Download demo exam in physics. Changes in the exam in physics
Specification
control measuring materials for carrying out
in 2018 of the main state exam in PHYSICS
1. Purpose of CMM for OGE - to assess the level of general education in physics of the 9th grade graduates of general educational organizations for the purpose of the state final certification of graduates. The results of the exam can be used when admitting students to specialized secondary school classes.
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 paper is determined on the basis of the Federal component of the state standard for basic general education in physics (Order of the Ministry of Education of Russia dated 05.03.2004 No. 1089 "On approval of the Federal component of state educational standards for primary general, basic general and secondary (complete) general education").
3. Approaches to the selection of content, development of the structure of CMM
The approaches to the selection of controlled elements of content used in the design of CMM variants provide the requirement for the functional completeness of the test, since in each variant the mastery of all sections of the physics course of the basic school is checked and for each section tasks of all taxonomic levels are proposed. At the same time, the content elements that are most important from a worldview point of view or the need for successful continuation of education are checked in the same version of CMM with tasks of different levels of complexity.
The structure of the KIM version provides verification 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 basic school physics course, mastering methodological knowledge and experimental skills, using educational tasks of texts of physical content, the use of knowledge in solving computational 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 USE) 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 the choice of students and performs two main functions: the final certification of graduates of the basic school and the creation of conditions for the differentiation of students when entering the specialized classes of secondary school. For these purposes, the CMM includes tasks of three levels of complexity. The fulfillment of tasks of the basic level of complexity makes it possible to assess the level of mastering the most significant content elements of the standard in physics of basic school and mastery of the most important types of activities, and the fulfillment of tasks of increased and high levels of complexity - the degree of the student's readiness to continue education at the next stage of study, taking into account the further level of study (basic or profile).
4. Connection of the exam model of the OGE with the KIM of the exam
The exam model of the OGE and KIM USE in physics are built on the basis of a single concept for assessing the educational achievements of students in the subject "Physics". Unified approaches are provided, first of all, by checking all types of activities formed in the framework of teaching the subject. In this case, similar work structures are used, as well as a single bank of job models. Continuity in the formation of various types of activity is reflected in the content of the tasks, as well as in the assessment system for tasks with a detailed answer.
Two significant differences can be noted between the exam model of the OGE and the KIM of the exam. So, the technological features of the USE do not allow for full control of the formation of experimental skills, and this type of activity is checked indirectly using specially designed tasks based on photographs. The OGE does not contain such restrictions, therefore, an experimental task was introduced into the work performed on real equipment. In addition, in the exam model of the OGE, a block for checking the techniques of working with various information of physical content is more widely represented.
5. Characteristics of the structure and content of CMM
Each version of 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 with a short answer in the form of one 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 choose two correct statements from the proposed list (multiple choice).
Part 2 contains four tasks (23-26), for which it is necessary to give a detailed answer. Task 23 is a hands-on exercise using laboratory equipment.
August 22, 2017In 2018, students will find 32 assignments in KIMs of the USE in physics. Recall that in 2017 the number of tasks was reduced to 31. An additional task will be a question on astronomy, which, by the way, is again introduced as a compulsory subject. It is not entirely clear, however, at the expense of what hours, but, most likely, physics will suffer. So, if in the 11th grade you do not have enough lessons, then the ancient science of the stars is probably to blame. Accordingly, you will have to prepare more on your own, because the volume of school physics will be extremely small in order to somehow pass the USE. But let's not talk about sad things.
The astronomy question is number 24 and ends the first test part. The second part, respectively, has shifted and now starts from number 25. In addition, no major changes were found. The same questions with a written short answer, assignments for establishing correspondences and multiple choice, and, of course, problems with a short and detailed answer.
The exam tasks cover the following sections of physics:
- Mechanics (kinematics, dynamics, statics, conservation laws in mechanics, mechanical vibrations and waves).
- Elements of astrophysics (Solar system, stars, galaxies and the universe)
Molecular physics (molecular kinetic theory, thermodynamics).
Electrodynamics and basics of SRT (electric field, direct current, magnetic field, electromagnetic induction, electromagnetic oscillations and waves, optics, basics of SRT).
The quantum physics (particle-wave dualism, physics of the atom and atomic nucleus).
Below you can familiarize yourself with the approximate tasks of the Unified State Examination in 2018 in a demo version from FIPI. And also get acquainted with the codifier and specification.
FIPI 2018 Early Unified State Exam in Physics with Answers and Solutions. answers to the early exam in physics 2018. options for the early exam in physics 2018 with answers
Answers
1. Answer: 12
In 0.5 seconds, the speed changed from 0 to 6 m / s
Acceleration projection \u003d
2. Answer: 0.25
According to the formula of friction force Ffr \u003d kN, where k is the coefficient of friction. k \u003d 1/4 \u003d 0.25. The graph shows that Ftr \u003d 0.25N. Hence k \u003d 0.25.
