Unified State Exam in Physics– an exam that is not included in the list of tests required for all graduates. Physics is chosen by potential engineering students. Moreover, each university sets its own bar – in prestigious educational institutions it can be very high. The graduate must understand this when starting preparation for the exam.Purpose of the exam– checking the level of knowledge and skills acquired during schooling, for compliance with the norms and standards specified in the program.

• The exam takes almost 4 hours - 235 minutes; this time must be correctly distributed between tasks in order to successfully complete all of them without wasting a single minute.
• You are allowed to bring a calculator with you as many complex calculations are required to complete the assignments. You can also take a ruler.
• The work consists of three parts, each has its own characteristics and consists of tasks of different levels of complexity.

Physics is a difficult subject, approximately every 15-1 take this exam every year to enter technical university. It is assumed that a graduate with such goals will not learn the subject “from scratch” in order to prepare for the Unified State Exam.
To successfully pass the test, you must:

• Start repeating the material in advance, approach the issue comprehensively;
• Actively apply theory into practice - solve many tasks of varying levels of complexity;
• Educate yourself;
• Pass online testing on questions from previous years.

C= 0.1 nF. By conductors with constant speed V= 2 m/s the conductive rod moves, which is in contact with the conductors. The entire system is in a uniform vertical magnetic field with induction IN D And E L= 40 cm, the total resistance of the circuit at this moment is R= 2 ohms, and current flows in the circuit with force I= 0.05 A. What is the charge of the capacitor at this moment? The inductance of the circuit is negligible. Express your answer in picoculombs.

Solution.

The movement of the rod will be uniform if an external force is applied to it where is the length of the jumper. Job external force will go to heat the circuit resistance and increase the energy of the capacitor.

where is the displacement of the rod. Let's find the power as the time derivative of the work done in each section. It is worth noting that the current strength shows how the charge changes per unit time, i.e.

Thus:

Answer: 2.

Answer: 2

Source: Physics training work 04/28/2017, version PH10503

Two straight conductors P 1 and P 2 are located in the same horizontal plane. Between their left ends there is a capacitor with a capacitance C= 0.2 nF. Along conductors at constant speed V= 4 m/s the conductive rod moves, which is in contact with the conductors. The entire system is in a uniform vertical magnetic field with induction IN= 0.15 T. At some point in time, the distance between points D And E, in which the rod touches the conductors, is equal to L= 20 cm, the total resistance of the circuit at this moment is R= 5 ohms, and current flows in the circuit with force I= 0.02 A. What is the charge of the capacitor at this moment? The inductance of the circuit is negligible. Express your answer in picoculombs.

Solution.

In order for the movement of the rod to be uniform, at the considered moment in time the external force must be equal work external force is used to release Joule heat and increase the energy of the capacitor:

where is the movement of the rod in a short time. Let's find the instantaneous power as the time derivative of the work. It is worth noting that the current strength shows how the charge changes per unit time, i.e.

Thus:

Answer: 4.

Answer: 4

Source: Physics training work 04/28/2017, version PH10504

The student assembled the electrical circuit shown in the figure. What energy will be released in the external part of the circuit when current flows for 10 minutes? (Express your answer in kJ. The necessary data is indicated in the diagram. Consider the ammeter ideal.)

Solution.

According to the Joule-Lenz law, the energy released over time when current flows through a resistance is equal to In the student’s diagram, resistors 2 Ohms and 4 Ohms are connected in series, which means their total resistance is equal to The current strength is 1 A. Thus, in the external circuit for 10 minutes will be allocated

Answer: 3.6.

A load resistor is connected to a current source with an emf of 4 V and internal resistance. What should it be equal for the source efficiency to be 50%? (Give your answer in ohms.)

Solution.

The source efficiency is defined as the ratio useful work(heat generated by the load) to the work done by the EMF: Both of these works are proportional to the time during which the current is passed, so we replace the ratio of work with the ratio of the corresponding powers: According to Ohm’s law for a complete circuit, the current strength in the circuit will be equal to The power of the source is equal to then the voltage per load, according to Ohm’s law for a section of the circuit, is equal to Therefore, the useful power is: Thus, in order for the efficiency to be equal to 50%, it is necessary that the equality be satisfied

Answer: 5.

Answer: 5

Alexey (St. Petersburg)

Good afternoon

Well, what other benefit is there from a resistance included in the circuit, only that it heats up?

