A physics contest question, a physics contest question?

Because it is a "light rope", it is not retractable, so in order to maintain the speed v to pull the rope head, the object "must" move horizontally to the right, and the velocity is also v

It's a catch-up question.

At the beginning, Xiao Wang and Xiao Zhang were separated by 300 6 4 = 200m, Xiao Wang ran faster than Xiao Zhang, and they were on the same road after Xiao Wang caught up with Xiao Zhang.

10 minutes.

The knowledge of the students is not very rich, so let me tell you a few key points:

1.The pressure of the liquid should be measured as p=density*g*h;

2.The density of water is the largest at 4 degrees Celsius, that is, when it changes from 0 to 4 degrees Celsius, the volume of water becomes smaller (because the density changes to the maximum), so h becomes smaller, but at this time you can't judge that the pressure p is getting smaller, because at the same time the density is also changing, and it is getting bigger, so you can't judge the change in pressure, but because the bottle is conical, the change in h is greater than the change in density, so the pressure becomes smaller.

By the same token, you can analyze the pressure change from 4 to 10 degrees Celsius.

*By inference, if the shape of the bottle changes, the bottom of the bottle is narrow, and the mouth of the bottle is wide, how does the pressure change? (It's easy.) ）

Water at 0 4 Abnormal expansion (heat contraction and cold tension) So...

Hope it helps.

Utilize the midpoint deformation method. The water expands abnormally at 0 4 (heat shrinkage and cold tension), and 4 to 10 resumes thermal expansion and cold contraction, so the density becomes larger before rising to 4, the liquid level height decreases, and the bottom area of the container after deformation becomes larger, and it is known by using p=f s that f does not change, s becomes larger, so the pressure p decreases; On the contrary, 4 to 10 will resume thermal expansion and cold contraction, and the liquid level will rise, and the result will increase. Therefore, the result of this problem is that the pressure change of water on the bottom of the Erlenmeyer flask is first smaller and then larger.

The height of the liquid level of the container after deformation remains the same as before deformation, so the pressure does not change. If the pressure change is analyzed, the container is changed back to the shape in the question, and the analysis is performed using f=ps. If you don't understand, he'll tell you about it.

Pick D. Ice is denser than water, and when water of the same mass solidifies into ice, it becomes larger in volume, so it can be pressed together. 100% right!

I'll give it to you. There are so many topics...

First question:

Pressure * volume = work.

The charged additional pressure of the sphere can be done using the "virtual work principle".

The electrostatic energy of a uniformly charged ball is: kq 2 2r, which represents the power operation, and the denominator of 2 is because the mutual electrostatic energy is double-counted.

Suppose the ball grows a little longer after stabilization, and the radius of the growth is r, then the corresponding change in energy is: E=kq 2 2(r+ r)-kq 2 2r=-(kq 2 2r 2)* r

The reduction of this potential energy is equivalent to the additional pressure external work: w=p v=p*4 r 2* r

The additional pressure is obtained from e=w, p=kq 2, 8 r 4

The formula is not easy to write, you read carefully and don't read it wrong)

The second question: the problem of elliptical orbit you have to experience and understand again, the elliptical motion of satellites, the earth is not in the center of the ellipse, but in a focal point of the ellipse, a is the major axis, b is the short axis, they should not be confused with r.

If you have any questions, send a message to ask.

As a competition question, this question has a certain degree of difficulty. I enlarged the cylinder and drew a diagram of the force. It's not hard to see that it's the equivalent of a moving pulley. The power arm is twice as large as the resistance arm and saves half the effort.

w total = f motion * s1 = 250 * s1

w has = f resistance * s2 = mg 2 * s2

By the mechanical efficiency formula:

w has w total.

De: Move*s1 f-resistant*s2

Find m=250kg

The bevel is long, but the force on both sides of the pulley is equal, and the force on the rope is equivalent w=2f*s

The principle is found in the work done by gravity -mgs sin30°=-mgs 2 buoyancy work 1000

500s×40%=250gs-mgs/2

2300=5m

m=460kg

The distance that the cylinder rolls is bevel long, but the distance that the hand pulls with force f is 2s, so the total work is w=f2s.

This cylinder is equivalent to a moving pulley.

Assuming that the height of the inclined plane, that is, the height of the object rising, is h, and the slope is 30 degrees, so the length of the inclined plane is 2h, then the mechanical efficiency is.