In 2018, graduates of grade 11 and secondary vocational education institutions will take the Unified State Exam 2018 in physics. The latest news regarding the Unified State Exam in Physics in 2018 is based on the fact that some changes, both major and minor, will be made to it.

What is the meaning of the changes and how many are there?

The main change related to the Unified State Examination in Physics compared to previous years is the absence of a multiple-choice test part. This means that preparation for the Unified State Exam must be accompanied by the student’s ability to give short or detailed answers. Consequently, it will no longer be possible to guess the option and score a certain number of points and you will have to work hard.

A new task 24 has been added to the basic part of the Unified State Exam in Physics, which requires the ability to solve problems in astrophysics. Due to the addition of No. 24, the maximum primary score increased to 52. The exam is divided into two parts according to difficulty levels: the basic part of 27 tasks, requiring a short or full answer. In the second part there are 5 advanced level tasks where you need to give a detailed answer and explain the process of your solution. One important caveat: many students skip this part, but even attempting these assignments can earn you one to two points.

All changes to the Unified State Examination in Physics are made with the aim of deepening preparation and improving the assimilation of knowledge in the subject. In addition, eliminating the test part motivates future applicants to accumulate knowledge more intensively and reason logically.

Exam structure

Compared to the previous year, the structure of the Unified State Exam has not undergone significant changes. 235 minutes are allotted for the entire work. Each task of the basic part should take from 1 to 5 minutes to solve. Problems of increased complexity are solved in approximately 5-10 minutes.

All CMMs are stored at the examination site and are opened during the test. The structure is as follows: 27 basic tasks test the examinee's knowledge in all areas of physics, from mechanics to quantum and nuclear physics. In 5 tasks of a high level of complexity, the student demonstrates skills in logical justification of his decision and the correctness of his train of thought. The number of initial points can reach a maximum of 52. They are then recalculated on a 100-point scale. Due to changes in the primary score, the minimum passing score may also change.

Demo version

A demo version of the Unified State Exam in Physics is already on the official FIPI portal, which is developing a unified state exam. The structure and complexity of the demo version is similar to the one that will appear on the exam. Each task is described in detail; at the end there is a list of answers to questions on which the student checks his solutions. Also at the end is a detailed breakdown for each of the five tasks, indicating the number of points for correctly or partially completed actions. For each task of high complexity you can get from 2 to 4 points, depending on the requirements and the extent of the solution. Tasks may contain a sequence of numbers that must be written down correctly, establishing correspondence between elements, as well as small tasks in one or two steps.

• Download demo: ege-2018-fiz-demo.pdf
• Download the archive with the specification and codifier: ege-2018-fiz-demo.zip

We wish you to successfully pass physics and enroll in your desired university, everything is in your hands!

Specification
control measuring materials for carrying out
in 2018 main state exam in PHYSICS

1. Purpose of CMM for OGE- to assess the level of general education training in physics of graduates of IX grades of general education organizations for the purpose of state final certification of graduates. The exam results can be used when admitting students to specialized classes in secondary schools.

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

2. Documents defining the content of CMM

The content of the examination work is determined on the basis of the Federal component of the state standard of basic general education in physics (order of the Ministry of Education of Russia dated 03/05/2004 No. 1089 “On approval of the Federal component of state educational standards of primary general, basic general and secondary (complete) general education”).

3. Approaches to content selection and CMM structure development

The approaches to the selection of controlled content elements used in the design of CMM variants ensure the requirement of functional completeness of the test, since in each variant the mastery of all sections of the basic school physics course is checked and tasks of all taxonomic levels are offered for each section. At the same time, the content elements that are most important from an ideological point of view or necessary for the successful continuation of education are tested in the same version of the CMM with tasks of different levels of complexity.

