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 different levels difficulty: basic, advanced and high. Tasks basic level, these are simple tasks that test the assimilation of the most important physical concepts, models, phenomena and laws. 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 2017, the tasks are 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.

Physics! For many modern schoolchildren, this sounds like something terrible, incomprehensible and of no practical interest. However, the development of science, technology, information technologies is the result of discoveries in this field of science. Therefore, choose as USE exam physics is necessary for most school graduates. In addition, the guys need to remember that physics is the science of nature, i.e. about what surrounds us. Whether you are studying a theory or solving a problem, you always need to imagine how this or that process occurs in real life.

The USE in Physics has been taken by graduates since 2003. Over the past 14 years, the structure of the Unified State Examination has undergone a lot of changes, and the next year 2017 will not be an exception. Let's take a look at some of them.

In 2017, the exam program remains unchanged. The encoder remains the same.

Big changes coming in part 1 USE options in physics. Part 2 will be fully preserved in its current form (3 tasks with a short answer + 5 tasks with a detailed solution).

What will change in part 1?

Of the options will go away completely tasks with a choice of answers (1 out of 4) - 9 tasks.

The number of tasks with a short answer and tasks where you need to choose 2 correct answers out of 5 will increase. The total number of tasks in part 1 is 23 tasks (was 24).

Tasks for sections in part 1 are distributed in almost the same way as before:

• Mechanics - 7 tasks
• Molecular physics- 5 tasks
• Electrodynamics - 6 tasks
• The quantum physics- 3 tasks (was 4)
• Methodology - 2 tasks

Inside the section, tasks will be arranged depending on their form. In task 13, this may not coincide with the sequence of presentation of the material.

The structure of the exam in physics in 2017

Total score for part 1: 10 + 7 + 9 + 4 + 2 = 32
Total score in Part 2: 3 + 5×3 = 18
The total sum of primary scores in the variant: 32 + 18 = 50 (as it is now).

Examples of problem solving

Sample task 13

Two long straight conductors perpendicular to the plane of the figure carry equal currents in opposite directions. How is the induction vector of the magnetic field of the conductors at point A directed (to the right, to the left, up, down, towards us, away from us)?

Sample task 19

Indicate the number of protons and the number of neutrons in the nucleus of the polonium isotope 214 84 Po

Good luck on your exam!

As it became known at the All-Russian parent meeting, KIM Unified State Examination in Physics, Chemistry and Biology will no longer contain multiple choice test items. This was emphasized in his speech by the Minister of Education of the Russian Federation O.Yu. Vasiliev, and full list changes appeared among the official documents of the exam.

1. KIM USE 2017 in physics has undergone significant changes.

The structure of the 1st part of tasks has been completely changed. In the first part, there are now 23 tasks instead of last year's 24.

In KIM USE 2016 in physics, tasks 1, 2, 8, 9, 13, 14, 19, 20, 23, 24 provided the examinee with 4 answer options, among which it was necessary to choose the correct one. There will be no more jobs like this. Judging by the published draft of the “Demonstration version” of the USE 2017 in physics, the main emphasis is on tasks. Moreover, the answer indicates the unit of measurement, which, on the one hand, should help the examinee solve the problem (by the unit of measurement, you can find the formula), and on the other hand, complicate the search for the correct answer if the unit of measurement is an order of magnitude higher or lower than that obtained with the standard solution .

As the compilers assure, neither the distribution of the main sections, nor the concept of testing the knowledge acquired at school, nor the final maximum number of points have been changed.

2. KIM USE 2017 in chemistry has been significantly changed

The changes affected the structure, the total number of tasks, the school for evaluating individual tasks, the maximum number of primary points.

Let's take a closer look.

The changes in the structure are as follows:

1. Tasks 1-3 have been completely changed. Several chemical elements, which must be characterized from different angles. The answer must be two numbers.
2. In KIM USE in chemistry for 2016, tasks 4, 5, 7, 8, 9, 10, 12-17, 19-23 are replaced by tasks that must contain several numbers in the answer.
3. Increased the number to search for matches from two columns. If in 2016 there were 9 of them (Nos. 27-35), now their number has been increased to 11 - Nos. 5, 10-12, 18, 19, 22-26.
4. Tasks for calculating indicators chemical reactions with the recording of the number, 3 remained - Nos. 27, 28, 29.
5. Part 2 remained unchanged in terms of the number of tasks.