3. Answer: 1.8
4. Answer: 0.5
According to the potential energy formula
Ep \u003d kx 2/2, because the maximum energy is needed Ep max \u003d kA 2/2
following. times at x \u003d -A through t \u003d T / 2 \u003d 0.5 (s)
5. Answer: 13
1) Body impulse P \u003d mv, 0 seconds impulse is 20 * 0 \u003d 0, in 20 seconds impulse is 20 * 4 \u003d 80 (correct)
2) in the time interval from 60 to 100 seconds, the average speed module is (0-4) / 2 \u003d 2 m / s, therefore, the body has passed 2 * 40 \u003d 80 meters (incorrect)
3) The resultant of all forces acting on the body is equal to F \u003d ma and since m \u003d 20 kg, and a \u003d 1/5, we obtain F \u003d 4 N (true)
4) the acceleration modulus in the time interval from 60 to 80 s is a \u003d dV / dt \u003d 1/20, the acceleration modulus in the time interval from 80 to 100 s hfdty 3/20. Less than 3 times (incorrect)
5) decreased by 90 times (incorrect)
6. Answer: 33
A body thrown horizontally from a height H moves horizontally uniformly (without acceleration) with speed. Time t depends on the height H as (the initial rate of fall is 0). The altitude does not change, therefore, the time remains the same.
There is no acceleration of movement, i.e. is equal to 0, and therefore will not change.
7. Answer: 14
8. Answer: 40
According to the ideal gas formula PV \u003d 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 \u003d R2T 0, i.e. Pressure remains the same P 2 \u003d 40kPa
9. Answer: 6
The graph shows that the process under study is isochoric. Since the volume of gas did not change, the gas did not perform work. Therefore, according to the first law of thermodynamics, the internal energy of a 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 \u003d 3 / 2vRT, because v and R - remain unchanged, then U 2 / U 1 \u003d 400/200 \u003d 2.
It turns out 2 times.
11. Answer: 15
1) The relative humidity of the air is defined as
where p is the partial pressure of water vapor; p H - saturated steam pressure (tabular 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 lower than on Wednesday. (right)
2) (wrong)
3) The partial pressure of water vapor is the pressure of this particular 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 the observer
14. Answer: 9
15. Answer: 80
16. Answer: 24
17. Answer: 31
Lorentz force modulus: 3) will not change
The orbital period of the 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)
Analysis of tasks 28 (part 2, qualitative problem)
Analysis of tasks 29 (part 2)
Secondary general education
Preparing for the exam-2018: analysis of the demo in physics
We bring to your attention an analysis of the USE tasks in physics from the 2018 demo version. The article contains explanations and detailed algorithms for solving problems, as well as recommendations and links to useful materials that are relevant in preparation for the exam.
USE-2018. Physics. Thematic training tasks
The edition contains:
assignments of different types on all topics of the 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 the topics, and for students: training tasks will allow you to systematically prepare for the exam when passing each topic.
A stationary point body begins to move along the axis Ox... The figure shows a graph of the dependence of the projection a xacceleration of this body from time to time t.
Determine which way the body went in the third second of the movement.
Answer: _________ m.
Decision
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 of 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 \u003d 2 s to t 2 \u003d 4 s, acceleration projection is zero. Consequently, the projection of the resultant force in this area, according to Newton's second law, is also zero. We determine the nature of the movement in this area: the body moved evenly. 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. Using the definition of acceleration, we write the equation for the projection of velocity V x = V 0x + a x t; since the body was initially at rest, the velocity projection by the end of the second second became
Then the path traversed by the body in a third second
Answer: 8 m.
Figure: 1
On a smooth horizontal surface there are two bars connected by a light spring. To a bar with a mass m\u003d 2 kg apply a constant force equal in modulus F\u003d 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 bar moves with an acceleration of 1 m / s 2.
Answer: _________ N.
Decision
Horizontally on a body with a mass m \u003d 2 kg two forces act, this is the force F\u003d 10 N and the elastic force from the side of the spring. The resultant of these forces imparts acceleration to the body. 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 control \u003d ma (2)
Let us express from formula (2) the elastic force modulus F control \u003d F – ma (3)
Substitute the numerical values \u200b\u200binto formula (3) and get, F control \u003d 10 N - 2 kg 1 m / s 2 \u003d 8 N.
Answer: 8 N.
Assignment 3
A body weighing 4 kg, located on a rough horizontal plane, was told along it a speed of 10 m / s. Determine the modulus of work performed 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.
Decision
The body is acted upon by the force of gravity, the reaction force of the support, the friction force that creates the braking acceleration. The body was initially given a speed of 10 m / s. Let's write 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)
Projected onto the axis X: –F tr \u003d - ma; F tr \u003d ma; (3) We need to determine the modulus of work of the friction force by the time when the speed becomes two times less, i.e. 5 m / s. Let's write down the formula for calculating the work.