The structure of the KIM version ensures testing of all types of activities provided for by the Federal component of the state educational standard (taking into account the restrictions imposed by the conditions of mass written testing of students’ knowledge and skills): mastering the conceptual apparatus of a primary school physics course, mastering methodological knowledge and experimental skills, using educational tasks of physical texts, application of knowledge in solving calculation problems and explaining physical phenomena and processes in situations of a practice-oriented nature.

The task models used in the examination work are designed for the use of blank technology (similar to the Unified State Examination) and the possibility of automated verification of part 1 of the work. The objectivity of checking tasks with a detailed answer is ensured by uniform assessment criteria and the participation of several independent experts evaluating one work.

The OGE in physics is an exam of students' choice and performs two main functions: the final certification of primary school graduates and the creation of conditions for differentiating students when entering specialized classes of secondary school. For these purposes, the CMM includes tasks of three levels of complexity. Completing tasks of a basic level of complexity allows you to assess the level of mastery of the most significant content elements of the standard in primary school physics and mastery of the most important types of activities, and completing tasks of increased and high levels of complexity - the degree of preparedness of the student to continue education at the next stage of education, taking into account the further level of study of the subject (basic or profile).

4. Connection of the OGE exam model with the Unified State Exam KIM

The examination model of the OGE and the KIM Unified State Examination in Physics are built on the basis of a unified concept for assessing the educational achievements of students in the subject “Physics”. Unified approaches are ensured, first of all, by checking all types of activities formed within the framework of teaching the subject. In this case, similar work structures are used, as well as a single bank of task models. Continuity in the formation of various types of activities is reflected in the content of tasks, as well as in the system for assessing tasks with a detailed answer.

It is possible to note two significant differences between the exam model of the OGE and the KIM Unified State Examination. Thus, the technological features of the Unified State Exam do not allow for full control of the development of experimental skills, and this type of activity is tested indirectly using specially designed tasks based on photographs. Carrying out the OGE does not contain such restrictions, so an experimental task was introduced into the work, performed on real equipment. In addition, in the examination model of the OGE, a block on testing techniques for working with a variety of physical information is more widely represented.

5. Characteristics of the structure and content of CMM

Each version of the CMM consists of two parts and contains 26 tasks that differ in form and level of complexity (Table 1).

Part 1 contains 22 tasks, of which 13 tasks require a short answer in the form of a single number, eight tasks that require a short answer in the form of a number or a set of numbers, and one task with a detailed answer. Tasks 1, 6, 9, 15 and 19 with a short answer are tasks to establish the correspondence of positions presented in two sets, or tasks to select two correct statements from the proposed list (multiple choice).

Part 2 contains four tasks (23-26), for which you need to provide a detailed answer. Task 23 is a practical task that uses laboratory equipment.

On the eve of the academic year, demo versions of the KIM Unified State Exam 2018 in all subjects (including physics) have been published on the official website of the FIPI.

This section presents documents defining the structure and content of the KIM Unified State Exam 2018:

Demonstration versions of control measurement materials of the Unified State Exam.
- codifiers of content elements and requirements for the level of training of graduates of general education institutions for conducting a unified state exam;
- specifications of control measuring materials for the Unified State Exam;

Demo version of the Unified State Exam 2018 in physics tasks with answers

Changes in the Unified State Exam KIM in 2018 in physics compared to 2017

The codifier of content elements tested on the Unified State Exam in Physics includes subsection 5.4 “Elements of Astrophysics”.

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

Duration of the Unified State Exam 2018 in physics

235 minutes are allotted to complete the entire examination work. The approximate time to complete 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 Unified State Examination

Each version of the examination paper consists of two parts and includes 32 tasks, differing in form and level of difficulty.

Part 1 contains 24 short answer questions. Of these, 13 tasks require the answer to be written in the form of a number, a word or two numbers, 11 tasks require matching 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 you need to provide a detailed answer.