The meaning of the task here is as follows. There is internal resistance, and there is load. Heat is released both here and there, but only the heat from the load is beneficial (this resistor, conditionally, can be moved somewhere and used to heat something), but a heating source is of no benefit :)

A resistor with resistance is connected to a current source with EMF and internal resistance. If you connect this resistor to a current source with EMF and internal resistance, how many times will the power released in this resistor increase?

Solution.

According to Ohm's law for a complete circuit, the current flowing through the resistor is equal to Therefore, the power released in the resistor is: Thus, since a we conclude that the power released in the resistor will increase by 4 times. Please note that the dependence of this quantity on the internal resistance of the source is less trivial.

Answer: 4.

W I

Solution.

W expressed in terms of current I and coil inductance L:

By choosing any convenient point on the graph, for example (3; 90), we find L:

The correct answer is listed at number 2.

The figure shows the energy dependence W magnetic field of the coil from the force I current flowing through it. The inductance of this coil is

Solution.

Coil magnetic field energy W expressed in terms of current I and coil inductance L:

By choosing any convenient point on the graph, for example (2; 20), we find L:

The correct answer is listed under number 1.

Flows through a conductor of constant cross-section D.C. with a force of 1 nA. On average, how many electrons pass through the cross section of this conductor in 0.72 µs?

Solution.

q t:

Answer: 4500.

Answer: 4500

Source: Physics training work 02/16/2017, version PH10303

A constant current of 1 nA flows through a conductor of constant cross-section. On average, how many electrons pass through the cross section of this conductor in 0.24 µs?

Solution.

Current strength shows how much charge q passes through the cross section of the conductor in time t:

where is the charge of the electron, is the number of electrons.

Let's find the number of electrons from here:

Answer: 1500.

Answer: 1500

Source: Physics training work 02/16/2017, version PH10304

Closed loop area S made of thin wire placed in a magnetic field. The plane of the contour is perpendicular to the vector of the magnetic induction of the field. Current oscillations with amplitude occur in the circuit if the magnetic induction of the field changes over time in accordance with the formula where Electrical resistance of the circuit What is the area of ​​the circuit?

Solution.

Therefore, the amplitude of the EMF is equal to The amplitude of the EMF is related to the amplitude of the current and resistance by Ohm's law: From here we obtain for the area of ​​the circuit

Source: Unified State Exam in Physics 06/06/2013. Main wave. Far East. Option 1.

A closed loop of thin wire is placed in a magnetic field. The plane of the contour is perpendicular to the vector of the magnetic induction of the field. Its contour area electrical resistance Ohm. The magnetic induction of the field changes over time in accordance with the formula where T, What is the amplitude of current oscillations in the circuit?

Solution.

According to Faraday's law, when the magnetic field flux through a circuit changes, a induced emf, equal Flow through the circuit is equal to Thus, the induced emf changes with time according to the law

Therefore, the amplitude of the EMF is equal to The amplitude of the EMF is related to the amplitude of the current and resistance by Ohm’s law: From here we get for the circuit resistance

Source: Unified State Exam in Physics 06/06/2013. Main wave. Far East. Option 3.

A closed loop of thin wire is placed in a magnetic field. The plane of the contour is perpendicular to the vector of the magnetic induction of the field. Loop area Current oscillations with amplitude occur in the loop if the magnetic induction of the field changes over time in accordance with the formula where What is the electrical resistance of the loop? R?

Solution.

According to Faraday's law, when the magnetic field flux through a circuit changes, an induced emf appears in the circuit equal to The flux through the circuit is equal to Thus, the induced emf changes with time according to the law

Preparation for the OGE and the Unified State Exam

Average general education

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

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

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

Preparing for the Unified State Exam in Physics: examples, solutions, explanations

Lebedeva Alevtina Sergeevna, physics teacher, 27 years of work experience. Certificate of Honor Ministry of Education of the Moscow Region (2013), Gratitude from the Head of Voskresensky municipal district(2015), Certificate of the President of the Association of Teachers of Mathematics and Physics of the Moscow Region (2015).

