Specification
control measuring materials
for holding the unified state exam in 2018
in PHYSICS

1. Appointment of KIM USE

The Unified State Examination (hereinafter referred to as the USE) is a form of objective assessment of the quality of training of persons who have mastered educational programs secondary general education, using tasks of a standardized form (control measuring materials).

The USE is conducted in accordance with Federal Law No. 273-FZ of December 29, 2012 “On Education in the Russian Federation”.

Control measuring materials allow you to establish the level of development by graduates of the Federal component of the state educational standard secondary (complete) general education in physics, basic and specialized levels.

The results of the unified state exam in physics are recognized educational organizations middle vocational education and educational organizations of higher professional education as the results of entrance examinations in physics.

2. Documents defining the content of KIM USE

3. Approaches to the selection of content, the development of the structure of the KIM USE

Each version of the examination paper includes controlled content elements from all sections school course physics, while for each section tasks of all taxonomic levels are offered. The most important content elements from the point of view of continuing education in higher educational institutions are controlled in the same variant by tasks of different levels of complexity. The number of tasks for a particular section is determined by its content content and in proportion to the study time allotted for its study in accordance with an exemplary program in physics. Various plans according to which are constructed exam options, are built on the principle of content addition so that, in general, all series of variants provide diagnostics for the development of all content elements included in the codifier.

The priority in the design of CMM is the need to verify the types of activities provided for by the standard (taking into account the limitations in the conditions of mass written testing of knowledge and skills of students): mastering the conceptual apparatus of a physics course, mastering methodological knowledge, applying knowledge in explaining physical phenomena and solving problems. Mastering the skills to work with information of physical content is checked indirectly when using various methods of presenting information in texts (graphs, tables, diagrams and schematic drawings).

The most important activity in terms of successful continuation of education at the university is problem solving. Each option includes tasks for all sections different levels difficulties that allow you to check the ability to apply physical laws and formulas both in typical educational situations and in non-traditional situations that require sufficient manifestation high degree 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 evaluation criteria, the participation of two independent experts evaluating one work, the possibility of appointing a third expert and the existence of an appeal procedure.

The Unified State Examination in Physics is an exam of the choice of graduates and is designed to differentiate when entering higher educational establishments. For these purposes, tasks of three levels of complexity are included in the work. Completing tasks basic level complexity allows you to assess the level of development of the most significant content elements of the course of physics high school and mastery of the most important species 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 points in physics, which confirms that the graduate has mastered the program of secondary (complete) general education in physics, is set based on the requirements for mastering the basic level standard. The use of tasks of increased and high levels complexity allows you to assess the degree of readiness of the student to continue education at the university.

4. The structure of KIM USE

Each version of the examination paper consists of two parts and includes 32 tasks that differ in form and level of complexity (Table 1).

Part 1 contains 24 short answer tasks. Of these, 13 tasks with a record of the answer in the form of a number, a word or two numbers. 11 matching and multiple choice tasks in which the answers must be written as a sequence of numbers.

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

The Unified State Examination is one of the most discussed topics in the Russian pedagogical community. Future graduates and teachers who are to prepare students for the USE are already wondering what the USE in physics will be like in the coming 2018 and whether we should expect any global changes in the structure examination papers or test format. Physics has always stood apart, and the exam in it is traditionally considered much more difficult than in other school subjects. At the same time, the successful passing of the exam in physics is a ticket to most technical universities.

At the moment, there is no official information about the adoption of any significant changes to the structure of the USE in 2018. The Russian language and mathematics remain compulsory, and physics is included in an extensive list of subjects that graduates can choose for themselves additionally, focusing on the requirements of the university they plan to enter.

In 2017, 16.5% of all 11th graders in the country chose physics. Such popularity of the subject is not accidental. Physics is necessary for everyone who plans to enter engineering specialties or connect their lives withIT-technologies, geology, aviation and many other areas that are popular today.

Launched by the Minister of Education and Science Olga Vasilyeva back in 2016, the process of modernizing the final certification procedure is actively continuing, from time to time information about possible innovations leaks into the media, such as:

1. Expansion of the list of subjects required for delivery by disciplines: physics, history and geography.
2. Introduction of a unified integrated examination in the natural sciences.

While discussions are underway on the proposals made, current high school students should thoroughly prepare for passing the most relevant USE bundle - mathematics profile level+ physics.

Is it worth specifying that mainly students of profile classes with in-depth study subjects of the mathematical cycle.

The structure of the examination paper in physics in 2018

The main session of the USE in the 2017-2018 academic year is planned from 05/28/18 to 07/09/18, but specific test dates for each subject have not yet been announced.

In 2017, exam papers have changed significantly compared to 2016.

Changes in the exam in physics in 2018

Tests have been completely removed from the tasks, leaving the possibility of a thoughtless choice of answer. Instead, students were offered tasks with a short or detailed answer. It is safe to say that in the 2017-2018 academic USE year in physics will not differ much in the structure and volume of assignments from last year. which means that:

• 235 minutes will be allotted to complete the work;
• in total, graduates will have to cope with 32 tasks;
• I block of the Unified State Examination (27 tasks) - tasks with a short answer, which can be represented by an integer, decimal or a numerical sequence;
• Block II (5 tasks) - tasks that require a similar description of the train of thought in the process of solving and justifying decisions taken based on physical laws and regularities;
• the minimum passing threshold is 36 points, which is equivalent to 10 correctly solved tasks from block I.

It is the last five tasks, from 27 to 31, that are the most difficult at the Unified State Exam in physics, and many students turn in work with empty fields in them. But there is very important nuance- if you read the rules for evaluating these tasks, it will become clear that by writing a partial explanation of the task and showing the correct direction of the train of thought, you can get 1 or 2 points, which many lose just like that, not reaching the full answer and not writing anything in the solution.

To solve most of the problems of their course of the subject "physics", not only a good knowledge of the laws and an understanding of physical processes, but also a good mathematical background is necessary, and therefore it is worth asking the question of expanding and deepening knowledge long before the upcoming USE 2018.

The ratio of theoretical and practical tasks in the examination papers is 3: 1, which means that in order to successfully pass, first of all, you need to master the basic physical laws and know all the formulas from the school course of mechanics, thermodynamics, electrodynamics, optics, as well as molecular, quantum and nuclear physics.

You should not count on cheat sheets and various other tricks. Using notepads with formulas, calculators and others technical means than so many students sin at school control work, is not allowed on the exam. Remember that compliance with this rule is monitored not only by observers, but also by the tireless eyes of video cameras located in such a way as to notice every questionable movement of the examinee.

You can prepare for the exam in physics by contacting an experienced teacher, or by repeating the school curriculum on your own again.

