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Early exam in physics with a solution. Preparation for the exam in physics: examples, solutions, explanations. Unified State Exam

As in the past year, in 2017 there are two "streams" of the unified state examination - an early period (it takes place in the middle of spring) and the main one, which traditionally starts at the end of the school year, in the last days of May. The official draft timetable for the USE "spelled out" all the dates for passing exams in all subjects in both of these periods - including additional reserve days provided for those who for a good reason (illness, coincidence of exam dates, etc.) could not pass the USE within the specified time frame.

Schedule of the early period for the USE - 2017

In 2017, the early "wave" of the unified state examination starts earlier than usual. If last year the peak of the spring exam period fell on the last week of March, then this season the spring break period will be free from the Unified State Exam.

The main dates of the early period are from March 14 to March 24... Thus, by the beginning of the spring school holidays, many "early students" will have time to pass the tests. And this may turn out to be convenient: among the graduates who have the right to take the Unified State Exam in an early wave are the guys who will take part in Russian or international competitions and competitions in May, and during spring break they often leave for sports camps, profile shifts to camps, etc. etc. Shifting exams to an earlier date will allow them to use the latter to its fullest.

Additional (reserve) days the early period of the USE-2017 will be held from 3 to 7 April... At the same time, many will probably have to write exams in reserve terms: if no more than two subjects were submitted on the same day in the last year's schedule on the same day, then in 2017 most elective exams are grouped "by triplets."

Separate days are allocated for only three subjects: the compulsory exam in the Russian language for graduates and all future applicants, as well as mathematics and the oral part of the exam in foreign languages... At the same time, this year, the "early adopters" will hand over the "speaking" before the written part.

It is planned to distribute the March exams by dates as follows:

March 14th(Tuesday) - exam in mathematics (both basic and specialized level);

March 16(Thursday) - chemistry, history, computer science;

18th of March(Saturday) - USE in foreign languages (oral part of the exam);

20th of March(Monday) - Russian language exam;

March 22(Wednesday) - biology, physics, foreign languages (written exam);

March 24(Friday) - Unified State Exam, Literature and Social Studies.

There is a nine-day pause between the main and reserve days of the early period. All additional tests for "reservists" will take place in three days:

April 3(Monday) - chemistry, literature, computer science, foreign (speaking);

April 5(Wednesday) - foreign (in writing), geography, physics, biology, social studies;

7 april(Friday) - Russian, basic and.

As a rule, the bulk of those taking the exam ahead of schedule are graduates of previous years, as well as graduates of secondary specialized educational institutions (in colleges and vocational lyceums, the program high school usually "pass" in the first year of study). In addition, school graduates who, during the main period of the USE, will be absent for valid reasons (for example, to participate in Russian or international competitions or to be treated in a sanatorium), or intend to continue their education outside of Russia, may “shoot out” with exams early.

Graduates of 2017 can also, at their own discretion, choose the date for passing exams in those subjects for which the program has been completed in full. This is true primarily for those who are planning - the school course on this subject is read until grade 10, and early delivery one of the exams can reduce tension during the main period of the exam.

Schedule of the main period for passing the exam - 2017

The main period for passing the exam in 2017 starts on May 26, and by June 16, most of the graduates will have completed the exam epic. For those who could not pass the exam on time for a good reason or chose subjects that coincide in terms of delivery, there are reserve examination days from June 19... Like last year, the last day of the exam period will become a "single reserve" - on June 30 it will be possible to pass an exam in any subject.

At the same time, the schedule of exams for the main period of the USE-2017 is much less dense in comparison with early-term students, and, most likely, most graduates will be able to avoid "overlapping" exams.

Separate examination days are allocated for the delivery of compulsory subjects: the Russian language, mathematics of the basic and specialized level (students have the right to take one of these exams or both at once, therefore, in the schedule of the main period, they are traditionally spread over several days).

As in the previous year, a separate day has been set aside for the most demanded elective exam - social studies. And for passing the oral part of the exam in foreign languages, two separate days are allocated at once. In addition, a separate day is allocated for the less demanded for USE subject- geography. Perhaps this was done in order to spread all subjects of the natural science profile in the schedule, reducing the number of matches.

