Senior 1 Physics Contest Questions, High School Physics Contest Topics

I am an undergraduate student of Tsinghua University, I have almost forgotten all this knowledge, but I can provide you with a little idea on this topic, just for reference, first of all, do the intention to show that the center of the 4 balls can be connected to get a regular tetrahedral OABC, so that the center of the top ball is O, and the length of each side of the tetrahedron is the diameter of the ball, the lower 3 balls have no interaction, so there is no friction, and there is friction between the upper ball and the 3 balls and the size is equal. We need to find the direction of friction, when the two balls are in contact, the friction is along the tangent direction at the intersection of the two balls, then the friction force is outward along the tangent, that is, the direction of the perpendicular line of the three edges of the tetrahedron. We need to use the knowledge of solid geometry in the second year of high school, but even if we haven't learned it, it is not difficult for students with strong comprehension to follow the method.

If O is perpendicular to the triangle ABC and is in M, and OABC is a tetrahedron, then M is the center of the regular triangle ABC, connecting MA to form a triangle MAO, and M is perpendicular to Ao and is in N, and the friction between the ball O and the ball A is perpendicular to Ao outward, which is obviously in the Mn direction. Let the diameter of the sphere, i.e., the edge length of the tetrahedron be 2a, in the regular triangle abc, ao= 3 a, then am=a, ao2=am 2+om 2, find om= 2 a, tan angle oam= 2.

3 supporting forces of equal size support ball o did not fall, indicating that the 3 support forces offset the weight of the ball's gravity in the direction of oa, ob, oc. Continuing with the interaction between the two balls of OA as an example, the relative direction of motion of ball O and ball A is MN, so the motion in other directions does not need to be considered with OA as the reference frame. At the same time, we need friction to cancel out the component of gravity in the mn direction, let the support force (equal to the pressure in the opposite direction) be f, there is f=1 3mg*sinoam, this should not be difficult to understand, otherwise we should make up trigonometric functions.

The gravitational force is canceled out by 3 supporting forces, so each is 1 3, and f= 6 9 mg is obtained while the frictional force is equal to the pressure multiplied by the static friction coefficient x, i.e., f=f*x= 6 9 mg*x, the 3 frictional forces are equal in magnitude and different directions, and have universal applicability for 3 contact surfaces.

The components of friction and gravity in the mn direction are equal, i.e., f= 6 9 mg*x=1 3mgcosoam, which can be obtained as x= 2 2.

Obviously, this x value is a critical value, and the static friction coefficient must be greater than this value, but for an object with a rigid surface that is not sticky, the static friction factor cannot reach 1, so the range is 2 2 <1

Of course, this is only the friction coefficient between the ball, the range between the ball and the ground can also be found, the idea is that the friction between the ball and the ground is equal to the component of the support force in the horizontal direction, pay attention to the calculation of the pressure of the ball ABC on the ground to include the gravitational component of the ball O.

This question is very good, although the conditions are few, but a lot of important information can be obtained. I hope that comrades will not blindly draw conclusions and think that there is something wrong with the topic because they have few conditions or do not know how to do it, but think more about it, and maybe they will suddenly become clearer.

You can't understand what you want to say about the topic yourself?

Summary. Dear to you, this is not an example.

How to write example questions.

Dear to you, this is not an example.

Yes, but I don't understand the process of the answer, I hope you can give a better answer.

This side has already explained it to you step by step.

Hello, why is the interval equal to the wavelength divided by the angle?

Because of the formula.

In general, double-slit interference is not x=d d.?

The formula for pro-double slit interference is x=l d, and the units of δx, d, l, and in the formula for slit interference are all meters (m).

So why is this interference related to angle?

The light source is different. The magnitude of the light level, varies.

It may all be related to the angle.

1) Smooth and frictionless, then the acceleration of the ring sliding along the metal rod a=gcos 1 2at 2=x solution t= 2gxcos 2) When the combined external force of the metal ring is exactly equal to the centripetal force required for it to move in a uniform circular motion r, the ring will remain balanced, let the distance from the axis of symmetry at this time be r, then f centripetal = mr 2=m*4n 2 2r

Whereas, the ring receives the resultant external force f=mgcot

Synapsion: r=gcot 4n 2 2

(1) After the ball is shot out of point C, vt=r, 1 2gt2=r, the solution is v = root number gr 2, at point c, the force of the tube to the ball is f, the direction is downward, mg+f=mv2 r, f=-mg 2, so the force of the tube to the ball is mg 2, the direction is up, the force of the ball to the tube is mg 2, the direction is downward, (2) the height of the ball from point A is h, from a to C only gravity does work, 1 2mv2=mg(h-r), assuming that the ball exactly reaches point n, vt=r+3r, 1 2gt2=r, from the above 3 equations can be obtained h=5r, the maximum height of the ball from point a is 5r

Since I can't draw pictures on a computer, you read my description carefully. I don't know much about the value of your dynamic friction factor, so you can calculate it yourself after I find the answer.