Teachers who teach the subject in specialized lyceums give such simple but effective advice:

1. Do not try to memorize complex formulas, try to understand their nature. Knowing how the formula was derived, you can easily write it out in a draft, while thoughtless memorization is fraught with mechanical errors.
2. Start solving the problem by deriving the final expression in literal form and only then look for the answer mathematically.
3. "Stuff your hand." The more different types of tasks on the topic you solve, the easier it will be to cope with the tasks of the exam.
4. Start preparing for the exam in physics at least a year before the exam. This is not the kind of subject that you can take "impudently" and learn another in a month, even with the best tutors.
5. Do not get hung up on the same type of simple tasks. Tasks for 1-2 formulas are only 1 stage. Unfortunately, many teachers in schools simply do not go further, descending to the level of the majority of students or relying on the fact that students in the humanities classes will not choose a subject that is not their profile when passing the USE. Solve problems that combine laws from different branches of physics.
6. Once again, repeat the physical quantities and their transformation. When solving problems, be especially attentive to the format in which the data is presented and, if necessary, do not forget to bring them to the desired form.

Excellent assistants in preparing for the exam in physics will be trial versions of examination tasks, as well as tasks on various topics that can be easily found on the net today. First of all, this is the FIPI website, where the USE archive in physics for 2008-17 with codifiers is located.

For more information about the changes that have already taken place in the USE and how to prepare for the exam, see video interview with Marina Demidova, Head of the Federal Commission for the Development of Assignments and USE in physics:

In anticipation school year demo versions of KIM USE 2018 in all subjects (including physics) have been published on the official website of FIPI.

This section presents documents that determine the structure and content of KIM USE 2018:

Demonstration options for control measuring materials of the unified state exam.
- codifiers of content elements and requirements for the level of training of graduates of educational institutions for the unified state exam;
- specifications of control measuring materials for the unified state examination;

Demo version of the exam 2018 in physics assignments with answers

Physics demo version of the exam 2018	option+answer
Specification	download
Codifier	download

Changes in KIM USE in 2018 in physics compared to 2017

Subsection 5.4 "Elements of Astrophysics" is included in the codifier of content elements tested at the Unified State Examination in Physics.

One multiple choice task has been added to part 1 of the exam paper, testing elements of astrophysics. The content of task lines 4, 10, 13, 14 and 18 has been expanded. Part 2 has been left unchanged. Maximum score for the performance of all tasks of the examination paper increased from 50 to 52 points.

USE duration 2018 in Physics

235 minutes are allotted to complete the entire examination paper. Estimated time to complete the tasks of various parts of the work is:

1) for each task with a short answer - 3-5 minutes;

2) for each task with a detailed answer - 15–20 minutes.

Structure of KIM USE

Each version of the examination paper consists of two parts and includes 32 tasks that differ in form and level of complexity.

Part 1 contains 24 short answer tasks. Of these, 13 tasks with the answer written as a number, word or two numbers, 11 tasks for establishing correspondence and multiple choice, in which the answers must be written as a sequence of numbers.

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

Secondary general education

Getting ready for the Unified State Exam-2018: analysis of the demo version in physics

We bring to your attention an analysis of the tasks of the exam in physics from the demo version of 2018. The article contains explanations and detailed algorithms for solving tasks, as well as recommendations and links to useful materials that are relevant in preparing for the exam.

USE-2018. Physics. Thematic training tasks

The edition contains:
tasks of different types for all USE topics;
answers to all questions.
The book will be useful both for teachers: it makes it possible to effectively organize the preparation of students for the exam directly in the classroom, in the process of studying all topics, and for students: training tasks will allow you to systematically, when passing each topic, prepare for the exam.

A point body at rest begins to move along the axis Ox. The figure shows a projection dependency graph ax acceleration of this body with time t.

Determine the distance traveled by the body in the third second of motion.

Answer: _________ m.

Solution

Being able to read graphs is very important for every student. The question in the problem is that it is required to determine from the graph the dependence of the projection of acceleration on time, the path that the body has traveled in the third second of motion. The graph shows that in the time interval from t 1 = 2 s to t 2 = 4 s, acceleration projection is zero. Consequently, the projection of the resultant force in this area, according to Newton's second law, is also equal to zero. We determine the nature of the movement in this area: the body moved uniformly. The path is easy to determine, knowing the speed and time of movement. However, in the interval from 0 to 2 s, the body moved uniformly accelerated. Using the definition of acceleration, we write the velocity projection equation Vx = V 0x + a x t; since the body was initially at rest, then the velocity projection by the end of the second second became

Then the path traveled by the body in the third second

Answer: 8 m

Rice. 1

On a smooth horizontal surface lie two bars connected by a light spring. To a bar of mass m= 2 kg apply a constant force equal in modulus F= 10 N and directed horizontally along the axis of the spring (see figure). Determine the modulus of the elastic force of the spring at the moment when this bar moves with an acceleration of 1 m / s 2.

Answer: _________ N.

Solution

Horizontally on a body of mass m\u003d 2 kg, two forces act, this is the force F= 10 N and elastic force, from the side of the spring. The resultant of these forces imparts acceleration to the body. We choose a coordinate line and direct it along the action of the force F. Let's write down Newton's second law for this body.

Projected onto axis 0 X: F – F extr = ma (2)

We express from formula (2) the modulus of the elastic force F extr = F – ma (3)

Substitute the numerical values ​​into formula (3) and get, F control \u003d 10 N - 2 kg 1 m / s 2 \u003d 8 N.

Answer: 8 N.

Task 3

A body with a mass of 4 kg, located on a rough horizontal plane, was reported along it with a speed of 10 m / s. Determine the modulus of work done by the friction force from the moment the body begins to move until the moment when the speed of the body decreases by 2 times.

Answer: _________ J.

Solution

The force of gravity acts on the body, the reaction force of the support is the friction force that creates a braking acceleration. The body was initially reported with a speed equal to 10 m / s. Let's write down Newton's second law for our case.

Equation (1) taking into account the projection on the selected axis Y will look like:

N – mg = 0; N = mg (2)

In the projection on the axis X: –F tr = - ma; F tr = ma; (3) We need to determine the modulus of work of the friction force by the time when the speed becomes half as much, i.e. 5 m/s. Let's write a formula for calculating work.

A · ( F tr) = – F tr S (4)

To determine the distance traveled, we take the timeless formula:

S =	v 2 - v 0 2	(5)
	2a

Substitute (3) and (5) into (4)

Then the modulus of work of the friction force will be equal to:

Let's substitute numerical values

A(F tr) =		4 kg	((	5 m	) 2 – (10	m	) 2)	= 150 J
		2		With		With

Answer: 150 J

USE-2018. Physics. 30 practice exam papers

The edition contains:
30 training options for the exam
instructions for implementation and evaluation criteria
answers to all questions
Training options will help the teacher to organize preparation for the exam, and students to independently test their knowledge and readiness for the final exam.

