Physics is good! 10, physics is good!

Plane Mirror Imaging:

Steps: a. The perpendicular line of the plane mirror is made at the point S.

b. Intercept the image point, so that the distance to the mirror is equal to the distance from S to the mirror.

c. Draw an image point (the image is the same size as the object).

Note: The reflective surface of a mirror is smooth, and such a mirror is called a flat mirror.

Characteristics of plane mirror imaging: like a virtual image; The image is equal in size to the object; The corresponding points of the image and the object are connected perpendicular to the mirror; The distance between the image and the object and the mirror is equal; The image is the opposite of the left and right of the object.

Reflection of light: When light hits the surface of an object, a part of the light will be reflected back by the surface of the object, a phenomenon called reflection of light.

Several terms related to the reflection of light:

1) Normal: A straight line perpendicular to the incident point and the mirror.

2) Angle of incidence: the angle between the incident ray and the normal.

3) Reflection angle: the angle between the reflected ray and the normal.

Note: It is necessary to clarify the meaning of one point (point of incidence), two angles (angle of reflection, angle of incidence), and three lines (reflected ray, incident ray, normal).

3 The law of reflection of light: when light is reflected, the incident ray, the reflected ray, and the normal are in the same plane; The reflected and incident rays are located on either side of the normal; The angle of reflection is equal to the angle of incidence.

1) When questioning, don't say "the angle of incidence is equal to the angle of reflection". Since there is an angle of incidence followed by an angle of reflection, and the angle of reflection is determined by the angle of incidence, it should be said that "the angle of reflection is equal to the angle of incidence". (2) The reflection of light occurs at the point of incidence and not elsewhere.

3) The change of the amount of occurrence at the two angles is corresponding, that is, when the angle of incidence changes, the angle of reflection also changes, the angle of reflection increases when the angle of incidence increases, the angle of reflection decreases when the angle of incidence decreases, and when the angle of incidence is 0 degrees, the angle of reflection is also 0 degrees, that is, the reflected ray and the incident ray coincide, but the direction is opposite.

When making an optical path diagram, the light rays should be drawn as solid lines, and arrows should be used to indicate the direction of light propagation, the back of the mirror should be drawn with short diagonal lines, and the normal and other auxiliary lines should be drawn with dashed lines.

How to map plane mirror imaging? Method? Should be noted?? Detailed!! Make a perpendicular line to the plane mirror from the corresponding point of the object, measure the length of the perpendicular line from the point to the plane mirror, extend the perpendicular line, and intercept an equal distance on the other side to find the position of the image of the corresponding point.

How to plot the reflection of light?Method? Should be noted?? Detailed!! Crossing the point of incidence as a normal. Measure the angle of incidence and the angle of reflection, and measure the same angle from the normal on the other side of the normal to find the reflected ray.

To draw a convex lens, draw a vertical line, draw two arrows with sharp outwards at each end of the line segment, and draw both arrows inward with the concave lens.

Example: Convex lens (not a dotted line, but a line segment).

Pay attention to the position of the focal length and focus and the way the object is painted in real or virtual (note whether it is zoomed in or out, upright or upright).

Reflection of light: Note here that the angle of reflection and the angle of incidence are the angle between the ray and the normal, and the normal should be perpendicular to the reflecting object.

Plane mirror imaging mapping, pay attention to the virtual reality of the light (outer reality and inner void), to ensure that the object image is symmetrical with respect to the plane mirror;

The reflection of light is plotted, and also pay attention to the virtual reality of the light (outer reality and inner void), ensure that the incident ray and the emitted ray are symmetrical about finding symmetry, and don't forget the arrows! It's easy to get the angle of incidence wrong.

Using symmetry, light travels in a straight line.

The law of reflection of light, which is first made of incident light.

Flat mirror to make a drawing, go there and see!

The 8n planks are accelerated by friction.

a1=μm1g/m2=

x1=1/2 a1 t²=

75cm = because the slider slides out.

x2=x1+l=

and x2=1 2 a2 t

So a2=7

By Niu II. f-μm1g=m1a2

So f=8n

Hope it helps you ......

First of all, it is judged that the block will definitely accelerate under the action of constant force, and the friction between the block and the board must be sliding friction, so we know that the friction received by the block is the sliding friction, which can be listed as follows.

a=（f-f）/m

x=1/2at²

where f= mg x = 75cm t =

So f=7n

When you write, you only need to write those two formulas and an answer, and physics only needs to write like this, and the above analysis process is up to you.

wolanshuanba

The acceleration of a small object to slide out of the plank is.

a relative = (2s t) (1 2) = 3

By the friction of a small object, the acceleration of a wooden board is.

