How to do physics problems in the second year of junior high school? Physics questions in the second

If you want to learn physics well, you need to understand the internal logic of physics, and it is not advisable to memorize formulas.

If you want to really learn physics well, the key is to be interested and understanding.

It's best to have a good understanding of common physical models ... Have a full interest in physics ...

Physics in the second year of junior high school is actually very simple, but this is science, so you have to understand, you can't just memorize by rote, learning this boy has an advantage, I studied like this when I was in the second year of junior high school, and I could get a full score in the exam. Not bragging.

It is recommended that you take a good look at the relevant example problems before doing each problem, understand them thoroughly, and find out the idea of solving the problem. When you do the problem independently, follow the steps of the example problem solving, develop the habit of solving the problem systematically, and don't solve the problem with a hammer in the east and a stick in the west.

Make sure to get the problem in the future, and immediately have a rough way to solve the problem.

You should develop the habit of solving problems independently, even with the help of other students, you must do it again independently.

Don't engage in sea tactics.

Learn physics to see more, ask more, more hands-on, physics is a very experimental discipline, you not only have to watch the teacher do a variety of demonstration experiments, but also do a lot of experiments by yourself, some experiments can be done at home because of the simplicity, it is recommended that you find some popular science physics books to look at first, improve their understanding of physics, interest, do not rigidly memorize definitions, memorize formulas, set of formulas to solve problems.

I can only briefly say it, I hope it will help you a little bit.

In the second year of junior high school, physics mainly focuses on market concepts, memorizing formulas and drawings, and some physics names in books, which should also be mainly concepts.

When ABC moves in a uniform linear motion under the action of F1=6N in Figure A, A is in equilibrium as a uniform linear motion, and there are two possibilities, one is the balance of the force in the horizontal direction, (without any contact with the object). The second is that there is no force, so a is not subject to friction at this time, that is, f=0. The answer is yes.

The whole is only subjected to two forces, the tensile force f1 = 6n, and the sliding friction between the whole and the ground due to the uniform linear motion f = 6n. And then according to the sliding friction formula:

f=n, the coefficient of friction =f n=f (mA+mb+mc)g

6/（3+2+1）g

The answer is wrong.

The gravitational force of b and the supporting force on it are not a pair of equilibrium forces. Because B has A pressed on it, it is supported by a force greater than its own gravity.

When B is pulled to the right by F2 in uniform motion, as shown in the figure above, the rope pull force (red) to the left of A and C is equal, F. =f。*

The mutual friction between abs (blue) is equal f1 = f1 * = n * = mag = 3 10 .

The mutual friction between BC (green) is also equal: f2 = f2 * = (ma + mb) g = (3 + 2) 10

The answer is: a (.)

f2=f1+f2=3+5=8n。

c is acted upon by 3 forces. Red rally f. * To the left, f. *=f。=f1=3n, the green friction f2*=5n to the right, you can see the static friction f of the ground experienced by c = f2*-f. *=5-3=2n。

This question can also be done in the second year of junior high school, but this question can be more clear in high school.

In Figure A, although the three blocks move in a straight line at a uniform speed to the right, there is a leftward static friction between block A and block B, as well as between block B and block C, and the magnitude of friction is equal to F1, so it is false, and the answer can be selected by the elimination method as B

, gravity and support are opposite in the vertical direction, and block b is at rest in the vertical direction, so gravity and support are a pair of balanced forces, correct.

With , the friction factor is not given in the question, and the magnitude of f2 cannot be directly calculated, and the direction of the friction force can be obtained from the force analysis.

1 cubic meter = 1 * 10 to the 6th power cubic centimeter, that is, 500 centimeters 3 = m3;

Alcohol mass = alcohol density * volume = >3*

The density of the alcohol does not change depending on the poured part, so the density does not change;

The mass of the water = the density of the water * the volume of the bottle = >3*

The first one is empty.

The second one is empty. The third is empty.

m alcohol = alcohol · v =

=400（g）

Density is an intrinsic property of a substance that does not change due to changes in shape, volume, position, etc.

So, even pour off half of the alcoholDensity stillm water = water · v =

500（g）

The direction of the propeller discharge should be against the current, that is, opposite to the direction of the current. This counteracts the force of the ocean current on the column and maintains the balance of the column.

Assuming that the direction of the current is from left to right, then the force of the current on the column is to the right. To counteract this force, the propeller has to drain water to the left, i.e. against the direction of the current. In this way, the reaction force generated by the displacement of the propeller can be equal to but opposite to the force of the ocean current, so that the resultant force on the column is zero and the balance is maintained.

Common types of questions in physics include force calculations, simple circuit diagrams, and fluctuating sound (light). The question type of computation is generally based on Newton's question.

