100 points. Ladies and gentlemen, help with physics problems

This is the working principle of Kepler double-convex lenses, that is, two convex lenses are placed one in front of the other, and the distance between them is adjusted until the focal lengths of the two convex lenses overlap.

Regarding magnification and reduction, you need to know the distance between the object and the convex lens and the law of the imaging effect, knowing this law, the above problem is very obvious.

For a convex lens, if the object is placed at infinity, the object will be imaged at the focal point, inverted, and reduced;

If the object is placed between the infinite 2x focal length, the object will be imaged between the focal point and the 2x focal length, inverted, zoomed out;

If the object is placed at 2x focal length, the object will be imaged at 2x focal length, upside down, equal in size;

If the object is placed between the 2x focal length and the focal point, the object will be imaged between the 2x focal length and infinity;

All of the above are real images.

If the object is placed in focus, the object will not be imaged, and a parallel light will be emitted;

If the object is placed in focus, the object will become a virtual image, magnified, and upright.

Based on the above principles, you can draw two scenarios in your question and compare the magnification effect.

I'll try to explain:

The role of the objective is equivalent to bringing distant objects to the front, and this process will have to shrink the object, although we don't want to do this (no way, usually objects are outside the 2x focal length), but if the focal length of the objective is large, then the object will be closer to the position of the 2x focal length, so the effect of being reduced is smaller, and the focal length of the visible objective should be as large as possible. The function of the eyepiece is to magnify the image of the objective lens, in order to make the magnification effect obvious, we should try to choose a lens with a short focal length.

Now for the swap. You will find that the object will be farther away from the 2x focal length, and the zooming effect will be more obvious, and the magnification effect of the lens with a long focal length will not be strong, so the resulting image will not be significantly magnified, and the final effect will be that the image of the object will be very small.

This has to do with focal length, and you just have to take a closer look at the relationship between the image and the focal length.

If you put the larger focal length in front, then the smaller focal length will be inside the larger focal length, so the object you see wants to be bigger!

There's a formula. 1 focal length = 1 object distance + 1 image distance Generally speaking, the object distance is quite large, so 1 object distance is almost zero and is not counted.

So focal length is roughly positively correlated with image distance.

The image formed by the front of the large focal length is also large (because the image distance is large, the farther away it is, the more divergent) (compared to the other case).

And then it can be explained.

The volume of steel v=m p, steel = 237kg to the power of 3 kg|M3 to the power = m3

The mass of polypropylene plastic material m = p plastic * v = 3 kg |3rd power of m * m3 = 33kg

V Steel = M Steel P Steel = 237kg ( =

M polypropylene = P polypropylene v polypropylene = * = 33kg

m steel - m polypropylene = 237kg - 33kg = 204kg

In a series circuit, p1=(u (r1+r2)) 2*r1 p2=(u (r1+r2)) 2 *r2

Since R1>R2 so P1 is greater than P2 because I have the same resistance and more power.

pTotal p1+p2

P1>P2 is available

p1 in parallel circuit'=u^2/r1 p2'= U2R2P'P2 Total'P2'+p1'

Hence p'> p2'>p1'

Since p total p1+p2 u 2 (r1+r2)p1'=u 2 r1 it can be seen that p2'>p total, so p total'>p2'>p1'>P total》P1>P2, you can also understand it this way, the two resistors are connected in series, that is either the total resistance is larger, or it is larger than either one, so that the power is naturally small at the same voltage.

Tandem: p1 + p2 = p total.

Parallel: P1' +p2'= p and total.

Because the voltage u does not change, p=u*u r

So the smaller the r, the more power.

p and total p2'>p1'

R1 and R2 are regarded as one resistor, and their total resistance is the largest, so their total power is not as good as the power on R1 when they are connected in parallel.

p and total p2'>p1'>p total.

In a series circuit, p=i*i*r, the current is the same.

So the greater the resistance, the greater the power, so P always >> P1>P2

So there is: p and total "p2".'>p1'pTotal p1+p2

Series: i=u (r1+r2) p1=[u (r1+r2)] 2*r1 p2=[u (r1+r2)] 2*r2 p total = u 2 r1+r2

Parallel: P1 = U 2 R1 P2 = U 2 R2 R Total = R1 * R2 R1 + R2 P Total = U2 R Total.

There is a reminder that something is wrong.

It is assumed to be an ideal circuit.

The total power of the series circuit is the smallest, and the total power of the parallel circuit is the largest.

In the series circuit, the power is divided by the large resistance, and the power is divided by the small resistance of the parallel circuit.

= p and total. ' > p2' > p1'pTotal p1+p2

Don't know if you want the process?

Let the lotus leaf area be s;

The impulse of the lotus leaf is f;

The pressure on the lotus leaf is: p;

The average megalowattle velocity of raindrops is v=12m s;

Take the direction of raindrops falling as the square direction;

Then according to the momentum theorem: ft δmv

The volume of rainwater received by the lotus leaf in time t:

v=s*t*;

The quality of the rain that hits the lotus leaf area in t time: m * vs. t *

The lotus leaf is subjected to the silver impulse of the family band:

f=δmv/t

s*t*v*

The average pressure of the lotus leaf: p=f s

s*t*v*

So the answer is a;

Note: The rate of precipitation of 45 mm per hour is to be converted into SI units.

So it's done!

We'll know if we read the question carefully:

The masses of the two wooden blocks A and B are m A and M B respectively, connected together with a thin wire, there is a light spring that is compressed in the middle, and the wooden block B is placed on the level ground at this time (the elastic force of the string is f). The f here is not the elastic force of the spring, but the force f experienced by the thin wire!

Unlike elastic force f m armor a plus m armor g ---, this force is the force f experienced by the spring

Answer: (mA+mB) g f f f is the elastic force of the thin wire, which is equal to m A a

lz you didn't use the letter well, your answer is right, the second answer is also right, but the f in the second answer refers to the tension of the rope, not the elastic force of the spring, for the balance equation of the object A, you can find that M A A = f pull, the two equations are equivalent to together.