Federal news. Preparing for the exam in physics: examples, solutions, explanations Preparing for the exam in physics: examples, solutions, explanations

The number of participants in the USE in physics in 2018 (main day) was 150,650 people, including 99.1% of graduates of the current year. The number of participants in the exam is comparable to the previous year (155,281 people), but lower than the number in 2016 (167,472 people). In percentage terms, the number of participants in the USE in physics amounted to 23% of the total number of graduates, which is slightly lower than last year. A slight decrease in the number of students taking the USE in physics may be due to an increase in universities that accept informatics as an entrance test.

The largest number of USE participants in physics is noted in Moscow (10,668), the Moscow Region (6,546), St. Petersburg (5,652), the Republic of Bashkortostan (5,271) and the Krasnodar Territory (5,060).

The average USE score in physics in 2018 was 53.22, which is comparable to last year's score (53.16 test scores). The maximum test score was scored by 269 exam participants from 44 constituent entities of the Russian Federation, in the previous year there were 278 people with 100 points. The minimum USE score in physics in 2018, as in 2017, was 36 TB, but in primary scores it was 11 points, compared to 9 primary points in the previous year. The proportion of exam participants who did not pass the minimum score in 2018 was 5.9%, which is slightly higher than those who did not reach the minimum score in 2017 (3.79%).

Compared to the previous two years, the proportion of poorly trained participants slightly increased (21-40 tb). The share of high-scorers (61-100 TB) increased, reaching the maximum values ​​for three years. This allows us to talk about the strengthening of differentiation in the training of graduates and the growth in the quality of training of students studying the profile course of physics.

In 2018, the proportion of exam participants who scored 81-100 points was 5.61%, which is higher than in 2017 (4.94%). For the USE in physics, the range from 61 to 100 test scores is significant, which demonstrates the readiness of graduates to successfully continue their education in universities. This year, this group of graduates increased compared to the previous year and amounted to 24.22%.

More detailed analytical and methodological materials of the USE 2018 are available at the link.

Our website contains about 3000 tasks for preparing for the exam in physics in 2019. The general plan of the examination paper is presented below.

PLAN OF THE EXAMINATION WORK OF THE USE IN PHYSICS 2019

Designation of the level of difficulty of the task: B - basic, P - advanced, C - high.

Correspondence between the minimum primary scores and the minimum test scores of 2019. Order on amendments to Appendix No. 1 to the order of the Federal Service for Supervision in Education and Science.

THRESHOLD SCORE
By order of Rosobrnadzor, a minimum number of points is established, confirming the mastering of the main general educational programs of secondary (complete) general education by the participants in the exams in accordance with the requirements of the federal state educational standard of secondary (complete) general education. THRESHOLD FOR PHYSICS: 11 primary points (36 test points).

EXAM FORMS
You can download the forms in high quality from the link.

WHAT YOU CAN BRING WITH YOU TO THE EXAM

At the physics exam, the use of a ruler for plotting graphs, optical and electrical circuits is allowed; a non-programmable calculator that performs arithmetic calculations (addition, subtraction, multiplication, division, root extraction) and the calculation of trigonometric functions (sin, cos, tg, ctg, arcsin, arcos, arctg), and also does not perform the functions of a communication facility, database storage and not having access to data networks (including the Internet). .

Preparation for the OGE and the Unified State Examination

Secondary general education

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

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

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

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

Lebedeva Alevtina Sergeevna, teacher of physics, work experience 27 years. Diploma of the Ministry of Education of the Moscow Region (2013), Gratitude of the Head of the Voskresensky 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 complexity: basic, advanced and high. Basic level tasks are simple tasks that test the assimilation of the most important physical concepts, models, phenomena and laws. Advanced level tasks 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 for the application of one or two laws (formulas) on any of the topics of the school physics course. In work 4, 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 fulfillment of such tasks requires the application of knowledge from two three sections of physics at once, i.e. high level of training. This option is fully consistent with the demo version of the USE in 2017, the 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 from the graph the path traveled by the car in the time interval from 0 to 30 s.

Solution. The path traveled by the car in the time interval from 0 to 30 s is most simply defined 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 100 kg mass is lifted vertically upwards with a rope. The figure shows the dependence of the velocity projection V load on the axis directed upwards, from time t. Determine the modulus of the cable tension during the lift.