The stepped block has an outer pulley with a radius of 24 cm. Weights are suspended from the threads wound on the outer and inner pulleys as shown in the figure. There is no friction in the axis of the block. What is the radius of the inner pulley of the block if the system is in equilibrium?

Rice. 1

Answer: _________ see

Solution

According to the condition of the problem, the system is in equilibrium. On the image L 1 , shoulder strength L 2 shoulder of force Balance condition: the moments of forces rotating the bodies clockwise must be equal to the moments of forces rotating the body counterclockwise. Recall that the moment of force is the product of the modulus of force and the arm. The forces acting on the threads from the side of the loads differ by a factor of 3. This means that the radius of the inner pulley of the block differs from the outer one also by 3 times. Therefore, the shoulder L 2 will be equal to 8 cm.

Answer: 8 cm

Task 5

Oh, at different times.

Select from the list below two correct statements and indicate their numbers.

1. The potential energy of the spring at time 1.0 s is maximum.
2. The period of oscillation of the ball is 4.0 s.
3. The kinetic energy of the ball at time 2.0 s is minimal.
4. The amplitude of the ball oscillations is 30 mm.
5. The total mechanical energy of the pendulum, consisting of a ball and a spring, is at a minimum at 3.0 s.

Solution

The table shows data on the position of a ball attached to a spring and oscillating along a horizontal axis. Oh, at different times. We need to analyze this data and choose the right two statements. The system is a spring pendulum. At the point in time t\u003d 1 s, the displacement of the body from the equilibrium position is maximum, which means this is the amplitude value. by definition, the potential energy of an elastically deformed body can be calculated by the formula

Ep = k	x 2	,
	2

Where k- coefficient of spring stiffness, X- displacement of the body from the equilibrium position. If the displacement is maximum, then the speed at this point is zero, which means that the kinetic energy will be zero. According to the law of conservation and transformation of energy, potential energy should be maximum. From the table we see that the body passes half of the oscillation for t= 2 s, total oscillation in twice the time T= 4 s. Therefore statements 1 will be true; 2.

Task 6

A small piece of ice was lowered into a cylindrical glass of water to float. After some time, the ice completely melted. Determine how the pressure on the bottom of the glass and the water level in the glass have changed as a result of the melting of the ice.

1. increased;
2. decreased;
3. hasn't changed.

Write in table

Solution

Rice. 1

Problems of this type are quite common in different USE options. And as practice shows, students often make mistakes. Let's try to analyze this task in detail. Denote m is the mass of a piece of ice, ρ l is the density of ice, ρ w is the density of water, V pt is the volume of the immersed part of the ice, equal to the volume of the displaced liquid (volume of the hole). Mentally remove the ice from the water. Then a hole will remain in the water, the volume of which is equal to V pm, i.e. volume of water displaced by a piece of ice 1( b).

Let us write down the condition of ice floating Fig. 1( A).

Fa = mg (1)

ρ in V pm g = mg (2)

Comparing formulas (3) and (4) we see that the volume of the hole is exactly equal to the volume of water obtained from the melting of our piece of ice. Therefore, if we now (mentally) pour the water obtained from ice into the hole, then the hole will be completely filled with water, and the water level in the vessel will not change. If the water level does not change, then the hydrostatic pressure (5), which in this case depends only on the height of the liquid, will also not change. Therefore, the answer will be

USE-2018. Physics. Training tasks

The publication is addressed to high school students to prepare for the exam in physics.
The allowance includes:
20 training options
answers to all questions
USE answer forms for each option.
The publication will assist teachers in preparing students for the exam in physics.

A weightless spring is located on a smooth horizontal surface and is attached to the wall at one end (see figure). At some point in time, the spring begins to deform, applying an external force to its free end A and uniformly moving point A.

Establish a correspondence between the graphs of dependences of physical quantities on deformation x springs and these values. For each position in the first column, select the corresponding position from the second column and write in table

Solution

It can be seen from the figure for the problem that when the spring is not deformed, then its free end, and accordingly point A, are in a position with the coordinate X 0 . At some point in time, the spring begins to deform, applying an external force to its free end A. Point A moves uniformly. Depending on whether the spring is stretched or compressed, the direction and magnitude of the elastic force arising in the spring will change. Accordingly, under the letter A), the graph is the dependence of the elastic modulus on the deformation of the spring.

The graph under the letter B) is the dependence of the projection of the external force on the magnitude of the deformation. Because with an increase in the external force, the magnitude of the deformation and the elastic force increase.

Answer: 24.

Task 8

When constructing the Réaumur temperature scale, it is assumed that at normal atmospheric pressure, ice melts at a temperature of 0 degrees Réaumur (°R), and water boils at a temperature of 80°R. Find the average kinetic energy of the translational thermal motion of an ideal gas particle at a temperature of 29°R. Express your answer in eV and round to the nearest hundredth.

Answer: _______ eV.

Solution

The problem is interesting in that it is necessary to compare two temperature measurement scales. These are the Réaumur temperature scale and the Celsius temperature scale. The melting points of ice are the same on the scales, but the boiling points are different, we can get a formula for converting degrees Réaumur to degrees Celsius. This

Let's convert the temperature of 29 (°R) to degrees Celsius

We translate the result into Kelvin using the formula

T = t°C + 273 (2);

T= 36.25 + 273 = 309.25 (K)

To calculate the average kinetic energy of the translational thermal motion of particles of an ideal gas, we use the formula

Where k– Boltzmann constant equal to 1.38 10 –23 J/K, T – absolute temperature on the Kelvin scale. It can be seen from the formula that the dependence of the average kinetic energy on temperature is direct, that is, how many times the temperature changes, the average kinetic energy of the thermal motion of molecules changes so many times. Substitute the numerical values:

The result is converted to electron volts and rounded to the nearest hundredth. Let's remember that

1 eV \u003d 1.6 10 -19 J.

For this

Answer: 0.04 eV.

One mole of a monatomic ideal gas is involved in process 1–2, the graph of which is shown in VT-diagram. Determine for this process the ratio of the change in the internal energy of the gas to the amount of heat imparted to the gas.

Answer: ___________ .

Solution

According to the condition of the problem in process 1–2, the graph of which is shown in VT-diagram, one mole of a monatomic ideal gas is involved. To answer the question of the problem, it is necessary to obtain expressions for changing the internal energy and the amount of heat imparted to the gas. Isobaric process (Gay-Lussac law). The change in internal energy can be written in two forms:

For the amount of heat imparted to the gas, we write the first law of thermodynamics:

Q 12 = A 12+∆ U 12 (5),

Where A 12 - gas work during expansion. By definition, work is

A 12 = P 0 2 V 0 (6).

Then the amount of heat will be equal, taking into account (4) and (6).