Let me start by listing the numbers: r1=r2=10cm, d=25cm, =30rad s, m=9kg, and the friction factor is =.v=4m/s,r3=15cm。

First of all, f pulls c along the direction of the axis, and c moves at a uniform speed, that is to say, the backward friction force experienced by c is equal to the magnitude of f!

Then we look at the ** of this friction. (Remember to draw).

First of all, A does a uniform circular motion, it has contact with C, so it will definitely have a relative sliding force on C when it rotates, so that A will have a sliding friction force against C, and the direction of this friction force is along the tangent direction of the circle, which is perpendicular to Ac, and points out of the triangle, and at the same time there is extrusion between A and C, so A will generate pressure on C, and this force will point to the center of the circle of C along Ac. In the same way, analyzing the force of B on C, we will get that the four forces are balanced by the gravity of C, because C is balanced by force in the vertical direction, otherwise it would move up or down.

Next is geometric analysis, and I can't do anything if I don't pass the math.

Vertically, the gravitational force of C is balanced by the four forces provided by A and B, so I will decompose these four forces into horizontal and vertical forces.

Let me let the friction of A to C be f, then the friction of B to C is also f, and the components of these two forces in the horizontal direction cancel each other out, and the force that balances gravity is the force in the numerical direction that the dust indicates them, which is fy=fcos60° (the direction of f is perpendicular to ab and points beyond the triangle). Tease oranges.

Let me let the pressure of ab on c be n, and the horizontal components of n generated by the same a and b will cancel each of them, and the force in the same numerical direction is: ny=ncos30° (the direction of n is on ac or bc, and points to the center of the circle of c).

Therefore, combined with the force balance, there will be: mg=2fy+2ny

We should also know that the f produced between a (or b) and c is produced by the extrusion and rotation of c by a (or b), so n is the factor of f production, so there will be: f = n

If you put the equations together, you will have it, mg=2 ncos60°+2ncos30°

With this equation you can find the size of n.

Then you think about it, when f pulls c, there is a friction force f' that creates a force that balances the force on c, and we know that the formation of friction must indicate that there is extrusion, so the extrusion between c and the other column causes f' to be generated.

Thus there is: f'=2 n (because there are two cylinders).

Then there is Newton's second law: f'=f. And so the answer comes out!

Note: Do you know if the answer is correct, can you tell me if you make a mistake?

Zhang Datong's gold medal road has a solution to this type of problem, which is pure, and there is a general solution in my **.

Zhang Datong's "The Road to Gold Medal Li" is best read by you, just on pages 3 and 4, and you can go down to the Internet to slow down.

First of all, the rate of increase in the absorption temperature v, according to the assumption, the heat absorbed per unit time is the heat generated by the resistance per unit time, that is, the power, so the heat generated per unit time is the power of the resistance, let the proportional coefficient be k, the voltage is u, and the temperature is a, the resistance is ra

So there is: u*u (ra)=kv

When a=0 degrees Celsius, there is u*U (ro)=kv0 V0 is 5k min at this time, and when A=30 degrees Celsius, there is U*U (r0(1+30a))=kv1 V1 at this time.

A comparison of the two equations gives us the temperature coefficient a=-1 300

The spring is only in contact with the block, not connected, and only accelerates the two pieces, after which it separates. Momentum is conserved horizontally during block acceleration.

According to the fact that they can pass the orbit to the fixed point, their speed at the top of the orbit can be found in the method of "Personality 10 Class Boss".

Then use the Conservation of Energy to find their velocities at the bottom end of the arc orbit, respectively. (The "Boss of Personality Class 10" didn't think about this, and it was right to add this.) If the top velocity is used, the horizontal momentum is not guaranteed, because the title does not say that they arrive at the same time, nor does it say that the orbit gives them equal horizontal impulses).

Since there is no friction, this velocity is the velocity at which the block and spring are separated.

Conservation of momentum analysis is then used to cut the rope ——— the block to separate from the spring during this process.

Momentum is a vector quantity, for the problem mentioned in the "Problem Supplement", because the direction of the orbital elastic force when the spring is doing work is upward, and the whole system is not affected by external forces in the horizontal direction, so the momentum in the horizontal direction is conserved, so the usable momentum is conserved.