Physics masters, please come in, it s in a hurry. Senior 1 Physics Questions

Let the total distance be 2s

Average speed = 2s (s v1 +s v2) = v1v2 (v1+v2)v1v2 (v1+v2) -2*v2 =-(v1v2+2v2) (v1+v2) 0

So it's impossible for him to average 2v2 in the whole process

Let the total distance be s

For the first half of the journey t1 s 2v1

The second half of the journey t2 s 2v2

The average speed of the whole journey is v s (t1 t2) 2v1v2 (v1 v2)v1v2 v1v2 v2 (v2 0).

v1v2／（v1＋v2）＜v2

v＜2v2

Use math. Solution: Let the total distance be 2

Column: t1=(1 v1)+(1 v2) t2=(2 2v2)=1 v2

T1 is the time taken for the first half of the distance to travel at a constant speed of V1 and the second half of the distance to travel at a constant speed of V2.

t2 is the time taken with an average speed of 2v2.

Obviously, t1 ≠ t2 (if t1 = t2, then 1 v1 = 0), so his average speed in the whole course cannot reach 2v2.

Please use the method of counter-proof in mathematics, it is easier.

Assumption: The average speed reaches 2v2.

Let t1 denote the time spent in the first half of the journey, and t2 denote the time spent in the second half of the journey.

Then: the length of the whole process s=2v2*(t1+t2).

Half-length s 2 = v2 * t2

So there is. 2v2*(t1+t2)=2*(v2*t2) divide both sides of the equation by 2v2, then:

t1+t2=t2

t1=0 comes to the absurd conclusion that t1=0.

Therefore, the assumption is wrong.

This proves that it is impossible to achieve an average speed of 2v2 in the whole process.

vAverage s ((s 2) v1+(s 2) v2)=2v1*v1 (v1+v2)=2v2*[v1 (v1+v2)].

and [v1 (v1+v2)] <1

So the average speed can't reach v2

According to the condition of the question, the actual displacement in this ** time is l 4 meters, and the velocity value of the middle moment of this time is equal to the average velocity in this time, that is, the velocity at the middle moment of ** time is v l t 4 m s

From v 2 2*a*s 20 2 2*a*40 , a 5 m s 2

The analysis of human force is done by orthogonal decomposition method, mg*sin37 degrees *mg*cos37 degrees m*a

Acceleration a g*(sin37 degrees *cos37 degrees)5 10*(, the kinetic friction factor is 1 8

4 is the velocity of the phantom midpoint. 2as=v*2, a=5m s*2

From the bovine law, a=gsin37*- gcos37*, 0 75 can be obtained

1) Before pushing the M2 firmly, the eggplant is in equilibrium.

m1＋m2）gsinθ=k1δl1

It is found that δL1=(M1 M2) is used to push GSIN K1 hard before fibrillation, and the deformation of the two springs is equal when the first spring is elongated, and the second spring is compressed.

For m1 there is k1δl+k2δl=m1gsin solution, and δl=m1gsin k1+k2

Therefore, the upward distance of m1 is δl1-δl=(m1 m2)gsin k1-=m1gsin k1+k2

The distance traveled on m2 is δl1+δl2=(m1 sells shirt m2)gsin k1+m2gsin k2

2) When the spring above is in the original length, the balance equation for the overall column of m1 m2 can be obtained.

f=（m1＋m2）gsinθ

The first question is calculated with the knowledge of energy.

The work done by gravity is finally converted into the gravitational potential energy of the chain.

e=mgh=gh

So 12=30*h where h is the distance from b to the center of gravity.

H=so we can know that the distance from a to the center of gravity is.

Then the gravitational potential energy that the work done is converted is.

Of course, there is another solution here, that is, the total work of lifting the entire chain is the sum of the work done from both ends respectively.

It can also be obtained directly with 30*1-12=18.

Question 2. Directly use the formula of energy.

for b is immovable.

Then the energy reduced by a is the total energy loss.

That's 1 2 mgr

Question 1: Known: w=12j g=30n Let h be the distance h of the center of gravity from the b terminus h'is the distance of the center of gravity from the a-end.

Find :w' solution: H=12 30=h'=

Hence w'= A: .Question 2:

Because the B-ball is fixed, only the A-ball goes from the right end to the lower end.

