Download kim ege in physics. Preparation for the exam in physics: examples, solutions, explanations
Specification
control measuring materials
for holding a unified state exam in 2017
in PHYSICS
1. The purpose of the KIM exam
The Unified State Exam (hereinafter - the USE) is a form of an objective assessment of the quality of training for people who have completed educational programs of secondary general education, using standardized tasks (control measurement materials).
The exam is carried out in accordance with Federal Law dated December 29, 2012 No. 273-ФЗ “On Education in the Russian Federation”.
Control measuring materials make it possible to establish the level of mastering by graduates of the Federal component of the state educational standard of secondary (complete) general education in physics, basic and specialized levels.
The results of the unified state examination in physics are recognized by educational organizations of secondary vocational education and educational organizations of higher professional education as the results of entrance tests in physics.
2. Documents defining the content of KIM USE
3. Approaches to the selection of content, the development of the structure of KIM USE
Each version of the examination work includes controlled content elements from all sections of the school physics course, with assignments of all taxonomic levels offered for each section. The most important from the point of view of continuing education in higher educational institutions, the content elements are controlled in the same version by tasks of different difficulty levels. The number of tasks in a particular section is determined by its substantive content and is proportional to the training time allocated for its study in accordance with the approximate program in physics. The various plans for constructing the examination options are built on the principle of meaningful additions so that in general all series of options provide diagnostics for the development of all the content elements included in the codifier.
The priority in the design of KIM is the need to verify the types of activities stipulated by the standard (taking into account restrictions in the conditions of mass written verification of students' knowledge and skills): mastering the conceptual apparatus of a physics course, mastering methodological knowledge, applying knowledge to explain physical phenomena and solve problems. The mastery of skills in working with information on physical content is verified indirectly using various methods of presenting information in texts (graphs, tables, diagrams and schematic drawings).
The most important type of activity from the point of view of successful continuation of education at a university is problem solving. Each option includes tasks in all sections of different difficulty levels, allowing you to test the ability to apply physical laws and formulas both in typical educational situations and in non-traditional situations requiring a sufficiently high degree of independence when combining well-known action algorithms or creating your own task execution plan .
The objectivity of checking tasks with a detailed answer is ensured by unified assessment criteria, the participation of two independent experts evaluating one work, the possibility of appointing a third expert and the presence of an appeal procedure.
The Unified State Exam in Physics is an exam of the choice of graduates and is intended to differentiate between admission to higher educational institutions. For these purposes, tasks of three difficulty levels are included in the work. Fulfillment of tasks of a basic level of complexity allows assessing the level of mastering the most significant substantive elements of a high school physics course and mastering the most important activities.
Among the tasks of the basic level, tasks whose content corresponds to the standard of the basic level are distinguished. The minimum number of exam points in physics, confirming the graduate's mastery of the program of secondary (complete) general education in physics, is established on the basis of the requirements for mastering the standard of the basic level. The use of tasks of increased and high levels of complexity in the examination work makes it possible to assess the degree of preparedness of a student for continuing education at a university.
4. The structure of KIM USE
Each version of the examination paper consists of 2 parts and includes 32 tasks that vary in form and level of difficulty (table 1).
Part 1 contains 24 tasks, of which 9 tasks with choosing and writing down the number of the correct answer and 15 tasks with a short answer, including tasks with self-recording the answer in the form of a number, as well as tasks for establishing correspondence and multiple choice, in which answers are necessary write as a sequence of numbers.
Part 2 contains 8 tasks, united by a common activity — solving problems. Of these, 3 tasks with a short answer (25-27) and 5 tasks (28-32), for which you need to provide a detailed answer.
Unified State Exam 2017 Physics Typical Test Tasks of Lukashev
M .: 2017 - 120 s.
Typical physics tests contain 10 variants of task sets drawn up taking into account all the features and requirements of the Unified State Exam in 2017. The purpose of the manual is to provide readers with information about the structure and content of control measuring materials 2017 in physics, as well as the degree of difficulty of the tasks. The collection contains answers to all test cases, as well as solutions to the most complex problems in all 10 variants. In addition, examples of forms used on the exam. The team of authors - experts of the federal subject commission of the exam in physics. The manual is addressed to teachers to prepare students for the physics exam, and high school students - for self-study and self-control.
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CONTENT
Instructions for completing work 4
OPTION 1 9
Part 1 9
Part 2 15
OPTION 2 17
Part 1 17
Part 2 23
OPTION 3 25
Part 1 25
Part 2 31
OPTION 4 34
Part 1 34
Part 2 40
OPTION 5 43
Part 1 43
Part 2 49
OPTION 6 51
Part 1 51
Part 2 57
OPTION 7 59
Part 1 59
Part 2 65
OPTION 8 68
Part 1 68
Part 2 73
OPTION 9 76
Part 1 76
Part 2 82
OPTION 10 85
Part 1 85
Part 2 91
ANSWERS. EXAMINATION EVALUATION SYSTEM
WORKS ON PHYSICS 94
To perform rehearsal work in physics, 3 hours 55 minutes (235 minutes) are given. The work consists of 2 parts, including 31 tasks.
In tasks 1-4, 8-10, 14, 15, 20, 24-26, the answer is an integer or a final decimal fraction. Write down the number in the answer field in the text of the work, and then transfer the following sample to answer form No. 1. The units of measurement of physical quantities need not be written.
The answer to tasks 27-31 includes a detailed description of the entire course of the task. In the answer form No. 2, indicate the number of the task and write down its complete solution.
In calculations, it is allowed to use a non-programmable calculator.
All USE forms are filled with bright black ink. You can use gel, capillary or fountain pens.
When completing tasks, you can use the draft. Draft entries are not taken into account when evaluating work.
The points you earned for completed assignments are cumulative. Try to complete as many tasks as possible and score the most points.
Preparation for the exam and exam
Secondary general education
Line of the teaching materials of A. V. Grachev. Physics (10-11) (base, deep)
Line of the teaching materials of A. V. Grachev. Physics (7-9)
Line of teaching materials of A.V. Peryshkin. Physics (7-9)
Preparation for the exam in physics: examples, solutions, explanations
We analyze the tasks of the exam in physics (Option C) with a teacher.Lebedeva Alevtina Sergeevna, physics teacher, work experience 27 years. Certificate of honor from the Ministry of Education of the Moscow Region (2013), Letter of Appreciation from the Head of the Voskresensky Municipal District (2015), Diploma from the President of the Association of Teachers of Mathematics and Physics of the Moscow Region (2015).
The paper presents tasks of different difficulty levels: basic, advanced and high. Basic level tasks are simple tasks that test the mastery of the most important physical concepts, models, phenomena and laws. Tasks of a higher level are aimed at testing the ability to use the concepts and laws of physics to analyze various processes and phenomena, as well as the ability to solve problems with the application of one or two laws (formulas) for any of the topics of the school physics course. In work 4, the tasks of Part 2 are tasks of a high level of complexity and test the ability to use the laws and theories of physics in an altered or new situation. The fulfillment of such tasks requires the application of knowledge from two three sections of physics at once, i.e. high level of training. This option is fully consistent with the demonstration version of the exam in 2017, the tasks are taken from the open bank of tasks of the exam.
The figure shows a graph of the dependence of the velocity module on time t. Determine on the schedule the path traveled by the car in the time interval from 0 to 30 s.
Decision. The path traveled by the car in the time interval from 0 to 30 s is most easily defined as the area of \u200b\u200bthe trapezoid, the basis of which is the time intervals (30 - 0) \u003d 30 s and (30 - 10) \u003d 20 s, and the height is the speed v \u003d 10 m / s, i.e.
| S = | (30 + 20) from | 10 m / s \u003d 250 m. |
| 2 |
Answer. 250 m.
A weight of 100 kg is lifted vertically with a cable. The figure shows the dependence of the velocity projection V load on the axis directed upwards from time to time t. Determine the tensile modulus of the cable during the lift.
Decision. According to the graph of the dependence of the projection of speed v load on an axis directed vertically upward from time to time t, you can determine the projection of the acceleration of the load
| a = | ∆v | = | (8 - 2) m / s | \u003d 2 m / s 2. |
| ∆t | 3 s |
The load is affected by: gravity directed vertically downward and cable tension force directed along the cable vertically upward see fig. 2. We write the basic equation of dynamics. We use the second law of Newton. The geometric sum of the forces acting on the body is equal to the product of the mass of the body and the acceleration communicated to it.
+ = (1)
We write the equation for the projection of vectors in the frame of reference associated with the earth, the OY axis is directed up. The projection of the tension force is positive, since the direction of the force coincides with the direction of the OY axis, the projection of gravity is negative, since the force vector is opposite to the OY axis, the projection of the acceleration vector is also positive, since the body moves upward with acceleration. We have
T – mg = ma (2);
from the formula (2) the modulus of the tension force
T = m(g + a) \u003d 100 kg (10 + 2) m / s 2 \u003d 1200 N.
