Senior 1 Physics Questions. Four methods are needed!

Method 1. It is a uniform deceleration for 3 seconds.

s = t · average speed = 3 * [(12 + 0) 2] = 18 > distance in 24 seconds of the method.

s=vt-a*t*t*1/2=18>

Method 3: Counterproof.

If enough or just right.

According to. s≥vt-a*t*t*1/2

a (while a = 4<

The counter-evidence is not established.

The fourth method.

Disproof. Acceleration = 4

Time = 3 Distance =

vt/2=s

v= is different from the question.

The driving time after braking t=12 4=3s, and 4s has stopped.

1) Displacement s=m > m, so it was crossed.

2) 0--12x12=--2x4xs s=18 m > m, so crossed.

3) The average speed in the time t after braking = 12 2 = 6m s displacement s = 6x3 = 18 m > m, so it was crossed.

4) The average speed within the time t after braking = 12 2 = 6m s Assuming that it stops just at the stop line, the average speed is 3 = m s < 6m s, so the stop line is crossed.

The force analysis of the object and the trolley as a whole shows that the gravity of Cha's slip vertically down = 1000N, the support force is perpendicular to the inclined plane, and f is parallel to the inclined plane. In the direction parallel to the inclined plane, f-1 2*1000=ma, i.e., f-500=ma. The force analysis of the isolated body method is done on the object, the gravity is 800n vertically downward, the support force n is vertically upward, and the static friction force f is horizontally to the right, and in order to make the object accelerate along the inclined plane direction, n must have a quantity n left in addition to balancing with gravity', this n'The resultant force with the horizontal direction f causes the object to move upwards with uniform acceleration along the inclined plane.

Analyze the force vector diagram, tg30=n'f, i.e., f=sqrt(3)*n'。The nuclear scale reads 1000n, so 1000=800+n'，n'=200，f=200*sqrt(3)。f and n'The resultant force = sqrt(f2+n'2) = 400n, so object acceleration = 5.

The overall acceleration is equal, so f=ma+500=100*5+500=1000n.

Decomposing the acceleration a horizontally and vertically, there is:

…2 points) take the whole car, object, and scale as the research object, and the force analysis obtains:

f-mgsinθ＝ma ……2 points).

f= mgsinθ+ma =1000n………1 point)2) ....Rock scattered hail .........2 points).

According to Newton's third law of the rough sail, the static friction of an object against the scale is , and the direction is ...... horizontally to the leftDig (1 point) ma

mgnfa

1. a=(30-0)/

f+mg=ma

f=ma-mg=

2.The speed is 0 at the highest point and there is no resistance. Therefore, it is only affected by gravity, and the acceleration is g=10m s 2

3.When falling, the resistance is upward, at this time mg-f=ma1a1=(mg-f) m=(

Here the drag force is calculated according to the average drag force, in fact the velocity is small at the beginning, and the drag force is also small, when the velocity reaches a certain value, the drag force is balanced with gravity, and the object falls at a uniform speed.

Solution:1Suppose the average resistance is f, which can be known from the momentum theorem:

f+mg）t=mv

Substituting data can be solved:

f=1n2.Since air resistance is related to velocity, when there is no velocity, it will not be affected by air resistance, therefore, at the highest point, since the velocity is 0, its acceleration is the acceleration due to gravity g=10m s

3.When the ball falls, the air resistance and gravity are in opposite directions, so the acceleration is magnitude of:

a=（mg-f）/m=8m/s²

1. Because a=g+f m, so f=(a-g)m and because a=v t=12(m s), f=(12-10) 2. Because the velocity of the ball is 0 at the highest point, it is not affected by air resistance, and the acceleration is g=10m s

3. Because the air resistance is unchanged, f=g-f, so a'=f/m=(5-1)÷

v=at yields a=12m s

It can then be obtained by f=mg+f=ma.

f=ma-mg=1n

Second question: Simpler a=12m s

The third question: f-he=mg-f=ma

a=8m s can be obtained

(1) Set the average resistance level f, for the ball there is 0-v=at, so a=12m s, for the ball, there is: g+f=ma, so f=1n, and the direction is vertical downward.

2) The ball is stationary at its highest point and is only subject to gravity, so the acceleration a'=g=10m s, the direction is vertically downward.

3) When the ball falls, the resistance is vertically upward, and there is: g-f=ma'', so a''=8m/s²

a 1.The wooden block is subjected to the left friction force given by the plank and is mg in magnitude. 2.

From Newton's third law, it can be seen that the wooden board is subjected to the frictional force of the wooden block to the right, and the magnitude is also mg of magnitude. 3.Because the plank is stationary, it is subject to a ground friction force to the right, which should be equal to mg

Draw a force analysis diagram...

Wood block m is subject to a frictional force f [umg] of wood block m, in the same way that wood board mm is subject to the anti-friction force of wood block m "[umg], if wood board m is to be stationary, wood board m must be subject to ground friction f [umg].

Note: This is not related to the quality of the planks. 】

The friction of the wooden board is the same as that of the wooden block, so it is a

The plank is subjected to only two forces in the horizontal direction, one is the friction given by the wood block and the other is the friction given by the ground, and these two forces are balanced.

The wooden block slides on the wooden board, and the wooden board is subjected to the sliding friction given by the wooden block, and the magnitude is mg, while the wooden board is at rest, and its force is balanced, so the static friction given by the wooden board is mg in the direction opposite to the friction given by the wooden block.

First, the gravitational force on the object is divided into a force of 25n downwards along the inclined plane and a pressure on the inclined plane by orthogonal decomposition, because the object moves upward at a uniform speed along the inclined surface, the force is balanced, and because it is a thrust of 30n, it is f-resistant+

f gravitational component = 30n so the frictional force experienced by the object is 5n, because the force is acting on each other, so the friction force on the inclined side of the slider is 5n

(1) Kinetic energy geography for small block columns.

mgh-q=1 2mv q=

2) Equations for the conservation of momentum for small blocks and long wooden boards.

mv=2mv'Gotta v'=1m/s

3) Newton's second law on long planks.

f=ma is solved by knowledge of kinematics.

a= therefore f=ma=

1), the use of mechanical energy conservation.

m s, conserved with momentum.

3), Newton's second law.

lz you are right, a uniform acceleration linear motion with zero initial velocity has a displacement ratio of 1:3:5 in the same time

7...It is also true for the movement of free fall. The distance between 12, 23, and 34 in the figure is not in this ratio, so the 1 position is definitely not the release position, nor is it the upper edge of the brick.

Let the time t take from the release position o to 1Then the 12 spacing d is equal to o to 2 minus the distance from o to 1 d = g(t + t) 2 2-gt 2 2

T is solved and substituted into the displacement formula to obtain the height of the release position.

Do you understand?

Follow the diagram you gave me and I can tell you where the ball is from"1"The reason for the upper surface of the brick above the position began to fall is that there is an important conclusion in the linear motion of uniform variable speed in the high school stage, "the difference of adjacent equal time displacement is a fixed value", and the figure shows that the displacement from 1 to 2 is 2d, the displacement from 2 to 3 is 3d, the displacement from 3 to 4 is 4d, and the displacement from 4 to 5 is 5d, so it can be known that the displacement difference in the adjacent t time is d, then the 1 position is one d less than 2d, and the starting position can only be there. I don't know how to make it clear?