Q 12 = P 0 2 V 0 + 3P 0 · V 0 = 5P 0 · V 0 (7)

Let's write the relation:

Answer: 0,6.

The handbook contains in full the theoretical material on the course of physics, necessary for passing the exam. The structure of the book corresponds to the modern codifier of the content elements in the subject, on the basis of which the examination tasks are compiled - control and measuring materials (CMM) of the Unified State Examination. The theoretical material is presented in a concise, accessible form. Each topic is accompanied by examples of exam tasks corresponding to the USE format. This will help the teacher organize preparation for a single state exam and for students to independently test their knowledge and readiness to pass the final exam.

A blacksmith forges an iron horseshoe weighing 500 g at a temperature of 1000°C. Having finished forging, he throws the horseshoe into a vessel of water. There is a hiss, and steam rises from the vessel. Find the mass of water that evaporates when a hot horseshoe is immersed in it. Consider that the water is already heated to boiling point.

Answer: _________

Solution

To solve the problem, it is important to remember the heat balance equation. If there are no losses, then heat transfer of energy occurs in the system of bodies. As a result, the water evaporates. Initially, the water was at a temperature of 100 ° C, which means that after the hot horseshoe is immersed, the energy received by the water will go immediately to vaporization. We write the heat balance equation

With and · m P · ( t n - 100) = lm in 1),

Where L is the specific heat of vaporization, m c is the mass of water that has turned into steam, m p is the mass of the iron horseshoe, With g is the specific heat capacity of iron. From formula (1) we express the mass of water

When recording the answer, pay attention to what units you want to leave the mass of water.

Answer: 90

One mole of a monatomic ideal gas is involved in a cyclic process, the graph of which is shown in TV- diagram.

Select two correct statements based on the analysis of the presented graph.

1. Gas pressure in state 2 is greater than gas pressure in state 4
2. The gas work in section 2–3 is positive.
3. In section 1–2, the gas pressure increases.
4. In section 4–1, a certain amount of heat is removed from the gas.
5. The change in the internal energy of the gas in section 1–2 is less than the change in the internal energy of the gas in section 2–3.

Solution

This type of task tests the ability to read graphs and explain the presented dependence of physical quantities. It is important to remember how dependency graphs look for isoprocesses in different axes, in particular R= const. In our example on TV The diagram shows two isobars. Let's see how the pressure and volume will change at a fixed temperature. For example, for points 1 and 4 lying on two isobars. P 1 . V 1 = P 4 . V 4 , we see that V 4 > V 1 means P 1 > P 4 . State 2 corresponds to pressure P 1 . Consequently, the gas pressure in state 2 is greater than the gas pressure in state 4. In section 2–3, the process is isochoric, the gas does no work, it is equal to zero. The assertion is incorrect. In section 1-2, the pressure increases, also incorrect. Just above we showed that this is an isobaric transition. In section 4–1, a certain amount of heat is removed from the gas in order to maintain the temperature constant when the gas is compressed.

Answer: 14.

The heat engine works according to the Carnot cycle. The temperature of the refrigerator of the heat engine was increased, leaving the temperature of the heater the same. The amount of heat received by the gas from the heater per cycle has not changed. How did the efficiency of the heat engine and the work of the gas per cycle change?

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

1. increased
2. decreased
3. hasn't changed

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

Solution

Heat engines operating on the Carnot cycle are often found in assignments on the exam. First of all, you need to remember the formula for calculating the efficiency factor. Be able to record it through the temperature of the heater and the temperature of the refrigerator

in addition to be able to write the efficiency through the useful work of the gas A g and the amount of heat received from the heater Q n.

We carefully read the condition and determined which parameters were changed: in our case, we increased the temperature of the refrigerator, leaving the temperature of the heater the same. Analyzing formula (1), we conclude that the numerator of the fraction decreases, the denominator does not change, therefore, the efficiency of the heat engine decreases. If we work with formula (2), we will immediately answer the second question of the problem. The work of the gas per cycle will also decrease, with all the current changes in the parameters of the heat engine.

Answer: 22.

negative charge - qQ and negative- Q(see picture). Where is it directed relative to the picture ( right, left, up, down, towards the observer, away from the observer) charge acceleration - q in this moment of time, if only charges act on it + Q And –Q? Write your answer in word(s)

Solution

Rice. 1

negative charge - q is in the field of two fixed charges: positive + Q and negative- Q, as shown in the figure. in order to answer the question of where the acceleration of the charge is directed - q, at the moment when only +Q and - charges act on it Q it is necessary to find the direction of the resulting force, as a geometric sum of forces According to Newton's second law, it is known that the direction of the acceleration vector coincides with the direction of the resulting force. The figure shows a geometric construction to determine the sum of two vectors. The question arises why the forces are directed in this way? Recall how similarly charged bodies interact, they repel each other, the Coulomb force of the interaction of charges is the central force. the force with which oppositely charged bodies attract. From the figure, we see that the charge is q equidistant from fixed charges whose moduli are equal. Therefore, modulo will also be equal. The resulting force will be directed relative to the figure down. Charge acceleration will also be directed - q, i.e. down.

Answer: Down.

The book contains materials for the successful passing of the exam in physics: brief theoretical information on all topics, tasks of different types and levels of complexity, problem solving advanced level difficulties, answers and evaluation criteria. Students do not have to search Additional information on the Internet and buy other benefits. In this book, they will find everything they need to independently and effectively prepare for the exam. The publication contains tasks of various types on all topics tested at the exam in physics, as well as solving problems of an increased level of complexity. The publication will provide invaluable assistance to students in preparing for the exam in physics, and can also be used by teachers in organizing the educational process.

Two resistors connected in series with a resistance of 4 ohms and 8 ohms are connected to a battery, the voltage at the terminals of which is 24 V. What thermal power is released in a resistor of a smaller rating?

Answer: _________ Tue.

Solution

To solve the problem, it is desirable to draw a series connection diagram of resistors. Then remember the laws of the series connection of conductors.

The scheme will be as follows:

Where R 1 = 4 ohm, R 2 = 8 ohms. The voltage at the battery terminals is 24 V. When the conductors are connected in series, the current strength will be the same in each section of the circuit. The total resistance is defined as the sum of the resistances of all resistors. According to Ohm's law for the circuit section we have:

To determine the thermal power released on a resistor of a smaller rating, we write:

P = I 2 R\u003d (2 A) 2 4 Ohm \u003d 16 W.

Answer: P= 16 W.

A wire frame with an area of ​​2 · 10–3 m 2 rotates in a uniform magnetic field around an axis perpendicular to the magnetic induction vector. The magnetic flux penetrating the frame area changes according to the law

Ф = 4 10 –6 cos10π t,

where all quantities are expressed in SI. What is the modulus of magnetic induction?

Answer: ________________ mT.