The Unified State Exam is one of the most discussed topics in the Russian pedagogical community. Future graduates and teachers who will have to prepare students to take the Unified State Exam are already today asking themselves what the Unified State Exam in physics will be like in the coming 2018 and whether we should expect any global changes in the structure of examination papers or the format of testing. 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, successfully passing the Unified State 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 Unified State Exam 2018. The Russian language and mathematics remain compulsory, and physics is included in the extensive list of subjects that graduates can additionally choose for themselves, focusing on the requirements of the university they plan to enroll in.

In 2017, 16.5% of all 11th graders in the country chose physics. This popularity of the subject is not accidental. Physics is necessary for everyone who plans to enter engineering professions or connect their lives withIT-technology, 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 leaks to the media about possible innovations, such as:

1. Expanding the list of subjects required to pass the following disciplines: physics, history and geography.
2. Introduction of a unified integrated exam in natural sciences.

While discussions are underway on the proposals made, current high school students should thoroughly prepare to take the most relevant combination of the Unified State Exam – profile-level mathematics + physics.

Is it worth clarifying that mainly students of specialized classes with in-depth study of mathematical subjects will feel confident in this area.

Structure of the examination paper in physics in 2018

The main session of the Unified State Exam 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 changed significantly compared to 2016.

Changes in the Unified State Examination in Physics in 2018

Tests have been completely removed from the assignments, leaving the possibility of mindlessly choosing an answer. Instead, students are offered tasks with short or extended answers. It is safe to say that in the 2017-2018 academic year the Unified State Exam in Physics will not differ much in the structure and volume of tasks 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;
• Block I of the Unified State Exam (27 tasks) – tasks with a short answer, which can be represented by an integer, a decimal fraction or a numerical sequence;
• Block II (5 tasks) – tasks that require a similar description of the train of thought in the decision process and justification of the decisions made 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 problems, from 27 to 31, that are the most difficult on the Unified State Exam in Physics and many schoolchildren pass the work with empty fields in them. But there is a very important nuance - if you read the rules for assessing 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, without reaching the full answer and without writing anything down in the decision.

To solve most of the problems in their physics course, they need not only a good knowledge of the laws and understanding of physical processes, but also good mathematical preparation, and therefore they should ask themselves the question of expanding and deepening their knowledge long before the upcoming Unified State Exam 2018.

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

You shouldn’t count on cheat sheets and various other tricks. The use of notebooks with formulas, calculators and other technical means, which many students do during school tests, is unacceptable during the exam. Remember that compliance with this rule is monitored not only by observers, but also by the tireless eyes of video cameras positioned in such a way as to notice every dubious movement of the examinee.

You can prepare for the Unified State Exam in Physics by contacting an experienced teacher, or by repeating the school curriculum on your own.

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

1. Don't try to memorize complex formulas, try to understand their nature. Knowing how the formula is derived, you can easily write it down in a draft, while mindless memorization is fraught with mechanical errors.
2. When solving a problem, start 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 problems on a topic you solve, the easier it will be to cope with the Unified State Examination tasks.
4. Start preparing for the Unified State Exam in Physics at least a year before the exam. This is not a subject that you can take “offhand” and learn in another month, even when studying with the best tutors.
5. Don't get hung up on the same type of simple tasks. Problems with 1-2 formulas are only stage 1. Unfortunately, many teachers in schools simply do not go further, descending to the level of the majority of students or counting on the fact that students in humanities classes will not choose a subject that is not their core when taking the Unified State Exam. Solve problems that combine laws from different branches of physics.
6. Review physical quantities and their transformations again. When solving problems, be especially attentive to the format in which the data is presented and, if necessary, do not forget to convert them to the desired form.

Excellent helpers in preparing for the Unified State Exam in physics will be trial versions of exam tasks, as well as tasks on various topics, which today can be easily found on the Internet. First of all, this is the FIPI website, where the archive of the Unified State Exam in Physics for 2008-17 with codifiers is located.

