Is it possible to prepare for the exam in physics on your own, having only access to the Internet? There is always a chance. About what to do and in what order, tells the author of the textbook “Physics. Complete course preparation for the exam ”I. V. Yakovlev.

Self-preparation for the exam in physics begins with the study of theory. Without this, it is impossible to learn how to solve problems. It is necessary first, taking any topic, thoroughly understand the theory, read the relevant material.

Take the topic "Newton's Law". You need to read about inertial reference systems, learn that forces add up in vectors, how vectors are projected onto an axis, how this can work in a simple situation - for example, on an inclined plane. It is necessary to learn what the friction force is, how the sliding friction force differs from the static friction force. If you do not distinguish between them, then most likely you will be mistaken in the corresponding task. After all, tasks are often given in order to understand certain theoretical points, therefore, the theory must be dealt with as clearly as possible.

For a complete mastering of the course of physics, we recommend you the textbook IV Yakovlev "Physics. A full course of preparation for the Unified State Exam ”. You can purchase it or read the materials online on our website. The book is written in simple and understandable language. It is also good in that the theory in it is grouped precisely according to the points of the USE codifier.

And then you have to take on tasks.
First step. To begin with, take the simplest problem book, and this is Rymkevich's book. You need to solve 10-15 problems on the chosen topic. In this collection, the tasks are quite simple, in one or two steps. You will understand how to solve problems on this topic, and at the same time you will remember all the formulas that are needed.

When you are preparing for the Unified State Exam in Physics on your own, you do not need to specially cram formulas and write cheat sheets. All this is effectively perceived only when it comes through solving problems. Rymkevich's problem book, like no other, meets this primary goal: to learn how to solve simple tasks and at the same time learn all the formulas.

Second phase. It's time to move on to training specifically for tasks of the exam... It is best to prepare using the wonderful manuals edited by Demidova (Russian tricolor on the cover). These collections are of two types, namely, collections of standard options and collections of thematic options. It is recommended to start with thematic options. These collections are structured as follows: first, there are options only for mechanics. They are arranged in accordance with the structure of the exam, but the tasks in them are only in mechanics. Then - the mechanics is fixed, thermodynamics is connected. Then - mechanics + thermodynamics + electrodynamics. Then optics, quantum physics are added, after which 10 full-fledged versions of the exam are given in this manual - on all topics.
Such a manual, which includes about 20 thematic options, is recommended as a second step after Rymkevich's problem book for those who independently prepare for the exam in physics.

For example, it can be a collection
“Unified State Exam Physics. Thematic examination options ". M.Yu. Demidova, I.I. Nurminsky, V.A. Mushrooms.

Similarly, we use collections in which typical examination options are selected.

Stage three. If time permits, it is highly desirable to reach the third step. This is training for the tasks of Phystech, a higher level. For example, the book of problems by Bakanina, Belonuchkin, Kozel (publishing house "Education"). The tasks of such collections seriously exceed USE level... But in order to successfully pass the exam, you need to be ready a couple of steps higher - for a variety of reasons, up to the banal self-confidence.

Don't be limited only USE grants... After all, it is not a fact that the tasks will be repeated on the exam. There may be problems that were not previously encountered in the collections of the exam.

How to allocate time when self-preparation for the exam in physics?
What to do when you have one year and 5 big topics: mechanics, thermodynamics, electricity, optics, quantum and nuclear physics?

The maximum amount - half of the entire preparation time - should be devoted to two topics: mechanics and electricity. These are the dominant topics, the most difficult ones. Mechanics is taught in grade 9 and students are considered to know it best. But actually it is not. The mechanics tasks are as difficult as possible. And electricity is a difficult topic in itself.
Thermodynamics and molecular physics is a fairly simple topic. Of course, there are also pitfalls here. For example, schoolchildren have a poor understanding of saturated couples. But on the whole, experience shows that there are no such problems as in mechanics and electricity. Thermodynamics and molecular physics at the school level is a simpler section. And the main thing is that this section is autonomous. It can be studied without mechanics, without electricity, it is on its own.

The same can be said about optics. Geometric optics is simple - it boils down to geometry. You have to learn the basics about thin lenses, the law of refraction, and that's it. Wave optics (interference, light diffraction) is present in the USE in minimal amounts. Compilers of variants do not give any difficult problems in the exam on this topic.

And there remains quantum and nuclear physics. Schoolchildren are traditionally afraid of this section, and in vain, because it is the simplest of all. The last problem from the final part of the exam - on the photoelectric effect, light pressure, nuclear physics - is easier than others. You need to know the Einstein equation for the photoelectric effect and the law of radioactive decay.