Solution

The magnetic flux changes according to the law

Ф = 4 10 –6 cos10π t,

where all quantities are expressed in SI. You need to understand what magnetic flux is in general and how this value is related to the magnetic induction modulus B and frame area S. Let's write the equation in general view to understand what quantities are included in it.

Φ = Φ m cosω t(1)

Remember that before the cos or sin sign there is an amplitude value of a changing value, which means Φ max \u003d 4 10 -6 Wb, on the other hand, the magnetic flux is equal to the product of the magnetic induction modulus and the circuit area and the cosine of the angle between the normal to the circuit and the magnetic induction vector Φ m = IN · S cosα, the flux is maximum at cosα = 1; express the modulus of induction

The answer must be written in mT. Our result is 2 mT.

Answer: 2.

The section of the electrical circuit is a series-connected silver and aluminum wires. Through them flows a constant electricity with a force of 2 A. The graph shows how the potential φ changes in this section of the circuit when it is displaced along the wires by a distance x

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

1. The cross-sectional areas of the wires are the same.
2. Cross-sectional area of ​​silver wire 6.4 10 -2 mm 2
3. Cross-sectional area of ​​silver wire 4.27 10 -2 mm 2
4. A thermal power of 2 W is released in the aluminum wire.
5. Silver wire produces less thermal power than aluminum wire.

Solution

The answer to the question in the problem will be two correct statements. To do this, let's try to solve a few simple problems using a graph and some data. The section of the electrical circuit is a series-connected silver and aluminum wires. A constant electric current of 2 A flows through them. The graph shows how the potential φ changes in this section of the circuit when it is displaced along the wires by a distance x. The specific resistances of silver and aluminum are 0.016 μΩ m and 0.028 μΩ m, respectively.

The wires are connected in series, therefore, the current strength in each section of the circuit will be the same. The electrical resistance of the conductor depends on the material from which the conductor is made, the length of the conductor, the cross-sectional area of ​​\u200b\u200bthe wire

R =	ρ l	(1),
	S

where ρ is the resistivity of the conductor; l- conductor length; S- cross-sectional area. It can be seen from the graph that the length of the silver wire L c = 8 m; aluminum wire length L a \u003d 14 m. Voltage on the section of silver wire U c \u003d Δφ \u003d 6 V - 2 V \u003d 4 V. Voltage in the section of aluminum wire U a \u003d Δφ \u003d 2 V - 1 V \u003d 1 V. By condition, it is known that a constant electric current of 2 A flows through the wires, knowing the voltage and current strength, we determine electrical resistance according to Ohm's law for the circuit section.

It is important to note that the numerical values ​​must be in the SI system for calculations.

Correct statement 2.

Let's check the expressions for power.

P a = I 2 · R a(4);

P a \u003d (2 A) 2 0.5 Ohm \u003d 2 W.

Answer:

The reference book contains in full the theoretical material on the course of physics, which is necessary for passing the exam. The structure of the book corresponds to the modern codifier of the content elements in the subject, on the basis of which the examination tasks are compiled - control and measuring materials (CMM) of the Unified State Examination. The theoretical material is presented in a concise, accessible form. Each topic is accompanied by examples of exam tasks corresponding to the USE format. This will help the teacher to organize preparation for the unified state exam, and students to independently test their knowledge and readiness for the final exam. At the end of the manual, answers are given to tasks for self-examination, which will help schoolchildren and applicants to objectively assess the level of their knowledge and the degree of preparedness for the certification exam. The manual is addressed to senior students, applicants and teachers.

A small object is located on the main optical axis of a thin converging lens between the focal length and twice the focal length from it. The object is brought closer to the focus of the lens. How does this change the image size and optical power of the lens?

For each quantity, determine the appropriate nature of its change:

1. increases
2. decreases
3. does not change

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

Solution

The object is located on the main optical axis of a thin converging lens between the focal and double focal lengths from it. The object begins to be brought closer to the focus of the lens, while the optical power of the lens does not change, since we do not change the lens.

D =	1	(1),
	F

Where F is the focal length of the lens; D is the optical power of the lens. To answer the question of how the image size will change, it is necessary to build an image for each position.

Rice. 1

Rice. 2

We built two images for two positions of the subject. It is obvious that the size of the second image has increased.

Answer: 13.

The figure shows a DC circuit. The internal resistance of the current source can be neglected. Establish a correspondence between physical quantities and formulas by which they can be calculated ( - EMF of the current source; R is the resistance of the resistor).

For each position of the first column, select the corresponding position of the second and write in table selected numbers under the corresponding letters.

Solution

Rice.1

By the condition of the problem, we neglect the internal resistance of the source. The circuit contains a constant current source, two resistors, resistance R, each and key. The first condition of the problem requires determining the current strength through the source with the key closed. If the key is closed, then the two resistors will be connected in parallel. Ohm's law for a complete circuit in this case will look like:

Where I- current strength through the source with the key closed;

Where N- the number of conductors connected in parallel, with the same resistance.

– EMF of the current source.

Substitute (2) in (1) we have: this is the formula under the number 2).

According to the second condition of the problem, the key must be opened, then the current will flow through only one resistor. Ohm's law for a complete circuit in this case will be of the form:

Solution

Let's write down the nuclear reaction for our case:

As a result of this reaction, the law of conservation of charge and mass number is fulfilled.

Z = 92 – 56 = 36;

M = 236 – 3 – 139 = 94.

Therefore, the charge of the nucleus is 36, and the mass number of the nucleus is 94.

New directory contains all the theoretical material for the course of physics required to pass the unified state exam. It includes all elements of the content, checked by control and measuring materials, and helps to generalize and systematize the knowledge and skills of the school physics course. The theoretical material is presented in a concise and accessible form. Each topic is accompanied by examples. test items. Practical tasks correspond to the USE format. Answers to the tests are given at the end of the manual. The manual is addressed to schoolchildren, applicants and teachers.

Period T The half-life of the potassium isotope is 7.6 min. Initially, the sample contained 2.4 mg of this isotope. How much of this isotope will remain in the sample after 22.8 minutes?

Answer: _________ mg.

Solution

The task is to use the law of radioactive decay. It can be written in the form

Where m 0 is the initial mass of the substance, t is the time it takes for a substance to decay T- half life. Let's substitute numerical values

Answer: 0.3 mg.

A beam of monochromatic light falls on a metal plate. In this case, the phenomenon of the photoelectric effect is observed. The graphs in the first column show the dependences of the energy on the wavelength λ and light frequency ν. Establish a correspondence between the graph and the energy for which it can determine the presented dependence.

For each position in the first column, select the corresponding position from the second column and write in table selected numbers under the corresponding letters.

