Urgent! Urgent! Urgent! Physics Problem High Reward !

1.Camera: U 2F 2F V F into a handstand Zoom out of the real image.

Projector: 2F u f v 2F into a handstand to magnify the real image.

Magnifying glass: u f v u upright magnifying virtual image.

2.This is actually very easy, you draw the light screen, the light fixture base (a long enough straight line), the object, the convex lens, the main optical axis, and follow the steps I said in order (remind with a pencil, because objects and so on have to move constantly).

Draw a ray parallel to the main optical axis along the top of the object, and then draw a ray through the center of light, because its propagation direction is unchanged, so the two rays converge through the convex lens, and then present to the optical screen through the main optical axis, and the line between the two points is the image of the object.

2) (3) (5) (6) are all done according to what I have discussed above, remember that the data must be accurate, otherwise the error will not be conclusive!

Let's talk about (4).

Basically, it is still carried out as above, but after the drawing, the two rays of light cannot converge after the convex lens, so how does the human eye form vision? This is due to the fact that after the light rays are reversed and extended, they form an enlarged image on the same side of the object according to the reversible light path, and it is not made of actual light, it is a virtual image, and you will find that the image is upright with the object!

The knowledge of junior 2 is forgotten, it seems to be related to the virtual image and the real image, and it should be easy to understand if you make a light path diagram.

1) (2) is a clear image at u>2f,2f>v>f; Because the optical path is reversible, 2f >u>f, v>2f

4) U>F is an upright magnified virtual image, so v>u, the farther away from the convex lens, the larger the virtual image.

5) U=F, the optical path is parallel, and the light rays have no intersection point, so they are not imaging.

Density is equal to mass divided by volume.

Draw g cm to calculate.

It's 250 15=

So the density is grams per cubic centimeter.

Move to the moon and the density remains the same.

Density = mass volume = 250 grams 15 cubic centimeters = mouth calculation, please check the landlord again).

Next question, same answer.

Fifty thirds of a gram per cubic centimeter.

The density sent to that is constant.

Density is equal to mass divided by volume. You mean that gravity is different on Earth and on the Moon, right? Density is only related to mass, not gravity.

Therefore, no matter how much is cut, no matter how much is put into the **, no matter whether the shape has changed or not, as long as it is still the metal, as long as it is still in the original metal state, the density will not change.

50 3 grams per cubic centimeter, moved to the moon the mass does not change, the volume does not change, the density does not change.

The small wooden block a is subjected to a constant force to the right and a frictional force to the left, f = 10-5 * 10 * a = 1

The long plank b is subjected to the frictional force to the right, f = 5*10* a = s = 1 2at 2 s1 = s2 + 1

1 2at 2=1 2at 2+1 t=2 The displacement of block A to the horizontal plane is s1=1 2at 2+1=2, and the displacement of block b to the horizontal plane is s2=1 2at 2=1

a:2m b:1m

If AB accelerates together, the force on B is 20|3 n>5n, they will not accelerate together.

The acceleration of a is 1m|The acceleration of s 2,b is that the relative slip is 1m, and the time is 2s, and then the results are calculated separately.

Explain that the first question and the second question are the second different questions, the first question is a uniform speed, and the second question is not a uniform speed, 1) the whole method: the person and the chair are regarded as a whole, then there is 2t=g person + g object to get t=400 n

2) The method is the same, but it is used, f together = ma, when the whole is as follows: 2t-g human-g thing = (m person + m object) a, and t=440n

For people: t+fn-g people = m (people) a, the acceleration of people and the whole system is the same, so a in the two equations is the same. The solution yields fn=330n

It's an interesting question.