In the version of the exam in physics there are 5 problems where you need to write a detailed solution. The peculiarity of the exam in physics is that the complexity of the problem does not increase with the growth of the number. You never know what problem will be difficult in the exam in physics. Sometimes mechanics are difficult, sometimes thermodynamics. But traditionally the task in quantum and nuclear physics is the simplest.

You can prepare for the exam in physics on your own. But if there is even the slightest opportunity to contact a qualified specialist, then it is better to do it. Schoolchildren, preparing for the exam in physics on their own, run the risk of losing a lot of points on the exam, simply because they do not understand the strategy and tactics of preparation. The specialist knows which way to go, but the student may not know this.

We invite you to our exam preparation courses in physics. A year of classes means mastering a physics course at a level of 80-100 points. Success in preparing for the exam!

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Preparation for the exam and exam

The average general education

UMK line A.V. Grachev. Physics (10-11) (basic, advanced)

UMK line A.V. Grachev. Physics (7-9)

UMK line A.V. Peryshkin. Physics (7-9)

Preparation for the exam in physics: examples, solutions, explanations

Lebedeva Alevtina Sergeevna, physics teacher, work experience 27 years. Certificate of Merit from the Ministry of Education of the Moscow Region (2013), Letter of Gratitude from the Head of the Resurrection municipal district(2015), Diploma of the President of the Association of Teachers of Mathematics and Physics of the Moscow Region (2015).

The work presents tasks of different levels of difficulty: basic, advanced and high. Tasks basic level, these are simple tasks that test the assimilation of the most important physical concepts, models, phenomena and laws. Tasks of an advanced level are aimed at testing the ability to use the concepts and laws of physics to analyze various processes and phenomena, as well as the ability to solve problems on the application of one or two laws (formulas) for any of the topics of the school physics course. In work 4, the tasks of part 2 are tasks of a high level of complexity and test the ability to use the laws and theories of physics in a changed or new situation. The implementation of such tasks requires the application of knowledge from two three sections of physics at once, i.e. high level of training. This option totally coincides demo version USE 2017, tasks are taken from the open bank of USE tasks.

The figure shows a graph of the dependence of the speed module on time t... Determine the path covered by the car in the time interval from 0 to 30 s.

Solution. The distance traveled by a car in the time interval from 0 to 30 s is easiest to define as the area of ​​a trapezoid, the bases of which are the time intervals (30 - 0) = 30 s and (30 - 10) = 20 s, and the height is the speed v= 10 m / s, i.e.

Answer. 250 m.

A load weighing 100 kg is lifted vertically upward using a rope. The figure shows the dependence of the speed projection V load on the upward axle from time t... Determine the modulus of the cable tension during the ascent.

Solution. According to the graph of the dependence of the projection of speed v load on an axle directed vertically upward, from time t, you can define the projection of the acceleration of the load

The load is influenced by: the force of gravity directed vertically downward and the tension force of the rope directed vertically upward along the rope, see fig. 2. Let us write down the basic equation of dynamics. Let's use Newton's second law. The geometric sum of the forces acting on the body is equal to the product of the body's mass by the acceleration imparted to it.

+ = (1)

Let us write the equation for the projection of the vectors in the frame of reference connected with the earth, the OY axis is directed upwards. The projection of the tensile force is positive, since the direction of the force coincides with the direction of the OY axis, the projection of the gravity is negative, since the force vector is oppositely directed to the OY axis, the projection of the acceleration vector is also positive, so the body moves with acceleration upward. We have

T – mg = ma (2);

from formula (2) the modulus of the tensile force

T = m(g + a) = 100 kg (10 + 2) m / s 2 = 1200 N.

Answer... 1200 N.

The body is dragged along a rough horizontal surface at a constant speed, the modulus of which is 1.5 m / s, applying force to it as shown in figure (1). In this case, the modulus of the sliding friction force acting on the body is 16 N. What is the power developed by the force F?

Solution. Imagine a physical process specified in the problem statement and make a schematic drawing indicating all the forces acting on the body (Fig. 2). Let's write down the basic equation of dynamics.

Tr + + = (1)

Having chosen a frame of reference associated with a fixed surface, we write down the equations for the projection of vectors onto the selected coordinate axes. According to the condition of the problem, the body moves uniformly, since its speed is constant and equal to 1.5 m / s. This means that the acceleration of the body is zero. Two forces act horizontally on the body: sliding friction force tr. and the force with which the body is dragged. The projection of the friction force is negative, since the force vector does not coincide with the direction of the axis NS... Force projection F positive. We remind you that to find the projection, we lower the perpendicular from the beginning and end of the vector to the selected axis. With this in mind, we have: F cosα - F tr = 0; (1) express the projection of the force F, this is F cosα = F tr = 16 N; (2) then the power developed by the force will be equal to N = F cosα V(3) We make a substitution, taking into account equation (2), and substitute the corresponding data into equation (3):

N= 16 N 1.5 m / s = 24 W.