Solution

It is useful to recall the definition of the photoelectric effect. This is the phenomenon of the interaction of light with matter, as a result of which the energy of photons is transferred to the electrons of matter. Distinguish between external and internal photoelectric effect. In our case, we are talking about the external photoelectric effect. When under the action of light, electrons are ejected from a substance. The work function depends on the material from which the photocathode of the photocell is made, and does not depend on the frequency of light. The energy of the incident photons is proportional to the frequency of the light.

E= h v(1)

where λ is the wavelength of light; With is the speed of light,

Substitute (3) into (1) We get

Let's analyze the resulting formula. Obviously, as the wavelength increases, the energy of the incident photons decreases. This type of dependence corresponds to the graph under the letter A)

Let's write the Einstein equation for the photoelectric effect:

hν = A out + E to (5),

Where hν is the energy of the photon incident on the photocathode, A vy – work function, E k is the maximum kinetic energy of photoelectrons emitted from the photocathode under the action of light.

From formula (5) we express E k = hν – A out (6), therefore, with an increase in the frequency of the incident light the maximum kinetic energy of photoelectrons increases.

red border

ν cr =	A exit	(7),
	h

this is the minimum frequency at which the photoelectric effect is still possible. The dependence of the maximum kinetic energy of photoelectrons on the frequency of the incident light is reflected in the graph under the letter B).

Answer:

Determine the ammeter reading (see figure) if the error direct measurement current is equal to the division value of the ammeter.

Answer: (____________________±___________) A.

Solution

The task tests the ability to record the readings of the measuring device, taking into account the specified measurement error. Let's determine the scale division value With\u003d (0.4 A - 0.2 A) / 10 \u003d 0.02 A. The measurement error according to the condition is equal to the scale division, i.e. Δ I = c= 0.02 A. We write the final result as:

I= (0.20 ± 0.02) A

It is necessary to assemble an experimental setup with which you can determine the coefficient of sliding friction of steel on wood. To do this, the student took a steel bar with a hook. Which two items from the list of equipment below should be additionally used to conduct this experiment?

1. wooden lath
2. dynamometer
3. beaker
4. plastic rail
5. stopwatch

In response, write down the numbers of the selected items.

Solution

In the task, it is required to determine the coefficient of sliding friction of steel on wood, therefore, to conduct the experiment, it is necessary to take a wooden ruler and a dynamometer from the proposed list of equipment to measure the force. It is useful to recall the formula for calculating the modulus of sliding friction force

fck = μ · N (1),

where μ is the coefficient of sliding friction, N is the reaction force of the support, equal in modulus to the weight of the body.

Answer:

The handbook contains detailed theoretical material on all topics tested by the USE in physics. After each section, multi-level tasks are given in the form of the exam. For the final control of knowledge at the end of the reference book are given training options corresponding to the exam. Students do not have to search for additional information on the Internet and buy other manuals. In this guide, they will find everything they need to independently and effectively prepare for the exam. The reference book is addressed to high school students to prepare for the exam in physics. The manual contains detailed theoretical material on all topics tested by the exam. After each section, examples of USE tasks and a practice test are given. All questions are answered. The publication will be useful to teachers of physics, parents for the effective preparation of students for the exam.

Consider a table containing information about bright stars.

Star name	Temperature, TO	Weight (in solar masses)	Radius (in solar radii)	Distance to the star (holy year)
Aldebaran		5
Betelgeuse

Select two statements that match the characteristics of the stars.

1. The surface temperature and radius of Betelgeuse indicate that this star belongs to the red supergiants.
2. The temperature on the surface of Procyon is 2 times lower than on the surface of the Sun.
3. The stars Castor and Capella are at the same distance from the Earth and, therefore, belong to the same constellation.
4. The star Vega belongs to the white stars of spectral class A.
5. Since the masses of the Vega and Capella stars are the same, they belong to the same spectral class.

Solution

Star name	Temperature, TO	Weight (in solar masses)	Radius (in solar radii)	Distance to the star (holy year)
Aldebaran
Betelgeuse
			2,5

In the task, you need to choose two true statements that correspond to the characteristics of the stars. The table shows that Betelgeuse has the lowest temperature and the largest radius, which means that this star belongs to red giants. Therefore, the correct answer is (1). To correctly choose the second statement, it is necessary to know the distribution of stars by spectral types. We need to know the temperature interval and the color of the star corresponding to this temperature. Analyzing the table data, we conclude that (4) will be the correct statement. The star Vega belongs to the white stars of spectral class A.

A 2 kg projectile flying at a speed of 200 m/s breaks into two fragments. The first fragment of mass 1 kg flies at an angle of 90° to the original direction with a speed of 300 m/s. Find the speed of the second fragment.

Answer: _______ m/s.

Solution

At the moment of projectile burst (Δ t→ 0), the effect of gravity can be neglected and the projectile can be considered as a closed system. According to the law of conservation of momentum: the vector sum of the momenta of the bodies included in a closed system remains constant for any interactions of the bodies of this system with each other. for our case we write:

- projectile speed; m- the mass of the projectile before rupture; is the speed of the first fragment; m 1 is the mass of the first fragment; m 2 – mass of the second fragment; is the speed of the second fragment.

Let's choose the positive direction of the axis X, coinciding with the direction of the projectile velocity, then in the projection onto this axis we write equation (1):

mv x = m 1 v 1x + m 2 v 2x (2)

According to the condition, the first fragment flies at an angle of 90° to the original direction. The length of the desired momentum vector is determined by the Pythagorean theorem for a right triangle.

p 2 = √p 2 + p 1 2 (3)

p 2 = √400 2 + 300 2 = 500 (kg m/s)

Answer: 500 m/s.

When compressing an ideal monatomic gas at constant pressure, external forces did 2000 J of work. How much heat was transferred by the gas to the surrounding bodies?

Answer: _____ J.

Solution

A challenge to the first law of thermodynamics.

Δ U = Q + A sun, (1)

Where Δ U – change in the internal energy of the gas, Q- the amount of heat transferred by the gas to the surrounding bodies, A sun - work external forces. According to the condition, the gas is monatomic and it is compressed at a constant pressure.

A sun = - A g(2),

Q = Δ U– A sun = Δ U+ A r =	3	pΔ V + pΔ V =	5	pΔ V,
	2		2

Where pΔ V = A G

Answer: 5000 J

A plane monochromatic light wave with a frequency of 8.0 · 10 14 Hz is incident along the normal onto a diffraction grating. A converging lens with a focal length of 21 cm is placed parallel to the grating behind it. The diffraction pattern is observed on the screen in the rear focal plane of the lens. The distance between its main maxima of the 1st and 2nd orders is 18 mm. Find the lattice period. Express your answer in micrometers (µm) rounded to the nearest tenth. Calculate for small angles (φ ≈ 1 in radians) tgα ≈ sinφ ≈ φ.

