Two high school physics questions! Two high school physics questions! Urgent

One. 1.The work done by gravity is equal to the amount of change in gravitational potential energy. The direction of gravity is the same as the direction of the ball's motion, doing positive work, w=gh=mgl

2.The pulling force is always perpendicular to the direction of velocity and no work is done.

3。If the ball is not told the speed of reaching point O, the work done by the external force cannot be calculated.

Two. 1.Because of the constant velocity, the tensile force is opposite to the gravity component force along the inclined plane downward, and the work done by the tensile force is equal to the product of the displacement and force along the direction of the force, w=fs=50*10*4=2000j

Negative work is done in the opposite direction of gravity.

3。The object is pulled by the frictional force of three forces, where the work done by friction is counterbalanced by the equal magnitude of the work done by the pulling force, so the work done is -1000j

Choose 3 for both questions

1: If the light has gravity, it should fall vertically, and the pull of the light rope should be horizontal, 2: The same goes for it.

The gravitational force of the ball does the work w=mgh=mgl

2) When the ball moves, the direction of velocity is perpendicular to the direction of the tensile force of the line, and the tensile force does not do work.

3) The total work done by the external force on the ball is equal to the work done by gravity mgl2Since the upward motion of the object is uniform, there is no change in kinetic energy. Hence the total work done by the respective forces on the object is 0; The work done by human tension is the same as the work done by gravity, but gravity does negative work, and their values are mglsin30=2000j

- - The first one is that you're playing m·s -right?

The height of the orbit from the ground is equal to the multiple of the radius of the earth, then the radius of the orbit is also a multiple of the radius of the earth, that is, let the mass of the artificial satellite be m

By <2 t> m <> then t = 4 * land.

The acceleration by gravity on the Earth's surface is = gm ground r gm ground = r

Bringing in the above equation gives t =4* so t = 20000 s

2 From p = fv at the start of the process analysis f maximum f as the velocity increases f begins to decrease when f is as large as the drag force.

v reaches the maximum of 60m s as mentioned in the title

In this case, f=ff, i.e., ff=p, v=900000, 60=15000n

Question 2: On the whole, the aircraft is subjected to the positive work done by its own engine and the negative work done by ground friction.

Positive work w1 = pt = 900000 * 15 j = 13500000j

The negative work done by the final kinetic energy before take-off ek=9000000j so frictional force wf=w1-ek=4500000j

After answering, I don't know how to ask me if I am.

First of all, assuming that the total distance is s, then s=vt1. On the way, braking to normal constant speed, went through three processes, that is, from v0 to 0,; Stop t2 time, and then accelerate from 0 to v0, so that the three processes, then t3-t2 is two sections, uniform reduction and uniform addition time.

According to v=at; Therefore, the acceleration of this uniform reduction and acceleration process is a=2v0 (t3-t2); According to s s = note 2 for the square of t);

Therefore, the total amount of uniform reduction and acceleration of the two stages is s1=v0(t3-t2) 2; If you still want to arrive at t1 time, then first use s-s1 to get the distance of uniform motion, and then divide it by the uniform motion time t1-(t2+t3), then after calculating vo=2vt1 (2t1-3t2-t3).

Forehead... After graduating from college for so many years, when I saw something like this, I thought it was fun and did it, and the questions were relatively simple.

Total displacement s=vt1

The second time from A to B, it is divided into a process of uniform motion and a process from the beginning of deceleration to the end of acceleration. Since the time does not change, it is still t1, therefore, the uniform motion time is t1-t3, and the displacement of the uniform process is s1=v0(t1-t3).

The deceleration time t is also set', acceleration time t'', then there is t'+t''+t2=t3, so, t'+t''=t3-t2

Deceleration displacement s2=', acceleration displacement s3='',s2+s3='+t'')s=s1+s2+s3, which can be solved: v0=vt1 (

Hello, do I need to provide you with a high school physics problem?

Questions. <>

Okay, wait a minute.

Questions. The mass of the slider and the skateboard is 1 kg, and the question is the acceleration of the skateboard?

Hello, the acceleration is 6

a g sin37° g 10 sin37° question. <>

It seems that there are fewer conditions.

1) Is it a uniform deceleration when braking?

2) Is the acceleration during braking the same as the acceleration during reacceleration?

There's an important formula for this problem, haven't you memorized yet? Still won't work? The formula is the displacement formula for the average velocity.