Answer. 24 watts

The load, fixed on a light spring with a stiffness of 200 N / m, makes vertical vibrations. The figure shows a plot of the dependence of the displacement x cargo from time to time t... Determine what the mass of the load is. Round your answer to the nearest whole number.

Solution. A spring loaded weight vibrates vertically. According to the graph of the dependence of the displacement of the load NS from time t, we define the period of fluctuations of the load. The oscillation period is T= 4 s; from the formula T= 2π we express the mass m cargo.

Answer: 81 kg.

The figure shows a system of two lightweight blocks and a weightless rope, with which you can balance or lift a load weighing 10 kg. Friction is negligible. Based on the analysis of the above figure, select two correct statements and indicate their numbers in the answer.

1. In order to keep the load in balance, you need to act on the end of the rope with a force of 100 N.
2. The block system shown in the figure does not give a power gain.
3. h, you need to stretch out a section of rope with a length of 3 h.
4. In order to slowly raise the load to a height hh.

Solution. In this task, it is necessary to recall simple mechanisms, namely blocks: a movable and fixed block. The movable block gives a twofold gain in strength, whereby the section of the rope must be pulled twice as long, and the stationary block is used to redirect the force. In operation, simple mechanisms of winning do not give. After analyzing the problem, we immediately select the necessary statements:

1. In order to slowly raise the load to a height h, you need to pull out a section of rope with a length of 2 h.
2. In order to keep the load in balance, you need to act on the end of the rope with a force of 50 N.

Answer. 45.

An aluminum weight, fixed on a weightless and inextensible thread, is completely immersed in a vessel with water. The cargo does not touch the walls and bottom of the vessel. Then an iron weight is immersed in the same vessel with water, the weight of which is equal to the weight of the aluminum weight. How will the modulus of the tension force of the thread and the modulus of the force of gravity acting on the load change as a result?

1. Increases;
2. Decreases;
3. Doesn't change.

Solution. We analyze the condition of the problem and select those parameters that do not change during the study: these are the body mass and the liquid into which the body is immersed on threads. After that, it is better to perform a schematic drawing and indicate the forces acting on the load: the tension force of the thread F control directed upward along the thread; the force of gravity directed vertically downward; Archimedean force a acting on the submerged body from the side of the liquid and directed upwards. According to the condition of the problem, the mass of the loads is the same, therefore, the modulus of the gravity force acting on the load does not change. Since the density of cargo is different, the volume will also be different.

The density of iron is 7800 kg / m 3, and the density of aluminum is 2700 kg / m 3. Hence, V f< V a... The body is in equilibrium, the resultant of all forces acting on the body is zero. Let's direct the coordinate axis OY up. The basic equation of dynamics, taking into account the projection of forces, is written in the form F control + F a – mg= 0; (1) Express the pulling force F control = mg – F a(2); Archimedean force depends on the density of the fluid and the volume of the submerged part of the body F a = ρ gV p.h.t. (3); The density of the liquid does not change, and the volume of the iron body is less V f< V a, therefore, the Archimedean force acting on the iron load will be less. We draw a conclusion about the modulus of the thread tension force, working with equation (2), it will increase.

Answer. 13.

Block weight m slides off a fixed rough inclined plane with an angle α at the base. The block acceleration modulus is a, the speed modulus of the bar increases. Air resistance is negligible.

Establish a correspondence between physical quantities and formulas with which they can be calculated. For each position of the first column, select the corresponding position from the second column and write down the selected numbers in the table under the corresponding letters.

B) Coefficient of friction of the bar on an inclined plane

3) mg cosα

Solution. This task requires the application of Newton's laws. We recommend making a schematic drawing; indicate all the kinematic characteristics of the movement. If possible, depict the acceleration vector and vectors of all forces applied to the moving body; remember that the forces acting on the body are the result of interaction with other bodies. Then write down the basic equation of dynamics. Choose a frame of reference and write down the resulting equation for the projection of the vectors of forces and accelerations;

Following the proposed algorithm, we will make a schematic drawing (Fig. 1). The figure shows the forces applied to the center of gravity of the bar and the coordinate axes of the reference system associated with the surface of the inclined plane. Since all forces are constant, the movement of the bar will be equally variable with increasing speed, i.e. the acceleration vector is directed towards the movement. Let's choose the direction of the axes as shown in the figure. Let's write down the projections of the forces on the selected axes.

Let's write down the basic equation of dynamics:

Tr + = (1)

Let us write this equation (1) for the projection of forces and acceleration.