Solution

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. Let us express from equation (1) the period of the diffraction grating

Rice. 1

According to the condition of the problem, we know the distance between its main maxima of the 1st and 2nd order, we denote it as Δ x\u003d 18 mm \u003d 1.8 10 -2 m, light wave frequency ν \u003d 8.0 10 14 Hz, focal length of the lens F\u003d 21 cm \u003d 2.1 10 -1 m. We need to determine the period of the diffraction grating. On fig. 1 shows a diagram of the path of rays through the grating and the lens behind it. On the screen located in the focal plane of the converging lens, a diffraction pattern is observed as a result of the interference of waves coming from all the slits. We use formula one for two maxima of the 1st and 2nd order.

d sinφ 1 = kλ(2),

If k = 1, then d sinφ 1 = λ (3),

write similarly for k = 2,

Since the angle φ is small, tgφ ≈ sinφ. Then from Fig. 1 we see that

Where x 1 is the distance from the zero maximum to the maximum of the first order. Similarly for the distance x 2 .

Then we have

grating period,

because by definition

Where With\u003d 3 10 8 m / s - the speed of light, then substituting the numerical values ​​we get

The answer was presented in micrometers, rounded to tenths, as required in the problem statement.

Answer: 4.4 µm.

Based on the laws of physics, find the reading of an ideal voltmeter in the circuit shown in the figure, before closing the key to and describe the changes in its readings after closing the key K. Initially, the capacitor is not charged.

Solution

Rice. 1

The tasks in Part C require the student to provide a full and detailed answer. Based on the laws of physics, it is necessary to determine the readings of the voltmeter before closing the key K and after closing the key K. Let us take into account that initially the capacitor in the circuit is not charged. Let's consider two states. When the key is open, only the resistor is connected to the power supply. The voltmeter reading is zero, since it is connected in parallel with the capacitor, and the capacitor is not initially charged, then q 1 = 0. The second state is when the key is closed. Then the readings of the voltmeter will increase until they reach the maximum value, which will not change with time,

Where r is the internal resistance of the source. Voltage across the capacitor and resistor, according to Ohm's law for the circuit section U = I · R will not change over time, and the voltmeter readings will stop changing.

A wooden ball is tied with a thread to the bottom of a cylindrical vessel with a bottom area S\u003d 100 cm 2. Water is poured into the vessel so that the ball is completely immersed in the liquid, while the thread is stretched and acts on the ball with a force T. If the thread is cut, the ball will float and the water level will change to h \u003d 5 cm. Find the tension in the thread T.

Solution

Rice. 1		Rice. 2

Initially, a wooden ball is tied with a thread to the bottom of a cylindrical vessel with a bottom area S\u003d 100 cm 2 \u003d 0.01 m 2 and completely immersed in water. Three forces act on the ball: the force of gravity from the side of the Earth, - the Archimedes force from the side of the liquid, - the force of the tension of the thread, the result of the interaction of the ball and the thread. According to the balance condition of the ball in the first case geometric sum of all forces acting on the ball, must be equal to zero:

Let's choose the coordinate axis OY and point it up. Then, taking into account the projection, equation (1) can be written:

Fa 1 = T + mg (2).

Let's write the force of Archimedes:

Fa 1 = ρ V 1 g (3),

Where V 1 - the volume of the part of the ball immersed in water, in the first it is the volume of the entire ball, m is the mass of the ball, ρ is the density of water. The equilibrium condition in the second case

Fa 2 = mg(4)

Let's write out the force of Archimedes in this case:

Fa 2 = ρ V 2 g (5),

Where V 2 is the volume of the part of the sphere immersed in the liquid in the second case.

Let's work with equations (2) and (4) . You can use the substitution method or subtract from (2) - (4), then Fa 1 – Fa 2 = T, using formulas (3) and (5) we obtain ρ · V 1 g– ρ · V 2 g= T;

ρg( V 1 – V 2) = T (6)

Given that

V 1 – V 2 = S · h (7),

Where h= H 1 - H 2; we get

T= ρ g S · h (8)

Let's substitute numerical values

Answer: 5 N.

All the information necessary for passing the exam in physics is presented in visual and accessible tables, after each topic there are training tasks for knowledge control. With the help of this book, students will be able to improve their knowledge in the shortest possible time, remember all the most important topics in a matter of days before the exam, practice completing assignments in the USE format and become more confident in their abilities. After repeating all the topics presented in the manual, the long-awaited 100 points will be much closer! The manual contains theoretical information on all topics tested at the exam in physics. After each section, training tasks of different types with answers are given. A visual and accessible presentation of the material will allow you to quickly find the information you need, eliminate gaps in knowledge and as soon as possible repeat a large amount of information. The publication will assist high school students in preparing for lessons, various forms current and intermediate control, as well as for preparing for exams.

Task 30

In a room with dimensions of 4 × 5 × 3 m, in which the air has a temperature of 10 ° C and a relative humidity of 30%, a humidifier with a capacity of 0.2 l / h was turned on. What will be the relative humidity of the air in the room after 1.5 hours? Saturated water vapor pressure at 10 °C is 1.23 kPa. Consider the room as a hermetic vessel.

Solution

When starting to solve problems for vapors and humidity, it is always useful to keep in mind the following: if the temperature and pressure (density) of the saturating vapor are given, then its density (pressure) is determined from the Mendeleev-Clapeyron equation. Write down the Mendeleev-Clapeyron equation and the relative humidity formula for each state.

For the first case at φ 1 = 30%. The partial pressure of water vapor is expressed from the formula:

Where T = t+ 273 (K), R is the universal gas constant. We express the initial mass of the vapor contained in the room using equations (2) and (3):

During the time τ of the humidifier operation, the mass of water will increase by

Δ m = τ · ρ · I, (6)

Where I– performance of the humidifier according to the condition, it is equal to 0.2 l / h = 0.2 10 -3 m 3 / h, ρ = 1000 kg / m 3 - the density of water. Substitute formulas (4) and (5) in (6)

We transform the expression and express

This is the desired formula for the relative humidity that will be in the room after the operation of the humidifier.

Substitute the numerical values ​​and get the following result

Answer: 83 %.

On horizontally arranged rough rails with negligible resistance, two identical rods of mass m= 100 g and resistance R= 0.1 ohm each. The distance between the rails is l = 10 cm, and the coefficient of friction between the rods and the rails is μ = 0.1. Rails with rods are in a uniform vertical magnetic field with induction B = 1 T (see figure). Under the action of a horizontal force acting on the first rod along the rail, both rods move translationally uniformly at different speeds. What is the speed of the first rod relative to the second? Ignore the self-inductance of the circuit.