The work presents tasks different levels Difficulty: basic, advanced and high. Quests basic level, This simple tasks, checking the assimilation of the most important physical concepts, models, phenomena and laws. Quests higher level aimed at testing the ability to use concepts and laws of physics for analysis various processes and phenomena, as well as the ability to solve problems using one or two laws (formulas) on any of the topics school course physics. In work 4 tasks of part 2 are tasks high level complexity and test the ability to use the laws and theories of physics in modified or new situation. Completing such tasks requires the application of knowledge from two or three sections of physics at once, i.e. high level of training. This option is fully consistent demo version Unified State Exam 2017, tasks taken from open bank Unified State Exam assignments.

The figure shows a graph of the speed modulus versus time t. Determine from the graph the distance traveled by the car in the time interval from 0 to 30 s.

Solution. The path traveled by a car in the time interval from 0 to 30 s can most easily be defined as the area of ​​a trapezoid, the bases of which are the time intervals (30 – 0) = 30 s and (30 – 10) = 20 s, and the height is the speed v= 10 m/s, i.e.

Answer. 250 m.

A load weighing 100 kg is lifted vertically upward using a cable. The figure shows the dependence of the velocity projection V load on the axis directed upwards, as a function of time t. Determine the modulus of the cable tension force during the lift.

Solution. According to the velocity projection dependence graph v load on an axis directed vertically upward, as a function of time t, we can determine the projection of the acceleration of the load

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

+ = (1)

Let's write the equation for the projection of vectors in the reference system associated with the earth, directing the OY axis upward. 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 the gravity force is negative, since the force vector is opposite to the OY axis, the projection of the acceleration vector is also positive, so the body moves with upward acceleration. We have

T – mg = ma (2);

from formula (2) tensile force modulus

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

Answer. 1200 N.

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

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

Tr + + = (1)

Having chosen a reference system associated with a fixed surface, we write the equations for the projection of vectors onto the selected coordinate axes. According to the conditions 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: the sliding friction force tr. and the force with which the body is dragged. The projection of the friction force is negative, since the force vector does not coincide with the direction of the axis X. Projection of force F positive. We remind you that to find the projection, we lower the perpendicular from the beginning and end of the vector to the selected axis. Taking this into account we have: F cosα – F tr = 0; (1) let us express the projection of force F, This F cosα = F tr = 16 N; (2) then the power developed by the force will be equal to N = F cosα V(3) Let’s make a replacement, taking into account equation (2), and substitute the corresponding data into equation (3):

N= 16 N · 1.5 m/s = 24 W.

Answer. 24 W.

A load attached to a light spring with a stiffness of 200 N/m makes vertical oscillation. The figure shows a graph of the displacement dependence x load from time to time t. Determine what the mass of the load is. Round your answer to a whole number.

Solution. A mass on a spring undergoes vertical oscillations. According to the load displacement graph X from time to time t, we determine the period of oscillation of the load. The period of oscillation is equal to T= 4 s; from the formula T= 2π let's express the mass m cargo

Answer: 81 kg.

The figure shows a system of two light blocks and a weightless cable, with which you can keep in balance or lift a load weighing 10 kg. Friction is negligible. Based on the analysis of the above figure, select twotrue statements and indicate their numbers in your answer.

1. In order to keep the load in balance, you need to act on the end of the rope with a force of 100 N.
2. The block system shown in the figure does not give any gain in strength.
3. h, you need to pull out a section of rope length 3 h.
4. To slowly lift a load to a height hh.

Solution. In this problem, it is necessary to remember simple mechanisms, namely blocks: a movable and a fixed block. The movable block gives a double gain in strength, while the section of the rope needs to be pulled twice as long, and the fixed block is used to redirect the force. In work, simple mechanisms of winning do not give. After analyzing the problem, we immediately select the necessary statements:

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

Answer. 45.

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

1. Increases;
2. Decreases;
3. Doesn't change.

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

The density of iron is 7800 kg/m3, and the density of aluminum cargo is 2700 kg/m3. Hence, V and< V a. The body is in equilibrium, the resultant of all forces acting on the body is zero. Let's direct the OY coordinate axis upward. We write the basic equation of dynamics, taking into account the projection of forces, in the form F control + F a – mg= 0; (1) Let us express the tension force F control = mg – F a(2); Archimedean force depends on the density of the liquid and the volume of the immersed part of the body F a = ρ gV p.h.t. (3); The density of the liquid does not change, and the volume of the iron body is smaller V and< V a, therefore the Archimedean force acting on the iron load will be less. We conclude about the modulus of the tension force of the thread, working with equation (2), it will increase.