On the OY axis: the projection of the support reaction force is positive, since the vector coincides with the direction of the OY axis N y = N; the projection of the friction force is zero since the vector is perpendicular to the axis; the projection of gravity will be negative and equal mg y= – mg cosα; acceleration vector projection a y= 0, since the acceleration vector is perpendicular to the axis. We have N – mg cosα = 0 (2) from the equation we express the force of the reaction acting on the bar, from the side of the inclined plane. N = mg cosα (3). Let's write projections onto the OX axis.

On the OX axis: force projection N equal to zero, since the vector is perpendicular to the OX axis; The projection of the friction force is negative (the vector is directed in the opposite direction relative to the selected axis); the projection of gravity is positive and equal to mg x = mg sinα (4) from a right triangle. Acceleration projection positive a x = a; Then we write equation (1) taking into account the projection mg sinα - F tr = ma (5); F tr = m(g sinα - a) (6); Remember that the friction force is proportional to the normal pressure force N.

A-priory F tr = μ N(7), we express the coefficient of friction of the bar on the inclined plane.

We select the appropriate positions for each letter.

Answer. A - 3; B - 2.

Task 8. Oxygen gas is in a 33.2 liter vessel. The gas pressure is 150 kPa, its temperature is 127 ° C. Determine the mass of gas in this vessel. Express your answer in grams and round to the nearest whole number.

Solution. It is important to pay attention to the conversion of units to the SI system. We convert the temperature to Kelvin T = t° С + 273, volume V= 33.2 l = 33.2 · 10 -3 m 3; We translate the pressure P= 150 kPa = 150,000 Pa. Using the ideal gas equation of state

express the mass of the gas.

Be sure to pay attention to the unit in which you are asked to write down the answer. It is very important.

Answer. 48 g

Task 9. An ideal monatomic gas in the amount of 0.025 mol adiabatically expanded. At the same time, its temperature dropped from + 103 ° С to + 23 ° С. What kind of work did the gas do? Express your answer in Joules and round to the nearest whole number.

Solution. First, the gas is a monoatomic number of degrees of freedom i= 3, secondly, the gas expands adiabatically - this means without heat exchange Q= 0. Gas does work by decreasing internal energy. Taking this into account, we write the first law of thermodynamics in the form 0 = ∆ U + A G; (1) express the work of the gas A r = –∆ U(2); The change in the internal energy for a monatomic gas can be written as

Answer. 25 J.

The relative humidity of a portion of air at a certain temperature is 10%. How many times should the pressure of this portion of air be changed in order for its relative humidity to increase by 25% at a constant temperature?

Solution. Questions related to saturated steam and air humidity are most often difficult for schoolchildren. Let's use the formula to calculate the relative humidity

According to the condition of the problem, the temperature does not change, which means that the saturated vapor pressure remains the same. Let us write down formula (1) for two states of air.

φ 1 = 10%; φ 2 = 35%

Let us express the air pressure from formulas (2), (3) and find the pressure ratio.

Answer. The pressure should be increased by 3.5 times.

The hot substance in the liquid state was slowly cooled in a constant power melting furnace. The table shows the results of measurements of the temperature of a substance over time.

Choose from the list provided two statements that correspond to the results of the measurements carried out and indicate their numbers.

1. The melting point of the substance under these conditions is 232 ° C.
2. In 20 minutes. after the start of measurements, the substance was only in a solid state.
3. The heat capacity of a substance in a liquid and solid state is the same.
4. After 30 min. after the start of measurements, the substance was only in a solid state.
5. The crystallization process of the substance took more than 25 minutes.

Solution. As the substance cooled, its internal energy decreased. The temperature measurement results allow you to determine the temperature at which the substance begins to crystallize. As long as the substance passes from a liquid to a solid state, the temperature does not change. Knowing that the melting point and crystallization temperature are the same, we choose the statement:

1. The melting point of the substance under these conditions is 232 ° C.

The second true statement is:

4. After 30 minutes. after the start of measurements, the substance was only in a solid state. Since the temperature at this point in time is already below the crystallization temperature.

Answer. 14.

V isolated system body A has a temperature of + 40 ° C, and body B has a temperature of + 65 ° C. These bodies are brought into thermal contact with each other. After some time, thermal equilibrium has come. How did the temperature of body B and the total internal energy of body A and B change as a result?

For each value, determine the corresponding change pattern:

1. Increased;
2. Decreased;
3. Hasn't changed.

Write down the selected numbers for each physical quantity in the table. The numbers in the answer may be repeated.

Solution. If in an isolated system of bodies there are no energy transformations except for heat exchange, then the amount of heat given off by bodies, the internal energy of which decreases, is equal to the amount of heat received by bodies, the internal energy of which increases. (According to the law of conservation of energy.) In this case, the total internal energy of the system does not change. Problems of this type are solved based on the heat balance equation.

where ∆ U- change in internal energy.