A · ( F tr) \u003d - F tr S (4)
To determine the distance traveled, let's take the timeless formula:
| S = | v 2 - v 0 2 | (5) |
| 2a |
Substitute (3) and (5) in (4)
Then the modulus of the friction force will be equal to:
Substitute numerical values
| A(F tr) \u003d | 4 kg | (( | 5 m | ) 2 – (10 | m | ) 2) | \u003d 150 J | |
| 2 | with | with |
Answer: 150 J.
USE-2018. Physics. 30 training options for exam papers
The edition contains:
30 training options for the exam
instruction for implementation and evaluation criteria
answers to all tasks
The training options will help the teacher organize preparation for the exam, and the students will 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 block axis. What is the radius of the block's inner pulley if the system is in equilibrium?
Figure: one
Answer: _________ see.
Decision
According to the problem statement, the system is in equilibrium. On the picture L 1, shoulder strength L 2 shoulder of force Equilibrium condition: the moments of forces rotating bodies clockwise should 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 per shoulder. The forces acting on the thread from the side of the weights differ 3 times. This means that the radius of the inner pulley of the block differs from the outer one also 3 times. Hence the shoulder L 2 will be equal to 8 cm.
Answer:8 cm.
Assignment 5
Oh, at different points in time.
From the list below, select twocorrect statements and indicate their numbers.
- The potential energy of the spring at the moment of time 1.0 s is maximum.
- The period of oscillation of the ball is 4.0 s.
- The kinetic energy of the ball at time 2.0 s is minimal.
- The vibration amplitude of the ball is 30 mm.
- The total mechanical energy of the pendulum, consisting of a ball and a spring, at the moment of time 3.0 s is minimal.
Decision
The table shows 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 select two statements correctly. The system is a spring-loaded pendulum. At a moment in time t \u003d 1 s, the displacement of the body from the equilibrium position is maximum, so this is the amplitude value. by definition, the potential energy of an elastically deformed body can be calculated by the formula
| E p = 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 velocity 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 vibration t \u003d 2 s, full oscillation in twice as long T \u003d 4 s. Therefore, statements 1 will be true; 2.
Assignment 6
A small piece of ice was dropped into a cylindrical glass with water. After a while, the piece of 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.
- increased;
- decreased;
- has not changed.
Write in table
Decision
 |
 |
Figure: one |
|
Problems of this type are quite common in different versions of the exam. And as practice shows, students often make mistakes. We will try to analyze this task in detail. We denote m - mass of a piece of ice, ρ l - ice density, ρ in - water density, V pmt - the volume of the submerged part of the ice, equal to the volume of the displaced liquid (the volume of the hole). Let's mentally remove ice from the water. Then a hole will remain in the water, the volume of which is V pht, i.e. the volume of water displaced by a piece of ice Fig. one( b).
Let us write down the condition of ice floating in Fig. one( and).
F a = mg (1)
ρ in V pht 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 the water obtained from ice into the hole, 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, also does not change. Hence the answer would be
USE-2018. Physics. Training tasks
The publication is addressed to high school students to prepare for the exam in physics.
The manual includes:
20 training options
answers to all tasks
USE answer forms for each option.
The publication will assist teachers in preparing students for the Unified State Exam in Physics.
The weightless spring is 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 correspondence between graphs of dependences of physical quantities on deformation xsprings and these values. For each position of the first column, select the corresponding position from the second column and write in table
Decision
From the figure to the problem it is seen that when the spring is not deformed, then its free end, and, accordingly, point A are in the 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. In this case, point A moves evenly. 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 modulus of the elastic force on the deformation of the spring.
The graph under the letter B) is the dependence of the projection of the external force on the amount of deformation. Because with an increase in the external force, the amount of deformation and the elastic force increase.
Answer: 24.
Assignment 8
When constructing the Reaumur temperature scale, it is assumed that at normal atmospheric pressure ice melts at 0 degrees Reaumur (° R), and water boils at 80 ° R. Find what is the average kinetic energy of translational thermal motion of an ideal gas particle at a temperature of 29 ° R. Express your answer in eV and round to hundredths.
Answer: ________ eV.
Decision
The problem is interesting because it is necessary to compare two scales for measuring temperature. These are the Reaumur temperature scale and the Celsius scale. The melting points of ice are the same on the scales, and the boiling points are different, we can get a formula for converting from Reaumur degrees to Celsius degrees. it
Convert temperature 29 (° R) to degrees Celsius
We convert the obtained result to Kelvin using the formula
T = t° C + 273 (2);
T \u003d 36.25 + 273 \u003d 309.25 (K)
To calculate the average kinetic energy of the 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. The formula shows that the dependence of the average kinetic energy on temperature is straight, that is, how many times the temperature changes, how many times the average kinetic energy of the thermal motion of molecules changes. Substitute the numerical values:
The result is converted into electron volts and rounded to the nearest hundredth. Recall that
1 eV \u003d 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-chart. 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: ___________.
