Senior 3 physics questions ask the experts to help answer them

Let the time be t, the horizontal distance of the ball from the inclined plane is s, and the ball leaves the inclined plane to move to the maximum distance (you don't mean not to fall on the inclined plane, then fall to the lowest part of the inclined plane) The height of the fall is h, then: s=vt; h=gt*t/2；tanθ=h/s;So t=2vtan g

2.The velocity at which the ball leaves the inclined plane and lands is v, i.e., the combined velocity (see **), and the change in momentum = mv-mv

3.The work done by gravity w=mgh; There is no h here, so considering the increment of kinetic energy, the increase in kinetic energy is completely obtained by the work done by gravity. So the work done by gravity = final kinetic energy - initial kinetic energy; w=(m*v*v-m*v*v)/2

4.The average power done by gravity p=w t, brings in the results obtained above.

You also didn't say that the inclined plane is placed on the ground, if the ball starts to move at the horizontal speed of v, the ball is doing parabolic motion, and it will definitely not move on the inclined plane, if there is a maximum distance, isn't it that the ball happens to move to the bottom of the inclined plane, I didn't upload it to you, the ball below is virtual, I didn't see it, saying that it is just a statement that the ground is true, so can it be changed to this now?

Upstairs is teaching in chaos...

To prove that I'm not talking nonsense!

I'm here to answer that question.

In fact, the most important thing in this question is to find his critical position.

So what is a critical position?

Borrow his picture.

It's actually very simple! That is, when the direction of velocity is also parallel to the inclined plane, and this is the time when the maximum distance is reached, (I've done this before, and I'm right!). Then there's the question of how to calculate, according to the formula of flat throwing motion, the ratio of velocity is twice the ratio of displacement, (I believe you know) and then according to the mathematical geometry, you are trying, I believe you can make it!

If you don't understand, you can ask me!

When a group of hydrogen atoms in the ground state is irradiated by a certain monochromatic light, it can only emit three monochromatic lights, A, B and C, indicating that a group of hydrogen atoms transitions to the excited state of n=3 when they are irradiated by a certain monochromatic light.

The wavelength of methyl light is the shortest, eA=e3-e1=

The wavelength of propylene light is the longest, epropylene = e3-e2 =

The length of the seat belt is 5 m, so the displacement of the tension process from the fall to the straightening of the seat belt is 5 m.

The velocity v=10m s is obtained by v = 2gs, and then the safety belt cushion makes the worker suspended, and the impulse theorem is obtained vertically upward.

f-mg）t=mv

The average impulse f=1100n is obtained

r=mv bq v, the larger r is.

t=2*pi*m bq is not rate-independent.

As long as you turn the same angle, the time is the same. I choose a CD

But you have to know that there is a component of the Q inclined plane on P at the beginning, so that it can be balanced, in fact, after the second due to the reduction of the strong spring elastic force, the acceleration of P is less than the acceleration of Q, and at the beginning of this second is the process of the acceleration brought by the spring to P is greater than the acceleration brought by F to Q.

Whoever said that at the beginning of the pq as the whole A=F (M1+M2), it is just a moment, and I personally think that the breakthrough point is in seconds.

Looking at the instantaneous resultant external force of q afterburner alone is f, q acceleration is indeed f q, and the force of this p on q is equal to the force of q strict inclination downward, but this is not a complete expression. The functional relation of the force on q should be the supporting force of f+p - the force of q downwards along the inclined plane. The initial f is the smallest, and with the elongation of the spring, the supporting force of p to q gradually decreases, and finally becomes 0, so it is necessary to gradually increase f to maintain a uniform acceleration motion, so the subsequent constant force is the maximum value of f.

The specific expression should be calculated in detail on paper, which is a little busy, and I can tell you if you need it.

The moment the force is applied, the support force of P to Q also changes immediately. Otherwise, q has acceleration and p has zero acceleration, and the two are separated.