Answer. 13.

A block of mass m slides off a fixed rough inclined plane with an angle α at the base. The acceleration modulus of the block is equal to a, the modulus of the block's velocity increases. Air resistance can be neglected.

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

B) Coefficient of friction between a block and an inclined plane

3) mg cosα

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

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

Let's write down the basic equation of dynamics:

Tr + = (1)

Let's write it down given equation(1) for the projection of forces and acceleration.

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

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

By definition F tr = μ N(7), we express the coefficient of friction of the block on the inclined plane.

We select the appropriate positions for each letter.

Answer. A – 3; B – 2.

Task 8. Oxygen gas is in a vessel with a volume of 33.2 liters. The gas pressure is 150 kPa, its temperature is 127° C. Determine the mass of the gas in this vessel. Express your answer in grams and round to the nearest whole number.

Solution. It is important to pay attention to the conversion of units to the SI system. Convert temperature to Kelvin T = t°C + 273, volume V= 33.2 l = 33.2 · 10 –3 m 3 ; We convert the pressure P= 150 kPa = 150,000 Pa. Using the equation of state ideal gas

Let's express the mass of gas.

Be sure to pay attention to which units are asked to write down the answer. This is very important.

Answer.'48

Task 9. An ideal monatomic gas in an amount of 0.025 mol expanded adiabatically. At the same time, its temperature dropped from +103°C to +23°C. How much work has been done by the gas? Express your answer in Joules and round to the nearest whole number.

Solution. Firstly, the gas is monatomic number of degrees of freedom i= 3, secondly, the gas expands adiabatically - this means without heat exchange Q= 0. The gas does work by decreasing internal energy. Taking this into account, we write the first law of thermodynamics in the form 0 = ∆ U + A G; (1) let us express the gas work A g = –∆ U(2); We write the change in internal energy for a monatomic gas 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%?

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

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

φ 1 = 10%; φ 2 = 35%

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

Answer. The pressure should be increased by 3.5 times.

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

Select from the list provided two statements that correspond to the results of the measurements taken and indicate their numbers.

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

Solution. Since the substance cooled, it internal energy decreased. The results of temperature measurements allow us to determine the temperature at which a substance begins to crystallize. While the substance passes from liquid state into a solid, the temperature does not change. Knowing that the melting temperature 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 correct statement is:

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

Answer. 14.

IN isolated system body A has a temperature of +40°C, and body B has a temperature of +65°C. These bodies were brought into thermal contact with each other. After some time it came thermal equilibrium. How did the temperature of body B and the total internal energy of bodies A and B change as a result?

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

1. Increased;
2. Decreased;
3. Hasn't changed.

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

Solution. If in an isolated system of bodies no energy transformations occur other than heat exchange, then the amount of heat given off 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. Problems of this type are solved based on the heat balance equation.

where ∆ U– change in internal energy.

In our case, as a result of heat exchange, the internal energy of body B decreases, which means 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 an electromagnet has a speed of perpendicular to the vector magnetic field induction as shown in the figure. Where is the Lorentz force acting on the proton directed relative to the drawing (up, towards the observer, away from the observer, down, left, right)

Solution. 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 the four fingers of the left hand along the velocity vector, for a positively charged particle, the vector should enter perpendicularly into the palm, thumb set aside by 90° shows the direction of the Lorentz force acting on the particle. As a result, we have that the Lorentz force vector is directed away from the observer relative to the figure.

Answer. from the observer.

Tension modulus electric field in a flat air capacitor with a capacity of 50 μF is equal to 200 V/m. The distance between the capacitor plates is 2 mm. Why charge equals capacitor? Write your answer in µC.

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

Where d– distance between the plates.

Let's express the voltage U=E d(4); Let's substitute (4) into (2) and calculate the charge of the capacitor.

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

Please pay attention to the units in which you need to write the answer. We received it in coulombs, but present it in µC.

Answer. 20 µC.

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

1. Increases
2. Decreases
3. Doesn't change
4. Record the selected numbers for each answer in the table. The numbers in the answer may be repeated.

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

Where n 2 – absolute indicator refraction of glass, the medium where the light goes; n 1 – absolute refractive index of the first medium, from where the light is coming. For air n 1 = 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 indicator refraction. The speed of light propagation in glass is slower. Please note that we measure angles 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. This will not change the refractive index of glass.

