An 8th grade physics problem, thank you, I want it tonight, and the bounty can be increased in 5 min

Known: v1=340m s; v2=5200m s to find: the length of the steel pipe l=?

Solution: According to v=l t, we get: t=l v

So, t1 = l v1 and t2 = l v2

t1-t2=2

That is, l v1-l v2=2, that is, l 340-l 5200=2 solution: l

The velocity transferred in iron minus the velocity in air multiplied by time.

The air propagation velocity is v1, and the steel pipe propagation velocity is v2 (v1v1*(t+2)=l

v2*t=l

The solution is: l=2*v1*v2 (v2-v1), t=2*v1 (v2-v1).

Known: v1=340m s; v2=5200m s to find: the length of the steel pipe l=?

Solution: According to the theme and common sense, the first sound should come from the iron pipe, and the second sound should come from the air.

According to v=l t, t=l v

So, t1 = l v1 and t2 = l v2

t1-t2=2

That is, l v1-l v2=2, that is, l 340-l 5200=2 solution: l

Answer: Slightly.

Set the length of the straight iron pipe to l Ah, ah, ah, a lot, the answer is meters.

Let the length of the iron pipe be L

2s = l v empty - l v iron.

2s=l 340m s-l 5000m sl 170m s-l 2500m s-1s=0 to get l=

Let the length of the iron pipe be l:l v air - l v iron = 2

Before, it would. It's been 20 years. I forgot.

Slide at a constant speed f2=

Uniform velocity rise f pull = 30n, this is the positive solution.

1. Analyze Figure A.

Gravity = support force (pressure) i.e.: mg = f

f1 = friction i.e.: f1 = f = 3n friction = 1 5 times the pressure, i.e.: f = 1 5f

In summary: mg=15n

2. Analyze Figure B.

Both forces in the horizontal direction are f2 in magnitude

Downward gravity in the vertical direction = 15n

The two contact surfaces give it two upward frictional forces, and if the sum of these two frictional forces is 15N, then the wooden block A slides at a uniform speed.

2f=mgf=1/5f2

It can be obtained: f2=

3. If the wooden block A slides up at a uniform speed, it is subject to a vertical downward gravity of 15N, and the friction of the two downward strokes totals 15N

Then the additional tensile force: t=mg+2f=30n

Figure A: When moving at a uniform velocity, the frictional force experienced by the object is f=3n. So the pressure is: 3*5=15n

f=f1=un

3=15uu=1/5=

mg=15 B figure: The weight of the wooden block cannot be ignored, otherwise there is no force on the wooden board, and the wooden block will not slide at a uniform speed. Let the mass of the wooden block be m and the afterburner needs to be f2, then:

mg=(uf2)/2

f2=2mg/u=2mg/

In Figure A, the wooden block moves in a straight line at a uniform velocity, so the two forces are balanced in the horizontal direction, g=f, =1 5, f=3n, so the gravity of the wooden block g=15n

In Figure B, the gravity of the wooden block must be considered, otherwise there is no need to do this problem. And this question is very loose, the wooden board against the wall should be fixed to the wall according to his meaning, otherwise the friction between the two should be considered.

Now, the problem can be looked at like this: the friction between two planks and the block caused by pressure is balanced by the gravity of the block, 2 f=g, =1 5, g=15n, so, f=.

The second question, give the upward pull force of the wooden block, then the friction force is the resistance, f=g+2 f, g=15n, =1 5, f=, so, f=30n.

Drawing, force analysis.

The friction force between the wooden block and the wooden board is the multiple of the pressure between them, and when the friction coefficient = a is placed on the wooden board, it can be moved at a uniform speed with a tensile force of 3n, and there is f = g = 3n so the gravitational force of a a is 15n

If two identical planks are used to clamp A in the vertical direction, in addition to its own gravity, when A slides at a uniform speed, it also has a long-lasting friction force due to the extrusion of the planks at both ends, that is, 2f=2(f)=g=15

f=yes, I don't know, I'm asking the squad leader to drop! It should be a drop.