Solution

Rice. 1

The task is complicated by the fact that two rods are moving and it is necessary to determine the speed of the first relative to the second. Otherwise, the approach to solving problems of this type remains the same. Change magnetic flux penetrating circuit leads to the emergence of induction EMF. In our case, when the rods move at different speeds, the change in the flux of the magnetic induction vector penetrating the circuit over the time interval Δ t is determined by the formula

ΔΦ = B · l · ( v 1 – v 2) Δ t (1)

This leads to the appearance of an EMF of induction. According to Faraday's law

By the condition of the problem, we neglect the self-induction of the circuit. According to Ohm's law for a closed circuit for the current that occurs in the circuit, we write the expression:

Current-carrying conductors in a magnetic field are affected by the Ampère force and the modules of which are equal to each other, and are equal to the product of the current strength, the module of the magnetic induction vector and the length of the conductor. Since the force vector is perpendicular to the direction of the current, then sinα = 1, then

F 1 = F 2 = I · B · l (4)

The braking force of friction still acts on the rods,

F tr = μ m · g (5)

by condition it is said that the rods move uniformly, which means that the geometric sum of the forces applied to each rod is equal to zero. Only the Ampere force and the friction force act on the second rod. Therefore, F tr = F 2 , taking into account (3), (4), (5)

Let us express from here the relative speed

Substitute the numerical values:

Answer: 2 m/s.

In an experiment to study the photoelectric effect, light with a frequency of ν = 6.1 · 10 14 Hz falls on the cathode surface, as a result of which a current appears in the circuit. Current dependency graph I from voltage U between the anode and cathode is shown in the figure. What is the power of the incident light R, if on average one out of 20 photons incident on the cathode knocks out an electron?

Solution

By definition, current is physical quantity numerically equal to the charge q passing through the cross section of the conductor per unit time t:

I =	q	(1).
	t

If all the photoelectrons knocked out of the cathode reach the anode, then the current in the circuit reaches saturation. The total charge passing through the cross section of the conductor can be calculated

q = N e · e · t (2),

Where e is the electron charge modulus, N e– the number of photoelectrons knocked out of the cathode in 1 s. According to the condition, one out of 20 photons incident on the cathode knocks out an electron. Then

Where N f is the number of photons incident on the cathode in 1 s. The maximum current in this case will be

Our task is to find the number of photons incident on the cathode. It is known that the energy of one photon is equal to E f = h · v, then the power of the incident light

After substituting the corresponding quantities, we obtain the final formula

P = N f · h · v =

20 · I max h USE-2018. Physics (60x84/8) 10 practice exam papers to prepare for the unified state exam The attention of schoolchildren and applicants is offered a new textbook in physics for USE preparation, which contains 10 options for training examination papers. Each option is compiled in full accordance with the requirements of the unified state exam in physics, includes tasks of different types and levels of complexity. At the end of the book, answers are given for self-examination of all tasks. The proposed training options will help the teacher organize preparation for the unified state exam, and students will independently test their knowledge and readiness for the final exam. The manual is addressed to schoolchildren, applicants and teachers.

FIPI 2018 Early exam in physics with answers and solutions. answers to the early exam in physics 2018. options early exam in physics 2018 with answers

Answers

1. Answer: 12

For a time of 0.5 seconds, the speed changed from 0 to 6 m/s

Acceleration projection =

2. Answer: 0.25

According to the friction force formula Ftr = kN, where k is the coefficient of friction. k=1/4=0.25. The graph shows that Ftr=0.25N. Hence k = 0.25.

3. Answer: 1.8

4. Answer: 0.5

According to the potential energy formula

Ep=kx 2/2, since the maximum energy is needed Ep.max=kA 2/2

after. times at x=-A through t=T/2=0.5(s)

5. Answer: 13

1) Body momentum P=mv, 0 second momentum is 20*0=0, 20 second momentum is 20*4=80 (correct)
2) in the time span of 60 to 100 seconds module average speed equals (0-4)/2=2 m/s, therefore, the body traveled 2*40=80 meters (incorrect)
3) The resultant of all forces acting on the body is F=ma and since m=20 kg, and a=1/5, we get F=4 N (correct)
4) the acceleration module in the time interval from 60 to 80 s is equal to a=dV/dt=1/20, the acceleration module in the time interval from 80 to 100 s hfdty 3/20. Less than 3 times (incorrect)
5) decreased by 90 times (wrong)

6. Answer: 33

A body thrown horizontally from a height H moves in a horizontal direction uniformly (without acceleration) with a speed. Time t depends on the height H as (the initial falling velocity is 0). The height does not change, therefore, the time will remain the same.

There is no acceleration of movement, i.e. is 0 and therefore will not change.

7. Answer: 14

8 Answer: 40

According to the ideal gas formula PV=vRT

First T \u003d T 0, P 1 \u003d 40 * 10 3, v 1 \u003d 2 mol, V \u003d V 0

P 2 V 0 =R2T 0 , i.e. Pressure remains the same P 2 =40kPa

9. Answer: 6

It can be seen from the graph that the process under study is isochoric. Since the volume of the gas did not change, the gas did no work. Therefore, according to the first law of thermodynamics, internal energy gas is equal to the amount of heat received by the gas.

10 Answer: 2

The graph shows T 1 \u003d 200K, T 2 \u003d 400K

U=3/2vRT, since v and R remain unchanged, then U 2 /U 1 =400/200 = 2.

It turns out 2 times.

11 Answer: 15

1) Relative humidity is defined as

where p is the partial pressure of water vapor; p H is the saturated vapor pressure (table value depending only on temperature). Since the pressure p on Tuesday was less than on Wednesday, and the saturated vapor pressure remained unchanged (the temperature did not change), the relative humidity on Tuesday was less than on Wednesday. (right)
2) (wrong)
3) The partial pressure of water vapor is the pressure of this individual vapor in the atmosphere. Since on Tuesday this pressure was less than on Wednesday, and the temperature remained constant, the density of water vapor on Tuesday was less than on Wednesday. (wrong)
4) The saturated vapor pressure was the same on both days, since the temperature did not change. (Incorrect)

5) The concentration of water vapor molecules in the air on Tuesday was less than on Wednesday. (right)

12 Answer: 32

13. Answer: from an observer

14 Answer: 9

15 Answer: 80

16 Answer: 24

17 Answer: 31

Lorentz force modulus: 3) will not change

The period of revolution of an α-particle: 1) will increase

18 Answer: 23

19 Answer: 37

20 Answer: 2

21 Answer: 31

22. Answer: (3.0 ± 0.2) V

23 Answer: 24

24 Answer: 12

Analysis of tasks 1 - 7 (mechanics)

Analysis of tasks 8 - 12 (MKT and thermodynamics)

Analysis of tasks 13 - 18 (electrodynamics)

Analysis of tasks 19 - 24

Analysis of tasks 25 - 27 (part 2)

Debriefing 28 (Part 2, Qualitative Problem)

Debriefing 29 (Part 2)