Solution. According to the speed projection curve v load on an axis directed vertically upwards, from time t, you can determine the projection of the acceleration of the load

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

+ = (1)

Let's write down the equation for the projection of vectors in the reference frame associated with the earth, the OY axis will be directed upwards. The projection of the tension force is positive, since the direction of the force coincides with the direction of the OY axis, the projection of the gravity force is negative, since the force vector is opposite to the OY axis, the projection of the acceleration vector is also positive, so the body moves with acceleration upwards. We have

T – mg = ma (2);

from formula (2) the modulus of the tension 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 a force to it as shown in Figure (1). In this case, the module of the sliding friction force acting on the body is 16 N. What is the power developed by the force F?

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

Tr + + = (1)

Having chosen a reference system associated with a fixed surface, we write 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 X. 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 cos- F tr = 0; (1) express the force projection F, This F cosα = F tr = 16 N; (2) then the power developed by the force will be equal to N = F cosα V(3) Let's make a replacement, taking into account equation (2), and substitute the corresponding data in equation (3):

N\u003d 16 N 1.5 m / s \u003d 24 W.

Answer. 24 W.

A load fixed on a light spring with a stiffness of 200 N/m oscillates vertically. The figure shows a plot of the offset x cargo from time t. Determine what the weight of the load is. Round your answer to the nearest whole number.

Solution. The weight on the spring oscillates vertically. According to the load displacement curve X from time t, determine the period of oscillation 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 cable, with which you can balance or lift a load of 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 system of blocks shown in the figure does not give a gain in strength.
3. h, you need to pull out a section of rope with a length of 3 h.
4. To slowly lift a load to a height hh.

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

1. To slowly lift a 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 load does not touch the walls and bottom of the vessel. Then, an iron load is immersed in the same vessel with water, the mass of which is equal to the mass of the aluminum load. 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 of this?

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: this is the mass of the body and the liquid into which the body is immersed on the threads. After that, it is better to make a schematic drawing and indicate the forces acting on the load: the force of the thread tension F control, directed along the thread up; gravity directed vertically downward; Archimedean force a, acting from the side of the liquid on the immersed body and directed upwards. According to the condition of the problem, the mass of the loads is the same, therefore, the modulus of the force of gravity acting on the load does not change. Since the density of goods is different, the volume will also be different.

The density of iron is 7800 kg / m 3, and the aluminum load is 2700 kg / m 3. Hence, V and< Va. The body is in equilibrium, the resultant of all forces acting on the body is zero. Let's direct the coordinate axis OY up. We write the basic equation of dynamics, taking into account the projection of forces, in the form F ex + Fa – mg= 0; (1) We express the tension force F extr = mg – Fa(2); Archimedean force depends on the density of the liquid and the volume of the submerged part of the body Fa = ρ gV p.h.t. (3); The density of the liquid does not change, and the volume of the iron body is less V and< Va, so 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.

Bar mass m slides off a fixed rough inclined plane with an angle α at the base. The bar acceleration modulus is equal to a, the bar velocity modulus increases. Air resistance can be neglected.

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) The coefficient of friction of the bar on the 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 the 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 reference system and write down the resulting equation for the projection of force and acceleration vectors;

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 in the direction of motion. Let's choose the direction of the axes as shown in the figure. Let's write down the projections of forces on the selected axes.

Let us 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 reaction force of the support 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 to mgy= – 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 reaction force acting on the bar from the side of the inclined plane. N = mg cosα (3). Let's write down the projections on the OX axis.

On the OX axis: force projection N is 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. Positive acceleration projection 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 force of friction is proportional to the force of normal pressure 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. Gaseous oxygen is in a vessel with a volume of 33.2 liters. 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. Convert temperature to Kelvin T = t°С + 273, volume V\u003d 33.2 l \u003d 33.2 10 -3 m 3; We translate 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

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

Solution. First, the gas is monatomic number of degrees of freedom i= 3, secondly, the gas expands adiabatically - this means no heat transfer Q= 0. The gas does work by reducing the internal energy. With this in mind, we write the first law of thermodynamics as 0 = ∆ U + A G; (1) we express the work of the gas A g = –∆ U(2); We write the change in internal energy for a monatomic gas 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 most often cause difficulties for schoolchildren. Let's use the formula for calculating the relative humidity of the air

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

φ 1 \u003d 10%; φ 2 = 35%

We express the air pressure from formulas (2), (3) and find the ratio of pressures.

