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In fact, this kind of problem is free fall, so it is a linear motion with a uniform velocity of 0.

According to the title. Drawings can be drawn.

5)|_4)|_3)|_2)|_1)|According to the title, the distance between the fourth drop and the fifth drop is 1mNote, in the uniform speed movement, as long as you know any 3 quantities, all the remaining quantities can be obtained].

So we now know that his initial velocity = 0m s

Acceleration = acceleration due to gravity =

Displacement = 1m can be based on the formula.

s=vot+1/2at^2

Can you understand how to find the time t?

And then the title says that one drop falls at an equal time.

Explain that the time taken for each of his displacements is the same.

As you can see from the diagram, there are four displacements.

So let's use the t*4 I just calculated

Get the total time.

And then according to the formula s=vot+1 2at 2, you can bring it in.

So you can find the displacement, which is the height of the house.

Summary] The title of free fall gives two implicit conditions.

1.The initial velocity is 0 2Acceleration is the acceleration of gravity = so don't ignore it when doing the problem, and drawing can simplify your thinking process.

I'm sorry, because it's too late, because it's very tight, I hope you can understand.

Draw a V-T diagram, and the displacement ratio in a continuous equal time interval is an odd ratio, and it will come out at a glance. 16m

s = gt^2/2

Let the time interval t between two drops of water

h = s1 = g(4t)^2/2 = 8gt^2s4 = gt^2/2 = 1

h = 8*2 = 16[m]

Let each drop of water be formed with a height of t and l

1/2gt^2=

The magnitude of friction is f= n

Moving at a constant speed along the horizontal plane, then the horizontal component of the tensile force is equal to the frictional force. i.e. fcos = f

Vertically, n=mg-fsin

From these formulas the magnitude f of the tensile force is obtained

mgμ/(cosθ

sinθ).

The work done by the tensile force is required to be in the vertical direction, without displacement, the work done is 0 in the horizontal direction, and the force is fcos *s=mg s (1tan ), and the work is fcos *s=mg s (1tan ).

The force in the horizontal direction is FCOS

The force in the vertical direction is fsin

So the pressure on the ground is n=mg-fsin

w=fcosαs=ns

Analysis: The force analysis of object a when moving under the action of horizontal force can be obtained.

mgcos +fsin = n1, fcos -f1-mgsin = ma1, f1 = n1, synthesis can be solved: a1 = 4m s, the force f is withdrawn after moving 2 meters, at this time, the velocity is v=4m s, because when the force is removed, the object a has velocity, the cave is auspicious, the object a does an upward deceleration motion, to the speed is 0, this process can be matched by the force analysis of the object a, mgcos = n2, mgsin +f2 = ma2.

f2 = n2, the simultaneous solvable mega fight: a2 =, so the time required for deceleration to 0 is t2 = v a1 = 10 21s, and the deceleration is 0 displacement is s1 = 20 21m, so the total displacement of the object a on the inclined plane s=2 + 20 21 = 62 21 meters, and then accelerates downward along the inclined plane. To the low end of the inclined plane, the force analysis of the object is carried out.

mgcos = n3, mgsin -f3=ma3, f3 = n3, the simultaneous equation can be solved: a3=, therefore, the descent time is t3= [2s) (a3)], 310 189)s, therefore, the time for the problem is t=t2+t3=10 21+ (310 189) seconds.

Solution: Before returning to the core and dismantling the bottom, it is necessary to walk 2m

After F is removed, the component force of gravity 30N slides downward, and the component of closed combustion and gravity 40N produces a sliding friction car virtual force of 12N, so the combined external force is 18N, and it is downward along the inclined plane. The acceleration is, t

1. Decompose the thrust and gravity along the direction of the inclined plane and the perpendicular direction of the inclined plane, then the force in the direction of the inclined plane is Duan Daqing 80 (thrust) - 30 (gravity) - 30 (friction) = 20 (n).

2. From the work fs=1 2mv2, find the speed v=43, when there is no push grip strength, find the acceleration of the downward lift vertical a=(30-40*4, s=vt+1 2at2,2=, find t

Why can't I see the title?? **I can't see how to get this??

You have a problem with this question. Take a look:

1. What kind of arrows do you mean by arrows? The arrows of the resultant force? Or is it a circle (ball or ring?) Arrows moving in the direction of the silver? Different arrows, methods of analysis and conclusions are completely different!

