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. Tasks basic level, these are simple tasks that test the assimilation of the most important physical concepts, models, phenomena and laws. Tasks advanced level 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 high level 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 compatible demo version USE 2017, assignments taken from open bank USE assignments.

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 slips off the fixed rough inclined plane with 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) of 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 must enter the palm perpendicularly, 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 passed through it direct current, is equal to 120 J. How many times should 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 U m= 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.

Year. Officials did not pass by and USE in physics. 2017 will bring several innovations to this exam that may affect total score student performance and reveal the true picture of their knowledge.

The main amendment is the exclusion of the test part. It is worth noting that this innovation will occur not only in the physics exam, but also in many others (history, literature, chemistry).

The main changes in the USE-2017

A few months ago it became known that the deputies are seriously thinking about adding another discipline to the list of compulsory subjects submitted for the Unified State Examination. In total, their total number will increase to three.

Until 2017, students at the end took Russian language and mathematics, as well as additional items required for admission to a university for a particular specialty. Starting next year, for the title compulsory subject claims first.

Civil servants, from whose submission the above innovations were made, justify their act by the fact that at present too few students have a decent knowledge in the field of national and world history. Few of them are interested in the past and do not know how their ancestors lived and how they “built” the state. According to them, such a trend cannot be called positive, and if appropriate measures are not taken, very few worthy educated citizens will remain in the country in the near future.

What will change in the physics exam?

Let's look at the exam in physics. This subject will not receive special amendments. The only thing to pay attention to is the exclusion of the test block. In its place, they plan to put an oral and written method of answering. It is too early to talk about any specific details on this issue, exactly the same as about what can be included in the tasks submitted for the exam.

As for the cancellation of the test part, it is worth noting that the officials did not come to this decision overnight. For many months there were heated negotiations in Rosobrnadzor regarding the expediency of this amendment. All pros and cons were weighed and carefully discussed.

Ultimately, as we see oral part decided to implement in many final tests. The most important advantage of this approach to testing knowledge is the exclusion of guessing or, as the people say, the “poke method”. Simply put, now it will not be possible to rely on “maybe lucky” and put the answer at random. In turn, the written and oral answers of the student will be able to show the examiner his level of education, as well as his ability to learn.

Date of the exam

There is not much time left before the start of the tests, so you can already familiarize yourself with the official schedule. So, the exam in physics in 2017 will be held on the following dates:

• The early period is March 22 (Wednesday). Reserve day - April 5th.
• The main period is June 13 (Tuesday). Reserve day - June 20.

The importance of the exam in Russia in the future

Note that in the next few years the procedure for conducting the Unified State Exam in Russia will change radically. The test part will be removed in all subjects and this is not the limit.

Closer to 2022, Rosobrnadzor intends to expand the list of compulsory disciplines to four. Most likely, it will be a foreign language, because in our time, knowledge, for example, in English is incredibly valued and gives a chance to apply for a prestigious high-paying position.

In addition to English, it will be possible to take German, French and Spanish.

You can already guess what the future of education will be like in Russian Federation. At the moment, even a simple person can see that the world and the trends in it are changing every day. What was previously unimportant comes to the fore. In modern society, the ability to establish connections, as well as the manifestation of diplomacy, is incredibly valued.

For supporting business relations with people of another nation, fluency in several languages ​​is necessary. Only by communicating with a person in his native dialect, it will be possible to establish close, trusting relationships. Actually, for this, Russian schools already pay great attention to foreign languages and their study among students.

How to prepare for the exam

Given the fact that physics is a complex subject and cannot be on a par with the Russian language or literature, eleventh graders should devote a little more time to it than to the rest of the subject. This is due to the fact that understanding a particular topic can take a long time, and without understanding a good result on the exam, you can forget. In addition, if you want to enter a prestigious university, knowledge in the field of physics is extremely important.

It is worth noting that there is a category of people who claim that the USE will be canceled in 2017. No need to mislead yourself and others - there will be no cancellation! And for the next 5-6 years, one can only dream of such a thing. Besides, for what to exchange something like an exam? Despite all its severity, the USE is able to show the real level of knowledge and preparation of the student for adult student life.

Where do you get knowledge from?

You can prepare for the exam in physics according to the same principle by which you plan to prepare for other subjects. First of all, of course, you should pay attention to educational materials: books and reference books. While studying at school, the teacher is obliged to give a huge amount of knowledge, which can later be used on. The main thing is to listen carefully to the teacher, ask again and understand the essence of the material presented.

Stock up on a collection of basic physical formulas so that this part of the exam does not become intimidating for you. Another tool for preparing for the exam in physics can be a collection of problems. It prints various problems with solutions that can be used as training. Of course, there will be completely different tasks on the test, but having got your hand in solving physical problems, the examination paper will not seem so difficult to you.