Answer.

Copper jumper at a point in time t 0 = 0 begins to move at a speed of 2 m/s along parallel horizontal conducting rails, to the ends of which a 10 Ohm resistor is connected. The entire system is in a vertical uniform magnetic field. The resistance of the jumper and the rails is negligible; the jumper is always located perpendicular to the rails. The flux Ф 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 your answer.

1. By the time t= 0.1 s change in magnetic flux through the circuit is 1 mWb.
2. Induction current in the jumper in the range from t= 0.1 s t= 0.3 s max.
3. The module of the inductive emf arising in the circuit is 10 mV.
4. The strength of the induction current flowing in the jumper is 64 mA.
5. 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.

Solution. Using a graph of the dependence of the flux of the magnetic induction vector through the circuit on time, we will determine the areas where the flux F changes and where the change in flux is zero. This will allow us to determine the time intervals during which an induced current will appear in the circuit. True statement:

1) By the time t= 0.1 s change in magnetic flux through the circuit is equal to 1 mWb ∆Ф = (1 – 0) 10 –3 Wb; The module of the inductive emf arising in the circuit is determined using the EMR law

Answer. 13.

According to the graph of current versus time in electrical circuit, the inductance of which is 1 mH, determine the module Self-induced emf in the time interval from 5 to 10 s. Write your answer in µV.

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

EMF formula self-induction has the form

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

∆t= 10 s – 5 s = 5 s

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

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

Let's substitute numeric values into formula (2), we get

| Ɛ | = 2 ·10 –6 V, or 2 µV.

Answer. 2.

Two transparent plane-parallel plates are pressed tightly 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 equal to n 2 = 1.77. Establish a correspondence between physical quantities and their meanings. For each position in the first column, select the corresponding position from the second column and write down the selected numbers in the table under the corresponding letters.

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

Let's write down the law of refraction

Let's plot the approximate path of the beam through the plates. We use formula (1) for the boundaries 2–3 and 3–1. In response 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 are produced as a result of the reaction thermonuclear fusion

+ → x+ y;

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

+ → x + y;

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

Answer. 1 – α-particle; 2 – protons.

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 your answer to the nearest tenth.

Solution. The momentum of the second photon is greater than the momentum of the first photon according to the condition, which means it can be represented p 2 = p 1 + Δ p(1). Photon energy can be expressed in terms of photon momentum using following equations. This E = mc 2 (1) and p = mc(2), then

E = pc (3),

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

We round the answer to tenths and get 8.2.

Answer. 8,2.

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

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

1. Increased;
2. Decreased;
3. Hasn't changed.

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

Solution. Positron β - decay in the atomic nucleus occurs when a proton transforms into 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 transformation reaction 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 monochromatic light with a certain wavelength. In all cases, the light 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 you used diffraction grating with a smaller period, and then the number of the experiment in which a diffraction grating with a larger period was used.

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

d sinφ = kλ (1),

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

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

Answer. 42.

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

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

1. Will increase;
2. Will decrease;
3. It won't change.

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

Solution. It is important to remember on what values ​​the conductor resistance depends. The formula for calculating 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 wire of the same material, the same length, but different sizes cross section. The area is twice as small. Substituting into (1) we find that the resistance increases by 2 times, and the current decreases by 2 times, therefore, the voltage does not change.

Answer. 13.

Oscillation period mathematical pendulum on the Earth's surface 1, 2 times more period its vibrations on some planet. Why modulus is equal acceleration free fall on this planet? The influence of the atmosphere in both cases is negligible.

Solution. A mathematical pendulum is a system consisting of a thread whose dimensions are many more sizes the ball and the ball itself. Difficulty may arise if Thomson's formula for the period of oscillation of a mathematical pendulum is forgotten.

T= 2π (1);

l– length of the mathematical pendulum; g– free fall acceleration.

By condition

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

Answer. 14.4 m/s 2.

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

Solution. If you place a current-carrying conductor in a magnetic field, the field on the current-carrying conductor will act with an Ampere force. Let's write down the formula for the Ampere force modulus

F A = I LB sinα ;

F A = 0.6 N

Answer. F A = 0.6 N.

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

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

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

Then the current ratio

Answer. The current must be increased 7 times. You enter only the number 7 on the answer form.