The application of related concepts such as the one, two, and three laws and the resultant force, and the simple circuit diagram is calculated by applying basic concepts such as Ohm's law. For the fluctuation question, you need to understand the basic formulas of fluctuation, such as wavelength, wave velocity and frequency. When doing the questions, you need to pay attention to the requirements of the questions, pay attention to the units and symbols of the data, and do not be hasty and hasty.

Through repeated practice and consolidation of the basic knowledge and calculation methods of physics courses, the performance and level of physics learning can be effectively improved.

The study of physics in the second year of junior high school mainly includes mechanics, optics, heat, electricity, etc. In mechanics, students learn about the laws of motion of objects, Newton's three laws, and the synthesis and decomposition of forces. In optics, students will learn the definition of light, the propagation, reflection, and refraction of light.

In thermal, students learn about temperature, heat, heat transfer, and thermal expansion. In terms of electricity, students will learn the principles, circuit basics, capacitors, resistors, inductors and other common electrical components. In the second year of junior high school physics learning, a lot of observation and practice, theory and practice are required to cultivate students' hands-on ability and ability to solve practical problems, such as measuring experiments, designing experiments, etc.

Physics is a basic science, which is of great significance for cultivating students' scientific literacy.

The main learning content of Physics in the second year of junior high school includes the basic concepts of physics, kinematics, mechanics, energy conversion and mechanical waves. Studying physics can enable students to better understand the laws of nature and improve their thinking skills and scientific literacy when solving practical problems. For example, studying kinematics can help us better design the training of athletes in order to understand the trajectory and velocity changes of objects in space

See the figure below for details, click to enlarge:

First of all, mass = volume x density.

1. The mass is the same, the density of A is 2 times that of B, then the volume of A is 1 2 of B;

2. The volume is the same, the density of A is twice that of B, and the mass of A is also twice that of B.

These two questions are the same thing, mass m = v

From the formula, m is equal, the density and volume are inversely proportional, the density is large, the volume is small, the density is several times larger, the volume is several times smaller.

The volume is equal, the density is proportional to the mass, and the density is several times larger, and the mass is several times larger. ，

<> the total weight of the water level is kilograms, usually kilograms, then you can push out the water is 2 kilograms. The density of water with water is known to be one gram per cubic meter. Knowing the weight and density, then the volume can be found, and the volume found is the volume of water.

It is also the volume of the bottle, so the first question is found.

The solution to Problem 2 is to use an equation. Whether it's liquid or water, the volume they put in the bottle is the same, so use the volume as a medium to make their density compare with the mass.

The density of water is equal to the mass of the water on the density of the liquid. And the weight of the liquid and the weight of the bottle are also known in the question, so subtract the weight of the bottle from the weight of the liquid and the weight of the bottle, and the rest is the weight of the liquid. Then the equation can be easily listed.

First of all, according to the formula of density equal to mass divided by volume to calculate the volume of the bottle, the second question can be calculated in the same way, and the specific steps will be sent to you in the form of **.

The density of water is one cubic meter and one ton, and the cup can hold 2 kilograms, and the volume is cubic meters. Volume 2000ml. Density is equal to mass divided by volume, so density is equal to 800 kilograms per cubic meter.

The spring dynamometer reads equal to the force at the hook, and since the hook is being held in the diagram, the force at the hook is exactly equal to the gravitational force of the spring dynamometer.

B dynamometer is used upside down, the spring bears both its own weight and the weight of the cow hook yard, and the spring dynamometer has a self-weight of 1 New, so the B dynamometer indicates that the number is Ox.

A dynamometer is used normally, its own weight is borne by the iron frame, the spring of a dynamometer bears the weight of B dynamometer and the weight of two ox hook yards, so a dynamometer shows 1 ox + cow + cow = 2 n. So choose C

A dynamometer is used normally, therefore, there is no problem with the reading of a nail dynamometer.

B The dynamometer is used upside down, so that it will appear, and when there is no object hanging on the pull ring, the dynamometer will also have an indication. This indication is the gravitational force of the dynamometer itself.

It's just two fools, it's good to measure it directly, it's really stupid. Full Solution:

Solution: The first sound is propagated by the railroad track, and the second time is the air propagation Let the rail length s, then: s 340m s—s 5000m s 2s >s.

This is just a written expression, there are steps on the scratch paper: write the above formula in the form of a formula, the common multiple of 340 and 5000 is 85000, multiply 85000 on both sides at the same time, remove the denominator, and get: 233 s 17 s 170000, you can find s.

The first time I heard it was from the railroad tracks, and the second time it was from the air.

Let the distance of this section of the track be s

s (340m s) - s (5000m s) = 2 seconds.

can find s = meters.