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Chapter 1 Mechanical Energy.

1 If an object is capable of doing work, that object has energy (energy).

2 Kinetic energy: The energy that an object has due to motion is called kinetic energy.

3 The greater the velocity and mass of the moving object, the greater the kinetic energy.

4 Potential energy is divided into gravitational potential energy and elastic potential energy.

5 Gravitational potential energy: The energy that an object has due to being lifted high.

6 The more massive the object is and the higher it is lifted, the greater the gravitational potential energy.

7 Elastic potential energy: The energy possessed by an object due to elastic deformation.

8 The greater the elastic deformation of an object, the greater its elastic potential energy.

9 Mechanical energy: A general term for kinetic energy and potential energy. (Mechanical energy = kinetic energy + potential energy) unit is: joule 10 Kinetic energy and potential energy can be converted into each other. The ways are: kinetic energy, gravitational potential energy; Kinetic energy Elastic potential energy.

11 The mechanical energy in nature that can be used by humans in large quantities is wind energy and water energy.

I remember that at this time last year, we were still in the first round of review, and we were doing the school's "Learning Sea Navigation". In the process of doing the questions, I found that there were many loopholes. In just a short winter vacation, we did 10 sets of questions in "Restructuring 168 Sets", and now I think back that doing the questions in a short period of time is very beneficial to the improvement of ability.

I think your first round of review is about the same, now you are not busy with comprehensive questions, review what you have learned, and list the knowledge points of each chapter on Saturday night to test yourself. During the day, you must concentrate in class, you must think about the questions that the teacher says, and you must think about the questions that the teacher says. Reach the topic and the collective process to keep it in mind, if you don't understand, you can ask the teacher again, the teacher knows you best, think about what you have learned before going to bed.

Don't be too anxious, every day is the most important thing before the college entrance examination, a heavy burden: keep up with the teacher's pace, every day to have a harvest. Have fun learning, take it easy.

Hope it helps.

1. Let the height of the parachute fall be h, and the speed when the parachute is v then 2gh=v2 - 0

From the above two equations, we can get h=

The height of the athlete from the ground when leaving the aircraft = 125+

2) v= is obtained from 2gh=v2

T1 = is obtained from v=gt1

From 5=v at2 we get t2=

t1 + t2 =

2.If an object falls from a suspended airplane, if the falling motion of the object is regarded as a free fall motion, and the height of the falling object in the last 1 second before landing is measured to be 9 25 (g=10m s2) of the entire falling height. Seeking:

1) The suspended height of the helicopter.

2) Average speed in the last 1 second.

2. Let the speed of landing be v, then the velocity of the previous second is (v-g*1) and has 2*g*h= v2 0

2*g*(16 25 *h)= v-g*1)2 0 to get v=50(m s) h=125(m).

Average velocity = (v + (v - 10)) 2 = 45 (m s).

2.If an object falls from a suspended airplane, if the falling motion of the object is regarded as a free fall motion, and the height of the falling object in the last 1 second before landing is measured to be 9 25 (g=10m s2) of the entire falling height. Seeking:

1) The suspended height of the helicopter.

2) Average speed in the last 1 second.

The falling motion of the object is free fall, h=(1 2)gt 21-9 25)h=(1 2)gt 2

Free fall movement time t=

The suspended altitude of the helicopter h=

The average velocity in the last 1 second v=(9h25) 1 m s=

You know, the slope of v—t is the expression of acceleration, so the slope of is positive, and the acceleration direction is the same, both are positive; The slope of is negative, and in a straight line, indicating that the acceleration is in the same direction, both are negative, and the magnitude of the acceleration is also the same;

So it is inconsistent with acceleration a, it is consistent with acceleration a, it is inconsistent with acceleration a, it is inconsistent with acceleration a.

The ratio of the rate is 2:1, so we use the formula to define the period t=2 r v(2* *r v1) (2* *4r v2)= v2 4v1 because v1=2v2, so it is 1:8

Which formula do you use to find the period and get different results? Tell me I'll do the math.

The ratio of the velocity is wrong, gm r 2=v 2 r then v1 v2= (m1r2 m2r1)= 2:1(1)t=2 r v then t1 t2=r1v2 r2v1=1:4 2(2)gm r 2=r(2 t) 2 gets t= (4 2r 3 gm) then t1 t2= (r1 3m2 r2 3m1)=1:

4 2 [I don't know if I miscalculated or if I answered this question incorrectly].

Gravitational force provides the centripetal force.

gmm r = mv r---m is the mass of the earth v= (gm r) --v is inversely proportional to the square root of the radius. It has nothing to do with the "quality of the satellite". --A True C False.

t=2 r v=2 (r gm)--t is proportional to "3 2 roots of radius". It has nothing to do with the "quality of the satellite". --b is false, d is true.

Question 1: If you want to pull the plank out from under the small wooden block, you need to make the speed change of the wooden board faster than the speed change of the small wooden block, that is, the acceleration of the wooden board a1 > the acceleration of the small wooden block a2

When you can pull it out, a2=umg m=ug... 1)

At this point, a1=[ f-u(m+m)g-umg] m... 2)

Simultaneous equations (1) and (2) When a1 > a2, the plank can be pulled out from under the small wooden block.

Second question: the first method: from the beginning of the movement to the process of pulling out the plank, let s1 is the displacement of the plank relative to the ground, and s2 is the displacement of the small wooden block relative to the ground, in the whole process, there are: s1 = 1 2 * a1 * t1 2 ...1)

s2=1/2*a2*t2^2...2)

s1-s2=l...3)

a1=[ f-u(m+m)g-umg ]/m ，f=5(m+m)ug...4)

a2=umg/m...5)

Simultaneous equations are solvable.

Let the acceleration of the board relative to the ground in the positive direction be a1, the acceleration of the small wooden block relative to the ground a2, and the acceleration of the small wooden block relative to the wooden board is a.

Then a2=a+a1 (this is a vector addition operation), so a=a2-a1;

From the equations (4) and (5) of the first method, a = -3u(m+m)g m; The minus sign represents the direction to the left.

So l=1 2at 2;

The answer is t.