UM plate G = M plate A plate.

A plate = ug =

The acceleration of a small object is.

A = A relative + A plate = 1 + 3

Force f=ma+umg=1*(1+3)+

Key points of investigation: 1. The maximum force of the wooden board is affected by the dynamic friction factor, FD = the mass of the small slider 1kgx the gravitational acceleration gx the dynamic friction factor or 1N (see if g is still 10);

2. The slider leaves the plank and the driving distance of the slider is greater than that of the two under the condition of a certain external force;

3. If the constant force f of the slider is assumed to be xn, then the external force of the slider actually affected acceleration = f-fd =;

4. Use the acceleration and distance formula (distance s = v0·t + a·t 2) to solve x, where the actual is the same, and the distance difference is .

5. S slip = v0·t + a slide·t 2;Wood = v0·t + wood·t 2;s slip - s wood =

v0=0；t=；(（

x = when taken, if g takes 10, then x = 8n).

1 2 a1 t squared - 1 2 a2 t square = x....1f-f=ma1...2

f=ma2...3

From the three formulas, f=8n

The friction of the small object upstairs doesn't seem to be analyzed by the force, wow...

No matter how large the tensile force f is, the friction between the board and the small slider is a certain =, then the board accelerates under the friction force of 1n, and the small slider accelerates under the force of f-1n.

If the height of b is h, the radius of the bottom surface is r, and the density of the liquid is p, then the pressure on the side is.

pgh*(2rπ)/2

The pressure on the underside is.

pgh*(r^2*π)

The two equations are equal, and the solution is: h=r

That is, should the bottom area and the side area be equal? It doesn't feel like pressure but pressure, and the force should be uniform, and the pressure will not wait I don't know, and it doesn't matter so much, I know that the answer should be a.

Khan: Ignore me, .........

= Desert puzzle **********====b height h bottom radius a gravity index g

Side pressure: 1 2 density * g * h * 2 ah

Ground pressure: density * g * h * aa

The two forces are equal. So.

h:a=1:1

====Desert puzzle **********====

Average speed of 70 meters = (8 + 6) 2 = 7 meters in seconds.

Time to advance 70 meters = 70 7 = 10 seconds.

Acceleration = (6-8) 10=

So the distance after another 50 seconds = v0t+

If you don't understand, ask me again, hopefully.

Average velocity of 70 m = (8 + 6) 2 = 7 m s Time to advance 70 m = 70 7 = 10 s

Acceleration = (6-8) 10=

So the distance after another 50 seconds is s=vt at2 2=6 50 (

After the train shuts down the engine, it should do a uniform deceleration motion. From:2as=vt 2-v0 2 can get:

a=(6^2-8^2)/2*70=-9/70.The time that the train can also move is t=(0-6) (-9 70)=140 3 and less than 50s

Therefore, the distance traveled by the train after another 50s should be: s=(0-6 2) (-2*9 70)=140m

Taking the passenger as a reference, the relative speed of the oncoming train is 210 3 = 70 m s = 252 km h

When using the ground as a reference, the passenger speed is 120 km h

Therefore, when the ground is used as a reference, the speed of the oncoming train is 252-120=132km/h

210 3 = 70m s, which is the speed of the oncoming train relative to the passenger.

The speed of the passenger is.

So the speed of the oncoming train is.

Let the oncoming train speed be x km h, in fact, the problem is an encounter problem, assuming that the passenger is stationary, the oncoming train speed is (x+120)km h=(x+120) [(x+120).

Solve the equation x = 132 km h

Relative velocity 210 3 70

So the speed of the incoming train is 120-70 50

You can see it......

Let the bottom area of the cornerstone be s, then the total gravity g=1000000+1*s*2000*10

1000000+1*s*2000*10）/s≤7×10⁴

s≥20m^2

Solution: Let the base area be s, then.

The mass of the cornerstone is: m=pv=2000s kg

The cornerstone weight is: g=mg=20000s n

Because the pressure subjected does not exceed 70000pa, i.e

p=f s because:f=(1000000+20000s)n,p=70000paso:s=f p=((1000000+20000s) 70000 solution: s=20 m

So the base area of the cornerstone is at least 20 m