For more information about the changes that have already occurred in the Unified State Examination and how to prepare for the exam, see video interview with Marina Demidova, head of the Federal Commission for the development of tasks and conducting the Unified State Exam in Physics:

Secondary general education

Preparing for the Unified State Exam 2018: analysis of the demo version in physics

Unified State Exam 2018. Physics. Thematic training tasks

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

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

Determine the distance the body traveled in the third second of movement.

Answer: _________ m.

Solution

Knowing how to read graphs is very important for every student. The question in the problem is that it is required to determine, from the graph of the projection of acceleration versus time, the path that the body has traveled in the third second of movement. 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 if you know 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, the velocity projection at the end of the second second became

Then the distance traveled by the body in the third second

Answer: 8 m.

Rice. 1

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

Answer: _________ N.

Solution

Horizontally on a body of mass m= 2 kg two forces act, this is a force F= 10 N and the elastic force on the side of the spring. The resultant of these forces imparts acceleration to the body. Let's 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.

In projection onto axis 0 X: F – F control = ma (2)

Let us express from formula (2) the modulus of the elastic force F control = F – ma (3)

Let's substitute the numerical values ​​into formula (3) and get, F control = 10 N – 2 kg · 1 m/s 2 = 8 N.

Answer: 8 N.

Task 3

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

Answer: _________ J.

Solution

The body is acted upon by the force of gravity, the reaction force of the support, the friction force, which creates a braking acceleration. The body was initially given a speed of 10 m/s. Let's write down Newton's second law for our case.

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

N – mg = 0; N = mg (2)

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

A · ( F tr) = – F tr · S (4)

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

Let's substitute (3) and (5) into (4)

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

Let's substitute numerical values

Answer: 150 J.

Unified State Exam 2018. Physics. 30 practice versions of exam papers

The publication contains:
30 training options for the Unified State Exam
instructions for implementation and evaluation criteria
answers to all tasks
Training options will help the teacher organize preparation for the Unified State Exam, and students will independently test their knowledge and readiness to take 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 block axis. 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 conditions of the problem, the system is in equilibrium. On the image L 1, shoulder strength L 2nd arm of force Equilibrium condition: moments of forces rotating bodies clockwise must be equal to 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 loads differ by a factor of 3. This means that the radius of the inner pulley of the block differs from the outer one by 3 times. Therefore, the shoulder L 2 will be equal to 8 cm.

Answer: 8 cm

Task 5

Oh, at different points in time.

From the list below, select 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's oscillations is 30 mm.
5. The total mechanical energy of a pendulum consisting of a ball and a spring at time 3.0 s is minimal.

Solution

The table presents data on the position of a ball attached to a spring and oscillating along a horizontal axis Oh, at different points in time. We need to analyze this data and choose the right two statements. The system is a spring pendulum. At a moment in time t= 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 using the formula

Where k– spring stiffness coefficient, X– displacement of the body from the equilibrium position. If the displacement is maximum, then the speed at this point is zero, which means 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 goes through half of the oscillation in t= 2 s, complete oscillation takes twice as long 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 level of water in the glass changed as a result of the melting of the ice.

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

Write to table

Solution

Rice. 1

Problems of this type are quite common in different versions of the Unified State Exam. And as practice shows, students often make mistakes. Let's try to analyze this task in detail. Let's denote m– mass of a piece of ice, ρ l – density of ice, ρ в – density of water, V pcht – the volume of the submerged part of the ice, equal to the volume of the displaced liquid (volume of the hole). Let's mentally remove the ice from the water. Then there will be a hole in the water whose volume is equal to V pcht, i.e. volume of water displaced by a piece of ice Fig. 1( b).

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

F a = mg (1)

ρ in V p.m. 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 melting our piece of ice. Therefore, if we now (mentally) pour water obtained from ice into a 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

Unified State Exam 2018. Physics. Training tasks

The publication is addressed to high school students to prepare for the Unified State Exam in physics.
The benefit includes:
20 training options
answers to all tasks
Unified State Exam answer forms for each option.
The publication will assist teachers in preparing students for the Unified State Exam in Physics.