2. The friction factor actually refers to the coefficient of friction, right? During my studies, I used the word coefficient of friction, and the "factor" was difficult to understand and easy to lead to ambiguity. Or is your "factor" another combustion?

3. What is the dynamic friction coefficient of rod 1? It's too big, isn't it? I seem to have learned that the coefficient of friction cannot be greater than or equal to 1.

4. When rod 2 turns to a 60-degree angle to rod 1, the circle rolls down - judging by my analysis. The rolling type is carried out in a clockwise direction. So the far center is moving vertically downward, and the rest of the part is rolling downward. Which part of the question is the direction of motion?

So it can be said that this is an unsolvable problem.

The score given is too low, right?

Not imaging.

U represents the object distance, V represents the image distance, and F represents the focal length.

When u f, that is, when the object distance is less than one time the focal length, it becomes an upright virtual image with magnification. (magnifying glass) when u=f, that is, when the object distance is equal to one time the focal length, no image is imaging. This one is best to memorize.

When f u 2f, that is, the object is in.

When the focal length is between one and two times, it becomes a magnified inverted real image. When U=2F, that is, when the object distance is equal to twice the focal length, it is an inverted real image of equal size.

When U 2F, that is, when the object distance is greater than twice the focal length, it becomes a reduced inverted real image. (Camera) notation: double the focal length is divided into virtual and real, and double the focal length is divided into size.

When it is a real image, the object is close and far away, and it is like a large image. (The object distance decreases, the image distance increases, the image becomes larger), and the object is far away and small than the near image.

When it is a virtual image, the object is far away and far like and large, and the object is close and small.

Unable to image, see below

Object distance (u) Imaging properties Application.

u>2f Handstand Zoom out Real Image camera.

u=2f Handstand Equal Size Real Image Equal Large Image Measurement Focal Length.

2f>u>f Inverted magnification Real image Slide projectors, projectors, projectors.

u=f does not image searchlights, photocopiers.

u

When the switch S1 is closed and S2 is disconnected, the current is IA when the sliding vane of the rheostat R0 is at the A end, and the current is IB. at the B end

The ratio of the number of voltage representations is 4:3

ia*ra:ib*rb=4：3

The power ratio of R1 is 4:9

ia^2:ib^2=4:9---ia:ib=2:3

In summary, ra:rb=2:1 --

Disconnect S1, close S2, slide vane at B, R2 and R0 have a total power of 9W

Let the circuit current at this point be ib' ,ib'^2)*(r2+rb)=9---

When the sliding vane is at a, set the current to ia'

ia'^2)*r2=2---

ia':ib'=(r2+rb):(r2+2*rb)--

The above four forms are combined, and the IA is eliminated'、ib'Get.

7*r2^2+r2*rb-8*rb^2=0

The solution is r2=rb or r2=(-8 7)*rb (rounded).

The power of r0 at a is (ia'^2）*ra=4*rb/r2=4w

PS: My process is a bit complicated, let's see an easier way.

If the power of R2 is used, the relationship between R2 and Rb can also be obtained, but the value is huge and meaningless.

The main purpose of this question is to find the relationship between R2 and Rb, and there are only a few formulas used in it, which is meaningless.

The power supply voltage is constant, when the switch S1 is closed, S2 is disconnected, and the sliding blade of the sliding rheostat R0 moves from A to B, the ratio of the voltmeter sequence is 4:3, and the power ratio of R1 is 4:9, keep the sliding vane at B, disconnect S1, close S2, then the total power of R2 and R0 is 9W, and then move the sliding blade P to A, the power of R2 is 2W, and the power of R0 is found at this time.

Answer: (Draw four equivalent circuit diagrams when looking at the problem).

Set: The currents in the four cases are: i1, i2, i3, i4

R1 is unchanged. p1/p2=（i1²·r1）/（i2²·r1）=i1²/i2²=4/9

i1/i2=2/3

ua/ub=（i1·ra）/（i2·rb）=4/3

ra/rb=2/1 ra=2rb---1)

and R2 is connected in series with RA.

p2’/pa=r2/ra---2）

The power supply voltage does not change.

p2+pb）/(p2'+pa)=(r2+ra)/(r2+rb)--3)

Substituting (1), (2), (P2+PB)=9W, P2'=2W into (3) obtains:

7r2²+r2·ra-8ra²=0---4）

Solve equation (4).

r2=-8rb 7 (discarded if it does not fit the topic).

r2=rbra=2r2

R2 is connected in series with RA.

pa=4w

Picture in**? The phone can't see it.