An electrical circuit consists of two light bulbs, two diodes and a turn of wire connected as shown in the figure. (A diode only allows current to flow in one direction, as shown at the top of the picture.) Which of the bulbs will light up if the north pole of the magnet is brought closer to the coil? Explain your answer by indicating what phenomena and patterns you used in your explanation.

Solution. Magnetic induction lines come out of north pole magnet and diverge. When the magnet approaches magnetic flux increases through a turn of wire. According to Lenz's rule, the magnetic field created induced current coil should be directed to the right. According to the gimlet rule, the current should flow clockwise (as viewed from the left). The diode in the second lamp circuit passes in this direction. This means the second lamp will light up.

Answer. The second lamp will light up.

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

Solution. Let's make an explanatory drawing.

– Thread tension force;

– Reaction force of the bottom of the vessel;

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

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

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

F a = Slρ in g (2)

Let's consider the moments of forces relative to the point of suspension of the spoke.

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

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

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

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

Let's substitute the numerical data and get that

F d = 0.025 N.

Answer. F d = 0.025 N.

Cylinder containing m 1 = 1 kg nitrogen, during strength testing exploded at temperature t 1 = 327°C. What mass of hydrogen m 2 could be stored in such a cylinder at a temperature t 2 = 27°C, having a fivefold safety margin? Molar mass nitrogen M 1 = 28 g/mol, hydrogen M 2 = 2 g/mol.

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

Where V– volume of the cylinder, T 1 = t 1 + 273°C. According to the condition, hydrogen can be stored at pressure p 2 = p 1 /5; (3) Considering that

We can express the mass of hydrogen by working directly with equations (2), (3), (4). Final formula has the form:

After substituting numeric data m 2 = 28 g.

Answer. m 2 = 28 g.

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

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

For amplitude (maximum) values ​​we write

and from equation (2) we express

Let's substitute (4) into (3). As a result we get:

I = I m (5)

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

I= 4.0 mA.

Answer. I= 4.0 mA.

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

Solution. Let's make an explanatory drawing

α is the angle of incidence of the beam;

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

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

According to the law of refraction of light

Consider the rectangular ΔADB. In it AD = h, then DB = AD

We get the following expression:

Let's substitute the numerical values ​​into the resulting formula (5)

Answer. 1.63 m.

In preparation for the Unified State Exam, we invite you to familiarize yourself with work program in physics for grades 7–9 to the UMK line Peryshkina A.V. And advanced level work program for grades 10-11 for teaching materials Myakisheva G.Ya. The programs are available for viewing and free downloading to all registered users.

Changes in Unified State Exam assignments in physics for 2019 no year.

Structure of Unified State Examination tasks in physics-2019

The examination paper consists of two parts, including 32 tasks.

Part 1 contains 27 tasks.

• In tasks 1–4, 8–10, 14, 15, 20, 25–27, the answer is an integer or finite number decimal.
• The answer to tasks 5–7, 11, 12, 16–18, 21, 23 and 24 is a sequence of two numbers.
• The answer to tasks 19 and 22 are two numbers.

Part 2 contains 5 tasks. The answer to tasks 28–32 includes detailed description the entire progress of the task. The second part of the tasks (with a detailed answer) is assessed by an expert commission on the basis of.

Unified State Exam topics in physics that will be included in the exam paper

1. Mechanics(kinematics, dynamics, statics, conservation laws in mechanics, mechanical vibrations and waves).
2. Molecular physics (molecular kinetic theory, thermodynamics).
3. Electrodynamics and fundamentals of SRT(electric field, direct current, magnetic field, electromagnetic induction, electromagnetic vibrations and waves, optics, fundamentals of SRT).
4. Quantum physics and elements of astrophysics(wave-particle dualism, atomic physics, physics atomic nucleus, elements of astrophysics).

Duration of the Unified State Exam in Physics

The entire examination work will be completed 235 minutes.

Estimated time to complete tasks various parts work is:

1. for each task with a short answer – 3–5 minutes;
2. for each task with a detailed answer – 15–20 minutes.

What you can take for the exam:

• A non-programmable calculator is used (for each student) with the ability to calculate trigonometric functions(cos, sin, tg) and ruler.
• The list of additional devices and devices, the use of which is permitted for the Unified State Examination, is approved by Rosobrnadzor.

Important!!! do not rely on cheat sheets, tips and use technical means(phones, tablets) during the exam. Video surveillance at the Unified State Exam 2019 will be strengthened with additional cameras.