Thus, in timetable of the exam there are two pairs and one "three" subjects, exams for which will be taken simultaneously:

chemistry, history and informatics;

foreign languages and biology,

literature and physics.

The exams must be passed on the following dates:

26 of May(Friday) - geography,

May 29(Monday) - Russian,

May 31(Wednesday) - history, chemistry, computer science and ICT,

2 June(Friday) - profile mathematics,

June 5th(Monday) - social studies;

June 7(Wednesday) -,

the 9th of June(Friday) - written foreign, biology,

June 13(Tuesday) - literature, physics,

June 15th(Thursday) and June 16(Friday) - foreign oral.

Thus, for graduation evenings, most schoolchildren will prepare "with a clear conscience", having already passed all the planned exams and received results in most subjects. Those who were ill during the main examination period, chose subjects that matched in terms of time, received a "bad" in Russian or mathematics, were removed from the exam, or faced technical or organizational difficulties during the exam (for example, a lack of additional forms or a power outage), will take exams on reserve dates.

Reserve days will be distributed as follows:

June 19(Monday) - computer science, history, chemistry and geography,

June 20(Tuesday) - physics, literature, biology, social studies, written foreign,

21st of June(Wednesday) - Russian,

June, 22(Thursday) - basic math,

June 28(Wednesday) - mathematics at a specialized level,

June 29(Thursday) - oral foreign,

30 June(Friday) - all items.

Can there be changes in the timetable for passing the exam

The draft official timetable for the exam is usually published at the beginning of the academic year, is discussed, and the final approval of the timetable for the exam takes place in the spring. Therefore, changes are possible in the USE schedule for 2017.

However, for example, in 2016 the project was approved without any changes and the actual dates of the exams completely coincided with those announced in advance - both in the early and in the main wave. So chances are high that the 2017 schedule will also be adopted unchanged.

When preparing for the USE, graduates are better off using options from official sources of information support for the final exam.

To understand how you need to perform the exam work, you should first of all familiarize yourself with the demos of the KIM USE in physics of the current year and with the options for the USE of the early period.

On May 10, 2015, in order to provide graduates with an additional opportunity to prepare for the unified state exam in physics, the FIPI website publishes one version of the CMM used for the USE in the early 2017 period. These are real options from the exam held on April 7, 2017.

Early versions of the exam in physics 2017

Demonstration version of the exam 2017 in physics

Option task + answers	variant + otvet
Specification	download
Codifier	download

Demo versions of the exam in physics 2016-2015

Physics	Download option
2016	version of the exam 2016
2015	variant EGE fizika

Changes in the KIM USE in 2017 compared to 2016

The structure of part 1 of the examination paper has been changed, part 2 is left unchanged. Tasks with a choice of one correct answer were excluded from the examination work and tasks with a short answer were added.

When making changes to the structure of the examination work, the general conceptual approaches to the assessment of educational achievements have been preserved. Including, the maximum score for completing all the tasks of the examination work remained unchanged, the distribution of maximum points for tasks of different levels of complexity and the approximate distribution of the number of tasks by sections of the school physics course and methods of activity were preserved.

A complete list of questions that can be controlled at the unified state exam in 2017 is given in the codifier of content elements and requirements for the level of training of graduates educational organizations for the 2017 unified state exam in physics.

The purpose of the demo version of the USE in physics is to enable any USE participant and the general public to get an idea of the structure of future CMMs, the number and form of tasks, and the level of their complexity.

The above criteria for assessing the performance of tasks with a detailed answer included in this option give an idea of the requirements for the completeness and correctness of recording a detailed answer. This information will allow graduates to develop a strategy for preparing and passing the exam.

Approaches to the selection of content, the development of the structure of the KIM USE in physics

Each version of the examination work includes tasks that check the mastery of controlled content elements from all sections of the school physics course, while for each section tasks of all taxonomic levels are proposed. The content elements most important from the point of view of continuing education in higher educational institutions are controlled in the same version by assignments of different levels of complexity.