From the question, we can know that = object weight mg = 15n

When placed in the vertical direction: 2 plates have 2 friction forces, 2f=mg, f=, and then calculate the pressure as fn=

Pull up: F-Mg-2F=0

F-pull = 30n

The weight of the wooden block is not counted cuowu, but the weight of the wooden board is not counted!! Death, let's take a look at the next question. Common sense, the teacher will not say that the wooden block is not counted, it is not a question!!

f=umg=f1 u=f n=1 5 get m=

Slide at a constant rate, 2f1=mg, f1=uf2 to get f2=

Slide up at a constant speed, and the pulling force f=mg+2f2=2mg=30n (because f2=uf2 =f1).

With a tensile force of 3n, it can be made to move in a straight line at a uniform speed f1=f=3n

Block A slides at a constant speed (block weight is not counted) f2=f=3n

If you make the block a slide up at a uniform speed f3=f=3n

If the weight of the wooden block is not considered, the vertical is the same as the horizontal need 3N, and the vertical is two sides, so it is 6N.

From Figure A, it is calculated that the wooden block g = 15 N, (1) when sliding at a uniform speed, the sum of the two frictional forces is equal to the wooden block g, so the friction of one surface is, the pressure is, (2) the pulling force is the downward friction plus the wooden block g, that is, 30n

I hope it helps you.,And the question is a wooden board that doesn't weigh it.。。。 If the wooden block does not weigh ... What is that...

by the friction of two planks.

That is, 6n, but the wooden block has gravity - 15n

Gravity, friction and tension can rise at a uniform speed when the three are balanced.

Hence the tensile force is 15-6=9n

Tell me how many answers you have.

Strange question: Why is there friction in front of the wooden block, regardless of the weight of the block? When horizontally? Still want 3n?

Figure A: When moving at a uniform velocity, the frictional force experienced by the object is f=3n. So the pressure is: 3*5=15n at this time the pressure = gravity, so mg=15n

Figure B: The weight of the wooden block cannot be ignored, otherwise the wooden board will not be strengthened, and the wooden block will not slide at a uniform speed. Let the mass of the wooden block be m and the afterburner needs to be f2, then:

mg=2*(uf2)

f2=mg/2u=mg/

Let the mass of the block be m and the required tensile force is f, then: f=mg+2uf2=15n+15n=30n

Solution: According to "the friction between the block and the plank is 1 5 of the pressure between them. It is known that when placed on a horizontal plank, it can be made to move in a straight line at a uniform speed with a tensile force of 3N" to obtain: f=1 5mg=3(1).

According to "two identical planks clamping a in the vertical direction, one of them is against the wall, as shown in Figure B, and F is applied to the other block to keep the wooden block A sliding at a uniform speed", we get: f=1 5*fn*2=mg(2).

To make a slide up at a uniform speed, then: f=f+mg=2mg(3) synthesis(1)(2)(3) can get the answer.

This kid is too cunning, and he already has a lot of correct answers, how can you people be blinded for 200 wealth, will he give you wealth so easily? The product of exam-oriented education is really helpless......Boy, let me tell you, the constant speed of the descent is 15N, and the constant speed upward force is 30N, and others don't answer the ......In answer is a fool.

This topic? Does the block weigh less? Where does the friction on the horizontal plank come from?

From p=ui and i=u r, r=u p is obtained to find the resistance of the lamp:

r1=u²/p1=(6v)²/3w=12ω，r2=u²/p2=(6v)²/9w=4ω；

Find the total resistance in series from r=r1+r2:

r=r1+r2=12ω+4ω=16ω；

Find the current of the series circuit by i=u r:

i=u/r=8v/16ω=；

The actual power of the two lamps is obtained by p=ui and i=u r, respectively

p'=i²r1=(，p''=i²r2=(。

R1=U1 Pi=(6V) 3W=12 R2=U2 P2=(6V) 9W=4 Total resistance R=R1+R2=16 when connected to the 8V power supply, the current is I=U R=8V 16 = actual power P1'=i²r1=(

p2'=i²r2=(

Within 1 times the focal length, the closer the object is to the focal point, the closer the image is to the lens, the smaller the size of the image is outside the focal length of 1 times the focal length, the real image, the closer the object is to the focal point, the farther the image is from the lens, the larger the size of the image, the size of the two focal points (the virtual and real refers to the image, the size refers to the magnification, reduction) The distant image of the object inside one focal point becomes larger, and the distant image of the object outside the object becomes smaller in one focal point.