Unified State Exam scores in physics

• 1 point - for 1-4, 8, 9, 10, 13, 14, 15, 19, 20, 22, 23, 25, 26, 27 tasks.
• 2 points - 5, 6, 7, 11, 12, 16, 17, 18, 21, 24.
• 3 points - 28, 29, 30, 31, 32.

Total: 52 points(maximum primary score).

What you need to know when preparing tasks for the Unified State Exam:

• Know/understand the meaning of physical concepts, quantities, laws, principles, postulates.
• Be able to describe and explain physical phenomena and properties of bodies (including space objects), experimental results... give examples practical use physical knowledge
• Distinguish between hypotheses and scientific theory, draw conclusions based on experiment, etc.
• Be able to apply acquired knowledge when making decisions physical problems.
• Use acquired knowledge and skills in practical activities and everyday life.

Where to start preparing for the Unified State Exam in Physics:

1. Study the theory required for each task.
2. Train in test tasks in physics, developed on the basis of the Unified State Exam. On our website, tasks and options in physics will be updated.
3. Manage your time correctly.

We wish you success!

The website “I will pass the GIA: I will solve the OGE and the Unified State Exam” by Dmitry Gushchin in 2019 will again be in the top chart of the best educational resources. Where does this confidence come from? It's all about a unique online trainer, with the help of which any user can “train” himself in any USE subject and OGE. Web resource available training work in Russian language, social studies, basic and specialized mathematics, chemistry and others school subjects. To thoroughly understand the structure of the portal, read our special review!

The resource can be found at sdamgia.ru. The site is divided into two parts: Unified State Exam and OGE. IN section of the Unified State Exam You can test your knowledge in 15 subjects:

1. Mathematics (basic);
2. Mathematics (profile);
3. Russian language;
4. Physics;
5. Chemistry;
6. English language;
7. Literature;
8. Informatics;
9. German;
10. Social science;
11. Story;
12. French;
13. Geography;
14. Biology;
15. Spanish.

The other section includes slightly fewer subjects - 14. Because on Dmitry Gushchin’s website, mathematics in the OGE part has no division. This is due to the fact that 9th graders, during the general state exam They just take mathematics, without dividing into “base” and “profile”.

Let's look at an online trainer using several items as an example.

“I WILL SOLVE the Unified State Exam”: Russian language

The page contains 15 training options. If the user clicks on any of them, a block of options with tasks will be loaded and a timer will start, which will count down exactly as much time as defined by the specification for the Unified State Exam in the Russian language. Namely – 3 hours 30 minutes.

Besides, practice test everyone can create it themselves. For example, if you activate the “Tasks B” option, the system will compose a training version consisting of 24 questions, without including task No. 25 (this is block C). Well, and vice versa - you can choose “Tasks C”. In this case, question No. 25 will be loaded. The system selects all options randomly.

“I WILL SOLVE the Unified State Exam”: physics

The same 15 options. The subject can solve any of them within the allotted time. After the time has passed, the system will check and grade the work done.

And, in addition to the ability to create a specialized test, the page contains an assembly training questions by task blocks. For example, if you click on the item “Kinematics, Newton’s laws”, sub-items such as “Friction Force”, “Newton’s Second Law”, “Circular Movements”, etc. will be loaded. By checking any of them and selecting the corresponding buttons below, the subject will receive a training test tailored to the solution certain types physical tasks.

“I WILL SOLVE the Unified State Exam”: social studies

Everything is absolutely exactly the same. The differences are only in small details.

• First section offers tasks for preparing for the social studies exam with sample solutions. Those. If you enter the number of a CMM solved in the past in the appropriate field, the program will load both the testing and measuring material itself and the options for its solution.
• Second section– these are the already familiar 15 training answers. If you are not satisfied with the quality of the tests presented, you can always use the archive of options.
• In the third section, unlike a similar block in social studies, 16, rather than 4 additional items are presented to create a specialized test.
• Fourth section– “Catalog of tasks” – there are 31 topics and 6 additional questions. Based on them, you can also create unique test papers.

Afterword

You can also get acquainted with chemistry in Dmitry Gushchin’s “SOLVE the Unified State Exam 2019”, foreign languages, biology. The online trainer has equally well selected training options for all items. The only resource that the user will need to use the simulator is personal time!