In our case, as a result of heat exchange, the internal energy of body B decreases, which means that the temperature of this body decreases. The internal energy of body A increases, since the body has received the amount of heat from body B, then its temperature will increase. The total internal energy of bodies A and B does not change.

Answer. 23.

Proton p, flown into the gap between the poles of the electromagnet, has a speed perpendicular to the magnetic induction vector, as shown in the figure. Where is the Lorentz force acting on the proton directed relative to the figure (up, towards the observer, from the observer, down, left, right)

Solution. The magnetic field acts on a charged particle with the Lorentz force. In order to determine the direction of this force, it is important to remember the mnemonic rule of the left hand, not to forget to take into account the particle charge. We direct four fingers of the left hand along the velocity vector, for a positively charged particle, the vector should enter the palm perpendicularly, the thumb set at 90 ° shows the direction of the Lorentz force acting on the particle. As a result, we have that the Lorentz force vector is directed away from the observer relative to the figure.

Answer. from the observer.

The modulus of the electric field strength in a 50 μF flat air capacitor is 200 V / m. The distance between the capacitor plates is 2 mm. What is the charge of a capacitor? Write down the answer in μC.

Solution. Let's convert all units of measurement to the SI system. Capacitance C = 50 μF = 50 · 10 -6 F, distance between plates d= 2 · 10 –3 m. The problem deals with a flat air capacitor - a device for accumulating electric charge and electric field energy. From the formula for electrical capacity

where d Is the distance between the plates.

Express the tension U= E d(4); Substitute (4) in (2) and calculate the capacitor charge.

q = C · Ed= 50 · 10 –6 · 200 · 0.002 = 20 μC

Pay attention to the units in which you need to write the answer. We got it in pendants, but we represent it in μC.

Answer. 20 μC.

The student conducted an experiment on the refraction of light, presented in the photograph. How does the angle of refraction of light propagating in glass and the refractive index of glass change with increasing angle of incidence?

1. Is increasing
2. Decreases
3. Does not change
4. Write down the selected numbers for each answer in the table. The numbers in the answer may be repeated.

Solution. In tasks of this kind, we remember what refraction is. This is a change in the direction of propagation of a wave when passing from one medium to another. It is caused by the fact that the wave propagation speeds in these media are different. Having figured out from which medium to which light it propagates, we write the law of refraction in the form

where n 2 - the absolute refractive index of the glass, the medium where the light goes; n 1 is the absolute refractive index of the first medium from which the light is coming. For air n 1 = 1. α is the angle of incidence of the beam on the surface of the glass semi-cylinder, β is the angle of refraction of the beam in the glass. Moreover, the angle of refraction will be less than the angle of incidence, since glass is an optically denser medium - a medium with great indicator refraction. The speed of propagation of light in glass is slower. Please note that the angles are measured from the perpendicular restored at the point of incidence of the ray. If you increase the angle of incidence, then the angle of refraction will also increase. The refractive index of the glass will not change from this.

Answer.

Copper jumper at a point in time t 0 = 0 begins to move at a speed of 2 m / s along parallel horizontal conductive rails, to the ends of which a 10 Ohm resistor is connected. The entire system is in a vertical uniform magnetic field. The resistance of the lintel and rails is negligible, the lintel is always perpendicular to the rails. The flux Ф of the magnetic induction vector through a circuit formed by a jumper, rails and a resistor changes over time t as shown in the graph.

Using the graph, select two correct statements and include their numbers in the answer.

1. By the point in time t= 0.1 s, the change in magnetic flux through the circuit is 1 mVb.
2. Induction current in the jumper in the range from t= 0.1 s t= 0.3 s max.
3. The EMF modulus of the induction arising in the circuit is 10 mV.
4. The strength of the induction current flowing in the jumper is 64 mA.
5. To maintain the movement of the bulkhead, a force is applied to it, the projection of which on the direction of the rails is 0.2 N.

Solution. According to the graph of the dependence of the flux of the magnetic induction vector through the circuit on time, we determine the sections where the flux Ф changes, and where the flux change is zero. This will allow us to determine the time intervals in which the induction current will occur in the circuit. Correct statement:

1) By the time t= 0.1 s the change in magnetic flux through the circuit is equal to 1 mWb ∆F = (1 - 0) · 10 –3 Wb; The EMF modulus of induction arising in the circuit is determined using the EMR law

Answer. 13.

According to the graph of the dependence of the current strength on time in an electric circuit, the inductance of which is 1 mH, determine the EMF modulus of self-induction in the time interval from 5 to 10 s. Write down the answer in μV.

Solution. Let's translate all the quantities into the SI system, i.e. the inductance of 1 mH is converted into H, we get 10 –3 H. The current shown in the figure in mA will also be converted into A by multiplying by 10 –3.