If the focal point is used as a reference point, the closer the object is to the focal point, the farther the image is from the lens——, the size of the image becomes larger

The imaging law of convex lens refers to the fact that the object is placed outside the focus and becomes an inverted real image on the other side of the convex lens, and there are three types of real images: reduction, equal size, and magnification. The smaller the object distance, the larger the image distance, and the larger the real image. The object is placed in focus, and the virtual image is magnified upright on the same side of the convex lens.

The smaller the object distance, the smaller the image distance, and the smaller the virtual image.

The first is near, and the second is large.

We divide the total distance by the planned average speed to get 90km divided by 60km h= which is the target hour reached.

The first half of the road is 45km, and the average speed is 50kmh, which can be obtained from the first half of the road 45km divided by 50km h=

So the second half of the road has to be done within 45km divided by the average speed of 75km h for the second half of the journey

90km / 60km/h=

90km/2=45km

45km / 50km/h=

Then if you want to reach the place at the scheduled time, then the remaining time is 45km

So the average speed of the rest of the road is 75 km h

t=s v=,t1=s 2v1=, then the second half of the journey takes t2=,v=s 2t2=75km h

1) The total volume of the stone = the total volume of the bottle - the total volume of the water in the bottle. Volume of water = mass of water Density of water = 2*10 -4.

So the total volume of the stone = 1*10 -4

2) The density of the stone = the mass of the stone The volume of the stone = cubic meters.

1. There is a layer of oil above the red soup, and the density of the oil is smaller than that of water, so it floats on the water, and when it is heated, because the oil layer is on top, it hinders the evaporation of the water below, and evaporation is an endothermic process, so that the red part boils faster because the evaporation absorbs less heat; Consommé, on the other hand, is caused by the direct evaporation of water, which absorbs and takes away a lot of heat, resulting in delayed boiling.

It corresponds to 4 squares, and 82 corresponds to 80 squares, so 2 82 occupies a total of 76 squares.

That is, every 80 corresponds to 76 squares.

Therefore, the temperature of each cell is: 80 76 20 19 ( ) The actual temperature indicated by the "0" grid is: 2 4 20 19 that is, the starting point is:

y=20x/

When x=26:

y=20×26/19－

When x=y:

x=20x/

That is: x 19 = x

1. Due to the oil layer on the side of the red soup halogen, it is difficult for the soup to evaporate before boiling (the boiling point of the oil is more than 200 degrees Celsius, and the water vapor condenses into a liquid in the oil layer), therefore, although the red soup halo and the clear soup halogen are heated the same, the external heat release is much more than the clear soup halogen, therefore, the red soup halogen boils before the clear soup halone!

Let y=ax+b, substitute y=2, x=4 and y=82, x=80 to get it.

a=20/19，b=-42/19

i.e. y=20 19x-42 19---1)2) substituting x=26 into Eq. 1).

y=20 19*26-42 19=478 19=degrees3) When the indication of this thermometer is equal to the actual temperature, that is, y=x, then there is x=20 19x-42 19

Solve the above equations.

y=x=42 degrees.

1.The flame on both sides of the pot body is the same, and the heat absorbed by the soup stock is about the same in the same time However, because there is a layer of oil above the liquid surface on one side of the red soup, the evaporation is slowed down and less heat is taken away The quality, main ingredients and initial temperature of the soup on both sides are almost the same, so the temperature of the red soup rises quickly and boils first

2: (1) Because the scale is uniform, the actual temperature increase is the same when the thermometer increases by the same indication According to this, it can be tentatively written that y=ax (a is a constant) But according to the actual situation described in the question, when x=0 y cannot also be 0, let y=b at this time, so the relationship between y and x should be y=ax+b in form

Since y1=2 is due to x1=4; When x1=80 and y1=82, these two sets of data are substituted into the formula.

4℃×a+b=2℃

80℃×a+b=82℃

Solving this simultaneous equation yields a= b=

That is: y=2) x=26 to get y=25, which means that when the thermometer shows 26, the actual temperature is 25

3) Let y=x be substituted and x= will be solved