Preparation for the OGE and the Unified State Exam

Secondary 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

We analyze the tasks of the Unified State Exam in physics (Option C) with the teacher.

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

The work 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. Advanced level tasks are aimed at testing the ability to use concepts and laws of physics to analyze various processes and phenomena, as well as the ability to solve problems using one or two laws (formulas) on 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 a changed 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 fully corresponds to the demo version of the Unified State Exam 2017; the tasks are taken from the open bank of Unified State Exam tasks.

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.

S = (30 + 20) With 10 m/s = 250 m.
2

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

a = ∆v = (8 – 2) m/s = 2 m/s 2.
∆t 3 s

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 a 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 the 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 undergoes vertical oscillations. 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 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


= T ; m = T 2 ; m = k T 2 ; m= 200 N/m (4 s) 2 = 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 keep in balance or lift a load weighing 10 kg. Friction is negligible. Based on the analysis of the above figure, select two true 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 make a schematic drawing and indicate the forces acting on the load: thread tension F control, directed upward along the thread; gravity directed vertically downwards; 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.

V = m .
p

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

4) sinα – a
g cosα

Solution. This task requires the application of Newton's laws. We recommend making a schematic drawing; indicate all kinematic characteristics of movement. 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 us write this 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 in the opposite direction 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.

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

μ = F tr = m(g sinα – a) = tgα – a (8).
N mg cosα g cosα

We select the appropriate positions for each letter.

Answer. A – 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 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 ideal gas equation of state

Let's express the mass of the gas.

Be sure to pay attention to which units are asked to write down the answer. It 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 that the saturated vapor pressure 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.

P 2 = φ 2 = 35 = 3,5
P 1 φ 1 10

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. In 20 minutes. 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. As the substance cooled, its internal energy decreased. The results of temperature measurements allow us to determine the temperature at which a substance begins to crystallize. While a substance changes from liquid to 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 an 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, thermal equilibrium occurred. 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 physical quantity in the table. 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.

∆U = ∑ n ∆U i = 0 (1);
i = 1

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 the electromagnet, has a speed perpendicular to the magnetic field induction vector, 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, the thumb set at 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.

The modulus of the electric field strength 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. What is the charge on the 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

sinα = n 2 ,
sinβ n 1

Where n 2 – absolute refractive index of glass, the medium where the light goes; n 1 is the absolute refractive index of the first medium from which the light comes. 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 high refractive index. 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.


Using the graph of current versus time in an electrical circuit whose inductance is 1 mH, determine the self-inductive emf module 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. We will also convert the current shown in the figure in mA to A by multiplying by 10 –3.

The formula for self-induction emf 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.

We substitute numerical 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 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 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

sinβ = sin50 = 0,4327 ≈ 0,433
1,77

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 thermonuclear fusion reaction

+ → x+ y;

Solution. In all 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). The energy of a photon can be expressed in terms of the momentum of the photon using the 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:

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

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 of monochromatic light with a specific 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. Indicate first the number of the experiment in which 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.

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. The angular directions to the maxima of the 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, with a wire of the same metal and the same length, but having half the cross-sectional area, and half the current was passed 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. Will not 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 cross-sectional area. 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.

The period of oscillation of a mathematical pendulum on the surface of the Earth is 1.2 times greater than the period of its oscillation on a certain planet. What is the magnitude of the acceleration due to gravity 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 much larger than the dimensions of 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- acceleration of gravity.

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.

A straight conductor 1 m long carrying a current of 3 A is 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

W m = LI 2 (1);
2

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

I 1 2 = 2W 1 ; I 2 2 = 2W 2 ;
L L

Then the current ratio

I 2 2 = 49; I 2 = 7
I 1 2 I 1

Answer. The current strength 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 emerge from the north pole of the magnet and diverge. As the magnet approaches, the magnetic flux through the coil of wire increases. In accordance with Lenz's rule, the magnetic field created by the inductive current of the coil must 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 is the 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 the Archimedean force modulus 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

NL cosα + Slρ in g (L – l )cosα = SLρ a g L cosα (7)
2 2

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:

F d = [ 1 Lρ a– (1 – l )lρ in ] Sg (8).
2 2L

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 of 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). The final formula looks like:

m 2 = m 1 M 2 T 1 (5).
5 M 1 T 2

After substituting numeric data m 2 = 28 g.

Answer. m 2 = 28 g.

In an ideal oscillatory circuit, the amplitude of current fluctuations in the inductor is 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

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

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

sinβ = sinα (3)
n 2

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

tgβ = h tgβ = h sinα = h sinβ = h sinα (4)
cosβ

We get the following expression:

AC = 2 DB = 2 h sinα (5)

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 of 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.