You can start going to a tutor, as well as study on the Internet on your own. Now there are a lot of online resources with which you can understand how the physics exam will actually take place.

Any of your efforts will once again prove that at this stage of life, the main thing for you is studying and you will do everything to make it successful!

Video news, demos

About 318,000 participants passed the USE in social science during the main period, more than 155,000 participants took the USE in physics, and more than 41,000 participants took the USE in literature. The average scores in all three subjects in 2017 are comparable to the results of the previous year.

The number of USE participants who failed to overcome the established minimum threshold in subjects decreased: in social science to 13.8% from 17.5% last year, in physics - to 3.8% from 6.1%, in literature - to 2 .9% from 4.4% a year earlier.

“The average scores are comparable to the results of the previous year, which indicates the stability of the exam and the objectivity of the assessment. It is important that the number of those who have not overcome the minimum thresholds is decreasing. This is largely due to the competent work with USE results when they are analyzed and used in the work of institutes for advanced training of teachers. In a number of regions, the project “I will pass the Unified State Examination” gave very serious results,” said Sergey Kravtsov, head of Rosobrnadzor.

Thanks to the use of scanning technology for participants' works at the examination points, the results of the USE in social science, literature and physics were processed ahead of schedule established by the schedule for issuing results. Graduates will be able to find out their result a day earlier.

• 0-35 points - score " 2 »,
• 36-52 points - score " 3 »,
• 53-67 points - score " 4 »,
• 68 and above points - score " 5 »;

The distribution of points for each task of the exam in physics

• 1 point- for 1-4, 8, 9, 10, 13, 14, 15, 19, 20, 22, 23, 25, 26, 27 tasks.
• 2 points - 5, 6, 7, 11, 12, 16, 17, 18, 21, 24.
• 3 points- 28, 29, 30, 31, 32 tasks with a detailed answer are evaluated according to the criteria from FIPI (see below)

Total: 52 points.

• The minimum USE test score is 36
• The minimum primary USE score is 11

Criteria for evaluating tasks with a detailed answer

1 point

• A solution corresponding to one of the following cases is presented. The correct answer to the question of the task is given, and an explanation is given, but it does not indicate two phenomena or physical laws necessary for a complete correct explanation.
• OR All the phenomena and laws, patterns necessary for explaining are indicated, but the existing reasoning aimed at obtaining an answer to the task question has not been completed.
• OR All the phenomena and laws necessary for explanation, regularities are indicated, but the existing reasoning leading to the answer contains errors.
• OR Not all the phenomena and laws, patterns necessary for explaining are indicated, but there are correct arguments aimed at solving the problem.

0 points

Maximum score - 3

3 points -

1. the law of conservation of momentum, the law of conservation of mechanical energy, Newton's second law for the motion of a body in a circle; it is taken into account that at the upper point the tension force of the thread vanishes);
2. (with the exception of the designations of the constants specified in the CMM variant and the designations used in the condition of the problem);
3. the necessary mathematical transformations and calculations were carried out, leading to the correct numerical answer (the solution “in parts” with intermediate calculations is allowed);
4. the correct answer is presented indicating the units of measurement of the desired value.

2 points

• All the necessary provisions of the theory, physical laws, regularities are correctly recorded, and the necessary transformations are carried out. But there are the following disadvantages.
• OR In the solution, extra entries that are not included in the solution (possibly incorrect) are not separated from the solution (not crossed out; not enclosed in brackets, frame, etc.).
• OR Errors were made in the necessary mathematical transformations or calculations, and (or) the transformations/calculations were not completed.
• OR .

1 point

• Only provisions and formulas are presented that express the physical laws, the application of which is necessary to solve the problem, without any transformations with their use, aimed at solving the problem, and the answer.
• OR The solution lacks ONE of the original formulas necessary to solve the problem (or the statement underlying the solution), but there are logically correct transformations with the available formulas aimed at solving the problem.
• OR ONE from the initial formulas necessary to solve the problem (or in the statement underlying the solution), an error was made, but there are logically correct transformations with the available formulas aimed at solving the problem

0 points

• All decision cases that do not meet the above criteria for scoring 1, 2, 3 points

Maximum score - 3

3 points - A complete solution is provided, including the following elements:

1. the provisions of the theory and physical laws are written down, the laws, the application of which is necessary to solve the problem in the chosen way (in this case: expression for the Archimedes force, the relationship between mass and density, the Mendeleev-Clapeyron equation, the condition for lifting the ball);
2. all newly introduced in the solution letter designations of physical quantities are described
3. the correct answer is presented indicating the units of measurement of the desired value

2 points

• Records corresponding to item II are incomplete or missing.
• AND (OR)
• AND (OR)
• AND (OR) Clause IV is missing or in error

1 point- Entries corresponding to one of the following cases are presented.