Therefore, the change in gravitational potential energy is 1 2 mgr

1, 18j

2, 1/2mgr

Just a reminder, your fraction notation is wrong, not 2 r, it should be r 2

If the roof angle is 2x, the speed of raindrops falling on the roof is 0 (this is easy to calculate). Raindrops slip from the roof at the time tThe acceleration due to gravity is g

Raindrop acceleration a = cosxg, 1 2cosxt2 = l 2sinx (t2 is the square of t)" t = l 2sin2x under the root number

When sin2x has a maximum value, t is the smallest. That is, 2x=90, sin2x=1, so when the top angle of the roof is 90 degrees, the falling raindrops can flow through the roof in the shortest time.

So that A does not slide off B, i.e., one of the critical conditions is that A slides to the right end of B with the same velocity and from then on they have the same acceleration.

A object is only subject to friction, acceleration a1 = f1 m1 = g = 2m s b is subjected to left friction and right tension f at critical conditions, a2 = (f-f1) m2 = 2f-4

At this time, A1=A2, that is, 2=2F-4, the maximum F=3N can be obtained, the second critical condition is that A first decelerates to the right, B accelerates to the right, and after the speed of AB is equal, A accelerates to the right, and the co-velocity and acceleration are equal at the leftmost end of B, and the minimum F=1N can be solved in the same way

There is a1=ug=2m s 2 for objects

Let the external force acting on the plank be f

then a2=(f+um1g) m2

Let the object not slip off the board at the elapsed time t, the distance the object passes is s1, the distance the board passes is s2, their velocity is v, and the object 2a1s1=4 2-v 2 v=4-a1t to the plank 2a2s2=v 2 v=a2t

There is s1-s2=1 for objects and planks

The results of the above joint solutions can be.

1.If you want to deal with the ground, he falls diagonally, and he does a flat throwing motion, that is, the egg seen by the person on the ground is a flat throwing motion, because he has a horizontal initial velocity like a train when it first falls.

2.If you want to be a train, it is a horizontal deceleration relative to the train (because the egg and the train want to stand in the horizontal direction, and the train accelerates forward evenly, which is equivalent to the egg decelerating backwards), and the vertical direction is free falling.

If it is thrown in the same direction as the direction of movement of the train, the direction of motion is the same as that of the train, and if the direction of the speed of the throw is opposite to the direction of the train, it is the same as the greater of the train or the direction of the throw.

Because the horizontal direction of the egg is not forced, there is an initial velocity, so the horizontal motion is a uniform linear motion, and the velocity is the vector sum of the throwing speed and the speed of the train;

In the vertical direction, only by gravity, the initial velocity is 0 and the acceleration is g, and the linear motion is uniformly accelerated;

The trajectory is the part of the parabola with the opening pointing downward.

Flat throwing motion. The initial velocity is the velocity when dropped from the train, and the downward acceleration is g.

The direction of motion is the same as that of the train.

No, because it also has a vertical downward gravitational acceleration in the same amount of time, although the displacement is the same in the horizontal direction, but in the vertical direction, it is 1:3:5:

7。。。because it has acceleration in the vertical direction (1:3:

5:7。。。This part can be seen by looking up the data, which is the displacement ratio of the initial velocity of 0 to the accelerated motion in the same time.

a represents the acceleration of the applied force, and x1 represents the displacement of the first ten seconds 1 2*a*t 2=x1

f-umg=ma is solved to f=15n2)a2 to denote the acceleration after the force is removed.

umg=ma2

v*2=2(a2)x x=25mv=at

First use the kinematics theorem s=1 2at2 to calculate the acceleration, and then use the ox two theorem, that is, f-f friction = ma to know the friction force to calculate f. The second question is to calculate the velocity at the end of 10 seconds, use s = v end 2 - v beginning 2 2a and then f friction = -ma the direction of attention to find the acceleration and then use the kinematic formula to find the final distance.

First of all, f (friction) = fn = g = 50 so f = f - f = ma

Because 1 2 at 2=x=50

So the solution is a=1m s 2

So f=ma+f=5+10=15n

The second question is too long, I'll go to your space.