The number of tasks for a particular section is determined by its content and proportional to the study time allotted for its study in accordance with the approximate physics program. The various plans, according to which the examination variants are constructed, are built according to the principle of substantive addition so that, in general, all series of variants provide diagnostics of the development of all the content elements included in the codifier.

Each option includes tasks for all sections of different levels of complexity, allowing you to test the ability to apply physical laws and formulas both in typical educational situations and in unconventional situations that require a sufficiently high degree of independence when combining known action algorithms or creating your own task execution plan ...

The objectivity of checking tasks with a detailed answer is ensured by uniform assessment criteria, the participation of two independent experts evaluating one work, the possibility of appointing a third expert and the presence of an appeal procedure. The Unified State Examination in Physics is an exam of the choice of graduates and is intended for differentiation in admission to higher education institutions.

For these purposes, the work includes tasks of three levels of complexity. Completing tasks of a basic level of complexity allows you to assess the level of mastering the most significant content elements of a secondary school physics course and mastering the most important types of activities.

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 scores in physics, which confirms that a graduate has mastered the program of secondary (complete) general education in physics, is established 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 makes it possible to assess the degree of a student's preparedness for continuing education at a university.

Preparation for the exam and exam

Secondary general education

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

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

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

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

We disassemble USE assignments in physics (Option C) with a teacher.

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

The work presents tasks of different levels of difficulty: basic, advanced and high. Basic level tasks are simple tasks that test the assimilation of the most important physical concepts, models, phenomena and laws. Tasks of an advanced level are aimed at testing the ability to use the concepts and laws of physics to analyze various processes and phenomena, as well as the ability to solve problems on the application of one or two laws (formulas) for any of the topics of the school physics course. In work, 4 tasks of part 2 are tasks high level difficulties and tests 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 exam 2017, tasks are taken from the open bank of USE tasks.

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

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

S =	(30 + 20) with	10 m / s = 250 m.
	2

Answer. 250 m.

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

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

a =	∆v	=	(8 - 2) m / s	= 2 m / s 2.
	∆t		3 sec

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

+ = (1)

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

T – mg = ma (2);

from formula (2) modulus of tensile force

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

Answer... 1200 N.

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

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

Tr + + = (1)

Having chosen a frame of reference associated with a fixed surface, we write down the equations for the projection of vectors on 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: the sliding friction force tr. and the force with which the body is dragged. The projection of the friction force is negative, since the force vector does not coincide with the direction of the axis NS... Force projection F positive. We remind you that to find the projection, we lower the perpendicular from the beginning and end of the vector to the selected axis. With this in mind, we have: F cosα - F tr = 0; (1) express the projection of the force F, this is F cosα = F tr = 16 N; (2) then the power developed by the force will be equal to N = F cosα V(3) We make a substitution, taking into account equation (2), and substitute the corresponding data into equation (3):

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

Answer. 24 watts

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

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

=	T	;	m	=	T 2	; m = k	T 2	; m= 200 H / m	(4 s) 2	= 81.14 kg ≈ 81 kg.
	2π		k		4π 2		4π 2		39,438

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 weighing 10 kg. Friction is negligible. Based on the analysis of the above figure, select two correct statements and indicate their numbers in the answer.

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

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

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

Answer. 45.

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

Increases;
Decreases;
Doesn't change.

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

V =	m	.
	p

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

Answer. 13.

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

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

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

3) mg cosα

4) sinα -	a
	g cosα

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

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

Let's write down the basic equation of dynamics:

Tr + = (1)

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

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

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

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

μ =	F tr	=	m(g sinα - a)	= tgα -	a	(8).
	N		mg cosα		g cosα

We select the appropriate positions for each letter.

Answer. A - 3; B - 2.

Task 8. Oxygen gas 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. We convert the temperature to Kelvin T = t° С + 273, volume V= 33.2 l = 33.2 · 10 -3 m 3; We translate the pressure P= 150 kPa = 150,000 Pa. Using the ideal gas equation of state

express the mass of the gas.

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

Answer. 48 g

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

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

Answer. 25 J.