• OR
• OR IN ONE from the initial formulas necessary for solving this problem (or in the statement underlying the solution), an error was made, but there are logically correct transformations with the available formulas aimed at solving the problem

0 points

Maximum score - 3

3 points - A complete solution is provided, including the following elements:

1. the provisions of the theory and physical laws are written down, the laws, the application of which is necessary to solve the problem in the chosen way (in this case: formulas for calculating electrical capacity, charge and voltage for capacitors connected in series and in parallel, determination of electrical capacity, formula for the energy of a charged capacitor);
2. all newly introduced in the solution letter designations of physical quantities are described (with the exception of the designations of the constants specified in the CMM version, the designations used in the problem statement, and the standard designations of quantities used when writing physical laws);
3. the necessary mathematical transformations and calculations were carried out, leading to the correct numerical answer (the solution “in parts” with intermediate calculations is allowed);
4. correct answer provided

2 points

• All the necessary provisions of the theory, physical laws, regularities are correctly recorded, and the necessary transformations are carried out. But there are one or more of the following disadvantages.
• Records corresponding to item II are incomplete or missing.
• AND (OR) The solution contains extra entries that are not included in the solution (possibly incorrect), which are not separated from the solution (not crossed out; not enclosed in brackets, frame, etc.).
• AND (OR) Errors were made in the necessary mathematical transformations or calculations, and/or logically important steps were omitted in the mathematical transformations/calculations.
• AND (OR) Clause IV is missing or in error

1 point- Entries corresponding to one of the following cases are presented.

• Only provisions and formulas are presented that express physical laws, the application of which is necessary and sufficient for solving this problem, without any transformations with their use aimed at solving the problem.
• OR The solution lacks ONE of the original formulas necessary to solve this problem (or the statement underlying the solution), but there are logically correct transformations with the available formulas aimed at solving the problem.
• OR IN ONE

0 points

• All cases of decision that do not meet the above criteria for scoring 1, 2, 3 points.

Maximum score - 3

3 points - A complete solution is provided, including the following elements:

1. the provisions of the theory and physical laws are written down, the laws, the application of which is necessary to solve the problem in the chosen way (in this case: thin lens formula, condition of equality of angular velocities of the source and its image, linear velocity formula);
2. all newly introduced in the solution letter designations of physical quantities are described (with the exception of the designations of the constants specified in the CMM version, the designations used in the problem statement, and the standard designations of quantities used when writing physical laws);
3. the necessary mathematical transformations and calculations were carried out, leading to the correct numerical answer (the solution “in parts” with intermediate calculations is allowed);
4. correct answer provided

2 points

• All the necessary provisions of the theory, physical laws, regularities are correctly recorded, and the necessary transformations are carried out. But there are one or more of the following disadvantages.
• Records corresponding to item II are incomplete or missing.
• AND (OR) The solution contains extra entries that are not included in the solution (possibly incorrect), which are not separated from the solution (not crossed out; not enclosed in brackets, frame, etc.).
• AND (OR) Errors were made in the necessary mathematical transformations or calculations, and/or logically important steps were omitted in the mathematical transformations/calculations.
• AND (OR) Clause IV is missing or in error

1 point- Entries corresponding to one of the following cases are presented.

• Only provisions and formulas are presented that express physical laws, the application of which is necessary and sufficient for solving this problem, without any transformations with their use aimed at solving the problem.
• OR The solution lacks ONE of the original formulas necessary to solve this problem (or the statement underlying the solution), but there are logically correct transformations with the available formulas aimed at solving the problem.
• OR IN ONE from the initial formulas necessary for solving this problem (or in the statement underlying the solution), an error was made, but there are logically correct transformations with the available formulas aimed at solving the problem.

0 points

• All cases of decision that do not meet the above criteria for scoring 1, 2, 3 points.

Maximum score - 3

In 2017, the control measuring materials in physics will undergo significant changes.

The version of the examination paper will consist of two parts and will include 31 tasks. Part 1 will contain 23 short answer items, including self-recording items as a number, two numbers, or a word, as well as matching and multiple choice items, in which responses must be recorded as a sequence of numbers. Part 2 will contain 8 tasks united by a common activity - problem solving. Of these, 3 tasks with a short answer (24–26) and 5 tasks (29–31), for which it is necessary to provide a detailed answer.

The work will include tasks of three levels of difficulty. Basic level tasks are included in part 1 of the work (18 tasks, of which 13 tasks record the answer in the form of a number, two numbers or a word and 5 tasks for matching and multiple choice). Among the tasks of the basic level, tasks are distinguished, the content of which corresponds to the standard of the basic level. The minimum number of USE points in physics, which confirms that the graduate has mastered the program of secondary (complete) general education in physics, is set based on the requirements for mastering the basic level standard.