Option No. 3109295

Early Unified State Exam in Physics 2017, option 101

When completing tasks with a short answer, enter in the answer field the number that corresponds to the number of the correct answer, or a number, a word, a sequence of letters (words) or numbers. The answer should be written without spaces or any additional characters. Separate the fractional part from the whole decimal point. There is no need to write units of measurement. In problems 1–4, 8–10, 14, 15, 20, 25–27, the answer is a whole number or a finite decimal fraction. The answer to tasks 5–7, 11, 12, 16–18, 21 and 23 is a sequence of two numbers. The answer to task 13 is a word. The answer to tasks 19 and 22 are two numbers.


If the option is specified by the teacher, you can enter or upload answers to tasks with a detailed answer into the system. The teacher will see the results of completing tasks with a short answer and will be able to evaluate the downloaded answers to tasks with a long answer. The scores assigned by the teacher will appear in your statistics.


Version for printing and copying in MS Word

The figure shows a graph of the projection of the speed of the body v x from time.

Determination of the projection of the acceleration of this body a x in inter-va-le time from 15 to 20 s. The answer is in m/s 2.

Answer:

Cube mass M= 1 kg, laterally compressed by springs (see figure), rests on a smooth horizontal table. The first spring is compressed by 4 cm, and the second is compressed by 3 cm. Stiffness of the first spring k 1 = 600 N/m. What is the stiffness of the second spring? k 2? Express your answer in N/m.

Answer:

Two bodies are moving at the same speed. The kinetic energy of the first body is 4 times less than the kinetic energy of the second body. Determine the ratio of the masses of the bodies.

Answer:

At a distance of 510 m from the blue sky, workers are driving in piles using a piledriver. How long will it take from the moment when the observer sees the impact of the pile driver to the moment when he hears the sound of the impact? The speed of sound in air is 340 m/s. The answer is you-ra-zi-te in the village.

Answer:

The figure shows graphs of pressure dependence p from diving depth h for two liquids at rest: water and heavy liquid diiodomethane, at constant temperature.

Choose two true statements that agree with the graphs given.

1) If the pressure inside a hollow ball is equal to atmospheric pressure, then in water at a depth of 10 m the pressure on its surface from the outside and from the inside will be equal to each other.

2) The density of kerosene is 0.82 g/cm 3, a similar graph of pressure versus depth for kerosene will be between the graphs for water and diiodomethane.

3) In water at a depth of 25 m, pressure p 2.5 times more than atmospheric.

4) As the depth of immersion increases, the pressure in diiodomethane increases faster than in water.

5) The density of olive oil is 0.92 g/cm 3, a similar graph of pressure versus depth for oil will be between the graph for water and the x-axis (horizontal axis).

Answer:

A massive load suspended from the ceiling on a weightless spring performs free vertical vibrations. The spring remains stretched all the time. How do the potential energy of a spring and the potential energy of a load behave in a gravitational field when the load moves upward from its equilibrium position?

1) increases;

2) decreases;

3) does not change.

Answer:

A truck moving along a straight, horizontal road at high speed v, braked so hard that the wheels stopped rotating. Cargo weight m, coefficient of friction of wheels on the road μ . Forms A and B allow one to calculate the values ​​of physical quantities, har-rak-te-ri-zu-yu- movement of the load.

Establishment of correspondence between the form-mu-la-mi and fi-zi-che-ski-mi ve-li-chi-na-mi, the meaning of something -ry can be calculated using these formulas.

To each position of the first column under-be-ri-te corresponding to the position of the second column and behind - write in the table the selected numbers under the corresponding letters.

AB

Answer:

As a result of cooling of rarefied argon, its absolute temperature decreased by 4 times. How many times did the average kinetic energy of thermal motion of argon molecules decrease?

Answer:

The working body of a thermal machine during a cycle receives from the heating the amount of heat equal to Noe 100 J, and at the end of the day the work is 60 J. What is the efficiency of the heating machine? The answer is in %.

Answer:

The relative humidity of the air in a closed vessel with a piston is 50%. What will be the relative humidity of the air in the vessel if the volume of the vessel at a constant temperature is reduced by 2 times? Express your answer in %.

Answer:

The hot substance, initially in a liquid state, was slowly cooled. The heat sink power is constant. The table shows the results of measurements of the temperature of a substance over time.

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

1) The crystallization process of the substance took more than 25 minutes.

2) The specific heat capacity of a substance in liquid and solid states is the same.

3) The melting point of the substance under these conditions is 232 °C.

4) After 30 min. after the start of measurements, the substance was only in a solid state.

5) After 20 minutes. after the start of measurements, the substance was only in a solid state.

Answer:

On graphs A and B there are diagrams p−T And p−V for processes 1−2 and 3−4 (hyper-bo-la), carried out with 1 mole of helium. On the diagrams p- pressure, V– volume and T– absolute gas temperature. Establishment of the correspondence between the graphics and the approval of the images processes reflected in graphs. To each position of the first column under-be-ri-te corresponding to the position of the second column and behind - write in the table the selected numbers under the corresponding letters.