Answer. 1200 N.
The body is dragged along a rough horizontal surface with a constant speed, the modulus of which is 1.5 m / s, applying force to it as shown in Figure (1). In this case, the module of the sliding friction force acting on the body is 16 N. What is the power developed by the force? F?
Decision. Imagine the physical process specified in the condition of the problem and make a schematic drawing indicating all the forces acting on the body (Fig. 2). We write the basic equation of dynamics.
Tr + + \u003d (1)
Having chosen a reference system associated with a fixed surface, we write down the equations for the projection of vectors onto the selected coordinate axes. By the condition of the problem, the body moves uniformly, since its speed is constant and equal to 1.5 m / s. This means the acceleration of the body is zero. Two forces act horizontally on the body: sliding friction force tr. and the force with which the body is dragged. The projection of the friction force is negative, since the force vector does not coincide with the direction of the axis X. Power projection F positive. We remind you, to find the projection, lower the perpendicular from the beginning and end of the vector to the selected axis. With this in mind, we have: F cosα - F mp \u003d 0; (1) express the projection of force F, this is Fcosα \u003d F mp \u003d 16 N; (2) then the power developed by the force will be equal to N = Fcosα V (3) We make a replacement, taking into account equation (2), and substitute the corresponding data in equation (3):
N \u003d 16 N · 1.5 m / s \u003d 24 W.
Answer. 24 watts
The load, mounted on a light spring with a stiffness of 200 N / m, performs vertical vibrations. The figure shows a graph of the displacement x cargo from time to time t. Determine what the mass of the cargo is. Round the answer to an integer.
Decision. The load on the spring performs vertical oscillations. According to the graph of the dependence of the load displacement x from time t, we determine the period of oscillation of the cargo. The oscillation period is T \u003d 4 s; from the formula T \u003d 2π we express the mass m cargo.
| = | T | ; | m | = | T 2 | ; m = k | T 2 | ; m \u003d 200 N / m | (4 s) 2 | \u003d 81.14 kg ≈ 81 kg. |
| 2π | k | 4π 2 | 4π 2 | 39,438 |
Answer: 81 kg.
The figure shows a system of two light blocks and a weightless cable, with which you can hold in balance or lift a load of 10 kg. Friction is negligible. Based on the analysis of the figure, select twotrue statements and indicate in the answer their numbers.
- In order to keep the load in balance, you need to act on the end of the rope with a force of 100 N.
- The block system shown in the figure does not give a gain in strength.
- h, you need to stretch a piece of rope with a length of 3 h.
- In order to slowly raise the load to a height hh.
Decision. In this problem, it is necessary to recall simple mechanisms, namely blocks: a movable and fixed block. The movable block gives a gain in force twice, while the section of the rope must be pulled out twice as long, and the fixed block is used to redirect the force. In the work, simple winning mechanisms do not give. After analyzing the task, we immediately select the necessary statements:
- In order to slowly raise the load to a height h, you need to stretch a piece of rope 2 h.
- In order to keep the load in balance, you need to act on the end of the rope with a force of 50 N.
Answer. 45.
An aluminum load is completely immersed in a vessel with water, mounted on a weightless and inextensible thread. The load does not touch the walls and bottom of the vessel. Then, an iron load, the mass of which is equal to the mass of aluminum cargo, is immersed in the same vessel with water. How, as a result of this, the modulus of the tension force of the thread and the modulus of gravity acting on the load change?
- Is increasing;
- Decreases;
- Does not change.
Decision. We analyze the condition of the problem and select those parameters that do not change during the study: these are body mass and liquid into which the body is immersed on threads. After that, it is better to perform a schematic drawing and indicate the forces acting on the load: the thread tension force F upr directed along the thread up; gravity directed vertically downward; Archimedean force a acting from the liquid side to the submerged body and directed upwards. According to the condition of the problem, the mass of cargo is the same, therefore, the modulus of gravity acting on the cargo does not change. Since the density of goods is different, the volume will also be different
| V = | m | . |
| p |
The density of iron is 7800 kg / m 3, and aluminum cargo 2700 kg / m 3. Consequently, V well< V a. The body in equilibrium, the resultant of all the forces acting on the body, is zero. Point the OY coordinate axis up. The basic equation of dynamics taking into account the projection of forces is written as F control + F a – mg \u003d 0; (1) Express the tension force F control \u003d mg – F a (2); Archimedean force depends on the density of the liquid and the volume of the submerged part of the body F a = ρ gVpct (3); The density of the liquid does not change, and the body volume of iron is less V well< V a, therefore, the Archimedean force acting on the iron load will be less. We conclude that the modulus of the tension force of the thread, working with equation (2), it will increase.
Answer. 13.
Bar mass m slides from a fixed rough inclined plane with an angle α at the base. The bar acceleration modulus is a, the bar velocity modulus increases. Air resistance can be neglected.
Set the correspondence between the physical quantities and the formulas by which they can be calculated. For each position of the first column, select the corresponding position from the second column and write down the selected numbers in the table under the corresponding letters.
B) The coefficient of friction of the bar on an inclined plane
3) mg cosα
| 4) sinα - | a |
| gcosα |
Decision. This task requires the application of Newton's laws. We recommend making a schematic drawing; indicate all kinematic characteristics of the movement. If possible, depict the acceleration vector and the vectors of all the forces applied to the moving body; remember that the forces acting on the body are the result of interaction with other bodies. Then write the basic equation of dynamics. Select a reference system and write the resulting equation for the projection of the force and acceleration vectors;
Following the proposed algorithm, we make a schematic drawing (Fig. 1). The figure shows the forces applied to the center of gravity of the bar, and the coordinate axes of the reference system associated with the surface of an inclined plane. Since all forces are constant, the movement of the bar will be equally variable with increasing speed, i.e. acceleration vector is directed towards the movement. Choose the direction of the axes as shown in the figure. We write the projection of the forces on the selected axis.
We write the basic equation of dynamics:
Tr + \u003d (1)
We write this equation (1) for the projection of forces and acceleration.
On the OY axis: the projection of the support reaction force is positive, since the vector coincides with the direction of the OY axis N y = N; the projection of the friction force is zero since the vector is perpendicular to the axis; the projection of gravity will be negative and equal mg y= – mgcosα; acceleration vector projection a y \u003d 0, since the acceleration vector is perpendicular to the axis. We have N – mgcosα \u003d 0 (2) from the equation we express the reaction force acting on the bar, from the side of the inclined plane. N = mgcosα (3). We write the projections onto the OX axis.
OX Axis: Power Projection N equal to zero, since the vector is perpendicular to the axis OX; The projection of the friction force is negative (the vector is directed in the opposite direction relative to the selected axis); the projection of gravity is positive and equal to mg x = mgsinα (4) from a right triangle. Positive acceleration projection a x = a; Then we write equation (1) taking into account the projection mgsinα - F mp \u003d ma (5); F mp \u003d m(gsinα - a) (6); Remember that the force of friction is proportional to the force of normal pressure N.
By definition F mp \u003d μ N (7), we express the coefficient of friction of the bar on an inclined plane.
| μ = | F tr | = | m(gsinα - a) | \u003d tgα - | a | (8). |
| N | mgcosα | gcosα |
Choose the appropriate position for each letter.
Answer. A is 3; B - 2.
Task 8. Gaseous oxygen is in a vessel with a volume of 33.2 liters. The gas pressure is 150 kPa, its temperature is 127 ° C. Determine the mass of gas in this vessel. Express the answer in grams and round to the nearest whole number.
Decision. It is important to pay attention to the conversion of units to the SI system. The temperature is converted to Kelvin T = t° C + 273, volume V \u003d 33.2 l \u003d 33.2 · 10 –3 m 3; We translate pressure P \u003d 150 kPa \u003d 150,000 Pa. Using the ideal gas equation of state
express the mass of gas.
Be sure to pay attention to which unit is asked to record the answer. It is very important.
Answer. 48 g
Task 9. The ideal monatomic gas in the amount of 0.025 mol adiabatically expanded. At the same time, its temperature dropped from + 103 ° С to + 23 ° С. What work did the gas do? Express the answer in Joules and round to the nearest integer.
Decision. First, the gas is a monatomic number of degrees of freedom i \u003d 3, secondly, the gas expands adiabatically - this means without heat transfer Q \u003d 0. Gas does the work by reducing internal energy. With this in mind, we write the first law of thermodynamics in the form 0 \u003d ∆ U + A g; (1) express gas work A r \u003d –∆ U (2); The change in internal energy for a monatomic gas can be written as
Answer. 25 J.
The relative humidity of a portion of air at a certain temperature is 10%. How many times should the pressure of this portion of air be changed so that at a constant temperature its relative humidity increases by 25%?