The EMF formula of self-induction has the form

in this case, the time interval is given according to the condition of the problem

∆t= 10 s - 5 s = 5 s

seconds and according to the graph we determine the interval of current change during this time:

∆I= 30 · 10 –3 - 20 · 10 –3 = 10 · 10 –3 = 10 –2 A.

Substituting numerical values ​​into formula (2), we obtain

| Ɛ | = 2 · 10 –6 V, or 2 µV.

Answer. 2.

Two transparent plane-parallel plates are tightly pressed against each other. A ray of light falls from the air onto the surface of the first plate (see figure). It is known that the refractive index of the upper plate is equal to n 2 = 1.77. Establish a correspondence between physical quantities and their values. For each position of the first column, select the corresponding position from the second column and write down the selected numbers in the table under the corresponding letters.

Solution. To solve problems on the refraction of light at the interface between two media, in particular, problems on the transmission of light through plane-parallel plates, the following order of solution can be recommended: make a drawing indicating the path of rays going from one medium to another; at the point of incidence of the ray at the interface between the two media, draw a normal to the surface, mark the angles of incidence and refraction. Pay special attention to the optical density of the media under consideration and remember that when a light beam passes from an optically less dense medium to an optically denser medium, the angle of refraction will be less than the angle of incidence. The figure shows the angle between the incident ray and the surface, but we need the angle of incidence. Remember that the angles are determined from the perpendicular restored at the point of incidence. We determine that the angle of incidence of the beam on the surface is 90 ° - 40 ° = 50 °, the refractive index n 2 = 1,77; n 1 = 1 (air).

Let's write the law of refraction

Let's construct an approximate path of the ray through the plates. We use formula (1) for the boundaries 2–3 and 3–1. In the answer we get

A) The sine of the angle of incidence of the beam on the boundary 2–3 between the plates is 2) ≈ 0.433;

B) The angle of refraction of the ray when crossing the boundary 3–1 (in radians) is 4) ≈ 0.873.

Answer. 24.

Determine how many α - particles and how many protons are produced as a result of a thermonuclear fusion reaction

+ → x+ y;

Solution. In all nuclear reactions, the laws of conservation of electric charge and the number of nucleons are observed. Let us denote by x - the number of alpha particles, y - the number of protons. Let's make the equations

+ → x + y;

solving the system, we have that x = 1; y = 2

Answer. 1 - α -particle; 2 - proton.

The modulus of the momentum of the first photon is 1.32 · 10 –28 kg · m / s, which is 9.48 · 10 –28 kg · m / s less than the modulus of the momentum of the second photon. Find the energy ratio E 2 / E 1 of the second and first photons. Round your answer to tenths.

Solution. The momentum of the second photon is greater than the momentum of the first photon by the condition, it means that we can represent p 2 = p 1 + Δ p(1). The energy of a photon can be expressed in terms of the momentum of a photon using the following equations. it E = mc 2 (1) and p = mc(2), then

E = pc (3),

where E- photon energy, p- photon momentum, m - photon mass, c= 3 · 10 8 m / s - the speed of light. Taking into account formula (3), we have:

We round the answer to tenths and get 8.2.

Answer. 8,2.

The nucleus of the atom has undergone radioactive positron β - decay. How did the electric charge of the nucleus and the number of neutrons in it change as a result?

For each value, determine the corresponding change pattern:

1. Increased;
2. Decreased;
3. Hasn't changed.

Write down the selected numbers for each physical quantity in the table. The numbers in the answer may be repeated.

Solution. Positron β - decay in an atomic nucleus occurs during the transformation of a proton into a neutron with the emission of a positron. As a result, the number of neutrons in the nucleus increases by one, the electric charge decreases by one, and the mass number of the nucleus remains unchanged. Thus, the transformation reaction of the element is as follows:

Answer. 21.

In the laboratory, five experiments were carried out to observe diffraction using various diffraction gratings. Each of the gratings was illuminated with parallel beams of monochromatic light with a specific wavelength. In all cases, the light was incident perpendicular to the grating. In two of these experiments, the same number of main diffraction maxima were observed. First indicate the number of the experiment in which a diffraction grating with a shorter period was used, and then the number of the experiment in which a diffraction grating with a longer period was used.

Solution. Diffraction of light is the phenomenon of a light beam in the area of ​​a geometric shadow. Diffraction can be observed when on the path of the light wave there are opaque areas or holes in large and opaque obstacles, and the sizes of these areas or holes are commensurate with the wavelength. One of the most important diffraction devices is a diffraction grating. The angular directions to the maxima of the diffraction pattern are determined by the equation

d sinφ = kλ (1),

where d Is the period of the diffraction grating, φ is the angle between the normal to the grating and the direction to one of the maxima of the diffraction pattern, λ is the light wavelength, k- an integer called the order of the diffraction maximum. Let us express from equation (1)

When choosing pairs according to the experimental conditions, we first select 4 where a diffraction grating with a shorter period was used, and then the number of the experiment in which a diffraction grating with a long period was used is 2.

