Ask 1 difficult physics question if you are 1 high, and ask a high school physics question, which is

1) Let the velocity be v1 when it reaches A and v2 when it is d1 from the hole. It is known that v1d1 = v2d2, so v2 = v1d1 d2.

2) Since it is a uniform deceleration motion, the average velocity from d1 to d2 is (v1+v2) 2, so the time taken from A to B (d2-d1) ((v1+v2) 2) = 2(d2-d1) (v1+v2).

Correction: Thanks somero for the reminder, because d = a v (a = d1v1 = d2v2), so time t = d v = a v 2, the square of the velocity is inversely proportional to the time, not a uniform deceleration motion.

t1 = a/(v1^2), t2=a/(v2^2)

So: the time taken from A to B t2-t1 = a (1 v2 2 - 1 v1 2) = (d1v1) (1 v2 2 - 1 v1 2), and it is enough to substitute v2 of 1) solution. I won't go into details.

Obviously not a uniform deceleration movement.

From the inscription, S is proportional to 1 V, let the scale coefficient be A, then S is the y-axis, and the image of 1 V as the X-axis is a straight line through the origin, and it is easy to calculate V2=V1D1 D2, and the time from D1 to D2 is the trapezoidal area between 1 V1 and 1 V2 below the graph, that is, T=(D1+D2)(D2-D1) 2V1D1

Summary. When the volume decreases, the density of molecules increases, but the average kinetic energy of molecules is not necessarily large, so the number of molecules colliding with the unit area of the vessel wall per unit time does not necessarily increase. So a mistake.

When the temperature increases, the average kinetic energy of the molecules becomes larger, but the density of the molecules is not necessarily larger, so the number of molecules colliding with the unit area of the device wall per unit time does not necessarily increase.

It is not difficult to ask a high school physics question.

The core types of high school physics include projectile motion problems, circular motion problems, celestial motion excavation problems, locomotive chain closure start problems, and energy-centered comprehensive application problems. Physics should not only pay attention to the usual accumulation of learning, but also pay attention to maintaining a good attitude and skills when answering questions.

Click here to see it. Why c

Pro, the pressure depends on the number of molecules that collide with the unit area of the device wall and the average kinetic energy of the molecule per unit time, the pressure does not change as a sail, the temperature and volume change, and the molecular level belt changes the average kinetic energy of the slag and hail, then the number of molecules that collide with the unit area of the device wall per unit time must change Therefore, c is correct.

When the volume decreases, the density of molecules increases, but the average kinetic energy of molecules is not necessarily large, so the number of molecules colliding with the unit area of the device wall does not necessarily increase during the single disadvantage rapid position time. So a mistake. When the temperature increases, the average kinetic energy of the molecules becomes larger, but the density of the molecules is not necessarily larger, so the number of molecules that collide with the unit surface of the wall does not necessarily increase per unit time.

Isn't that possible with volume and temperature accelerating the kinetic energy of the molecule and decreasing the kinetic energy unchanged?

The kinetic energy of a is eventually completely converted into other energy, and as long as the loss is calculated, its initial kinetic energy can be found, and the velocity can be found.

Let the masses be m, the velocity v before the collision, the velocity v after the collision, and the velocity v during separation, then:

mv = mgl1 (a return process, v can be found ) m + m) v m + m) v = (2m) g (2l ) (the process of colliding to the separation process ab moving together, because the spring is still the original length, the elastic potential energy does not change, so it is not displayed, v can be found).

mv = 2mv (momentum is conserved during the collision, v can be found) mv0 - mv = mgl1 (the motion process of a before the collision can be found v0) Combining these equations, v0 can be solved

Without taking into account the loss of energy. It is obtained from the conservation of kinetic energy.

Since the collision of the two sliders at the beginning, the kinetic energy is converted into elastic potential energy, and later the elastic potential energy is converted into kinetic energy. So springs are used to confuse you,

Solution: Before the lower end of the stone enters the water, the two stones are regarded as a whole, and they work together to do free fall motion, that is: v=g·t (1) After the lower end of the stone enters the water, the upper end of the stone continues to do accelerated motion, that is: s=v·t gt 2 (2) from the title, s=3m, t=,g 10m s, from (2) obtain: v=

v=g·tt=∴h′=gt²/2=

h=h′＋s=

Draw an image of v t, set the time of the rock below t1, and the time of the top t2 bridge deck h

h-3=gt1^2/2

h=gt2^2/2

t2-t1=

The joint solution of the three formulas can obtain h

Because both stones start to fall freely from a standstill, so there is no tension in the rope during the fall, and we set the speed of the stone falling into the water is v1, and the speed of the stone falling into the water is v2

Get: v22-v12=2gs

v2=v1+gt

where g s t is known to be v2, and again:

v22-02=2g（h-s）

H=

Both stones are in free fall, take the gravitational acceleration g=10m s, and set the falling time of the stones at the lower end of the rope t, according to the title.

g((t + solution t=s)

The distance from the bridge deck to the water surface h=(gt 2)+3=m)).

I haven't done a physics problem for a long time, try it Solution: Let the height of the water surface from the bridge be h, and the time of the second stone falling into the water is t, then there is h= 1 2gt2 h—3 1 2g(t 0 1)2 From the second equation, the time t can be calculated, and then the distance will be calculated.

1. The work done to overcome its own gravity is 50*10*24=12000J, and the power = 12000 120=100W(Units are not included in the calculation).

2. The density of blood is about the same as that of water, and the work that the heart needs to do = 12000J 40% = 30000J

3. The heart can deliver 70cm3 of blood every 1 contraction, and the muscles do the work. The heart contracts (i.e., heart rate) 30000j 2 times per minute.

If each second is divided into three segments, it is 9 segments of the same time, and the displacement ratio of each segment is 1:3:5:

13:15:17, the total is 81 parts, and the displacement of each copy is s 81, then the time of the last 1 3 in the 2nd second is the 6th time, the displacement is 11s 81, and the time after 1 3 in the 3rd second is the 9th time, and the displacement is 17s 81.