As a result, there were fewer tasks (there were 40, and there were 34), but they became more difficult.

Tasks 9 and 17 are now evaluated not with 1 point, but with 2.

It was changed primary score- was 64, and became 60.

3. The largest number changes made to the exam 2017 in biology

All USE changes 2017 in biology were called optimization.

So, tests with several answers to choose from were excluded from CIMs. If we compare demo options 2016 from 2017, you can easily find the differences:

It can be noted that, despite the decrease in the number of tasks, the exam did not become easier because of this. All changes were made with a focus on complication, and a slight decrease in the initial score and an increase in the duration of the exam by half an hour only confirm this.

In our center for additional education Merlin, we have fully analyzed all the changes, made adjustments to the work programs and are already “at the start” for conducting preparation courses for the Unified State Examination 2017 in biology, chemistry and physics.

Changes in the Unified State Examination in Physics, adopted by Rosobrnadzor not so long ago, will come into force in 2017. The main innovation is the complete exclusion of the test part. Starting in 2017, this will also affect chemistry and biology.

USE-2017: main changes

Earlier it became known that almost certainly in 2017 the third exam will be added to the program for passing the Unified State Exam compulsory subject. Prior to this, there were two compulsory academic disciplines that served as a test of the knowledge of all schoolchildren without exception: the Russian language and mathematics. Since 2017, and recently, rumors about this have not subsided, history will be added to them.

The officials, on whose instructions the relevant amendments were made to the exam, indicate that at the present time many young people are not interested in the past and do not know how their ancestors lived, which, according to them, is very bad. So, they believe that it is necessary to know this, so now future students will also be tested for their knowledge in the context of the history of Russia and the world.

USE-2017 in physics: what will change?

Let's get back to physics. The exam in physics in 2017, as we have already said, will change only in that the test part will be left out. It will be replaced by oral and written. We have not yet received specific details on what exactly will change in the tasks.

The cancellation of the test part is the result of a long-term discussion of officials, during which they considered the pros and cons of making this or that decision. In the end, they agreed to approve a positive answer. One of the features of this approach, they believe, will be the complete elimination of the possible guessing of answers. At the same time, the oral and written answers of the applicant will clearly show his consistency and ability to learn.

Prospects for the exam in Russia in the near future

Soon the cancellation of tests will affect other subjects. In addition, we want to note that by 2022 Rosobrnadzor plans to include in passing the exam fourth compulsory subject. By then they will be foreign language. Among the proposed options for the delivery of this academic discipline to date, English, German, French and Spanish have been approved by officials.

It is not difficult to guess in what direction the development of education in the Russian Federation will be directed by such a course of affairs. Today, with the naked eye, you can see how quickly the world is changing, and one of the features of this process is the communication of people representing the interests of the most different states peace. To build close relationships with like-minded people who speak a different language, you need to learn the one that most people have in common. Actually, the four listed in the text earlier are just the same and are among them.

Preparation for the exam

Preparation in the subject of Physics should take a student a little more time, in contrast to how many hours a day he devotes to understanding chemistry and biology, the Russian language and mathematics. Yes, mathematics is somewhat similar to physics - and perhaps the main thing that unites them is formulas - but it must be taken, and physics - on demand - at will, in order to subsequently enter the appropriate university, which will certainly require a positive result achieved by the student in the exam.

I would like to immediately tell all skeptics who are among the people who firmly believe that the Unified State Examination will be canceled in 2017, that they are mistaken in their judgments. This will not happen, at least for another 5-6 years. And then, what will they exchange the exam for, huh? After all, this is the only test of knowledge, which, although strict, is at the same time indicative in many respects.

Where to get knowledge?

It will be necessary to prepare for the exam in physics using the following educational materials: books and reference books. School program gives the student a lot of what he needs to know in the first place, so you should not neglect it - you should listen carefully to the teacher and try to understand everything he says.