Answer. The pressure should be increased by 3.5 times.

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

Choose from the proposed list two statements that correspond to the results of measurements 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 the solid state.
3. The heat capacity of a substance in the liquid and solid state is the same.
4. After 30 min. after the start of measurements, the substance was only in the solid state.
5. The process of crystallization of the substance took more than 25 minutes.

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

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

The second correct statement is:

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

Answer. 14.

In an 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 is reached. 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 appropriate nature of the change:

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

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

Solution. If in an isolated system of bodies no energy transformations occur except for heat exchange, then the amount of heat given off by bodies whose internal energy decreases is equal to the amount of heat received by bodies whose internal energy 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 on the basis of the heat balance equation.

where ∆ U- change in internal energy.

In our case, as a result of heat transfer, 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 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 an electromagnet, has a speed perpendicular to the magnetic field 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, away from the observer, down, left, right)

Solution. A 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 charge of the particle. We direct the 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 aside by 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 flat air capacitor with a capacity of 50 μF is 200 V/m. The distance between the capacitor plates is 2 mm. What is the charge on the capacitor? Write your answer in µC.

Solution. Let's convert all units of measurement to the SI system. Capacitance C \u003d 50 μF \u003d 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 electric capacitance formula

Where d is the distance between the plates.

Let's Express the Tension U= E d(4); Substitute (4) in (2) and calculate the charge of the capacitor.

q = C · Ed\u003d 50 10 -6 200 0.002 \u003d 20 μC

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

Answer. 20 µC.

The student conducted the 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. Doesn't change
4. Record the selected numbers for each answer in the table. Numbers in the answer may be repeated.

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

Where n 2 - the absolute refractive index of glass, the medium where the light goes; n 1 is the absolute refractive index of the first medium where the light comes from. For air n 1 = 1. α is the angle of incidence of the beam on the surface of the glass half-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 a high refractive index. The speed of light propagation in glass is slower. Please note that the angles are measured from the perpendicular restored at the point of incidence of the beam. If you increase the angle of incidence, then the angle of refraction will also increase. The refractive index of glass will not change from this.

Answer.

Copper jumper at time t 0 = 0 starts moving 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 jumper and the rails is negligible, the jumper is always perpendicular to the rails. The flux Ф of the magnetic induction vector through the circuit formed by the jumper, rails and resistor changes over time t as shown in the chart.

Using the graph, select two true statements and indicate their numbers in your answer.

1. By the time t\u003d 0.1 s, the change in the magnetic flux through the circuit is 1 mWb.
2. Induction current in the jumper in the range from t= 0.1 s t= 0.3 s max.
3. The module of the EMF of induction that occurs in the circuit is 10 mV.
4. The strength of the inductive current flowing in the jumper is 64 mA.
5. To maintain the movement of the jumper, 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 flow of the magnetic induction vector through the circuit on time, we determine the sections where the flow Ф changes, and where the change in the flow is zero. This will allow us to determine the time intervals in which the inductive current will occur in the circuit. Correct statement:

1) By the time t= 0.1 s the change in the magnetic flux through the circuit is 1 mWb ∆F = (1 - 0) 10 -3 Wb; The EMF module of induction that occurs in the circuit is determined using the EMP law

Answer. 13.

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

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

The self-induction EMF formula 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 schedule we determine the interval of current change during this time:

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

We substitute numerical values ​​into formula (2), we obtain

| Ɛ | \u003d 2 10 -6 V, or 2 μV.

Answer. 2.

Two transparent plane-parallel plates are tightly pressed against each other. A beam 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 passage of light through plane-parallel plates, the following order of solution can be recommended: make a drawing indicating the path of rays coming from one medium to another; at the point of incidence of the beam at the interface between 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 beam and the surface, and 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 build an approximate path of the beam through the plates. We use formula (1) for the 2–3 and 3–1 boundaries. In response 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 beam when crossing the boundary 3–1 (in radians) is 4) ≈ 0.873.

Answer. 24.

