I know how to parse a physics problem in my first year of high school, but there is one thing I can

The <> area should be equal to t2>t1, and the acceleration time should be long and then uniform.

The acceleration will not be infinite, and the maximum friction it gives to the object is UMG, regardless of the speed of the conveyor belt.

There can be two scenarios for the acceleration of an object:

1. The speed of the conveyor belt is not large, the object has not moved to the B end, the speed is the same as the conveyor belt, and the object first accelerates the movement with ug, and then the speed is uniform.

2. The speed of the conveyor belt is relatively large, and when the object moves to the B end, the speed is still smaller than the speed of the conveyor belt, or just the same as the conveyor belt, and the object always accelerates with the acceleration of ug in this process.

Because the initial velocity of the workpiece is 0, and the friction between the workpiece and the conveyor belt will not exceed the sliding friction at most, so when the workpiece is just placed on A, it is a uniform acceleration motion with a muzzle velocity of 0, and the acceleration cannot exceed g, no matter how large the speed of the conveyor belt is, the maximum acceleration of the workpiece can only be g and not greater, so it is impossible to "give him an infinite speed". The acceleration calculated earlier 1m s 2 is the maximum sliding friction acceleration g, so the acceleration in the back is equal to the acceleration in the front.

Because in the problem, the acceleration of the object is provided by the frictional force, and the frictional force experienced by the object is a certain (by a, b are separated by l 10 m. The workpiece is placed on the conveyor belt without initial velocity from A, and several conditions can be calculated by transferring the workpiece to B after the elapsed time t 6 s), so the acceleration calculated in the front is the acceleration amount used in the back.

Combined with the view, it is easy to understand the answer.

1.Suppose the distance from the cave is l when the horn is sounded for the first time

The distance of V1T1 is advanced by the car when the sound of the echo bucket source is heard for the first time, and the distance of sound propagation can be calculated according to the position of the car horn and the sound of the echo state is 2L-V1T1

According to the speed of sound propagation, the distance of sound propagation can also be obtained as V2T1, >2L-V1T1=V2T1

2.When the horn is sounded for the second time, the rook advances v1(t1+t2), then the distance between the rook and the cave is l-v1(t1+t2).

When the echo is heard for the second time, the car advances the distance of V1T3 again, and the distance of the sound propagation is 2[L-V1(T1+T2)]-V1T3 according to the position of the car honking and hearing the echo.

According to the speed of sound propagation, it can also be obtained that the distance of sound propagation is V2T3.

2[l-v1(t1+t2)]-v1t3=v2t3

<> hee hee handwriting can be a little scribbled sedan chair Xiyan or wanton hope can be understood!

The answer is D, ABC is all wrong.

a.The full force of the object placed on the table is 0, but it is stationary, so A is wrong.

b.Assuming that the initial velocity of a object is 0, and the resultant force is 0, then at the beginning, its acceleration is greater than 0, but the velocity is 0;

Suppose that the initial velocity of object b is 100 and the resultant force is 0, then its acceleration is 0 and the velocity is 100, so b is wrong.

c.Suppose that an object with velocity v experiences a resultant force in the opposite direction to the velocity, then its acceleration is in the opposite direction to the velocity, so c is false. In addition, in the circular motion of a standard celestial body, the direction of the resultant force is perpendicular to the direction of velocity, which also indicates that c is wrong.

Because the small bulb emits light normally, it consumes (36 2 r)*t in one cycle.

There is a diode on the secondary side, which is a half-wave rectification, so it is only energized for half a cycle, so the current consumption is (u2 2 r)*t 2.

The two are equal, so there is a first step of analysis.

According to the radius of the bottom surface, the diameter is 4, and the radius is 2.

This results in a 3:4:5 right triangle.

According to the decomposition of the force 3 5g=4 5g*

Got =

This question should be understood as follows:

The circumference of the bottom surface of the cone on his side is meters, and the height of the cone is meters, and the cone is the shape of the cone that rolls down the surface of the cone at a uniform speed along the surface of the cone, that is, the sliding force is just equal to the friction force at this time. According to the data, it is not difficult to find that the radius of the cone is r=2, and the angle between the conical bus bar and the horizontal plane is , then mgsin = mgcos is substituted into sin = 3 5, and cos = 4 5 is obtained by =3 4=

For wooden blocks, the horizontal direction is subjected to a rightward tensile force f and a leftward frictional force f = m2*g. then f-f=m2*a2

For planks, if the horizontal direction is subjected to a frictional force to the right f= m2g, then f=m1a1

For acceleration, a1=a2 is possible, in which case f=(m1+m2)a. In B, A1>A2, then V1>V2, then the wooden block should be subjected to the right friction, the wooden board should be subjected to the left friction, and the wooden board will be evenly decelerated. Here there is a contradiction.

So for acceleration, a is possible, and b is impossible.

For speed, v2>v1, it is reasonable for the block to experience a frictional force to the left and the plank to the right friction to accelerate.

If v1>v2, the block should be subjected to the right friction, and the board should be subjected to the left friction, and the board will be evenly decelerated. Here there is a contradiction. So for velocity, c is possible, d is not.

Choose the AC [Handsome Wolf Hunting] team to answer for you.

Force Analysis:

For the wooden block, the horizontal direction is subjected to the right tensile force f, and the wooden board contact has a leftward friction force f2, the size is not determined, but the maximum is m2g, where is the friction coefficient (assuming that the dynamic and static friction coefficients are the same, are .