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

Establish a correspondence between the graphs of the 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

From the figure for the problem it is clear 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 by 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 generated in the spring will change. Accordingly, under the letter A) the graph is the dependence of the modulus of the elastic force on the deformation of the spring.

The graph under letter B) shows the dependence of the projection of the external force on the magnitude of the deformation. Because with increasing external force, the magnitude of deformation and 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 translational thermal motion of a particle of an ideal gas at a temperature of 29°R. Express your answer in eV and round to the nearest hundredth.

Answer: ________ eV.

Solution

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

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

Let's convert the result to Kelvin using the formula

T = t°C + 273 (2);

T= 36.25 + 273 = 309.25 (K)

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

Where k– Boltzmann constant equal to 1.38 10 –23 J/K, T– absolute temperature on the Kelvin scale. From the formula it is clear that the dependence of the average kinetic energy on temperature is direct, that is, the number of times the temperature changes, the number of times the average kinetic energy of the thermal motion of molecules changes. Let's substitute the numerical values:

Let's convert the result into electronvolts and round to the nearest hundredth. Let's remember that

1 eV = 1.6 10 –19 J.

For this

Answer: 0.04 eV.

One mole of a monatomic ideal gas participates in process 1–2, the graph of which is shown in VT-diagram. For this process, determine 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 conditions 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 the change in internal energy and the amount of heat imparted to the gas. The process is isobaric (Gay-Lussac's 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+Δ U 12 (5),

Where A 12 – gas work during expansion. By definition, work is equal to

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 reference book contains in full the theoretical material for the physics course required to pass the Unified State Exam. The structure of the book corresponds to the modern codifier of content elements in the subject, on the basis of which examination tasks are compiled - test and measuring materials (CMM) of the Unified State Examination. Theoretical material is presented in a concise, accessible form. Each topic is accompanied by examples of exam tasks that correspond to the Unified State Exam format. This will help the teacher organize preparation for the unified state exam, and students will independently test their knowledge and readiness to take 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. A hissing sound is heard and steam rises above 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: _________ g.

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 immersing the hot horseshoe, the energy received by the water will go directly to steam formation. Let us write the heat balance equation

With and · m P · ( t n – 100) = Lm in 1),

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

When writing down the answer, pay attention to the units in which you want to leave the mass of water.

Answer: 90

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

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

1. The gas pressure in state 2 is greater than the gas pressure in state 4
2. 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 what dependence graphs look like for isoprocesses in different axes, in particular R= const. In our example at TV The diagram shows two isobars. Let's see how pressure and volume 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 not do any work; it is zero. The statement is incorrect. In section 1–2, the pressure increases, which is also incorrect. We just showed above that this is an isobaric transition. In section 4–1, a certain amount of heat is removed from the gas in order to maintain a constant temperature as the gas is compressed.

Answer: 14.

The heat engine operates 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 gas work per cycle change?

For each quantity, determine the corresponding nature of the change:

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

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

Solution

Heat engines operating according to the Carnot cycle are often found in exam tasks. First of all, you need to remember the formula for calculating the efficiency factor. Be able to write it down using the temperature of the heater and the temperature of the refrigerator

in addition, be able to write down the efficiency through the useful work of 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 gas work per cycle will also decrease, with all 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 drawing ( right, left, up, down, towards the observer, away from the observer) charge acceleration – q in this moment in time, if only charges + act on it Q And –Q? Write the answer in word(s)

Solution

Rice. 1

Negative charge – q is in the field of two stationary charges: positive + Q and negative - Q, as shown in the figure. in order to answer the question of where the charge acceleration is directed - q, at the moment of time when only charges +Q and – 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 are the forces directed this way? Let us remember how similarly charged bodies interact, they repel, the force Coulomb force of 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 stationary charges whose moduli are equal. Therefore, they will also be equal in modulus. The resulting force will be directed relative to the drawing down. The charge acceleration will also be directed - q, i.e. down.