Decision. Questions related to saturated steam and air humidity often cause difficulties for students. We use the formula for calculating the relative humidity
According to the condition of the problem, the temperature does not change, which means that the saturated vapor pressure remains the same. We write formula (1) for two states of air.
φ 1 \u003d 10%; φ 2 \u003d 35%
We express the air pressure from formulas (2), (3) and find the pressure ratio.
| P 2 | = | φ 2 | = | 35 | = 3,5 |
| P 1 | φ 1 | 10 |
Answer. Pressure should be increased by 3.5 times.
A hot substance in a liquid state was slowly cooled in a melting furnace with a constant power. The table shows the results of measurements of the temperature of the substance over time.
Choose from the proposed list two statements that correspond to the results of measurements and indicate their numbers.
- The melting point of the substance under these conditions is 232 ° C.
- In 20 minutes. after the start of measurements, the substance was only in a solid state.
- The heat capacity of the substance in the liquid and solid state is the same.
- After 30 minutes after the start of measurements, the substance was only in a solid state.
- The process of crystallization of the substance took more than 25 minutes.
Decision. Since the substance was cooled, its internal energy decreased. The temperature measurement results allow you to determine the temperature at which the substance begins to crystallize. While the substance passes from a liquid to a solid state, the temperature does not change. Knowing that the melting point and crystallization temperature are the same, we choose the statement:
1. The melting point of the substance under these conditions is 232 ° C.
The second true statement is:
4. After 30 minutes after the start of measurements, the substance was only in a solid state. Since the temperature at this point in time, already below the crystallization temperature.
Answer.14.
In an isolated system, body A has a temperature of + 40 ° C, and body B has a temperature of + 65 ° C. These bodies brought into thermal contact with each other. After some time, thermal equilibrium came. How did the temperature of body B and the total internal energy of body A and B change as a result?
For each value, determine the corresponding nature of the change:
- Increased;
- Decreased;
- Not changed.
Record in the table the selected numbers for each physical quantity. The numbers in the answer can be repeated.
Decision. If in an isolated system of bodies no energy transformations take place except heat exchange, then the amount of heat given up by bodies whose internal energy decreases is equal to the amount of heat received by bodies whose internal energy increases. (According to the law of conservation of energy.) In this case, the total internal energy of the system does not change. Tasks of this type are solved on the basis of the heat balance equation.
| ∆U \u003d ∑ | n | ∆U i \u003d0 (1); |
| i = 1 |
where Δ U - change in internal energy.
In our case, as a result of heat transfer, the internal energy of body B decreases, which means that the temperature of this body decreases. The internal energy of body A increases, since the body received an amount of heat from body B, its temperature will increase. The total internal energy of bodies A and B does not change.
Answer. 23.
Proton p, flying into the gap between the poles of the electromagnet, has a speed perpendicular to the magnetic field induction vector, as shown in the figure. Where the Lorentz force acting on the proton is directed relative to the figure (up, to the observer, from the observer, down, left, right)
Decision. A magnetic field acts on a charged particle with the Lorentz force. In order to determine the direction of this force, it is important to remember the mnemonic rule of the left hand, do not forget to take into account the charge of the particle. We direct four fingers of the left hand along the velocity vector, for a positively charged particle, the vector should perpendicularly enter the palm of the hand, the thumb set aside 90 ° indicates the direction of the Lorentz force acting on the particle. As a result, we have that the Lorentz force vector is directed away from the observer relative to the figure.
Answer. from the observer.
The electric field strength modulus in a 50 μF flat air capacitor is 200 V / m. The distance between the plates of the capacitor is 2 mm. What is the capacitor charge? Write the answer in μC.
Decision. We will transfer all units of measure to the SI system. Capacitance C \u003d 50 μF \u003d 50 · 10 –6 F, the distance between the plates d \u003d 2 · 10 –3 m. The problem speaks of a flat air condenser - a device for the accumulation of electric charge and electric field energy. From the formula of electric capacity
where d - the distance between the plates.
Express stress U \u003d E d(4); We substitute (4) in (2) and calculate the capacitor charge.
q = C · Ed\u003d 50 · 10 –6 · 200 · 0.002 \u003d 20 μC
Please note in which units you want to write the answer. Received in pendants, but represent in μC.
Answer. 20 μC.
The student conducted an experiment on the refraction of light, presented in the photograph. How does the angle of refraction of light propagating in the glass and the refractive index of glass change with increasing angle of incidence?
- Is increasing
- Decreases
- Does not change
- Record the selected numbers for each answer in the table. The numbers in the answer can be repeated.
Decision. In tasks of such a plan, we recall what refraction is. This is a change in the direction of wave propagation as it passes from one medium to another. It is caused by the fact that the wave propagation velocities in these media are different. Having figured out from which medium into which light is distributed, we write the law of refraction in the form
| sinα | = | n 2 | , |
| sinβ | n 1 |
where n 2 - the absolute refractive index of glass, the medium where the light goes; n 1 - the absolute refractive index of the first medium from where the light comes. For air n 1 \u003d 1. α is the angle of incidence of the beam on the surface of the glass half cylinder, β is the angle of refraction of the beam in the glass. Moreover, the angle of refraction will be less than the angle of incidence, since glass is an optically denser medium - a medium with a large refractive index. The speed of light propagation in glass is less. Please note that the angles are measured from the perpendicular restored at the point of incidence of the beam. If you increase the angle of incidence, then the angle of refraction will increase. The refractive index of the glass from this will not change.
Answer.
Copper jumper at time t 0 \u003d 0 begins to move at a speed of 2 m / s along parallel horizontal conductive rails, to the ends of which a resistor with a resistance of 10 Ohms is connected. The entire system is in a vertical uniform magnetic field. The resistance of the jumper and rails is negligible, the jumper is always perpendicular to the rails. The flux Φ of the magnetic induction vector through the circuit formed by the jumper, rails, and resistor changes over time t as shown in the graph.
Using the graph, select two correct statements and indicate their numbers in the answer.
- By time t \u003d 0.1 s, the change in the magnetic flux through the circuit is 1 mVb.
- Induction current in jumper in the range from t \u003d 0.1 s t \u003d 0.3 s maximum.
- The EMF module of the induction arising in the circuit is 10 mV.
- The strength of the induction current flowing in the jumper is 64 mA.
- To maintain the movement of the jumper, a force is applied to it, the projection of which on the direction of the rails is 0.2 N.
Decision. According to the graph of the dependence of the flux of the magnetic induction vector through the circuit on time, we determine the areas where the flux Φ changes and where the flux change is zero. This will allow us to determine the time intervals in which the induction current will occur in the circuit. True statement:
1) By time t \u003d 0.1 s, the change in the magnetic flux through the circuit is 1 mVb ∆Ф \u003d (1 - 0) · 10 –3 Vb; The EMF module of the induction arising in the circuit is determined using the law of EMP
Answer. 13.
According to the graph of the current strength versus time in an electric circuit whose inductance is 1 mH, determine the self-induction EMF module in the time interval from 5 to 10 s. Write the answer in μV.
Decision. We transfer all quantities to the SI system, i.e. the inductance of 1 mH will be converted into HH, we get 10 –3 HH. The current strength shown in the figure in mA will also be converted to A by multiplying by 10 –3.
The self-induction EMF formula has the form
the time interval is given by the condition of the problem
∆t\u003d 10 s - 5 c \u003d 5 s
seconds and according to the schedule we determine the interval of current change during this time:
∆I\u003d 30 · 10 –3 - 20 · 10 –3 \u003d 10 · 10 –3 \u003d 10 –2 A.
Substitute the numerical values \u200b\u200bin the formula (2), we obtain
| Ɛ | \u003d 2 · 10 –6 V, or 2 μV.
Answer. 2.
Two transparent plane-parallel plates are tightly pressed against each other. A ray of light falls from the air onto the surface of the first plate (see figure). It is known that the refractive index of the upper plate is n 2 \u003d 1.77. Set the correspondence between physical quantities and their values. For each position of the first column, select the corresponding position from the second column and write down the selected numbers in the table under the corresponding letters.
Decision. To solve the problems of refraction of light at the interface between two media, in particular the problems of the passage of light through plane-parallel plates, the following solution procedure can be recommended: make a drawing indicating the path of the rays going from one medium to another; at the point of incidence of the beam at the interface between the two media, draw a normal to the surface, note the angles of incidence and refraction. Pay particular attention to the optical density of the media under consideration and remember that when a light beam passes from an optically less dense medium to an optically denser medium, the angle of refraction will be less than the angle of incidence. The figure shows the angle between the incident beam and the surface, and we need the angle of incidence. Remember that the angles are determined from the perpendicular restored at the point of incidence. We determine that the angle of incidence of the beam on the surface 90 ° - 40 ° \u003d 50 °, the refractive index n 2 = 1,77; n 1 \u003d 1 (air).