Decision
According to the condition of the problem in process 1–2, the graph of which is shown on 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 the 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
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 physics course required 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 exam. The theoretical material is presented in a concise, accessible form. Each topic is accompanied by examples of exam assignments that correspond to the USE format. This will help the teacher to organize the preparation for the unified state exam, and the students to independently check their knowledge and readiness for the final exam.
A blacksmith forges an iron horseshoe weighing 500 g at a temperature of 1000 ° C. When he finishes forging, he throws the horseshoe into a vessel of water. A hiss is heard and steam rises from the vessel. Find the mass of water that evaporates when a hot horseshoe is immersed in it. Assume that the water is already heated to the boiling point.
Answer: _________
Decision
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 immersion of the hot horseshoe, the energy received by the water will go directly to vaporization. Let us write the heat balance equation
with f m P · ( t n - 100) \u003d 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 of iron. From formula (1) we express the mass of water
When writing down the answer, pay attention to what units you want to leave the mass of water.
Answer: 90 g
One mole of a monatomic ideal gas participates in a cyclic process, the graph of which is shown in TV- diagram.
Please select twocorrect statements based on the analysis of the presented schedule.
- Gas pressure in state 2 is greater than gas pressure in state 4
- Gas work in section 2-3 is positive.
- In the section 1–2, the gas pressure increases.
- In section 4-1, a certain amount of heat is removed from the gas.
- The change in the internal gas energy in the section 1–2 is less than the change in the internal gas energy in the section 2–3.
Decision
This type of task tests the ability to read graphs and explain the presented dependence of physical quantities. It is important to remember how the dependence plots for isoprocesses look in different axes, in particular r \u003d const. In our example, on TVThe 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 one . 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 perform work, it is equal to zero. The statement is incorrect. In section 1-2, the pressure increases, also incorrect. We have just shown 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 the temperature constant when the gas is compressed.
Answer: 14.
The heat engine works according to the Carnot cycle. The temperature of the heat engine cooler was increased, leaving the heater temperature the same. The amount of heat received by the gas from the heater during the cycle did not change. How did the efficiency of the heat engine and the gas work per cycle change?
For each value, determine the corresponding change pattern:
- increased
- decreased
- hasn't changed
Write in table selected figures for each physical quantity. The numbers in the answer may be repeated.
Decision
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. Be able to record it through the temperature of the heater and the temperature of the refrigerator
in addition to be able to record 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, the temperature of the refrigerator was increased, leaving the heater temperature 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), then 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 - qQand negative - Q(see figure). Where is it directed relative to the figure ( to the right, left, up, down, towards the observer, from the observer) charge acceleration - q inthis moment of time, if only charges act on it + Q and –Q? Write the answer in a word (words)
Decision
Figure: one
Negative charge - q is in the field of two stationary charges: positive + Q and negative - Qas shown in the figure. in order to answer the question where the charge acceleration is directed - q, at a time when only charges + Q and - 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? Let's remember how similarly charged bodies interact, they are repelled, force The Coulomb force of interaction of charges is the central force. the force with which oppositely charged bodies are attracted. From the figure we see that the charge is q equidistant from stationary charges, the modules of which are equal. Therefore, the modulus will also be equal. The resulting force will be directed relative to the drawing way down.Charge acceleration will also be directed - q, i.e. way down.
Answer: Way down.
The book contains materials for successfully passing the exam in physics: brief theoretical information on all topics, assignments of different types and levels of difficulty, solving problems of an increased level of complexity, answers and assessment criteria. Students do not have to search the Internet for additional information and buy other manuals. In this book, they will find everything they need to prepare independently and effectively for the exam. The publication contains assignments of different types on all topics tested in the exam in physics, as well as solving problems of an increased level of complexity. The publication will provide invaluable assistance to students in preparation for the exam in physics, and can also be used by teachers in organizing the educational process.
Two series-connected resistors with a resistance of 4 Ohm and 8 Ohm 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 lower rating?
Answer: _________ Tue.
Decision
To solve the problem, it is advisable to draw a diagram of a series connection of resistors. Then remember the laws of the serial connection of conductors.
The scheme will be as follows:
Where R 1 \u003d 4 Ohm, R 2 \u003d 8 ohms. The voltage at the battery terminals is 24 V. When the conductors are connected in series on each section of the circuit, the current will be the same. 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 on a smaller resistor, we write:
P = I 2 R \u003d (2 A) 2.4 ohm \u003d 16 W.
Answer: P \u003d 16 W.
A wire frame with an area of \u200b\u200b2 · 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 \u003d 4 · 10 –6 cos10π t,
where all quantities are expressed in SI units. What is the modulus of magnetic induction?
Answer: ________________ mT.
