Physics in the first year of high school is so difficult, and there are several equations to be list

High school physics is the embodiment of equation thinking, that is, what is not known, and the connection between the previous process and the next process should be bravely set out, which is different from the step-by-step formula of junior high school physics. Real physics is about exploring the unknown, so using equation thinking now proves that you're already on your way to becoming a high-tech talent.

In fact, physics in the first year of high school is not difficult, analyze and draw a good situation, draw each force well, and remember the core point"Balance"Any object moving in a straight line at rest or at a uniform speed is balanced.

Analyzing a good situation and drawing a good force, physics is actually not difficult to learn.

I hope you can learn physics well.

Cultivating an interest in physics is very important, and your teachers are very rigid, right? If so, it's up to you to learn to have fun in doing problems.

It is recommended that you use the book "One Use Five-Year College Entrance Examination Three-Year Mock (High School Edition)" for three years.

There are a few little stories about physics underneath that will pique your interest.

It will make you fall in love with physics.

I'm using it too. There is also the book The Road to Success.

Hope it helps.

Encourage you to do so.

Because I'm also in the practical phase.

Hehe. It works great.

Hopefully your physics is getting better and better.

It seems difficult now, but when you finish high school and then do the problem, you don't have to list a few equations to solve the problem, and you probably have to wonder if you are doing it wrong.

If you think about the reasoning more, if you can't do it, you can use your hands to make a gesture, and at the beginning you understand the simple problems, and slowly the difficult ones can be solved.

What's your problem?

At the moment when the hanging basket rope is cut, the position relationship of ABC does not change, B and C are still supported on A, A and C are not separated, the distance between BC remains unchanged, and the spring elasticity does not change. b by gravity and spring support, balanced, acceleration is 0; c is subject to gravity, the elastic force of the spring (equal to the gravitational force of b), and the supporting force of a. For a certain period of time, C is pressed on A by the spring elastic force, and moves with A, and the acceleration is the same, set to A, and the AC force is N.

A is subject to gravity and n, mg ten n = ma, 6 mg ten 2 mg-n = 2 mA2 mg ten 2n

3n=6mg，n=2mg

a=3g。

Let the time elapsed for b to meet the lower end of a bar is t [i.e.].

t=, then the sum of the distances between a and b is.

H-1 at this time.

aThe displacement of the fall is .

ha=1/2gt²

b The upward displacement is.

hb=v0t-1/2gt²

ha+hb=h-1

1/2gt²+v0t-1/2gt²=h-1v0t=h-1

t=(h-1)/v0

Let the time elapsed for B to meet the lower end of the A rod.

t', [i.e.]

t'= then the sum of the distances between a and b is h at this time.

aThe displacement of the fall is .

ha=1/2gt'²

b The upward displacement is.

hb=v0t'-1/2gt'²

ha+hb=h

1/2gt'²+v0t'-1/2gt'²=hv0t'=h

t'=h/v0

By and , get.

5/6=(h-1)/h

h=6mv0=10m/s

Buses depart from bus stations every T0, indicating that the distance between each two buses is the same. So the problem translates into people and cars in the same direction for 3 minutes, and the reverse is 5 minutes. Let the speed of the vehicle v1 and the speed of the person v2, the distance between the two cars is s, and the distance between the two cars is s, and the distance between the two cars is s

v1+v2）*3=s

v1 -v2）*5=s

Here are the graphs and formulas, and the answer is up to you.

1.Analysis: The original speed of the locomotive v1 = 36 km h = 10 m s The locomotive travels to the end of the downhill slope, and the speed is v2 = 54 km h = 15 m s, which is defined by the acceleration formula a=v2-v1 t

The time taken by the locomotive to pass this downhill road can be obtained.

t=v2-v1/a=15-10/

2.Solution: The maximum speed v, the average speed of the first 5 min is v 2, and the average speed of the last 3 min is also v 2, so the total average speed is also v 2

s=vt/2

v=2s/t=2* km/h

The maximum speed of the car on this journey is 36 km h.

3.Analysis: Because the box moves at a constant speed.

So in the horizontal direction fcos = f

Friction force f= n= (g-fsin )

So fcos = (g-fsin).

f = g cos + sin = root number 3 2 + 120n

1 Initial velocity v1 10m s final velocity v2 15m s v2=v1+at solution t 25s