We write the law of refraction
| sinβ \u003d | sin50 | = 0,4327 ≈ 0,433 |
| 1,77 |
We construct an approximate beam path through the plates. We use formula (1) for the boundaries 2–3 and 3–1. In the answer we get
A) The sine of the angle of incidence of the beam on the boundary 2–3 between the plates is 2) ≈ 0.433;
B) The angle of refraction of the beam when crossing the boundary 3–1 (in radians) is 4) ≈ 0.873.
Answer. 24.
Determine how many α - particles and how many protons result from a fusion reaction
+ → x+ y;
Decision. For all nuclear reactions, the laws of conservation of electric charge and the number of nucleons are observed. Denote by x the number of alpha particles, y the number of protons. Let's make the equations
+ → x + y;
solving the system we have that x = 1; y = 2
Answer. 1 - α-particle; 2 - proton.
The momentum modulus of the first photon is 1.32 · 10 –28 kg · m / s, which is 9.48 · 10 –28 kg · m / s less than the momentum modulus of the second photon. Find the energy ratio E 2 / E 1 of the second and first photons. Round the answer to tenths.
Decision. The momentum of the second photon is greater than the momentum of the first photon under the condition means it can be imagined p 2 = p 1 + Δ p (1). The photon energy can be expressed in terms of the photon momentum using the following equations. it E = mc 2 (1) and p = mc (2) then
E = pc (3),
where E - photon energy, p - photon momentum, m - photon mass, c \u003d 3 · 10 8 m / s is the speed of light. Given formula (3), we have:
| E 2 | = | p 2 | = 8,18; |
| E 1 | p 1 |
We round the answer to tenths and get 8.2.
Answer. 8,2.
The nucleus of an atom has undergone radioactive positron β decay. How did the electrical charge of the nucleus and the number of neutrons in it change as a result of this?
For each value, determine the corresponding nature of the change:
- Increased;
- Decreased;
- Not changed.
Record in the table the selected numbers for each physical quantity. The numbers in the answer can be repeated.
Decision. Positron β - decay in the atomic nucleus occurs during the conversion of a proton to a neutron with the emission of a positron. As a result of this, the number of neutrons in the nucleus increases by one, the electric charge decreases by one, and the mass number of the nucleus remains unchanged. Thus, the reaction of the transformation of the element is as follows:
Answer. 21.
Five experiments were carried out in the laboratory to observe diffraction using various diffraction gratings. Each of the gratings was illuminated by parallel beams of monochromatic light with a specific wavelength. Light in all cases fell perpendicular to the grating. In two of these experiments, the same number of main diffraction maxima was observed. First indicate the number of the experiment in which the diffraction grating with a shorter period was used, and then the number of the experiment in which the diffraction grating with a longer period was used.
Decision. Diffraction of light is the phenomenon of a light beam in the region of a geometric shadow. Diffraction can be observed when there are opaque sections or holes in large and opaque to light obstacles in the path of the light wave, and the sizes of these sections or holes are commensurate with the wavelength. One of the most important diffraction devices is the diffraction grating. The angular directions to the maxima of the diffraction pattern are determined by the equation
dsinφ \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 Is an integer called the order of the diffraction maximum. Express from equation (1)
Selecting pairs according to the conditions of the experiment, we first select 4 where the diffraction grating with a shorter period was used, and then the number of the experiment in which the diffraction grating with a longer period was used - this is 2.
Answer. 42.
Current flows through the wirewound resistor. The resistor was replaced with another one, with a wire of the same metal and the same length, but with half the cross-sectional area, and half the current passed through it. How will the voltage across the resistor and its resistance change?
For each value, determine the corresponding nature of the change:
- Will increase;
- Will decrease;
- Will not change.
Record in the table the selected numbers for each physical quantity. The numbers in the answer can be repeated.
Decision. It is important to remember what values \u200b\u200bthe resistance of the conductor depends on. The formula for calculating the resistance is
ohm's law for a section of the circuit, from formula (2), we express the voltage
U = I R (3).
According to the conditions of the problem, the second resistor is made of a wire of the same material, the same length, but different cross-sectional area. The area is two times smaller. Substituting in (1) we get that the resistance increases by 2 times, and the current decreases by 2 times, therefore, the voltage does not change.
Answer. 13.
The oscillation period of a mathematical pendulum on the surface of the Earth is 1, 2 times the period of its oscillations on a planet. What is the free fall acceleration module on this planet? The influence of the atmosphere in both cases is negligible.
Decision. A mathematical pendulum is a system consisting of a thread whose dimensions are much larger than the sizes of the ball and the ball itself. Difficulty may arise if the Thomson formula for the oscillation period of a mathematical pendulum is forgotten.
T \u003d 2π (1);
l - the length of the mathematical pendulum; g - acceleration of gravity.
By condition
Express from (3) g n \u003d 14.4 m / s 2. It should be noted that the acceleration of gravity depends on the mass of the planet and the radius
Answer. 14.4 m / s 2.
A rectilinear conductor 1 m long, through which a current of 3 A flows, is located in a uniform magnetic field with induction IN \u003d 0.4 T at an angle of 30 ° to the vector. What is the modulus of the force acting on the conductor from the side of the magnetic field?
Decision. If a conductor with current is placed in a magnetic field, then the field on the conductor with current will act with an Ampere force. We write the formula for the Ampere force modulus
F A \u003d I LBsinα;
F A \u003d 0.6 N
Answer. F A \u003d 0.6 N.
The energy of the magnetic field stored in the coil when passing a direct current through it is 120 J. How many times do you need to increase the current flowing through the coil of the coil so that the stored magnetic field energy increases by 5760 J.
Decision. The energy of the magnetic field of the coil is calculated by the formula
| W m \u003d | Li 2 | (1); |
| 2 |
By condition W 1 \u003d 120 J, then W 2 \u003d 120 + 5760 \u003d 5880 J.
| I 1 2 = | 2W 1 | ; I 2 2 = | 2W 2 | ; |
| L | L |
Then the ratio of currents
| I 2 2 | = 49; | I 2 | = 7 |
| I 1 2 | I 1 |
Answer. The current must be increased by 7 times. In the answer form you enter only the number 7.
The electric circuit consists of two light bulbs, two diodes and a coil of wire connected as shown in the figure. (The diode transmits current in only one direction, as shown in the upper part of the figure). Which of the bulbs will light when the north pole of the magnet is brought closer to the loop? Explain the answer by indicating which phenomena and patterns you used in the explanation.
Decision. The lines of magnetic induction exit the north pole of the magnet and diverge. As the magnet approaches, the magnetic flux through the coil of wire increases. In accordance with the Lenz rule, the magnetic field created by the induction current of the coil must be directed to the right. According to the rule of the gimlet, the current should go clockwise (when viewed from the left). In this direction, passes a diode standing in the circuit of the second lamp. So, the second lamp will light up.
Answer. The second lamp will light up.
Long spoke aluminum L \u003d 25 cm and cross-sectional area S \u003d 0.1 cm 2 suspended on the threads of the upper end. The lower end rests on the horizontal bottom of the vessel into which water is poured. The length of the submerged part of the spoke l \u003d 10 cm. Find strength F, with which the spoke presses on the bottom of the vessel, if it is known that the thread is located vertically. The density of aluminum ρ a \u003d 2.7 g / cm 3 the density of water ρ b \u003d 1.0 g / cm 3. Acceleration of gravity g \u003d 10 m / s 2
Decision. Perform an explanatory drawing.
- Strength of thread tension;
- The reaction force of the bottom of the vessel;
a - Archimedean force acting only on the submerged part of the body, and applied to the center of the submerged part of the spoke;
- the force of gravity acting on the spoke from the side of the Earth and applied to the center of the entire spoke.
By definition, the weight of the spoke m and the module of Archimedean force are expressed as follows: m = SLρ a (1);
F a \u003d Slρ in g (2)
Consider the moments of force relative to the point of suspension of the spokes.
M(T) \u003d 0 is the moment of tension force; (3)
M(N) \u003d Nlcosα is the moment of reaction force of the support; (4)
Given the signs of moments, we write the equation
| Nlcosα + Slρ in g (L – | l | ) cosα \u003d SLρ a g | L | cosα (7) |
| 2 | 2 |
given that according to Newton’s third law, the reaction force of the bottom of the vessel is equal to the force F q with which the needle presses on the bottom of the vessel we write N = F q and from equation (7) we express this force:
| F d \u003d [ | 1 | Lρ a– (1 – | l | )lρ in] Sg (8). |
| 2 | 2L |
Substitute the numerical data and get that
F d \u003d 0.025 N.
Answer. Fd \u003d 0.025 N.