Taking people as the research object, then people are subjected to downward gravity mg, the support force of the chair to the person n, and the force f of the person pulling the rope, and there is a balance formula when the person moves at a uniform speed in the first question: mg=n+f=700 (ox); (1)

In the second question, when the acceleration a=1m per second, there is a balance formula: n+f=mg+ma=770 (N); (2)

When taking the chair as the research object, the chair is subjected to the downward gravitational force m*g, the pressure of the person on the chair n*, and the tensile force f of the rope, which also has a balance formula when moving at a uniform speed: f-n*=m*g=100 (Ox); (3)

In the second question, when the acceleration a=1m per second, there is a balanced formula: f-n*=m*g+m*a=110 (NN). (4)

where m=70kg, m*=10kg, n*=n

Solving simultaneous equations (1) and (3) yields: 2f = 800 (N), so at a constant velocity, the pulling force of the person pulling the rope is 400 N, and the pressure of the person on the chair is 300 N.

Solving simultaneous equations (2) and (4) obtains: 2f=880 (N), so when the acceleration a=1m per second, the pulling force of the person pulling the rope is 440 N, and the pressure of the person on the chair is: 440-n*=110, n*=330 (N).

Figure also. There is no figure on how to engage in force analysis.

Guess what.

In the second question, the reason why the athlete accelerates upwards is the combined force of gravity and the support force given by the chair, and the support force and the pressure given by the athlete are a pair of interaction forces.

The first question is that there is no ma.

(1) Perform force analysis on the node o of the rope pulling the hanging chair, and the gravity g1 of the downward hanging chair is subjected to the upward pulling force t1 and the pressure of the person on the hanging chair fn1 = g 2 (representing the gravity of the person)-t1, then t1 = g1 + fn = g1 + g2-t1 substituting the value to obtain t1 = 400

According to fn1=g2-t1 substituting the value, solve fn1=300 The first problem is solved: (2) The person and the chairlift are regarded as a whole, and the upward acceleration of the combined external force is one meter, and the tensile force of the rope is t2

You can list the formulas.

2t2-（g1+g2）=（m1+m2）a

The solution yields t2=440

Then the force analysis of the person is carried out by the upward pull of the rope, the upward support force of the chairlift, the gravity, the upward external force, and the acceleration is one meter per square second.

You can list the formulas.

t2+fn2-g2=m2 multiplied by a

The solution yields fn2=330

The second question is also easy to solve!

LZ give points.

Taking the heavy object as the research object, 100 x A = 120 x g - 100 x g is obtained to obtain a=(120-100)xg 100=2m s 2 direction upward.

m x = m x g - 120 x g, find m = 1200. It can lift up to 160kg when descending.

120*10-100*10=100*a, the solution is a=2, and the direction is downward.

The support force received by the weight is the force given to him by the athlete, the maximum is 120*10, and the weight is also subjected to a downward gravity force of 100*10, and the above relationship can be obtained according to the force balance.

In the same way, the second question: 10m+, the solution is m=96kg

Problems with overweight and weightlessness.

1.Do the acceleration movement first, and then do the uniform movement.

2.According to the title, f kv 2 = 1 4g, that is, when g = 100k constant velocity, the magnitude of gravity is equal to the magnitude of drag, f=g

f=g=kv＾2=100k

It can be seen that v=10m s

The process is too cumbersome to type, just talk about analysis.

1.Giving you the voltage and the power at that voltage will find the resistance no(p=u, square r)2Two resistors were calculated, and the rated current (U r) 3Two resistors, in series, multiplied by the smaller rated current, i.e. the total maximum voltage.

4.Know the current and the resistance of a at this time, and find the power.

1.Sound waves transmit the information and energy of the vibration of the sound source, while the sound transmission medium does not move forward with the waves.

2.The propagation of the sound-emitting body in the medium is called a mechanical wave, and all the dharma bodies are vibrating, but the vibrating object does not make a sound, and we call the sound-emitting object the sound source.

Energy direction No.

Vibration vibration is not necessarily vibration.

With a difference of 100 meters, the sound travels in seconds (because the speed of light is very fast, the time taken is about 0).

That's all it takes, 100

At 100 meters, the movement time of light is negligible, and the movement time of sound is seconds. 100 about 345 meters per second.