic points on the upper floor of the m m are all wrong.

P.S. Recommended answers, although the recommended ones are clearly stated.

But it's really wrong.

The integral of 1 t is LNT, then the product from the large mg product to mg is equal to ln(mg)-ln(mg)=ln(mg mg).

This category administrator doesn't even look at it clearly, so they recommend it casually.

Anonymous users2024-01-22

Set: Wood block volume: v, wood block mass: m, iron block volume: v, iron block mass: m, iron block density: .

The standard units used in the questions are all standard, so you can omit the units and write only the numeric values. So, the density of water is: 1000.

We see the wood and the iron as a whole. Depending on the title, this "whole" needs to be "floating" or "suspended" in water. And, when "floating", the volume that enters the water is at least v (i.e., the block is just submerged in the water); When "suspended", the volume that enters the water is v v.

According to the equilibrium conditions, the "gravitational force" of this "whole" is exactly equal to the "buoyant force" on it, that is, the total weight of this "whole" is exactly equal to the weight of the water it "displaces". Since the gravitational coefficient is constant, we only need to have equal "mass" between the two.

The quality of the whole "m m;

The whole "volume entering the water, i.e. the volume of the water discharged v,v v];

Overall "quality of discharged water 1000V, 1000(V V)].

We require: the quality of the whole The quality of the water discharged as a whole.

So, there is. 1000v ≤ m ＋ m ≤ 1000（v ＋ v）

In the above formula, v v m m .

We know that the volume of water discharged when a block of wood floats alone in water is , and the mass of this water is exactly equal to the mass of the block. So: m 1000

Substituting v, v, and m into the inequality above, we get:

1 ≤ m ≤ 1 ＋ m/

Solving the two inequalities separately gives the range of values for m is: [, 39 85] in kg.

The answer to this question has nothing to do with the size of the container, as long as the container is deep enough and the water is enough to ensure that the wood and iron blocks are completely submerged, even if there is water overflowing. Because wood and iron are only subject to gravity and buoyancy except for the interaction between them. Buoyancy is only related to the volume of water discharged by the object, that is, the volume of the object entering the water and the density of the water, and has nothing to do with the size of the container, the height of the water surface, and the shape of the object.