Cylinder containing m 1 \u003d 1 kg of nitrogen, when tested for strength exploded at a temperature t 1 \u003d 327 ° C. What mass of hydrogen m 2 could be stored in such a cylinder at a temperature t 2 \u003d 27 ° C, having a five-fold safety margin? Molar mass of nitrogen M 1 \u003d 28 g / mol, hydrogen M 2 \u003d 2 g / mol.
Decision. We write the equation of state of an ideal Mendeleev-Clapeyron gas for nitrogen
where V - cylinder volume T 1 = t 1 + 273 ° C. By condition, hydrogen can be stored at pressure p 2 \u003d p 1/5; (3) Given that
we can express the mass of hydrogen working immediately with equations (2), (3), (4). The final formula is:
| m 2 = | m 1 | M 2 | T 1 | (5). | ||
| 5 | M 1 | T 2 |
After numerical data substitution m 2 \u003d 28 g.
Answer. m 2 \u003d 28 g.
In an ideal oscillatory circuit, the amplitude of the current oscillations in the inductor I m \u003d 5 mA, and the amplitude of the voltage across the capacitor U m \u003d 2.0 V. At time t the voltage across the capacitor is 1.2 V. Find the current strength in the coil at that moment.
Decision. In an ideal oscillatory circuit, the energy of vibrations is conserved. For time t, the energy conservation law has the form
| C | U 2 | + L | I 2 | = L | I m 2 | (1) |
| 2 | 2 | 2 |
For amplitude (maximum) values, we write
and from equation (2) we express
| C | = | I m 2 | (4). |
| L | U m 2 |
We substitute (4) into (3). As a result, we get:
I = I m (5)
Thus, the current strength in the coil at time t is equal to
I \u003d 4.0 mA.
Answer. I \u003d 4.0 mA.
At the bottom of the reservoir 2 m deep lies a mirror. A ray of light, passing through the water, is reflected from the mirror and leaves the water. The refractive index of water is 1.33. Find the distance between the point where the beam enters the water and the point where the beam exits the water if the angle of incidence of the beam is 30 °
Decision. Let's make an explanatory drawing
α is the angle of incidence of the beam;
β is the angle of refraction of the beam in water;
AC - the distance between the point of entry of the beam into the water and the point of exit of the beam from the water.
According to the law of refraction of light
| sinβ \u003d | sinα | (3) |
| n 2 |
Consider a rectangular ΔADB. In it AD \u003d hthen DB \u003d AD
| tgβ \u003d htgβ \u003d h | sinα | = h | sinβ | = h | sinα | (4) |
| cosβ |
We get the following expression:
| AC \u003d 2 DB \u003d 2 h | sinα | (5) |
We substitute the numerical values \u200b\u200bin the resulting formula (5)
Answer. 1.63 m.
In preparation for the exam, we suggest you familiarize yourself with a work program in physics for grades 7–9 to the line of the CMC A. Peryshkina and advanced level work program for grades 10-11 to the teaching materials department Myakisheva G.Ya. Programs are available for viewing and free download to all registered users.
Preparation for the exam and exam
Secondary general education
Line of the teaching materials of A. V. Grachev. Physics (10-11) (base, deep)
Line of the teaching materials of A. V. Grachev. Physics (7-9)
Line of teaching materials of A.V. Peryshkin. Physics (7-9)
Preparation for the exam in physics: examples, solutions, explanations
We analyze the tasks of the exam in physics (Option C) with a teacher.Lebedeva Alevtina Sergeevna, physics teacher, work experience 27 years. Certificate of honor from the Ministry of Education of the Moscow Region (2013), Letter of Appreciation from the Head of the Voskresensky Municipal District (2015), Diploma from the President of the Association of Teachers of Mathematics and Physics of the Moscow Region (2015).
The paper presents tasks of different difficulty levels: basic, advanced and high. Basic level tasks are simple tasks that test the mastery of the most important physical concepts, models, phenomena and laws. Tasks of a higher level are aimed at testing the ability to use the concepts and laws of physics to analyze various processes and phenomena, as well as the ability to solve problems with the application of one or two laws (formulas) for any of the topics of the school physics course. In work 4, the tasks of Part 2 are tasks of a high level of complexity and test the ability to use the laws and theories of physics in an altered or new situation. The fulfillment of such tasks requires the application of knowledge from two three sections of physics at once, i.e. high level of training. This option is fully consistent with the demonstration version of the exam in 2017, the tasks are taken from the open bank of tasks of the exam.
The figure shows a graph of the dependence of the velocity module on time t. Determine on the schedule the path traveled by the car in the time interval from 0 to 30 s.
Decision. The path traveled by the car in the time interval from 0 to 30 s is most easily defined as the area of \u200b\u200bthe trapezoid, the basis of which is the time intervals (30 - 0) \u003d 30 s and (30 - 10) \u003d 20 s, and the height is the speed v \u003d 10 m / s, i.e.
| S = | (30 + 20) from | 10 m / s \u003d 250 m. |
| 2 |
Answer. 250 m.
A weight of 100 kg is lifted vertically with a cable. The figure shows the dependence of the velocity projection V load on the axis directed upwards from time to time t. Determine the tensile modulus of the cable during the lift.
Decision. According to the graph of the dependence of the projection of speed v load on an axis directed vertically upward from time to time t, you can determine the projection of the acceleration of the load
| a = | ∆v | = | (8 - 2) m / s | \u003d 2 m / s 2. |
| ∆t | 3 s |
The load is affected by: gravity directed vertically downward and cable tension force directed along the cable vertically upward see fig. 2. We write the basic equation of dynamics. We use the second law of Newton. The geometric sum of the forces acting on the body is equal to the product of the mass of the body and the acceleration communicated to it.
+ = (1)
We write the equation for the projection of vectors in the frame of reference associated with the earth, the OY axis is directed up. The projection of the tension force is positive, since the direction of the force coincides with the direction of the OY axis, the projection of gravity is negative, since the force vector is opposite to the OY axis, the projection of the acceleration vector is also positive, since the body moves upward with acceleration. We have
T – mg = ma (2);
from the formula (2) the modulus of the tension force
T = m(g + a) \u003d 100 kg (10 + 2) m / s 2 \u003d 1200 N.
Answer. 1200 N.
The body is dragged along a rough horizontal surface with a constant speed, the modulus of which is 1.5 m / s, applying force to it as shown in Figure (1). In this case, the module of the sliding friction force acting on the body is 16 N. What is the power developed by the force? F?
Decision. Imagine the physical process specified in the condition of the problem and make a schematic drawing indicating all the forces acting on the body (Fig. 2). We write the basic equation of dynamics.
Tr + + \u003d (1)
Having chosen a reference system associated with a fixed surface, we write down the equations for the projection of vectors onto the selected coordinate axes. By the condition of the problem, the body moves uniformly, since its speed is constant and equal to 1.5 m / s. This means the acceleration of the body is zero. Two forces act horizontally on the body: sliding friction force tr. and the force with which the body is dragged. The projection of the friction force is negative, since the force vector does not coincide with the direction of the axis X. Power projection F positive. We remind you, to find the projection, lower the perpendicular from the beginning and end of the vector to the selected axis. With this in mind, we have: F cosα - F mp \u003d 0; (1) express the projection of force F, this is Fcosα \u003d F mp \u003d 16 N; (2) then the power developed by the force will be equal to N = Fcosα V (3) We make a replacement, taking into account equation (2), and substitute the corresponding data in equation (3):
N \u003d 16 N · 1.5 m / s \u003d 24 W.
Answer. 24 watts
The load, mounted on a light spring with a stiffness of 200 N / m, performs vertical vibrations. The figure shows a graph of the displacement x cargo from time to time t. Determine what the mass of the cargo is. Round the answer to an integer.
Decision. The load on the spring performs vertical oscillations. According to the graph of the dependence of the load displacement x from time t, we determine the period of oscillation of the cargo. The oscillation period is T \u003d 4 s; from the formula T \u003d 2π we express the mass m cargo.
| = | T | ; | m | = | T 2 | ; m = k | T 2 | ; m \u003d 200 N / m | (4 s) 2 | \u003d 81.14 kg ≈ 81 kg. |
| 2π | k | 4π 2 | 4π 2 | 39,438 |
Answer: 81 kg.
The figure shows a system of two light blocks and a weightless cable, with which you can hold in balance or lift a load of 10 kg. Friction is negligible. Based on the analysis of the figure, select twotrue statements and indicate in the answer their numbers.
- In order to keep the load in balance, you need to act on the end of the rope with a force of 100 N.
- The block system shown in the figure does not give a gain in strength.
- h, you need to stretch a piece of rope with a length of 3 h.
- In order to slowly raise the load to a height hh.
Decision. In this problem, it is necessary to recall simple mechanisms, namely blocks: a movable and fixed block. The movable block gives a gain in force twice, while the section of the rope must be pulled out twice as long, and the fixed block is used to redirect the force. In the work, simple winning mechanisms do not give. After analyzing the task, we immediately select the necessary statements:
- In order to slowly raise the load to a height h, you need to stretch a piece of rope 2 h.