Determine how many α - particles and how many protons are obtained 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. Denote by x the number of alpha particles, y the number of protons. Let's make equations

+ → x + y;

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

Answer. 1 – α-particle; 2 - protons.

The momentum modulus of the first photon is 1.32 · 10 -28 kg m/s, which is 9.48 · 10 -28 kg m/s less than the momentum module 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 condition, so we can imagine p 2 = p 1 + ∆ p(1). The photon energy can be expressed in terms of the photon momentum using the following equations. This E = mc 2(1) and p = mc(2), then

E = pc (3),

Where E is the photon energy, p is the momentum of the photon, m is the mass of the photon, c= 3 10 8 m/s is 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 an atom has undergone radioactive positron β-decay. How did this change the electric charge of the nucleus and the number of neutrons in it?

For each value, determine the appropriate nature of the change:

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

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

Solution. Positron β - decay in the 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 an element is as follows:

Answer. 21.

Five experiments were carried out in the laboratory to observe diffraction using various diffraction gratings. Each of the gratings was illuminated by parallel beams of monochromatic light with a certain wavelength. The light in all cases was incident perpendicular to the grating. In two of these experiments, the same number of principal diffraction maxima were observed. Indicate first 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 into the region of a geometric shadow. Diffraction can be observed when opaque areas or holes are encountered in the path of a light wave in large and light-opaque barriers, and the dimensions 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 is an integer called the order of the diffraction maximum. Express from equation (1)

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

Answer. 42.

Current flows through the wire resistor. The resistor was replaced with another, 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 appropriate nature of the change:

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

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

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

Ohm's law for the circuit section, 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 different cross-sectional area. The area is twice as small. 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 greater than the period of its oscillation on some planet. What is the gravitational acceleration modulus on this planet? The effect of the atmosphere in both cases is negligible.

Solution. A mathematical pendulum is a system consisting of a thread, the dimensions of which are much larger than the dimensions of the ball and the ball itself. Difficulty may arise if the Thomson formula for the period of oscillation of a mathematical pendulum is forgotten.

T= 2π (1);

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

By condition

Express from (3) g n \u003d 14.4 m / s 2. It should be noted that the acceleration of free fall depends on the mass of the planet and the radius

Answer. 14.4 m / s 2.

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

Solution. If a current-carrying conductor is placed in a magnetic field, then the field on the current-carrying conductor will act with the Ampere force. We write the formula for the Ampère force modulus

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 120 J. How many times must the strength of the current flowing through the coil winding be increased in order for the energy of the magnetic field stored in it to increase by 5760 J.

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

By condition W 1 = 120 J, then W 2 \u003d 120 + 5760 \u003d 5880 J.

Then the current ratio

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

An electrical circuit consists of two light bulbs, two diodes, and a coil of wire connected as shown in the figure. (A diode only allows current to flow 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 coil? Explain your answer by indicating what phenomena and patterns you used in the explanation.

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

Answer. The second lamp will light up.

Aluminum spoke length L= 25 cm and cross-sectional area S\u003d 0.1 cm 2 is suspended on a thread by the upper end. The lower end rests on the horizontal bottom of the vessel in which water is poured. The length of the submerged part of the spoke l= 10 cm Find strength 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 in = 1.0 g / cm 3. Acceleration of gravity g= 10 m/s 2

Solution. Let's make an explanatory drawing.

– Thread tension force;

– Reaction force of the bottom of the vessel;

a is the 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 side of the Earth and is applied to the center of the entire spoke.

By definition, the mass 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 is the 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

given 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 needle presses on the bottom of the vessel we write N = F e and from equation (7) we express this force:

Plugging in the numbers, we get that

F d = 0.025 N.

Answer. F d = 0.025 N.

A bottle containing m 1 = 1 kg of nitrogen, when tested for strength exploded at a temperature t 1 = 327°C. What mass of hydrogen m 2 could be stored in such a cylinder at a temperature t 2 \u003d 27 ° C, with a fivefold margin of safety? Molar mass of nitrogen M 1 \u003d 28 g / mol, hydrogen M 2 = 2 g/mol.