In addition to the above educational materials, it will not be superfluous to resort to studying collections with formulas in order to test yourself for a sufficient amount of knowledge in this part of the exam.

Also, as you yourself understand, before the exam in physics in 2017, you just need to purchase collections of problems with 100% necessity. If solutions are already indicated in them, do not be alarmed, on the contrary, this will help you understand how to get the desired result in a particular task. In any case, there will be completely different tasks on the exam, requiring, quite possibly, a non-standard approach to their solution. Therefore, so to speak, to get a hand in this matter will obviously not be superfluous.

You can go to consultations, if there are any in your school, you can hire a tutor. And don't be shy about it. By this you show your readiness for learning and that you are determined to enter the university, where you have dreamed of studying since childhood.

On the official site"Federal Institute of Pedagogical Measurements"provides information on planned changes in the structureKIM USE 2017years, which will also affect the disciplines of the natural cycle.

Changes in KIM in physics in 2017 compared to KIM in 2016.

In 2017, big changes will occur in part 1 of the exam options in physics. Part 2 will be fully preserved in its current form (3 tasks with a short answer + 5 tasks with a detailed solution). In part 1 of the options, tasks with a choice of answers (1 out of 4) will completely disappear - 9 tasks. The number of tasks with a short answer and tasks where you need to choose 2 correct answers out of 5 will increase. The total number of tasks in part 1 is 23 tasks (was 24). Inside the section, tasks will be arranged depending on their form. In task 13, this may not coincide with the sequence of presentation of the material.

Changes in KIM in chemistry in 2017 compared to 2016.

The following changes have been made in the 2017 examination paper compared to the 2016 paper.

The structure of part 1 of the KIM has been fundamentally changed, due to which its greater compliance with the structure of the chemistry course itself has been achieved. The tasks included in this part of the work are grouped into separate thematic blocks. In each of these blocks there are tasks of both basic and elevated levels difficulties. Within each block, tasks are arranged in ascending order of that number. learning activities required for their implementation.

In the examination paper of 2017, the total number of tasks was reduced from 40 (in 2016) to 34. This is primarily due to the fact that the activity basis and the practice-oriented orientation of the content of all tasks of the basic level of complexity have been significantly strengthened, as a result of which the implementation of each of they require a systematic application of generalized knowledge. The change in the total number of tasks in the KIM USE in 2017 was carried out mainly due to a decrease in the number of those tasks, the implementation of which involved the use of similar types of activities.

The assessment scale (from 1 to 2 points) for completing tasks of the basic level of complexity, which test the assimilation of knowledge about the genetic relationship of inorganic and organic substances, has been changed (9 and 17). The initial total score for the performance of the work as a whole will be 60 points (instead of 64 points in 2016).

In general, the adopted changes in the examination work in 2017 are focused on increasing the objectivity of testing the formation of a number of important general educational skills, primarily such as: apply knowledge in the system, independently assess the correctness of the implementation of educational and educational and practical tasks, as well as combine knowledge about chemical objects with an understanding of the mathematical relationship between various physical quantities.

Changes in KIM in biology in 2017 compared to KIM in 2016.

The structure of the examination paper has been optimized.

1. Tasks with a short answer in the form of one digit corresponding to the number of the correct answer were excluded from the examination paper.
2. Reduced the number of tasks from 40 to 28.
3. Reduced maximum primary score from 61 in 2016 to 59 in 2017
4. The duration of the examination work has been increased from 180 to 210 minutes.
5. Part 1 includes new types of tasks that differ significantly in the types of learning activities: filling in the missing elements of a diagram or table, finding correctly indicated symbols in a figure, analyzing and synthesizing information, including those presented in the form of graphs, diagrams and tables with statistical data.

Changes in KIM by geography in 2017 compared to KIM in 2016.

There are no changes in the structure and content of KIM. The maximum score for completing tasks 3, 11, 14, 15 has been increased to 2. The maximum score for completing tasks 9, 12, 13, 19 has been reduced to 1. The maximum primary score has not changed.

Information source: http://fipi.ru/