Decision
The magnetic flux changes according to the law
Ф \u003d 4 · 10 –6 cos10π t,
where all quantities are expressed in SI units. You need to understand what a magnetic flux is in general and how this value is related to the modulus of magnetic induction B and frame area S... Let's write the equation in general form to understand what quantities are included in it.
Φ \u003d Φ m cosω t(1)
Remember that in front of 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 by the area of \u200b\u200bthe circuit and the cosine of the angle between the normal to the circuit and the magnetic induction vector Φ m \u003d IN · Scosα, the flux is maximum at cosα \u003d 1; express the modulus of induction
The answer must be recorded in mT. Our result is 2 mT.
Answer: 2.
The section of the electrical circuit consists of silver and aluminum wires connected in series. 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 displaced along the wires by a distance x
Using the graph, select twocorrect statements and indicate their numbers in the answer.
- The cross-sectional areas of the wires are the same.
- Cross-sectional area of \u200b\u200ba silver wire 6.4 · 10 –2 mm 2
- Cross-sectional area of \u200b\u200ba silver wire 4.27 · 10 –2 mm 2
- A thermal power of 2 W is generated in the aluminum wire.
- Silver wire produces less heat output than aluminum wire.
Decision
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 consists of silver and aluminum wires connected in series. 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 displaced along the wires by a distance x... The specific resistances of silver and aluminum are 0.016 μOhm · m and 0.028 μOhm · m, respectively.
The connection of the wires is serial, 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, the cross-sectional area of \u200b\u200bthe wire
| R = | ρ l | (1), |
| S |
where ρ is the specific resistance of the conductor; l - conductor length; S - cross-sectional area. The graph shows that the length of the silver wire L s \u003d 8 m; length of aluminum wire L a \u003d 14 m. Voltage on a section of silver wire U s \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. 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 determine the electrical resistance according to Ohm's law for a section of the circuit.
It is important to note that the numerical values \u200b\u200bmust be in SI for calculations.
Correct statement option 2.
Let's check the expressions for the cardinality.
P a \u003d 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 physics course required 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 exam. The theoretical material is presented in a concise, accessible form. Each topic is accompanied by examples of exam assignments that correspond to the USE format. This will help the teacher to organize the preparation for the unified state exam, and the students to independently check their knowledge and readiness for the final exam. At the end of the manual, answers to tasks for self-examination are given, which will help students and applicants to objectively assess their level of knowledge and the degree of preparedness for the certification exam. The manual is addressed to senior students, applicants and teachers.
A small object is positioned on the main optical axis of a thin converging lens between focal length and double focal length from it. The subject begins to move closer to the focus of the lens. How do the size of the image and the optical power of the lens change?
For each value, determine the corresponding character of its change:
- increases
- decreases
- does not change
Write in table selected figures for each physical quantity. The numbers in the answer may be repeated.
Decision
The object is located on the main optical axis of a thin converging lens between focal length and double focal length 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 - 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
Figure: 2
Constructed two images for two positions of the subject. 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 Is the resistance of the resistor).
For each position of the first column, select the corresponding position of the second and write in table the selected numbers under the corresponding letters.
Decision
Figure:1
By the condition of the problem, the internal resistance of the source is neglected. The circuit contains a constant current source, two resistors, resistance R, each and the key. The first condition of the problem requires determining the current through the source with a closed switch. 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 through the source with a closed switch;
where N - the number of conductors connected in parallel with the same resistance.
- EMF of the current source.
Substituting (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 only through one resistor. Ohm's law for a complete circuit in this case will be:
Decision
Let's write down the nuclear reaction for our case:
As a result of this reaction, the law of conservation of charge and mass numbers 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.
The new handbook contains all the theoretical material on the physics course required to pass the unified state exam. It includes all the elements of the content verified 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 of test items. Practical assignments correspond to the USE format. At the end of the manual you will find the answers to the tests. The manual is addressed to schoolchildren, applicants and teachers.
Period Tthe half-life of the potassium isotope is 7.6 minutes. 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.
Decision
The problem of using the law of radioactive decay. It can be written as
where m 0 - the initial mass of the substance, t - the time during which the substance disintegrates, T - half life. Substitute numerical values
Answer: 0.3 mg.
A beam of monochromatic light is incident 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 energy on wavelength λ and frequency of light ν. Establish a correspondence between the graph and the energy for which it can determine the presented dependence.
For each position of the first column, select the corresponding position from the second column and write in table the selected numbers under the corresponding letters.
Decision
It is useful to remember the definition of photoeffect. This is the phenomenon of the interaction of light with matter, as a result of which the energy of the photons is transferred to the electrons of the matter. Distinguish between external and internal photoelectric effect. In our case, we are talking about an external photoeffect. When, under the action of light, electrons are pulled out of the substance. The work function depends on the material from which the photocell photocathode is made, and does not depend on the light frequency. The energy of the incident photons is proportional to the frequency of the light.