Answeraba Rutherford began in 1909 to conduct the famous particle scattering experiment, the purpose of which was to confirm the correctness of the Thomson atomic model, but the results of the experiment became strong evidence to disprove the Thomson atomic model. On this basis, Rutherford proposed a model of the nucleus-like structure.

b Hertz discovered the photoelectric effect in 1887 and was the first to successfully explain the photoelectric effect. The effect of emitting electrons from a metal surface under the action of light irradiation, and the emitted electrons are called photoelectrons. Electrons can only be emitted when the wavelength of light is less than a certain critical value, that is, the limit wavelength, and the corresponding frequency of light is called the limit frequency.

The critical value depends on the metallic material, while the energy of the emitted electrons depends on the frequency of the light and is independent of the light intensity, which cannot be explained by the wave nature of the light.

c, the energy emitted when the dispersed nucleons form the nucleus is called the nucleus binding energy. The combination of atomic charges is called nuclear fusion, and it requires a lot of energy!

d.It is true that there is a mass loss, but the loss part is only a very small part of the total mass, and the mass number is still conserved.

The vector decomposition of v1 is divided into velocity along the ab bar and velocity perpendicular to it, and v1cos = vab

The vector decomposition of v2 is divided into velocity along the ab rod and velocity perpendicular to it, and v2sin = vab

So v1cos = v2sin = vab so v1 = v2sin cos = v2tan

Hello, do I need to provide you with a high school physics problem?

Questions. <>

Okay, wait a minute.

Questions. The mass of the slider and the skateboard is 1 kg, and the question is the acceleration of the skateboard?

Hello, the acceleration is 6

a g sin37° g 10 sin37° question. <>

The solution is as follows: (Additional note: 1. The car and the police car have the same acceleration, so the police car is used to calculate the acceleration.)

2. Because people do not have a great impact on the movement of the car, it is believed that the car stops at a uniform deceleration.

3. The time of the person from D to B is the sum of the reaction time of the car driver and the time from braking to B).

The object with a mass of 1kg is parked on the horizontal ground, and the maximum static friction between the object and the ground is 2N, and the dynamic friction factor is, and an external force f in the horizontal direction is applied to the object, then when f = 1N, the friction force between the object and the ground F1 = 1 N; When f=2n, the friction between the object and the ground f2= 2 n; When f=3n, the frictional force between the object and the ground f3= 2 n. (g takes 10m s2).

Analysis: Static friction is the generation of friction between the surfaces of another object in contact with it (in this case, the ground) when an object has a tendency to move, but is not moving (remains at rest). The maximum static friction is the maximum amount of static friction that can be generated between the contact surfaces of two objects.

Static friction can be any value greater than zero and less than the maximum static friction, and since it is in equilibrium (at rest), the static friction must be equal to the part of the external force that causes the object to move and form a pair of equilibrium forces. For example, in this problem, the maximum static friction between the object and the ground is 2n, therefore, the static friction can be 02n, the object is already in motion, and the sliding friction is f3 = n = mg = Note: Vertical direction at rest n

mg）。

The trajectory length is the displacement of the object in that second.

Let the velocity of the stone at the upper end of the trajectory be v0, then there is.

There is s=v0*t+1 2*gt 2, you can find the initial velocity of the object in this second v0 = this velocity is the last velocity of the front, v0 2 = 2gs1, the displacement of the front s1 = the above is calculated by g = 10m s 2.

Let the velocity of the stone at the upper end of the trajectory be v1, then there is.

h1=v1t+1 2*gt (1) It is known that h1=2cm=,t=,g=uniform deceleration, so the acceleration is negative) Let the distance between the upper end of the trajectory and the starting point be s, and the known initial velocity is 0, then there is: v1 -v0 =2gs (2).

From equation (1), v1 is substituted into equation (2) to obtain s

Let's figure it out for yourself

According to the meaning of the title, it can be seen that when the car just collides with the truck, its speed is equal to the speed of the truck, and then it is less than the speed of the truck, let the acceleration be a, and the time of collision is tthen there is v0-at=v1 and v0t-1 2a('2=v1t+65.

v0 is the speed of the original car, and v1 is the speed of the truck. Substituting the data to get a=? Do the math yourself.

Suppose the acceleration is a

162km/h=

72km/h=2m/s

The car is gone and the truck is gone, then the <=

at=< obtained by the above two formulas=t<=when t=, a is the smallest.