For the plank, the horizontal direction is subject to friction at the contact with the wooden block, the direction is to the right, and the size is equal to F2. The contact with the smooth horizontal plane is not clear, whether it is considered that the friction coefficient is 0?. Then there is no friction at the point of contact.

Problem analysis: 1. F is relatively small, less than the maximum friction m2g at the contact between the wooden block and the wooden board, and there is no friction at the contact between the wooden board and the smooth horizontal plane, then the wooden board and the wooden block move to the right together at the same speed and the same constant acceleration, and its value is f (m1+m2). This is shown in Figure A.

2. F is greater than the maximum friction force m2g at the contact between the wooden block and the wooden board, then the wooden block moves to the right with the acceleration of m2g (m1), and the wooden block moves to the right with the acceleration of the value (f- m2g) (m2) relative to the wooden board (note that it is relative to the wooden board), that is, the acceleration of the wooden block is greater than the acceleration of the wooden board. Then it is not possible in the acceleration diagram A and B, only in Figure C: it is possible that the velocity of the wooden block is greater than the velocity of the wooden board.

3. F= M2G, critical state, unstable motion (it may be the first article above, which is the state of Figure A, or the second article above, which is the state of Figure C). However, it is not possible to see the states of Figures b and d.

Therefore, the ones that may be consistent with the exercise are (a, c).

This kind of problem should pay attention to the relationship between f and f, when f "maximum friction m2g", f = f, not m2g.

It is easy to prove in mathematics that the angle theta is measured in radians, so when the angle theta is very small, the theta value is very close to its sinusoidal value, i.e.,

For very small teta, the approximation is as follows: sin(theta) = theta is replaced by the commonly used unit allpha (degrees, minutes, seconds).

allpha=(

By the way, for very small angles theta, there is an approximate equation: theta = sin (theta) = tan (theta).

In mathematics, if you draw a unit, it is easy to see that the above equation is approximately true, and the smaller the angle, the smaller the error.

When very small, sin

So here sin = , we know radian =

My physics teacher: 1. First of all, let's talk about the solution you use here: the "whole" here includes:

1) Slide bar and base with mass m; (2) monkeys with mass m; 2. In this whole or system, m is stationary, and m is accelerating decline; 3. Here a is the acceleration of the monkey, if you write it all, it should be: (m+m)g-fn=ma+ma'Since m is stationary, i.e. a. a'=0.So it's the formula above.

4. In fact, when I first started learning physics, this solution was very bad and did not do much good to students. So how should we think about it?

With the isolation method: first take the monkey as the research object: let the friction force of the rod to m upwards be f, then Newton's second law is: mg-f=ma;

Then take m as the object of study: m should be subjected to three forces: mg, f (the reaction force of m to m), and fnThe same goes for :

mg+f=fn

If you join the above two formulas, you can get FN

5. This solution seems to be very complicated, but in fact, it is very beneficial to understand the principles of physics and develop good thinking habits, and Newton's second and third laws are used here. If you start from the basics first, figure out the basic principles, and then use the overall method in the third year of high school, it is of course beneficial to improve the speed of problem solving. The so-called:

Ability = Basic knowledge + basic method.

Ok, I hope it helps you in your studies, and I wish you progress in your studies.

This "whole" is made up of two parts: the slider and the base m., the monkey m

a is the acceleration of the monkey.

If this "whole" is not easy to understand, it is completely possible to isolate it:

Taking the monkey as the research object and rolling together: using Newton's second law to obtain mg-f=ma, using the slide bar and the base as the research object: using the equilibrium condition to obtain mg+f=fn and eliminating f from the above two formulas, that is, (m+m)g-fn=maThat's easy to understand.

Therefore, we do not advocate the use of a "holistic" approach on this topic.

A is the acceleration of the monkey Sun Zijing, the overall method mentioned here is just to facilitate the fn into the calculation, in fact, it is still the monkey Qi Dan as the research object.

The teacher should have said that there are the following rules for uniform acceleration linear motion starting from rest:

The ratio of displacements of the previous second, the first two seconds, the first three seconds, and the first four seconds ,,, is 1:4:9:

16, the first second, the second second of the old answer, the third second, the fourth second of the ,,, of the displacement ratio is 1:3:5:

7, from which it can be seen that the ratio of the displacement of the first second to the fifth second is 1:9, and it is easy to calculate that the displacement in the fifth second is 18m

From the object that starts from rest to do a uniform acceleration linear motion, in the first second through a distance of 2 meters and s at to get a 4m s, and substitute t 5 into s5 50, the same way s4, so the answer is b. The result of the calculation is 18, 25 is wrong and 18 is not the same, and the calculation of multiple choice questions is like this.

x=1/2at^2。Hui Zheng buried this formula before you learned it, and then, with the value, you can find the acceleration aEqual to 4.

And then it's okay to use the formula once. 1/2*4*（5*5-4*4）=18。Got it.

Uniform acceleration of a linear moving object from a stationary start through the first second displacement: second displacement: third second displacement: . . . Nth second displacement = 1:3:5:. 2n-1

Your question is this rule, and the exam is used directly.

You take the first second and pass two meters, and at this scale, the fifth second is 18 meters.

Hope! is also a high side ridge of a middle school.

The uniform acceleration of the linear motion with an initial velocity of 0 has a displacement ratio of 1:3:5:7 in the same adjacent time

The displacement ridge in the first second is 2, then the displacement in the fifth second is 2x9 = 18 meters.