Solution. We write the equation of state of an ideal gas Mendeleev - Clapeyron for nitrogen

Where V- the volume of the balloon, T 1 = t 1 + 273°C. According to the condition, hydrogen can be stored at a pressure p 2 = p 1 /5; (3) Given that

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

After substituting numerical data m 2 = 28

Answer. m 2 = 28

In an ideal oscillatory circuit, the amplitude of current oscillations in the inductor I m= 5 mA, and the amplitude of the voltage across the capacitor Um= 2.0 V. At 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 energy of vibrations is conserved. 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

Let us substitute (4) into (3). As a result, we get:

I = I m (5)

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

I= 4.0 mA.

Answer. I= 4.0 mA.

There is a mirror at the bottom of a reservoir 2 m deep. A beam of light, passing through the water, is reflected from the mirror and exits 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 beam incidence angle;

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

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

According to the law of refraction of light

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

We get the following expression:

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

Answer. 1.63 m

In preparation for the exam, we invite you to familiarize yourself with work program in physics for grades 7–9 to the line of teaching materials Peryshkina A.V. And the working program of the in-depth level for grades 10-11 to the TMC Myakisheva G.Ya. Programs are available for viewing and free download to all registered users.

Analysis of the results of the state (final) certification

in the form of a unified state exam (USE)

graduates of MBOU "Secondary School No. 6" NMR RT

in physics in 2017

The Unified State Examination (hereinafter referred to as the USE) is a form of objective assessment of the quality of training of persons who have mastered the educational programs of secondary general education, using tasks in a standardized form (control measuring materials). The USE is conducted in accordance with Federal Law No. 273-FZ of December 29, 2012 “On Education in the Russian Federation”. Control measuring materials allow to establish the level of development by graduates of the Federal component of the state educational standard of secondary (complete) general education in physics, basic and profile levels.

The results of the unified state exam in physics are recognized by educational institutions of higher professional education as the results of entrance examinations in physics.

In preparation for the exam, all the work was aimed at organizing group work with students, with the aim of orienting the preparation of "weak" students to overcome the required minimum, and also with the aim of orienting the preparation of "strong" students to work out complex topics, analyze the criteria for checking tasks of high and high level. To increase the effectiveness of mastering the course of physics in the classroom, reference notes were used, containing a mandatory minimum of knowledge on a particular topic; used demo versions in her work; Also, in preparation for the Unified State Examination, it was planned to repeat the knowledge and skills formed during the study of the material in the primary and secondary schools. The main focus of the work was the organization of independent learning activities for the implementation of specific tasks with a written fixation of the results, their further analysis. When solving CIM tasks, students independently processed the information presented in the tasks, made conclusions and argued them.

Each version of the examination paper consists of two parts and includes 31 tasks that differ in form and level of complexity (Table 1).

Part 1 contains 23 tasks with a short answer. Of these, 13 tasks with a record of the answer in the form of a number, a word or two numbers, 10 tasks for establishing correspondence and multiple choice, in which answers must be written as a sequence of numbers.

Part 2 contains 8 tasks united by a common activity - problem solving. Of these, 3 tasks with a short answer (24–26) and 5 tasks (27–31), for which it is necessary to provide a detailed answer.

Table 1. Distribution of tasks of the examination work by parts of the work

In total, several plans are used to form the KIM USE 2017.

In part 1, to ensure a more accessible perception of information, tasks 1–21 are grouped based on the thematic assignment of tasks: mechanics, molecular physics, electrodynamics, quantum physics. In part 2, tasks are grouped depending on the form of presentation of tasks and in accordance with thematic affiliation.

In the examtook part in physics4 (22.2%) graduate.

Overcame the "threshold" in physics (the minimum number of points is 36) 4 out of 4 graduates (100% of the total number of those who took the exam in physics).

The maximum USE score was - 62 (Nikolaeva Anastasia).

USE in physics iselective exam and is designed to differentiate when entering higher education institutions. For these purposes, tasks of three levels of complexity are included in the work. Among the tasks of the basic level of complexity, tasks are distinguished, the content of which corresponds to the standard of the basic level. The minimum number of USE points in physics (36 points), confirming that a graduate has mastered a secondary general education program in physics, is set based on the requirements for mastering the basic level standard.

Table 2 - Sections and topics of the examination work of the exam in physics

The result of the completed tasks of the Unified State Examination in Physics by graduates of MBOU "Secondary School No. 6" NMR RT in 2017

Analyzing the completed tasks of part 1 (1-24) of the KIM USE in PHYSICS of various levels of complexity, it can be noted that more than half of the graduates successfully cope with the taskswith answer choicemechanics.