- In order to keep the load in balance, you need to act on the end of the rope with a force of 50 N.
Answer. 45.
An aluminum load is completely immersed in a vessel with water, mounted on a weightless and inextensible thread. The load does not touch the walls and bottom of the vessel. Then, an iron load, the mass of which is equal to the mass of aluminum cargo, is immersed in the same vessel with water. How, as a result of this, the modulus of the tension force of the thread and the modulus of gravity acting on the load change?
- Is increasing;
- Decreases;
- Does not change.
Decision. We analyze the condition of the problem and select those parameters that do not change during the study: these are body mass and liquid into which the body is immersed on threads. After that, it is better to perform a schematic drawing and indicate the forces acting on the load: the thread tension force F upr directed along the thread up; gravity directed vertically downward; Archimedean force a acting from the liquid side to the submerged body and directed upwards. According to the condition of the problem, the mass of cargo is the same, therefore, the modulus of gravity acting on the cargo does not change. Since the density of goods is different, the volume will also be different
| V = | m | . |
| p |
The density of iron is 7800 kg / m 3, and aluminum cargo 2700 kg / m 3. Consequently, V well< V a. The body in equilibrium, the resultant of all the forces acting on the body, is zero. Point the OY coordinate axis up. The basic equation of dynamics taking into account the projection of forces is written as F control + F a – mg \u003d 0; (1) Express the tension force F control \u003d mg – F a (2); Archimedean force depends on the density of the liquid and the volume of the submerged part of the body F a = ρ gVpct (3); The density of the liquid does not change, and the body volume of iron is less V well< V a, therefore, the Archimedean force acting on the iron load will be less. We conclude that the modulus of the tension force of the thread, working with equation (2), it will increase.
Answer. 13.
Bar mass m slides from a fixed rough inclined plane with an angle α at the base. The bar acceleration modulus is a, the bar velocity modulus increases. Air resistance can be neglected.
Set the correspondence between the physical quantities and the formulas by which they can be calculated. For each position of the first column, select the corresponding position from the second column and write down the selected numbers in the table under the corresponding letters.
B) The coefficient of friction of the bar on an inclined plane
3) mg cosα
| 4) sinα - | a |
| gcosα |
Decision. This task requires the application of Newton's laws. We recommend making a schematic drawing; indicate all kinematic characteristics of the movement. If possible, depict the acceleration vector and the vectors of all the forces applied to the moving body; remember that the forces acting on the body are the result of interaction with other bodies. Then write the basic equation of dynamics. Select a reference system and write the resulting equation for the projection of the force and acceleration vectors;
Following the proposed algorithm, we make a schematic drawing (Fig. 1). The figure shows the forces applied to the center of gravity of the bar, and the coordinate axes of the reference system associated with the surface of an inclined plane. Since all forces are constant, the movement of the bar will be equally variable with increasing speed, i.e. acceleration vector is directed towards the movement. Choose the direction of the axes as shown in the figure. We write the projection of the forces on the selected axis.
We write the basic equation of dynamics:
Tr + \u003d (1)
We write this equation (1) for the projection of forces and acceleration.
On the OY axis: the projection of the support reaction force is positive, since the vector coincides with the direction of the OY axis N y = N; the projection of the friction force is zero since the vector is perpendicular to the axis; the projection of gravity will be negative and equal mg y= – mgcosα; acceleration vector projection a y \u003d 0, since the acceleration vector is perpendicular to the axis. We have N – mgcosα \u003d 0 (2) from the equation we express the reaction force acting on the bar, from the side of the inclined plane. N = mgcosα (3). We write the projections onto the OX axis.
OX Axis: Power Projection N equal to zero, since the vector is perpendicular to the axis OX; The projection of the friction force is negative (the vector is directed in the opposite direction relative to the selected axis); the projection of gravity is positive and equal to mg x = mgsinα (4) from a right triangle. Positive acceleration projection a x = a; Then we write equation (1) taking into account the projection mgsinα - F mp \u003d ma (5); F mp \u003d m(gsinα - a) (6); Remember that the force of friction is proportional to the force of normal pressure N.
By definition F mp \u003d μ N (7), we express the coefficient of friction of the bar on an inclined plane.
| μ = | F tr | = | m(gsinα - a) | \u003d tgα - | a | (8). |
| N | mgcosα | gcosα |
Choose the appropriate position for each letter.
Answer. A is 3; B - 2.
Task 8. Gaseous oxygen is in a vessel with a volume of 33.2 liters. The gas pressure is 150 kPa, its temperature is 127 ° C. Determine the mass of gas in this vessel. Express the answer in grams and round to the nearest whole number.
Decision. It is important to pay attention to the conversion of units to the SI system. The temperature is converted to Kelvin T = t° C + 273, volume V \u003d 33.2 l \u003d 33.2 · 10 –3 m 3; We translate pressure P \u003d 150 kPa \u003d 150,000 Pa. Using the ideal gas equation of state
express the mass of gas.
Be sure to pay attention to which unit is asked to record the answer. It is very important.
Answer. 48 g
Task 9. The ideal monatomic gas in the amount of 0.025 mol adiabatically expanded. At the same time, its temperature dropped from + 103 ° С to + 23 ° С. What work did the gas do? Express the answer in Joules and round to the nearest integer.
Decision. First, the gas is a monatomic number of degrees of freedom i \u003d 3, secondly, the gas expands adiabatically - this means without heat transfer Q \u003d 0. Gas does the work by reducing internal energy. With this in mind, we write the first law of thermodynamics in the form 0 \u003d ∆ U + A g; (1) express gas work A r \u003d –∆ U (2); The change in internal energy for a monatomic gas can be written as
Answer. 25 J.
The relative humidity of a portion of air at a certain temperature is 10%. How many times should the pressure of this portion of air be changed so that at a constant temperature its relative humidity increases by 25%?
Decision. Questions related to saturated steam and air humidity often cause difficulties for students. We use the formula for calculating the relative humidity
According to the condition of the problem, the temperature does not change, which means that the saturated vapor pressure remains the same. We write formula (1) for two states of air.
φ 1 \u003d 10%; φ 2 \u003d 35%
We express the air pressure from formulas (2), (3) and find the pressure ratio.
| P 2 | = | φ 2 | = | 35 | = 3,5 |
| P 1 | φ 1 | 10 |
Answer. Pressure should be increased by 3.5 times.
A hot substance in a liquid state was slowly cooled in a melting furnace with a constant power. The table shows the results of measurements of the temperature of the substance over time.
Choose from the proposed list two statements that correspond to the results of measurements and indicate their numbers.
- The melting point of the substance under these conditions is 232 ° C.
- In 20 minutes. after the start of measurements, the substance was only in a solid state.
- The heat capacity of the substance in the liquid and solid state is the same.
- After 30 minutes after the start of measurements, the substance was only in a solid state.
- The process of crystallization of the substance took more than 25 minutes.
Decision. Since the substance was cooled, its internal energy decreased. The temperature measurement results allow you to determine the temperature at which the substance begins to crystallize. While the substance passes from a liquid to a solid state, the temperature does not change. Knowing that the melting point and crystallization temperature are the same, we choose the statement:
1. The melting point of the substance under these conditions is 232 ° C.
The second true statement is:
4. After 30 minutes after the start of measurements, the substance was only in a solid state. Since the temperature at this point in time, already below the crystallization temperature.
Answer.14.
In an isolated system, body A has a temperature of + 40 ° C, and body B has a temperature of + 65 ° C. These bodies brought into thermal contact with each other. After some time, thermal equilibrium came. How did the temperature of body B and the total internal energy of body A and B change as a result?
For each value, determine the corresponding nature of the change:
- Increased;
- Decreased;
- Not changed.
Record in the table the selected numbers for each physical quantity. The numbers in the answer can be repeated.
Decision. If in an isolated system of bodies no energy transformations take place except heat exchange, then the amount of heat given up by bodies whose internal energy decreases is equal to the amount of heat received by bodies whose internal energy increases. (According to the law of conservation of energy.) In this case, the total internal energy of the system does not change. Tasks of this type are solved on the basis of the heat balance equation.
| ∆U \u003d ∑ | n | ∆U i \u003d0 (1); |
| i = 1 |
where Δ U - change in internal energy.
In our case, as a result of heat transfer, the internal energy of body B decreases, which means that the temperature of this body decreases. The internal energy of body A increases, since the body received an amount of heat from body B, its temperature will increase. The total internal energy of bodies A and B does not change.
Answer. 23.