E= hν (1)
where λ is the wavelength of light; with - the speed of light,
Substitute (3) into (1) We obtain
We analyze the resulting formula. Obviously, with increasing wavelength, the energy of incident photons decreases. This type of dependence corresponds to the graph under the letter A)
Let's write down 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 out - 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 \u003d 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 \u003d | A out | (7), |
| h |
this is the minimum frequency at which the photo 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 readings (see figure) if the error in direct current measurement is equal to the division value of the ammeter.
Answer: (___________ ± ___________) A.
Decision
The task tests the ability to record the readings of the measuring device, taking into account the specified measurement error. Determine the scale division value with \u003d (0.4 A - 0.2 A) / 10 \u003d 0.02 A. The measurement error by the condition is equal to the division price, i.e. Δ I = c \u003d 0.02 A. The final result is written as:
I \u003d (0.20 ± 0.02) A
It is necessary to assemble an experimental setup with which it is possible to determine the coefficient of sliding friction of steel against wood. for this, the student took a steel bar with a hook. Which two items from the list of equipment below should be additionally used for this experiment?
- wooden lath
- dynamometer
- beaker
- plastic rail
- stopwatch
In response, write down the numbers of the selected items.
Decision
In the task, it is required to determine the coefficient of sliding friction of steel on wood, therefore, for 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 the sliding friction force
F ck = μ · N (1),
where μ is the coefficient of sliding friction, N - the reaction force of the support, equal in absolute value to the body weight.
Answer: |
The handbook contains detailed theoretical material on all topics tested by the USE in physics. After each section, there are different-level tasks in the form of the exam. For the final control of knowledge at the end of the handbook, training options are given that correspond to the USE. Students do not have to search the Internet for additional information and buy other manuals. In this guide, they will find everything they need to prepare for the exam independently and effectively. 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 covered by the exam. After each section, examples of USE tasks and a practice test are given. All tasks are answered. The publication will be useful for physics teachers, parents for effective preparation of students for the Unified State Exam.
Consider the table for bright stars.
Star name |
Temperature, |
Weight |
Radius |
Distance to the star |
Aldebaran |
5 |
|||
Betelgeuse |
||||
Please select twostatements that match the characteristics of the stars.
- The surface temperature and radius of Betelgeuse indicate that this star belongs to red supergiants.
- The temperature on the surface of Procyon is 2 times lower than on the surface of the Sun.
- The stars Castor and Capella are at the same distance from the Earth and therefore belong to the same constellation.
- The star Vega belongs to white stars of spectral type A.
- Since the masses of the stars Vega and Capella are the same, they belong to the same spectral type.
Decision
Star name |
Temperature, |
Weight |
Radius |
Distance to the star |
Aldebaran |
||||
Betelgeuse |
||||
| 2,5 |
||||
In the assignment, you need to choose two correct statements that correspond to the characteristics of the stars. The table shows that Betelgeuse has the lowest temperature and a large radius, which means that this star belongs to the red giants. Therefore, the correct answer is (1). To choose the second statement correctly, you need to know the distribution of stars by spectral type. We need to know the temperature range and the corresponding color of the star. Analyzing the data in the table, we conclude that the correct statement will be (4). The star Vega belongs to white stars of spectral type A.
A projectile weighing 2 kg, flying at a speed of 200 m / s, explodes into two fragments. The first fragment weighing 1 kg flies at an angle of 90 ° to the original direction at a speed of 300 m / s. Find the speed of the second shard.
Answer: _______ m / s.
Decision
At the moment the 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 between the bodies of this system. for our case, we write:
- projectile speed; m - mass of the projectile before rupture; Is the speed of the first fragment; m 1 - 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 Xcoinciding with the direction of the velocity of the projectile, 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 required impulse vector is determined by the Pythagorean theorem for a right-angled triangle.
p 2 = √p 2 + p 1 2 (3)
p 2 \u003d √400 2 + 300 2 \u003d 500 (kg m / s)
Answer: 500 m / s.
When an ideal monatomic gas was compressed at constant pressure, external forces performed work of 2000 J. What amount of heat was transferred by the gas to the surrounding bodies?
Answer: _____ J.
Decision
The problem for 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 - the work of external forces. By condition, the gas is monoatomic and it is compressed at constant pressure.
A sun \u003d - A r (2),
| Q = Δ U– A sun \u003d Δ U+ A r \u003d | 3 | pΔ V + pΔ V = | 5 | pΔ V, |
| 2 | 2 |
where pΔ V = A r
Answer: 5000 J.
A plane monochromatic light wave with a frequency of 8.0 · 10 14 Hz is incident along the normal onto the diffraction grating. A collecting lens with a focal length of 21 cm is placed in 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 tenths. Consider for small angles (φ ≈ 1 in radians) tgα ≈ sinφ ≈ φ.
