Find the physics questions in chapters 2 to 5 of Physics in Senior 1 Physics

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1 Chapter 1: Description of the Movement 71:58

2 Chapter 1: Description of the Movement (II) 70:26

8 Force Analysis 71:27

9 Fourth Section Synthesis of Forces 70:41

10 Section 5: Decomposition of Forces 69:44

12 Force Balancing Exercise 71:32

13 Section 5 Newton's Third Law 72:42

14 Newton's Second Law 70:21

15 Newton's Second Law 70:20

Physics teacher Zhou Wenkuo won a first in Shandong Province. 41:06

These chapters are also used as titles. It's simple.

Kinematics, or simple. The first chapter is purely conceptual, and the second chapter gives several formulas1 for velocity: vt = v0 + at

2 Displacement formula: s = v0t + at2

3 Velocity-displacement relation: -2as

4 Average Speed Formula: =

v0 + vt) ⑤

5 Displacement difference formula : s = at2

Formula description: (1) Except for the above formula, other formulas are only applicable to linear motion with uniform variable speed. (2) The formula refers to the fact that in a linear motion with uniform velocity, the value of the average velocity of a certain period of time is exactly equal to the velocity at the intermediate moment of this period, so that a connection between the average velocity and the velocity is established.

Chang rot 6The uniform acceleration of a linear motion with zero initial velocity holds true of the following law:

1).At the end of 1t seconds, at the end of 2t seconds, at the end of 3t seconds...The ratio of the velocity at the end of the nt second is: 1 : 2 : 3 : n

2).Within 1t seconds, 2t seconds, 3t seconds...The ratio of displacement in nt seconds is: 12 : 22 : 32 : n2

3).In the 1st t, 2nd, 3rd, etcThe ratio of displacement in the first second is: 1 : 3 : 5 : 2 n-1).

4).Within 1t seconds, 2 seconds, 3 seconds...The ratio of the average velocity in the first second is: 1 : 3 : 5 : 2 n-1).

From the first question f- mg=0 =8 40=second question, when m is placed at the far right end of the board (the upper surface is smooth, the block is stationary relative to the ground) to the plank f- (m+m)g=ma a=

l=v0t+1/2at^2

t1=t2= rounded.

The magnetic inductance lines are closed curves, so for an infinite plane or any closed surface, there are many magnetic inductance lines that are penetrated, and there are equal numbers of magnetic inductance lines that are worn out, so that the magnetic flux that penetrates the magnetic lines in a certain way is positive, then the magnetic flux that is penetrated in the opposite way is negative. Magnetic flux, on the other hand, is defined as "net magnetic flux", which means algebraic sum.

In this way, the smallest coil inside is energized, which is equivalent to a bar magnet, and all the magnetic inductance lines pass through the coil, while only some of the magnetic inductance lines pass through the outside of the coil in the opposite way. If the magnetic flux passing through the coil is positive, then the magnetic flux passing through the outside of the coil is negative, so the smaller the coil, the smaller the number of magnetic inductance wires that can pass through from the outside, and the greater the net magnetic flux.

Therefore, it is concluded that the smaller the radius of the coil, the greater the magnetic flux passing through.

The closer to the wire, the denser the magnetic inductance lines, which is the law of magnetic field distribution.

This problem is that the magnetic flux of a small area is large.

Solution: The displacement of the last second is 2m.

a=4m/s^2

So v=at=28m s displacement is.

From the idea that 1 2at 2 = s and the reversible idea of uniform variable velocity, it can be seen that a=4m s 2 then the initial velocity v of the object is a*t = 7*4 =28 m s and the displacement is 1 2*v*t =

Draw a v--t image.

We just finished talking, you see from s to the end of another line, through two triangles, that is, two displacements, so it is 2s