Proton p, flying into the gap between the poles of the electromagnet, has a speed perpendicular to the magnetic field induction vector, as shown in the figure. Where the Lorentz force acting on the proton is directed relative to the figure (up, to the observer, from the observer, down, left, right)
Decision. A magnetic field acts on a charged particle with the Lorentz force. In order to determine the direction of this force, it is important to remember the mnemonic rule of the left hand, do not forget to take into account the charge of the particle. We direct four fingers of the left hand along the velocity vector, for a positively charged particle, the vector should perpendicularly enter the palm of the hand, the thumb set aside 90 ° indicates the direction of the Lorentz force acting on the particle. As a result, we have that the Lorentz force vector is directed away from the observer relative to the figure.
Answer. from the observer.
The electric field strength modulus in a 50 μF flat air capacitor is 200 V / m. The distance between the plates of the capacitor is 2 mm. What is the capacitor charge? Write the answer in μC.
Decision. We will transfer all units of measure to the SI system. Capacitance C \u003d 50 μF \u003d 50 · 10 –6 F, the distance between the plates d \u003d 2 · 10 –3 m. The problem speaks of a flat air condenser - a device for the accumulation of electric charge and electric field energy. From the formula of electric capacity
where d - the distance between the plates.
Express stress U \u003d E d(4); We substitute (4) in (2) and calculate the capacitor charge.
q = C · Ed\u003d 50 · 10 –6 · 200 · 0.002 \u003d 20 μC
Please note in which units you want to write the answer. Received in pendants, but represent in μC.
Answer. 20 μC.
The student conducted an experiment on the refraction of light, presented in the photograph. How does the angle of refraction of light propagating in the glass and the refractive index of glass change with increasing angle of incidence?
- Is increasing
- Decreases
- Does not change
- Record the selected numbers for each answer in the table. The numbers in the answer can be repeated.
Decision. In tasks of such a plan, we recall what refraction is. This is a change in the direction of wave propagation as it passes from one medium to another. It is caused by the fact that the wave propagation velocities in these media are different. Having figured out from which medium into which light is distributed, we write the law of refraction in the form
| sinα | = | n 2 | , |
| sinβ | n 1 |
where n 2 - the absolute refractive index of glass, the medium where the light goes; n 1 - the absolute refractive index of the first medium from where the light comes. For air n 1 \u003d 1. α is the angle of incidence of the beam on the surface of the glass half cylinder, β is the angle of refraction of the beam in the glass. Moreover, the angle of refraction will be less than the angle of incidence, since glass is an optically denser medium - a medium with a large refractive index. The speed of light propagation in glass is less. Please note that the angles are measured from the perpendicular restored at the point of incidence of the beam. If you increase the angle of incidence, then the angle of refraction will increase. The refractive index of the glass from this will not change.
Answer.
Copper jumper at time t 0 \u003d 0 begins to move at a speed of 2 m / s along parallel horizontal conductive rails, to the ends of which a resistor with a resistance of 10 Ohms is connected. The entire system is in a vertical uniform magnetic field. The resistance of the jumper and rails is negligible, the jumper is always perpendicular to the rails. The flux Φ of the magnetic induction vector through the circuit formed by the jumper, rails, and resistor changes over time t as shown in the graph.
Using the graph, select two correct statements and indicate their numbers in the answer.
- By time t \u003d 0.1 s, the change in the magnetic flux through the circuit is 1 mVb.
- Induction current in jumper in the range from t \u003d 0.1 s t \u003d 0.3 s maximum.
- The EMF module of the induction arising in the circuit is 10 mV.
- The strength of the induction current flowing in the jumper is 64 mA.
- To maintain the movement of the jumper, a force is applied to it, the projection of which on the direction of the rails is 0.2 N.
Decision. According to the graph of the dependence of the flux of the magnetic induction vector through the circuit on time, we determine the areas where the flux Φ changes and where the flux change is zero. This will allow us to determine the time intervals in which the induction current will occur in the circuit. True statement:
1) By time t \u003d 0.1 s, the change in the magnetic flux through the circuit is 1 mVb ∆Ф \u003d (1 - 0) · 10 –3 Vb; The EMF module of the induction arising in the circuit is determined using the law of EMP
Answer. 13.
According to the graph of the current strength versus time in an electric circuit whose inductance is 1 mH, determine the self-induction EMF module in the time interval from 5 to 10 s. Write the answer in μV.
Decision. We transfer all quantities to the SI system, i.e. the inductance of 1 mH will be converted into HH, we get 10 –3 HH. The current strength shown in the figure in mA will also be converted to A by multiplying by 10 –3.
The self-induction EMF formula has the form
the time interval is given by the condition of the problem
∆t\u003d 10 s - 5 c \u003d 5 s
seconds and according to the schedule we determine the interval of current change during this time:
∆I\u003d 30 · 10 –3 - 20 · 10 –3 \u003d 10 · 10 –3 \u003d 10 –2 A.
Substitute the numerical values \u200b\u200bin the formula (2), we obtain
| Ɛ | \u003d 2 · 10 –6 V, or 2 μV.
Answer. 2.
Two transparent plane-parallel plates are tightly pressed against each other. A ray of light falls from the air onto the surface of the first plate (see figure). It is known that the refractive index of the upper plate is n 2 \u003d 1.77. Set the correspondence between physical quantities and their values. For each position of the first column, select the corresponding position from the second column and write down the selected numbers in the table under the corresponding letters.
Decision. To solve the problems of refraction of light at the interface between two media, in particular the problems of the passage of light through plane-parallel plates, the following solution procedure can be recommended: make a drawing indicating the path of the rays going from one medium to another; at the point of incidence of the beam at the interface between the two media, draw a normal to the surface, note the angles of incidence and refraction. Pay particular attention to the optical density of the media under consideration and remember that when a light beam passes from an optically less dense medium to an optically denser medium, the angle of refraction will be less than the angle of incidence. The figure shows the angle between the incident beam and the surface, and we need the angle of incidence. Remember that the angles are determined from the perpendicular restored at the point of incidence. We determine that the angle of incidence of the beam on the surface 90 ° - 40 ° \u003d 50 °, the refractive index n 2 = 1,77; n 1 \u003d 1 (air).
We write the law of refraction
| sinβ \u003d | sin50 | = 0,4327 ≈ 0,433 |
| 1,77 |
We construct an approximate beam path through the plates. We use formula (1) for the boundaries 2–3 and 3–1. In the answer we get
A) The sine of the angle of incidence of the beam on the boundary 2–3 between the plates is 2) ≈ 0.433;
B) The angle of refraction of the beam when crossing the boundary 3–1 (in radians) is 4) ≈ 0.873.
Answer. 24.
Determine how many α - particles and how many protons result from a fusion reaction
+ → x+ y;
Decision. For all nuclear reactions, the laws of conservation of electric charge and the number of nucleons are observed. Denote by x the number of alpha particles, y the number of protons. Let's make the equations
+ → x + y;
solving the system we have that x = 1; y = 2
Answer. 1 - α-particle; 2 - proton.
The momentum modulus of the first photon is 1.32 · 10 –28 kg · m / s, which is 9.48 · 10 –28 kg · m / s less than the momentum modulus of the second photon. Find the energy ratio E 2 / E 1 of the second and first photons. Round the answer to tenths.
Decision. The momentum of the second photon is greater than the momentum of the first photon under the condition means it can be imagined p 2 = p 1 + Δ p (1). The photon energy can be expressed in terms of the photon momentum using the following equations. it E = mc 2 (1) and p = mc (2) then
E = pc (3),
where E - photon energy, p - photon momentum, m - photon mass, c \u003d 3 · 10 8 m / s is the speed of light. Given formula (3), we have:
| E 2 | = | p 2 | = 8,18; |
| E 1 | p 1 |
We round the answer to tenths and get 8.2.
Answer. 8,2.
The nucleus of an atom has undergone radioactive positron β decay. How did the electrical charge of the nucleus and the number of neutrons in it change as a result of this?
For each value, determine the corresponding nature of the change:
- Increased;
- Decreased;
- Not changed.
Record in the table the selected numbers for each physical quantity. The numbers in the answer can be repeated.
Decision. Positron β - decay in the atomic nucleus occurs during the conversion of a proton to a neutron with the emission of a positron. As a result of this, the number of neutrons in the nucleus increases by one, the electric charge decreases by one, and the mass number of the nucleus remains unchanged. Thus, the reaction of the transformation of the element is as follows:
Answer. 21.
Five experiments were carried out in the laboratory to observe diffraction using various diffraction gratings. Each of the gratings was illuminated by parallel beams of monochromatic light with a specific wavelength. Light in all cases fell perpendicular to the grating. In two of these experiments, the same number of main diffraction maxima was observed. First indicate the number of the experiment in which the diffraction grating with a shorter period was used, and then the number of the experiment in which the diffraction grating with a longer period was used.