3 people out of 4 gave correct answers to tasks with a short answer (1).

The analysis data allow us to conclude that graduates are most successfully able to perform tasks of 2-4 basic levels of complexity, for which it is necessary to know/understand the lawuniversal gravitation, Hooke's law, as well as a formula for calculating the friction force.

Also, a high percentage of completion of task 5 of the basic level of complexity (3 people out of 4), which tested the assimilation of basic concepts on the topics “Rigid body equilibrium condition”, “Archimedes force”, “Pressure”, “Mathematical and spring pendulums”, “Mechanical waves and sound."

Task 7 was of an increased level of complexity, in which in different versions it was required to establish a correspondence between graphs and physical quantities, between physical quantities and formulas, units of measurement. However, more than half of the graduates successfully completed this task: 25% of the graduates scored 1 point, having made one mistake, and 50% scored the primary 2 points, having completed this task completely correctly.

Practically the same result was demonstrated by the graduates when completing task 6 of the basic level of complexity.

Bymolecular physics in part 1 of the KIM USE, 3 tasks were presented with the choice and recording of the number of the correct answer (8-10), for the correct implementation of which 1 point was given. All students coped with task 8, in the 9th task 1 person out of 4 made a mistake. In addition, 2 tasks with a short answer (11-12) are presented, these are assignments for establishing correspondence and multiple choice, in which answers must be written in the form sequences of numbers. The students showed the most successful performance when performing 11 tasks. In general, with tasks forgraduates did well in molecular physics.

Byelectrodynamics in part 1 of the KIM USE, 4 tasks were presented with the choice and recording of the number of the correct answer (13-16), for the correct implementation of which 1 point was given. In addition, 2 tasks with a short answer (17-18) are presented, these are tasks for establishing correspondence and multiple choice, in which the answers must be written in the form of a sequence of numbers.

The analysis data allow us to conclude that, in general, graduates completed tasks in electrodynamics much worse than similar tasks in mechanics and molecular physics.

The most difficult task for the graduates was task 13 of the basic level of complexity, in which their ideas aboutelectrization of bodies, the behavior of conductors and dielectrics in an electric field, the phenomenon of electromagnetic induction, interference of light, diffraction and dispersion of light.

The most successful graduates completed task 16 of the basic level of complexity, for which it is necessary to have an understanding of Faraday's law of electromagnetic induction, an oscillatory circuit, the laws of reflection and refraction of light, the course of rays in a lens (75 %).

Task 18 of an increased level of complexity, in which in different versions it was required to establish a correspondence between graphs and physical quantities, between physical quantities and formulas, units of measurement, the graduates performed no worse than a similar task in mechanics and molecular physics.

Byquantum physics in part 1 of the KIM USE, 3 tasks were presented with the choice and recording of the number of the correct answer (19-21), for the correct implementation of which 1 point was given. In addition, 1 task with a short answer (22) is presented. The highest percentage of completion (2 people out of 2) was in the case of task 20 of the basic level of complexity, which tested the knowledge of graduates on the topics "Radioactivity", "Nuclear reactions" and "Fission and fusion of nuclei".

Most of the students (3 out of 4) did not start and did not score primary points when completing tasks with a detailed answer (part C).

However, it is worth noting that there were no students who would have coped successfully (by 3 maximum points) with at least one task. This is explained by the fact that physics is studied at school at a basic level, and these tasks mainly involve specialized training in the subject.

Students showed an average level of preparation for the exam in physics. The presented data indicate that in part 1 of the KIM USE in physics, graduates performed much better tasks in mechanics and molecular physics than in electrodynamics and quantum physics.

Many students did not realize that the new criteria for assessing tasks require explanations for each formula for solving these problems.

Use the results of the analysis to prepare for the Unified State Examination - 2018.

To form in students the skills indicated in the education standard as the main goals in teaching physics:

Correctly explain physical phenomena;

Establish relationships between physical quantities;

Give examples of confirmation of fundamental laws and their consequences.