The use of tasks of increased and high levels of complexity in the examination work allows us to assess the degree of readiness of the student to continue education at the university. Advanced questions are divided between parts 1 and 2 of the examination paper: 5 short answer questions in part 1, 3 short answer questions and 1 long answer question in part 2. The last four problems of part 2 are tasks of a high level of difficulty.

Part 1 examination work will include two blocks of tasks: the first checks the development of the conceptual apparatus of the school physics course, and the second - the mastery of methodological skills. The first block includes 21 tasks, which are grouped based on thematic affiliation: 7 tasks in mechanics, 5 tasks in MKT and thermodynamics, 6 tasks in electrodynamics and 3 in quantum physics.

A group of tasks for each section begins with tasks with an independent formulation of the answer in the form of a number, two numbers or a word, then there is a multiple choice task (two correct answers out of five proposed), and at the end - tasks for changing physical quantities in various processes and on establishing a correspondence between physical quantities and graphs or formulas in which the answer is written as a set of two numbers.

Multiple choice and matching tasks are 2-point and can be constructed on any content elements in this section. It is clear that in the same version, all tasks related to one section will check different elements of the content and refer to different topics this section.

In the thematic sections on mechanics and electrodynamics, all three types of these tasks are presented; in the section on molecular physics- 2 tasks (one of them for multiple choice, and the other - either for changing physical quantities in processes, or for matching); in the section on quantum physics - only 1 task for changing physical quantities or matching. Particular attention should be paid to tasks 5, 11 and 16 for multiple choice, which assess the ability to explain the studied phenomena and processes and interpret the results of various studies presented in the form of tables or graphs. Below is an example of such a task in mechanics.

Attention should be paid to the change in the shape of individual task lines. Task 13 to determine the direction of vector physical quantities (Coulomb force, electric field strength, magnetic induction, Ampère force, Lorentz force, etc.) is proposed with a short answer in the form of a word. Wherein possible options answers are given in the text of the task. An example of such a task is shown below.

In the section on quantum physics, I would like to pay attention to task 19, which tests knowledge about the structure of an atom, an atomic nucleus, or nuclear reactions. This task has changed the presentation form. The answer, which is two numbers, must first be written in the proposed table, and then transferred to the answer form No. 1 without spaces and additional characters. Below is an example of such a task form.

At the end of Part 1, there will be 2 tasks of the basic level of difficulty, testing various methodological skills and related to different sections physics. Task 22, using photographs or drawings of measuring instruments, is aimed at testing the ability to record instrument readings when measuring physical quantities, taking into account the absolute measurement error. The absolute measurement error is specified in the text of the task: either as half of the division value, or as the division value (depending on the accuracy of the instrument). An example of such a task is shown below.

Task 23 tests the ability to choose equipment for the experiment according to a given hypothesis. In this model, the form of the task presentation has changed, and now it is a multiple choice task (two items out of five proposed), but it is estimated at 1 point if both elements of the answer are correctly indicated. Three different task models can be offered: a choice of two drawings that graphically represent the respective test setups; the choice of two rows in the table that describes the characteristics of the experimental setups, and the choice of the names of two pieces of equipment or instruments that are necessary to carry out the specified experiment. Below is an example of one of these tasks.

Part 2 work is devoted to problem solving. This is traditionally the most significant result of mastering a physics course. high school and the most demanded activity in the further study of the subject at the university.

In this part, KIM 2017 will have 8 different tasks: 3 computational tasks with independent recording of a numerical answer of an increased level of complexity and 5 tasks with a detailed answer, of which one is qualitative and four are computational.

At the same time, on the one hand, in different problems in one variant, the same not too significant meaningful elements are used, on the other hand, the application of fundamental conservation laws can occur in two or three problems. If we consider the "binding" of the topics of tasks to their position in the variant, then position 28 will always have a task in mechanics, position 29 - in MKT and thermodynamics, position 30 - in electrodynamics, and position 31 - mainly in quantum physics (if only material quantum physics will not be involved in the qualitative task at position 27).

The complexity of tasks is determined by both the nature of the activity and the context. In computational problems of an increased level of complexity (24–26), the use of the studied algorithm for solving the problem is assumed and typical learning situations are proposed that students encountered in the learning process and in which explicitly specified physical models are used. In these tasks, preference is given to standard formulations, and their selection will be carried out mainly with a focus on open bank assignments.

The first of the tasks with a detailed answer is a qualitative task, the solution of which is a logically structured explanation based on physical laws and patterns. For computational problems of a high level of complexity, it is necessary to analyze all stages of the solution, therefore they are offered in the form of tasks 28–31 with a detailed answer. Here, modified situations are used in which it is necessary to operate with a larger number of laws and formulas than in typical problems, to introduce additional justifications in the decision process or completely new situations that have not been encountered before in the educational literature and involve serious activity in the analysis of physical processes and independent choice of a physical model for solving the problem.