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

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

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

φ 1 = 10%; φ 2 = 35%

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

P 2	=	φ 2	=	35	= 3,5
P 1		φ 1		10

Answer. The pressure should be increased by 3.5 times.

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

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

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

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

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

The second true statement is:

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

Answer. 14.

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 has come. How did the temperature of body B and the total internal energy of body A and B change as a result?

For each value, determine the corresponding change pattern:

Increased;
Decreased;
Hasn't changed.

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

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

∆U = ∑	n	∆U i = 0 (1);
	i = 1

where ∆ U- change in internal energy.

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

Answer. 23.

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

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

Answer. from the observer.

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

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

where d Is the distance between the plates.

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

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

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

Answer. 20 μC.

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

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

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

sinα	=	n 2	,
sinβ		n 1

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

Answer.

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

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

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

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

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

Answer. 13.

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

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

The EMF formula of self-induction has the form

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

∆t= 10 s - 5 s = 5 s

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

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

Substituting numerical values into formula (2), we obtain

| Ɛ | = 2 · 10 –6 V, or 2 µV.

Answer. 2.

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

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

Let's write the law of refraction

sinβ =	sin50	= 0,4327 ≈ 0,433
	1,77

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

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

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

Answer. 24.

Determine how many α - particles and how many protons are produced by a thermonuclear fusion reaction

+ → x+ y;

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

+ → x + y;

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

Answer. 1 - α -particle; 2 - proton.

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

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

E = pc (3),

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

E 2	=	p 2	= 8,18;
E 1		p 1

We round the answer to tenths and get 8.2.

Answer. 8,2.

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

For each value, determine the corresponding change pattern:

Increased;
Decreased;
Hasn't changed.

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

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

Answer. 21.

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

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

d sinφ = kλ (1),

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

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

Answer. 42.

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

For each value, determine the corresponding change pattern:

Will increase;
Will decrease;
Will not change.

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

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

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

U = I R (3).

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

Answer. 13.

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

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

T= 2π (1);

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

By condition

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

Answer. 14.4 m / s 2.

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

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

F A = I LB sinα;

F A = 0.6 N

Answer. F A = 0.6 N.

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

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

W m =	LI 2	(1);
	2

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

I 1 2 =	2W 1	; I 2 2 =	2W 2	;
	L		L

Then the ratio of currents

I 2 2	= 49;	I 2	= 7
I 1 2		I 1

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

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

Solution. 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. According to Lenz's rule, the magnetic field created by the induction current of the loop must be directed to the right. According to the rule of the gimbal, the current should flow clockwise (when viewed from the left). A diode in the circuit of the second lamp passes in this direction. This means that the second lamp will light up.

Answer. The second lamp comes on.

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

Solution. Let's make an explanatory drawing.

- Thread tension force;

- Force of reaction of the bottom of the vessel;

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

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

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

F a = Slρ in g (2)

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

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

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

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

NL cosα + Slρ in g (L –	l	) cosα = SLρ a g	L	cosα (7)
	2		2

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

F d = [	1	Lρ a– (1 –	l	)lρ in] Sg (8).
	2		2L

Substitute numerical data and get that

F d = 0.025 N.

Answer. F d = 0.025 N.

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

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

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

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

m 2 =	m 1		M 2		T 1	(5).
	5		M 1		T 2

After substitution of numeric data m 2 = 28 g.

Answer. m 2 = 28 g.

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

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

C	U 2	+ L	I 2	= L	I m 2	(1)
	2		2		2

For the amplitude (maximum) values, we write

and from equation (2) we express

C	=	I m 2	(4).
L		U m 2

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

I = I m (5)

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

I= 4.0 mA.

Answer. I= 4.0 mA.

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

Solution. Let's make an explanatory drawing

α is the angle of incidence of the beam;

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

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

According to the law of refraction of light

sinβ =	sinα	(3)
	n 2

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

tgβ = h tgβ = h	sinα	= h	sinβ	= h	sinα	(4)
	cosβ

We get the following expression:

AC = 2 DB = 2 h	sinα	(5)

Substitute the numerical values into the resulting formula (5)

Answer. 1.63 m.

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

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