4. Use the laws of physics to analyze phenomena at the qualitative and computational levels.

5. Carry out calculations based on data presented in graphical or tabular forms.

Physics teacher __________________ / Mochenova O.V. /

The Unified State Examination in Physics is an exam of the choice of graduates and is designed to differentiate when entering higher educational institutions. For these purposes, tasks of three levels of complexity are included in the work. Completing tasks of a basic level of complexity allows assessing the level of mastering the most significant content elements of a high school physics course and mastering the most important activities. The use of tasks of increased and high levels of complexity in the USE makes it possible to assess the degree of student's preparation for continuing education at a university.

Each version of the examination paper consists of 2 parts and includes 32 tasks that differ in form and level of complexity (see table).

Part 1 contains 24 tasks, of which 9 tasks with the choice and recording the number of the correct answer and 15 tasks with a short answer, including tasks with self-recording the answer in the form of a number, as well as tasks for establishing correspondence and multiple choice, in which answers are required write as a sequence of numbers.

Part 2 contains 8 tasks united by a common activity - problem solving. Of these, 3 tasks with a short answer (25–27) and 5 tasks (28–32), for which it is necessary to provide a detailed answer.

Number of tasks

Maximum primary score

Percentage of maximum primary score

Job type

Distribution of tasks by topics

When developing the content of KIM, the need to check the assimilation of knowledge in the following sections of the physics course is taken into account:

• Mechanics(kinematics, dynamics, statics, conservation laws in mechanics, mechanical oscillations and waves);
• Molecular physics(molecular-kinetic theory, thermodynamics);
• Electrodynamics and fundamentals of SRT(electric field, direct current, magnetic field, electromagnetic induction, electromagnetic oscillations and waves, optics, fundamentals of SRT);
• The quantum physics(particle-wave dualism, physics of the atom, physics of the atomic nucleus)

The total number of tasks in the examination paper for each of the sections is approximately proportional to its content content and the study time allotted for the study of this section in the school physics course.

Distribution of tasks by difficulty level

The examination paper presents tasks of different levels of complexity: basic, advanced and high.

Basic level tasks are included in part 1 of the work (19 tasks, of which 9 tasks with the choice and recording of the number of the correct answer and 10 tasks with a short answer). These are simple tasks that test the assimilation of the most important physical concepts, models, phenomena and laws.

Tasks of an advanced level are distributed between the first and second parts of the examination paper: 5 tasks with a short answer in part 1, 3 tasks with a short answer and 1 task with a detailed answer in part 2. These tasks are aimed at testing the ability to use the concepts and laws of physics to apply one -two laws (formulas) on any of the topics of the school physics course.

The four tasks in Part 2 are advanced tasks and test the ability to use the laws and theories of physics in a changed or new situation. The fulfillment of such tasks requires the application of knowledge from two or three branches of physics at once, i.e., a high level of training.

Task difficulty levels

Grading system

The task with the choice and recording of the number of the correct answer is considered completed if the number of the answer recorded in form No. 1 matches the correct answer. Each of these tasks is worth 1 point.

A task with a short answer is considered completed if the answer recorded in form No. 1 matches the correct answer.

Assignments 3–5, 10, 15, 16, 21 of Part 1 and Assignments 25–27 of Part 2 are worth 1 point.

Items 6, 7, 11, 12, 17, 18, 22 and 24 of Part 1 are worth 2 points if both elements of the answer are correct; 1 point if an error was made in specifying one of the elements of the answer, and 0 points if two errors were made.

Answers to tasks with the choice and recording of the number of the correct answer and a short answer are processed automatically after scanning the answer forms No. 1.

A task with a detailed answer is evaluated by two experts taking into account the correctness and completeness of the answer. The maximum initial score for tasks with a detailed answer is 3 points. For each task, detailed instructions for experts are provided, which indicate what each point is set for - from zero to the maximum score. In the examination version, before each type of task, an instruction is offered that contains general requirements for the design of answers.

Exam duration and equipment

To complete the entire examination work is given 235 minutes. Estimated time to complete the tasks of various parts of the work is:

• for each task with a choice of answers - 2-5 minutes;
• for each task with a short answer - 3-5 minutes;
• for each task with a detailed answer - from 15 to 25 minutes.

used non-programmable calculator(for each student) with the ability to calculate trigonometric functions (cos, sin, tg) and a ruler. The list of additional devices and materials, the use of which is allowed for the exam, is approved by Rosobrnadzor.