Answer. 42.

Current flows through the wirewound resistor. The resistor was replaced with another one, with a wire of the same metal and the same length, but having half the cross-sectional area, and half the current was passed through it. How will the voltage across the resistor and its resistance change?

For each value, determine the corresponding change pattern:

1. Will increase;
2. Will decrease;
3. Will not change.

Write down the selected numbers for each physical quantity in the table. The numbers in the answer may be repeated.

Solution. It is important to remember on what values ​​the resistance of the conductor depends. The formula for calculating the resistance is

Ohm's law for a section of the circuit, from formula (2), we express the voltage

U = I R (3).

According to the condition of the problem, the second resistor is made of wire of the same material, the same length, but a different cross-sectional area. The area is half the size. Substituting in (1), we get that the resistance increases by 2 times, and the current decreases by 2 times, therefore, the voltage does not change.

Answer. 13.

The period of oscillation of a mathematical pendulum on the surface of the Earth is 1, 2 times longer than the period of its oscillation on a certain planet. What is the modulus of acceleration of gravity on this planet? The influence of the atmosphere in both cases is negligible.

Solution. A mathematical pendulum is a system consisting of a thread, the dimensions of which are many more sizes the ball and the ball itself. Difficulty can arise if Thomson's formula for the period of oscillation of a mathematical pendulum is forgotten.

T= 2π (1);

l- the length of the mathematical pendulum; g- acceleration of gravity.

By condition

Let us express from (3) g n = 14.4 m / s 2. It should be noted that the acceleration of gravity depends on the mass of the planet and the radius

Answer. 14.4 m / s 2.

A straight conductor 1 m long, through which a current of 3 A flows, is located in a uniform magnetic field with induction V= 0.4 T at an angle of 30 ° to the vector. What is the modulus of the force acting on the conductor from the side of the magnetic field?

Solution. If you place a conductor with current in a magnetic field, then the field on the conductor with current will act with the force of Ampere. We write the formula for the modulus of the Ampere force

F A = I LB sinα;

F A = 0.6 N

Answer. F A = 0.6 N.

The energy of the magnetic field stored in the coil when a direct current is passed through it is equal to 120 J. How many times must the current flowing through the coil winding be increased in order for the stored magnetic field energy to increase by 5760 J.

Solution. The magnetic field energy of the coil is calculated by the formula

By condition W 1 = 120 J, then W 2 = 120 + 5760 = 5880 J.

Then the ratio of currents

Answer. The current strength must be increased by 7 times. In the answer form, you enter only the number 7.

The electrical circuit consists of two light bulbs, two diodes and a coil of wire, connected as shown. (The diode only passes current in one direction, as shown at the top of the figure). Which of the bulbs will light up if the north pole of the magnet is brought closer to the loop? Explain the answer by indicating what phenomena and patterns you used when explaining.

Solution. The magnetic induction lines leave the north pole of the magnet and diverge. As the magnet approaches, the magnetic flux through the coil of wire increases. According to Lenz's rule, the magnetic field created by the induction current of the loop must be directed to the right. According to the rule of the gimbal, the current should flow clockwise (when viewed from the left). A diode in the circuit of the second lamp passes in this direction. This means that the second lamp will light up.

Answer. The second lamp comes on.

Aluminum spoke length L= 25 cm and cross-sectional area S= 0.1 cm 2 suspended on a thread at the upper end. The lower end rests on the horizontal bottom of a vessel into which water is poured. Length of the submerged spoke l= 10 cm. Find the force F, with which the needle presses on the bottom of the vessel, if it is known that the thread is located vertically. The density of aluminum ρ a = 2.7 g / cm 3, the density of water ρ b = 1.0 g / cm 3. Acceleration of gravity g= 10 m / s 2

Solution. Let's make an explanatory drawing.

- Thread tension force;

- Force of reaction of the bottom of the vessel;

a - Archimedean force acting only on the immersed part of the body, and applied to the center of the immersed part of the spoke;

- the force of gravity acting on the spoke from the Earth and is applied to the center of the entire spoke.

By definition, the weight of the spoke m and the modulus of the Archimedean force are expressed as follows: m = SLρ a (1);

F a = Slρ in g (2)

Consider the moments of forces relative to the suspension point of the spoke.

M(T) = 0 - moment of tension force; (3)

M(N) = NL cosα is the moment of the reaction force of the support; (4)

Taking into account the signs of the moments, we write the equation

considering that according to Newton's third law, the reaction force of the bottom of the vessel is equal to the force F d with which the spoke presses on the bottom of the vessel, we write N = F e and from equation (7) we express this force:

Substitute numerical data and get that

F d = 0.025 N.