Answer: Down.

The book contains materials for successfully passing the Unified State Exam in Physics: brief theoretical information on all topics, tasks of different types and levels of complexity, solving problems of an increased level of complexity, answers and assessment criteria. Students will not have to search for additional information on the Internet and buy other textbooks. 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 on the Unified State Exam in Physics, as well as solutions to problems of an increased level of complexity. The publication will provide invaluable assistance to students in preparing for the Unified State Exam in physics, and can also be used by teachers in organizing the educational process.

Two series-connected resistors with a resistance of 4 Ohms and 8 Ohms are connected to a battery whose terminal voltage is 24 V. What thermal power is released in the lower value resistor?

Answer: _________ Tue.

Solution

To solve the problem, it is advisable to draw a diagram of the series connection of resistors. Then remember the laws of 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 at each section of the circuit, the current will be the same. The total resistance is defined as the sum of the resistances of all resistors. According to Ohm's law, for a section of the circuit we have:

To determine the thermal power released by a resistor of a lower value, we write:

P = I 2 R= (2 A) 2 · 4 Ohm = 16 W.

Answer: P= 16 W.

A wire frame with an area of ​​2·10–3 m2 rotates in a uniform magnetic field around an axis perpendicular to the magnetic induction vector. The magnetic flux penetrating the frame area varies according to the law

Ф = 4 10 –6 cos10π t,

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

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 quantity is related to the magnetic induction module B and frame area S. Let's write the equation in general form to understand what quantities are included in it.

Φ = Φ m cosω t(1)

We remember that before the cos or sin sign there is an amplitude value of a changing value, which means Φ max = 4 10 –6 Wb. On the other hand, the magnetic flux is equal to the product of the magnetic induction module by the area of ​​the circuit and the cosine of the angle between the normal to the circuit and the magnetic induction vector Φ m = IN · S cosα, the flow is maximum at cosα = 1; let's express the induction modulus

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

Answer: 2.

The electrical circuit section consists of silver and aluminum wires connected in series. A direct electric current of 2 A flows through them. The graph shows how the potential φ changes in this section of the circuit when displaced along the wires by a distance x

Using the graph, select two true statements and indicate their numbers in your 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. The aluminum wire produces a thermal power of 2 W.
5. Silver wire produces less thermal power than aluminum wire

Solution

The answer to the question in the problem will be two true statements. To do this, let's try to solve a few simple problems using a graph and some data. The electrical circuit section consists of silver and aluminum wires connected in series. A direct electric current of 2 A flows through them. The graph shows how the potential φ changes in this section of the circuit when displaced along the wires by a distance x. The resistivities 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 a conductor depends on the material from which the conductor is made, the length of the conductor, and the cross-sectional area of ​​the conductor

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

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

Correct statement option 2.

Let's check the expressions for power.

P a = I 2 · R a(4);

P a = (2 A) 2 0.5 Ohm = 2 W.

The reference book contains in full the theoretical material for the physics course required to pass the Unified State Exam. The structure of the book corresponds to the modern codifier of content elements in the subject, on the basis of which examination tasks are compiled - test and measuring materials (CMM) of the Unified State Examination. Theoretical material is presented in a concise, accessible form. Each topic is accompanied by examples of exam tasks that correspond to the Unified State Exam format. This will help the teacher organize preparation for the unified state exam, and students will independently test their knowledge and readiness to take the final exam. At the end of the manual, answers to self-test tasks are provided that 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 high school students, applicants and teachers.

A small 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 move closer to the focus of the lens. How does the size of the image and the optical power of the lens change?

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

1. increases
2. decreases
3. does not change

Write to table selected numbers for each physical quantity. The 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.

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

Rice. 1

Rice. 2

We constructed two images for two positions of the object. Obviously 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– resistor resistance).

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