Physics questions in the second year of junior high school, who will help

Reasons: 1 Hundreds of thousands of people live in modern large cities, and a lot of heat is emitted by industrial production and transportation.

Abbreviation: Produces a lot of heat.

2. Due to the large heat capacity and small reflectivity of the building complex and asphalt pavement, a large amount of solar radiation heat is stored.

That is, the heat dissipation is not good.

80% more solar energy is stored than in the countrysideAt night, the city cools down slowly, so the first half of the summer months are still muggy. In addition, there is often an inversion layer over the city at night on a sunny day, which also hinders the diffusion of heat to the upper altitude.

The way to do it is: a reasonable building layout.

Plant trees to reduce the use of irrational fuels.

A large amount of heat emitted by industrial production and transportation.

The city's architectural ensemble and asphalt store the sun.

It can increase the green area.

Reduce the use of carbon-containing fuels.

Let's start by understanding the heat island effect

Because the asphalt road and cement pavement in the city have a greater "heat absorption capacity" than the soil and vegetation in the suburbs (it seems to be called the heat absorption rate or something), so that more heat is stored in the urban area, and the atmosphere radiates to the surrounding areas, so that the temperature in the urban area is higher than the temperature of the surrounding suburbs, and the high temperature urban area is therefore surrounded by low temperature suburbs, causing temperature differences, this phenomenon is called the urban heat island effect.

Your topic is the human activity that causes the heat island effect:

1: Urban ground buildings;

2: Industrial production counts.

I can't remember much of the second question, so you can sort it out by yourself;

3: On the one hand, it is recommended to expand the area of green vegetation (greening);

Also, you can. Hey, I can't think of it, old man.

Let the distance of the person from mountain A be s, then the distance from mountain B is (1000-s) The distance traveled by the sound is twice the distance, therefore:

2s 340 - 2(1000-s) 340=4 solution: s = 840m

1000-s=160m

So the distance of this person from mountain A is 840m, and the distance from mountain B is 160m

This person is from 840m to 160m from one mountain to another mountainLet the distance to mountain A be L1 and the distance from mountain B to l2, then l1+l2=1000 (1) then 2l1 340+2l can be solved by two tests.

Suppose that a person is at point A, the mountain on one side is at point B, the mountain on the other side is at point C, and the person is close to mountain B, assuming that the time it takes for sound to propagate from point A to point B is t1, and the time it takes for sound to propagate from point A to point C is t2, then he hears the first echo at 2t1 time after the sound is made, and hears the second echo at 2t2 time after the sound, according to the time interval between the two echoes is 4s, it can be obtained:

2t2-2t1=4s, i.e. t2-t1=2 (for ease of calculation, without units) This is the equation

Through the above analysis, it takes time t1 for sound to go from A to B, and time T2 to go from A to C, so the time it takes for sound to go from B to C is T1+T2, and the distance between Bc is 1000m, then there is: T1+T2=1000 340 This is the equation

The simultaneous equation sum , the solution of t1 = 8 17 t2 = 42 17 then the ab distance is 340 8 17 = 160m

AC distance is 340 42 17 = 840m

People are at point A, a mountain is at point B, and the other mountain is at point C, assuming that people are close to mountain B, and it takes time T1 to get the sound from A to point B, and it takes time T2 to go from point A to C, then the time it takes for the sound to go from B to C is he T1+T2, the first echo is heard at 2T1 time after the sound, and the second echo is heard at 2T2 time after the sound, and the interval between the two echoes is 4s, which can be obtained

2T2-2T1=4S, i.e. T2-T1=2 Equation From the above analysis, we know that the time it takes for sound to go from B to C is T1+T2, and the distance between BC is 1000m, then there is: T1+T2=1000 340 equation

The simultaneous equation sum , the solution of t1 = 8 17 t2 = 42 17 then the ab distance is 340 8 17 = 160m

AC distance is 340 42 17 = 840m

Let the time when the first echo is heard t, then there is 340m s (2t+4s)=2 1000m, and the solution is t=

The distance between the person and the mountain: d=vt=340m s The distance from the other mountain is: 1000m-320m=680m=

The two mountains can be set as A and B, and the position of the person is C. Let the time for sound to propagate between ACs is t1 and the distance to travel between BCs is t2, then the system of equations can be listed.

t2 - t1 = 4

1000=340 x( t1 + t2)1 2 solution: t1 = 16 17s, t2 = 84 17s, then the ac distance is t1 x v = 320m

The BC distance is T2 x V = 680m

After 4s, the difference is 4 340 = 1360mTherefore, the distance between people and the two mountains is 1360 2=680m, so the distance from the two mountains is (1000+680) 2=840m, (1000-680) 2=160m

Since there is a gap between the sounds that people hear, and the position where people stand is not in the middle of the two mountains, then it is assumed that the distance farther away from the person is S1, and the closer distance is S2. From the question, we can get: 2x(s1 v-s2 v)=4.

then s1-s2=680Then join the two equations with s1+s2=1000 to solve s1=

If the distance from one mountain is s1 and the distance from another is s2, then l s1 s2 is 1000 meters.

Then the time difference is δt (2*s2 v) (2*s1 v) and here s1 s2 is assumed

4 (2*s2 340) (2*s1 340) is coupled with 1000 s1 s2.

S1 160 m, S2 840 m.