Decision
The angular directions to the maxima of the diffraction pattern are determined by the equation
d Sinφ \u003d k Λ (1),
where d Is the period of the diffraction grating, φ is the angle between the normal to the grating and the direction to one of the maxima of the diffraction pattern λ is the light wavelength, k - an integer called the order of the diffraction maximum. Let us express from Eq. (1) the period of the diffraction grating
Figure: one
By 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, the frequency of the light wave ν \u003d 8.0 · 10 14 Hz, the 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. In fig. 1 shows a diagram of the path of rays through the grating and the lens behind it. A diffraction pattern is observed on the screen located in the focal plane of the collecting lens, as a result of the interference of waves coming from all the slits. Let's use formula one for two maxima of the 1st and 2nd order.
dsinφ 1 \u003d kλ (2),
if k \u003d 1, then dsinφ 1 \u003d 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 - distance from zero maximum to first order maximum. Likewise for distance x 2 .
Then we have
Diffraction grating period,
since by definition
where with \u003d 3 10 8 m / s - the speed of light, then substituting the numerical values \u200b\u200bwe get
The answer was presented in micrometers, rounded to tenths, as required in the problem statement.
Answer: 4.4 microns.
Based on the laws of physics, find the reading of an ideal voltmeter in the diagram 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.
Decision
Figure: one
Part C assignments require a complete and detailed answer from the student. 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's take into account that the capacitor in the circuit is not charged initially. Consider two states. When the switch is open, only a resistor is connected to the power supply. The voltmeter readings are zero, since it is connected in parallel with the capacitor, and the capacitor is not initially charged, then q 1 \u003d 0. Second state when the key is closed. Then the voltmeter readings will increase until they reach the maximum value, which will not change over time,
where r Is the internal resistance of the source. The voltage across the capacitor and resistor, according to Ohm's law for a section of the circuit U = I · Rwill 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 pulled and acts on the ball with force T... If the thread is cut, the ball will float, and the water level will change by h \u003d 5 cm. Find the thread tension T.
Decision
|
|
|
Figure: one |
Figure: 2 |
The original wooden ball is tied with a thread to the bottom of a cylindrical vessel with the area of \u200b\u200bthe bottom S \u003d 100 cm 2 \u003d 0.01 m 2 and is completely submerged in water. Three forces act on the ball: gravity from the side of the Earth, - the force of Archimedes from the side of the liquid, - the tension force of the thread, the result of the interaction of the ball and the thread. According to the condition of equilibrium of the ball in the first case, the geometric sum of all forces acting on the ball must be equal to zero:
Let's choose the coordinate axis OY and direct it up. Then, taking into account the projection, equation (1) is written:
F a 1 = T + mg (2).
Let's write down the strength of Archimedes:
F a 1 \u003d ρ V 1 g (3),
where V 1 - the volume of a part of the ball immersed in water, in the first it is the volume of the whole ball, m - ball mass, ρ - water density. Equilibrium condition in the second case
F a 2 \u003d mg (4)
Let's write down the strength of Archimedes in this case:
F a 2 \u003d ρ V 2 g (5),
where V 2 - the volume of the part of the ball 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 F a 1 – F a 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)
Considering that
V 1 – V 2 = S · h (7),
where h \u003d H 1 - H 2; get
T \u003d ρ g S · h (8)
Substitute numerical values
Answer: 5 N.
All the information necessary for passing the exam in physics is presented in clear and accessible tables, after each topic - training tasks to control knowledge. With the help of this book, students will be able to improve their knowledge in the shortest possible time, recall all the most important topics a few 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 become much closer! The manual contains theoretical information on all topics tested on the exam in physics. Each section is followed by training tasks of different types with answers. A clear and accessible presentation of the material will allow you to quickly find the necessary information, eliminate gaps in knowledge and quickly repeat a large amount of information. The publication will assist high school students in preparing for lessons, various forms of current and intermediate control, as well as to prepare for exams.
Task 30
In a room measuring 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 is the relative humidity in the room after 1.5 hours? The pressure of saturated water vapor at a temperature of 10 ° C is 1.23 kPa. Consider the room as an airtight vessel.
Decision
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 set, 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 \u003d 30%. We express the partial pressure of water vapor from the formula:
where T = t + 273 (C), R Is a universal gas constant. Let us express the initial mass of steam contained in the room using equations (2) and (3):
During the time τ of operation of the humidifier, the mass of water will increase by
Δ m = τ · ρ · I, (6)
where I– the productivity of the humidifier by the condition it is equal to 0.2 l / h \u003d 0.2 · 10 –3 m 3 / h, ρ \u003d 1000 kg / m 3 is the density of water. Let us substitute formulas (4) and (5) in (6)
Let's transform the expression and express
This is the sought-after formula for the relative humidity in the room after the humidifier is in operation.
Substitute the numerical values \u200b\u200band get the following result
Answer:83 %.