Decision. Diffraction of light is the phenomenon of a light beam in the region of a geometric shadow. Diffraction can be observed when there are opaque sections or holes in large and opaque to light obstacles in the path of the light wave, and the sizes of these sections or holes are commensurate with the wavelength. One of the most important diffraction devices is the diffraction grating. The angular directions to the maxima of the diffraction pattern are determined by the equation
dsinφ \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 Is an integer called the order of the diffraction maximum. Express from equation (1)
Selecting pairs according to the conditions of the experiment, we first select 4 where the diffraction grating with a shorter period was used, and then the number of the experiment in which the diffraction grating with a longer period was used - this is 2.
Answer. 42.
Current flows through the wirewound resistor. The resistor was replaced with another one, with a wire of the same metal and the same length, but with half the cross-sectional area, and half the current passed through it. How will the voltage across the resistor and its resistance change?
For each value, determine the corresponding nature of the change:
- Will increase;
- Will decrease;
- Will not change.
Record in the table the selected numbers for each physical quantity. The numbers in the answer can be repeated.
Decision. It is important to remember what values \u200b\u200bthe resistance of the conductor depends on. The formula for calculating the resistance is
ohm's law for a section of the circuit, from formula (2), we express the voltage
U = I R (3).
According to the conditions of the problem, the second resistor is made of a wire of the same material, the same length, but different cross-sectional area. The area is two times smaller. Substituting in (1) we get that the resistance increases by 2 times, and the current decreases by 2 times, therefore, the voltage does not change.
Answer. 13.
The oscillation period of a mathematical pendulum on the surface of the Earth is 1, 2 times the period of its oscillations on a planet. What is the free fall acceleration module on this planet? The influence of the atmosphere in both cases is negligible.
Decision. A mathematical pendulum is a system consisting of a thread whose dimensions are much larger than the sizes of the ball and the ball itself. Difficulty may arise if the Thomson formula for the oscillation period of a mathematical pendulum is forgotten.
T \u003d 2π (1);
l - the length of the mathematical pendulum; g - acceleration of gravity.
By condition
Express from (3) g n \u003d 14.4 m / s 2. It should be noted that the acceleration of gravity depends on the mass of the planet and the radius
Answer. 14.4 m / s 2.
A rectilinear conductor 1 m long, through which a current of 3 A flows, is located in a uniform magnetic field with induction IN \u003d 0.4 T at an angle of 30 ° to the vector. What is the modulus of the force acting on the conductor from the side of the magnetic field?
Decision. If a conductor with current is placed in a magnetic field, then the field on the conductor with current will act with an Ampere force. We write the formula for the Ampere force modulus
F A \u003d I LBsinα;
F A \u003d 0.6 N
Answer. F A \u003d 0.6 N.
The energy of the magnetic field stored in the coil when passing a direct current through it is 120 J. How many times do you need to increase the current flowing through the coil of the coil so that the stored magnetic field energy increases by 5760 J.
Decision. The energy of the magnetic field of the coil is calculated by the formula
| W m \u003d | Li 2 | (1); |
| 2 |
By condition W 1 \u003d 120 J, then W 2 \u003d 120 + 5760 \u003d 5880 J.
| I 1 2 = | 2W 1 | ; I 2 2 = | 2W 2 | ; |
| L | L |
Then the ratio of currents
| I 2 2 | = 49; | I 2 | = 7 |
| I 1 2 | I 1 |
Answer. The current must be increased by 7 times. In the answer form you enter only the number 7.
The electric circuit consists of two light bulbs, two diodes and a coil of wire connected as shown in the figure. (The diode transmits current in only one direction, as shown in the upper part of the figure). Which of the bulbs will light when the north pole of the magnet is brought closer to the loop? Explain the answer by indicating which phenomena and patterns you used in the explanation.
Decision. The lines of magnetic induction exit the north pole of the magnet and diverge. As the magnet approaches, the magnetic flux through the coil of wire increases. In accordance with the Lenz rule, the magnetic field created by the induction current of the coil must be directed to the right. According to the rule of the gimlet, the current should go clockwise (when viewed from the left). In this direction, passes a diode standing in the circuit of the second lamp. So, the second lamp will light up.
Answer. The second lamp will light up.
Long spoke aluminum L \u003d 25 cm and cross-sectional area S \u003d 0.1 cm 2 suspended on the threads of the upper end. The lower end rests on the horizontal bottom of the vessel into which water is poured. The length of the submerged part of the spoke l \u003d 10 cm. Find strength F, with which the spoke presses on the bottom of the vessel, if it is known that the thread is located vertically. The density of aluminum ρ a \u003d 2.7 g / cm 3 the density of water ρ b \u003d 1.0 g / cm 3. Acceleration of gravity g \u003d 10 m / s 2
Decision. Perform an explanatory drawing.
- Strength of thread tension;
- The reaction force of the bottom of the vessel;
a - Archimedean force acting only on the submerged part of the body, and applied to the center of the submerged part of the spoke;
- the force of gravity acting on the spoke from the side of the Earth and applied to the center of the entire spoke.
By definition, the weight of the spoke m and the module of Archimedean force are expressed as follows: m = SLρ a (1);
F a \u003d Slρ in g (2)
Consider the moments of force relative to the point of suspension of the spokes.
M(T) \u003d 0 is the moment of tension force; (3)
M(N) \u003d Nlcosα is the moment of reaction force of the support; (4)
Given the signs of moments, we write the equation
| Nlcosα + Slρ in g (L – | l | ) cosα \u003d SLρ a g | L | cosα (7) |
| 2 | 2 |
given that according to Newton’s third law, the reaction force of the bottom of the vessel is equal to the force F q with which the needle presses on the bottom of the vessel we write N = F q and from equation (7) we express this force:
| F d \u003d [ | 1 | Lρ a– (1 – | l | )lρ in] Sg (8). |
| 2 | 2L |
Substitute the numerical data and get that
F d \u003d 0.025 N.
Answer. Fd \u003d 0.025 N.
Cylinder containing m 1 \u003d 1 kg of nitrogen, when tested for strength exploded at a temperature t 1 \u003d 327 ° C. What mass of hydrogen m 2 could be stored in such a cylinder at a temperature t 2 \u003d 27 ° C, having a five-fold safety margin? Molar mass of nitrogen M 1 \u003d 28 g / mol, hydrogen M 2 \u003d 2 g / mol.
Decision. We write the equation of state of an ideal Mendeleev-Clapeyron gas for nitrogen
where V - cylinder volume T 1 = t 1 + 273 ° C. By condition, hydrogen can be stored at pressure p 2 \u003d p 1/5; (3) Given that
we can express the mass of hydrogen working immediately with equations (2), (3), (4). The final formula is:
| m 2 = | m 1 | M 2 | T 1 | (5). | ||
| 5 | M 1 | T 2 |
After numerical data substitution m 2 \u003d 28 g.
Answer. m 2 \u003d 28 g.
In an ideal oscillatory circuit, the amplitude of the current oscillations in the inductor I m \u003d 5 mA, and the amplitude of the voltage across the capacitor U m \u003d 2.0 V. At time t the voltage across the capacitor is 1.2 V. Find the current strength in the coil at that moment.
Decision. In an ideal oscillatory circuit, the energy of vibrations is conserved. For time t, the energy conservation law has the form
| C | U 2 | + L | I 2 | = L | I m 2 | (1) |
| 2 | 2 | 2 |
For amplitude (maximum) values, we write
and from equation (2) we express
| C | = | I m 2 | (4). |
| L | U m 2 |
We substitute (4) into (3). As a result, we get:
I = I m (5)
Thus, the current strength in the coil at time t is equal to
I \u003d 4.0 mA.
Answer. I \u003d 4.0 mA.
At the bottom of the reservoir 2 m deep lies a mirror. A ray of light, passing through the water, is reflected from the mirror and leaves the water. The refractive index of water is 1.33. Find the distance between the point where the beam enters the water and the point where the beam exits the water if the angle of incidence of the beam is 30 °
Decision. Let's make an explanatory drawing
α is the angle of incidence of the beam;
β is the angle of refraction of the beam in water;
AC - the distance between the point of entry of the beam into the water and the point of exit of the beam from the water.
According to the law of refraction of light
| sinβ \u003d | sinα | (3) |
| n 2 |
Consider a rectangular ΔADB. In it AD \u003d hthen DB \u003d AD
| tgβ \u003d htgβ \u003d h | sinα | = h | sinβ | = h | sinα | (4) |
| cosβ |
We get the following expression:
| AC \u003d 2 DB \u003d 2 h | sinα | (5) |
We substitute the numerical values \u200b\u200bin the resulting formula (5)
Answer. 1.63 m.
In preparation for the exam, we suggest you familiarize yourself with a work program in physics for grades 7–9 to the line of the CMC A. Peryshkina and advanced level work program for grades 10-11 to the teaching materials department Myakisheva G.Ya. Programs are available for viewing and free download to all registered users.