Answer. F d = 0.025 N.

A container containing m 1 = 1 kg nitrogen, exploded in strength test at temperature t 1 = 327 ° C. What is the mass of hydrogen m 2 could be stored in such a container at a temperature t 2 = 27 ° С, having a fivefold safety factor? Molar mass nitrogen M 1 = 28 g / mol, hydrogen M 2 = 2 g / mol.

Solution. Let us write the equation of state of the ideal gas of Mendeleev - Clapeyron for nitrogen

where V- the volume of the cylinder, T 1 = t 1 + 273 ° C. By condition, hydrogen can be stored at pressure p 2 = p 1/5; (3) Taking into account that

we can express the mass of hydrogen by working directly with equations (2), (3), (4). The final formula is:

After substitution of numeric data m 2 = 28 g.

Answer. m 2 = 28 g.

In an ideal oscillatory circuit, the amplitude of the current fluctuations in the inductor I m= 5 mA, and the amplitude of the voltage across the capacitor U m= 2.0 V. At the time t the voltage across the capacitor is 1.2 V. Find the current in the coil at this moment.

Solution. In an ideal oscillatory circuit, the vibration energy is stored. For the moment of time t, the energy conservation law has the form

For the amplitude (maximum) values, we write

and from equation (2) we express

Substitute (4) into (3). As a result, we get:

I = I m (5)

Thus, the current in the coil at the moment of time t is equal to

I= 4.0 mA.

Answer. I= 4.0 mA.

There is a mirror at the bottom of the reservoir 2 m deep. A ray of light, passing through the water, is reflected from the mirror and comes out of the water. The refractive index of water is 1.33. Find the distance between the point of entry of the beam into the water and the point of exit of the beam from the water, if the angle of incidence of the beam is 30 °

Solution. Let's make an explanatory drawing

α is the angle of incidence of the beam;

β is the angle of refraction of the ray in water;

AC is the distance between the point of entry of the beam into the water and the point of exit of the beam from the water.

According to the law of refraction of light

Consider a rectangular ΔADB. In it AD = h, then DВ = АD

We get the following expression:

Substitute the numerical values ​​into the resulting formula (5)

Answer. 1.63 m.

In preparation for the exam, we suggest that you familiarize yourself with a working program in physics for grades 7–9 for the line of the UMK Peryshkina A.V. and working program of an in-depth level for grades 10-11 for the educational complex Myakisheva G.Ya. The programs are available for viewing and free download for all registered users.

Physics is a fairly complex subject, so preparation for the USE in physics 2020 will take a fair amount of time. In addition to theoretical knowledge, the commission will check the ability to read circuit diagrams and solve problems.

Consider the structure of the examination paper

It consists of 32 tasks distributed over two blocks. For understanding, it is more convenient to arrange all the information in the table.

The whole theory of the exam in physics by sections

• Mechanics. This is a very large, but relatively simple section that studies the motion of bodies and the interactions between them occurring at the same time, including dynamics and kinematics, conservation laws in mechanics, statics, oscillations and waves of a mechanical nature.
• Molecular physics. In this topic, special attention is paid to thermodynamics and molecular kinetic theory.
• Quantum physics and components of astrophysics. These are the most difficult sections that cause difficulties both during study and during testing. But also, perhaps, one of the most interesting sections. Here knowledge is tested on such topics as physics of the atom and atomic nucleus, particle-wave dualism, astrophysics.
• Electrodynamics and special theory of relativity. Here you cannot do without studying optics, the basics of SRT, you need to know how an electric and magnetic field works, what a direct current is, what are the principles of electromagnetic induction, how electromagnetic oscillations and waves arise.

Yes, there is a lot of information, the volume is very decent. In order to successfully pass the exam in physics, you need to be very good at the entire school course in the subject, and it has been studied for five years. Therefore, it will not be possible to prepare for this exam in a few weeks or even a month. You need to start now so that you feel calm during the tests.

Unfortunately, the subject of physics causes difficulties for many graduates, especially for those who have chosen it as a major subject for admission to a university. Effective study this discipline has nothing to do with memorizing rules, formulas and algorithms. In addition, it is not enough to assimilate physical ideas and read as much theory as possible; you need to be proficient in mathematical technique. Often, poor mathematical training does not allow a student to pass physics well.

How do you prepare?

Everything is very simple: choose a theoretical section, read it carefully, study it, trying to understand all physical concepts, principles, postulates. After that, reinforce the preparation by solving practical problems on the chosen topic. Use online tests to test your knowledge, this will allow you to immediately understand where you are making mistakes and get used to the fact that a certain time is given to solve the problem. We wish you good luck!