On horizontally located rough rails with negligible resistance, two identical rods of mass can slide m \u003d 100 g and resistance R \u003d 0.1 ohm each. The distance between the rails is l \u003d 10 cm, and the coefficient of friction between the rods and the rails is μ \u003d 0.1. Rails with rods are in a uniform vertical magnetic field with induction B \u003d 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? Disregard the self-induction of the circuit.
Decision
Figure: one
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. A change in the magnetic flux of the penetrating circuit leads to the emergence of EMF induction. In our case, when the rods move at different speeds, the change in the flux of the magnetic induction vector penetrating the contour over the time interval Δ tdetermined by the formula
ΔΦ = B · l · ( v 1 – v 2) Δ t (1)
This leads to the emergence of EMF induction. According to Faraday's law
By the condition of the problem, the self-induction of the circuit is neglected. According to Ohm's law for a closed circuit for the current that occurs in the circuit, we write the expression:
On conductors with a current in a magnetic field, the Ampere force acts and the modules of which are equal to each other, and are equal to the product of the current strength, the modulus 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 \u003d 1, then
F 1 = F 2 = I · B · l (4)
The braking force of friction still acts on the rods,
F tr \u003d μ m · g (5)
according to the 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 \u003d F 2, taking into account (3), (4), (5)
Let us express from this the relative speed
Substitute the numerical values:
Answer: 2 m / s.
In an experiment to study the photoelectric effect, light with a frequency ν \u003d 6.1 · 10 14 Hz falls on the cathode surface, as a result of which a current arises in the circuit. Current dependence graph I from stresses U between the anode and the cathode is shown in the figure. What is the power of the incident light R, if, on average, one of 20 photons incident on the cathode knocks out an electron?
Decision
By definition, current strength is a physical quantity numerically equal to the charge qpassing through the cross-section of the conductor per unit of 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 passed through the cross section of the conductor can be calculated
q = N e · e · t (2),
where e - electron charge modulus, N e– the number of photoelectrons ejected from the cathode in 1 s. According to the condition, one of the 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 E f \u003d h · v, then the power of the incident light
After substitution of the corresponding values, we obtain the final formula
| P = N f · h · v = | 20 · I max h USE-2018. Physics (60x84 / 8) 10 training options for examination papers to prepare for the unified state exam A new textbook on physics for the preparation of the Unified State Exam is offered to the attention of schoolchildren and applicants, which contains 10 options for training examinations. 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 difficulty. At the end of the book, self-test answers are given for all tasks. The proposed training options will help the teacher organize preparation for the unified state exam, and the students independently test their knowledge and readiness for the final exam. The manual is addressed to schoolchildren, applicants and teachers. |
In 2018, graduates of grade 11 and institutions of secondary vocational education will take the Unified State Examination 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 introduced into it.
What is the meaning of the changes and how many there are
The main change related to the Unified State Exam in Physics, compared to previous years, is the absence of a test part with a choice of answers. This means that preparation for the exam should be accompanied by the student's ability to give short or detailed answers. Therefore, it will not 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 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 one of 27 tasks, which involves a short or complete answer. In the second part, there are 5 advanced tasks, where it is necessary to give a detailed answer and explain the course of your solution. One important caveat: many students skip this part, but even trying to complete these tasks can get from one to two points.
All changes in the exam in physics are made in order to deepen preparation and improve the assimilation of knowledge in the subject. In addition, the elimination of the test part motivates future applicants to accumulate knowledge more intensively and to reason logically.
Exam structure
Compared to the previous year, the structure of the USE has not undergone significant changes. All work is given 235 minutes. Each task of the basic part should be solved from 1 to 5 minutes. Problems of increased complexity are solved in about 5-10 minutes.
All CMMs are stored at the exam site, and autopsy is performed during the test. The structure is as follows: 27 basic tasks check whether the examinee has knowledge in all areas of physics, from mechanics to quantum and nuclear physics. In 5 tasks of a high level of difficulty, the student demonstrates skills in the logical justification of his decision and the correctness of his train of thought. The number of primary points can reach a maximum of 52. Then they are recalculated within a 100-point scale. Due to the change in the primary score, the minimum passing score may also change.
Demo version
The demo version of the Unified State Exam in Physics is already on the official portal of FIPI, which develops a unified state exam. The structure and complexity of the demo version is similar to the one that will appear on the exam. Each assignment is detailed, and at the end there is a list of answers to questions, for which the student checks against his solutions. Also at the end there is a detailed layout for each of the five tasks, indicating the number of points for correctly or partially performed actions. For each task of high complexity, you can get from 2 to 4 points, depending on the requirements and deployment of the solution. Assignments can contain a sequence of numbers that you need to correctly write down, establishing correspondence between elements, as well as small tasks in one or two steps.
- Download demo: ege-2018-fiz-demo.pdf
- Download archive with specification and codifier: ege-2018-fiz-demo.zip
We wish you to successfully pass physics and enter the desired university, everything is in your hands!