GRA-FI-KI APPROVAL

1) Work is done on the gas, while the gas gives off a certain amount of heat.

2) Gas receives the same amount of heat, while its internal energy does not come from me. yes.

3) Work is carried out over the gas, while its internal energy increases.

4) Gas receives the full amount of heat, while its internal energy increases -yes.
AB

Answer:

How is the Ampere force acting on conductor 1 from conductor 2 directed relative to the figure (to the right, left, up, down, towards the observer, away from the observer) (see figure), if the conductors are thin, long, straight, parallel to each other? ( I- current strength.) Write the answer in word(s).

Answer:

A direct current flows through a section of the circuit (see figure) I= 4 A. What current will be shown by an ideal ammeter connected to this circuit if the resistance of each resistor r= 1 Ohm? Express your answer in amperes.

Answer:

In an experiment to observe electromagnetic induction, a square frame made of one turn of thin wire is placed in a uniform magnetic field perpendicular to the plane of the frame. The magnetic field induction uniformly increases from 0 to the maximum value IN max per time T. In this case, an induced emf equal to 6 mV is excited in the frame. What induced emf will occur in the frame if T reduce by 3 times, and IN Reduce max by 2 times? Express your answer in mV.

Answer:

A single electric field is created equally across an extended horizontal plastic. The field strength lines to the right are vertically upward (see ri-su-nok).

From the list below, select two correct statements and indicate their numbers.

1) If to the point A place a test point-reactive charge, then a force will act on it from the side of the plate, in the right direction flaxen ver-ti-kal-but down.

2) The plastic has a negative charge.

3) Po-ten-tsi-al of the electric field at the point IN lower than at point WITH.

4) Field strength at a point A less than at point WITH.

5) Work of the electric field according to the relocation of the sample -but-th-za-rya-yes from the point A and to the point IN equal to zero.

Answer:

An electron moves in a circle in a uniform magnetic field. How will the Lorentz force acting on the electron and its period of revolution change if its kinetic energy is increased?

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

1) will increase;

2) will decrease;

3) will not change.

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

Answer:

The figure shows a DC circuit. Establish a correspondence between physical quantities and formulas by which they can be calculated ( ε – EMF of the current source, r– internal resistance of the current source, R– resistor resistance).

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

PHYSICAL QUANTITIES FORMULAS

A) current strength through the source with switch K open

B) current strength through the source with the key K closed

Answer:

Two monochromatic electromagnetic waves propagate in a vacuum. The energy of a photon of the first wave is 2 times greater than the energy of a photon of the second wave. Determine the ratio of the lengths of these electromagnetic waves.

Answer:

How will they change when β − − decay mass number of the nucleus and its charge?

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

1) will increase

2) will decrease

3) will not change

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

Answer:

Determination of the volt-meter (see ri-su-nok), if the error of the direct line is due to the measurement of the direct line -the same is equal to the price of the volt-meter. Indicate the answer in volts. In this way, meaning and sinfulness are merged, but without space.

Answer:

To conduct laboratory work to detect the dependence of the resistance of a conductor on its length, the student was given five conductors, the characteristics of which are indicated in the table. Which two of the following guides should a student take to conduct this study?

Duration of the physics exam - 3 hours 55 minutes
The work consists of two parts, including 31 tasks.
Part 1: tasks 1 - 23
Part 2: tasks 24 - 31.
In tasks 1–4, 8–10, 14, 15, 20, 24–26 the answer is
whole number or finite decimal fraction.
Answer to tasks 5–7, 11, 12, 16–18, 21 and 23
is a sequence of two digits.
The answer to task 13 is a word.
The answer to tasks 19 and 22 are two numbers.
The answer to tasks 27–31 includes
a detailed description of the entire progress of the task.
Minimum test score (on a 100-point scale) - 36

Demo version of the Unified State Exam 2020 in physics (PDF):

Unified State Exam

The purpose of the demonstration version of USE tasks is to enable any USE participant to get an idea of ​​the structure of the CMM, the number and form of tasks, and their level of complexity.
The given criteria for assessing the completion of tasks with a detailed answer, included in this option, give an idea of ​​the requirements for the completeness and correctness of recording a detailed answer.
To successfully prepare for passing the Unified State Exam, I propose to analyze solutions to prototypes of real tasks from the Unified State Exam.

When preparing for the Unified State Exam, it is better for graduates to use options from official sources of information support for the final exam.

To understand how to complete the exam work, you should first of all familiarize yourself with the demo versions of the KIM Unified State Examination in Physics of the current year and with the options for the Unified State Examination of the early period.

On May 10, 2015, in order to provide graduates with an additional opportunity to prepare for the Unified State Exam in Physics, one version of the KIM used for the Unified State Exam in the early period of 2017 is published on the FIPI website. These are real options from the exam conducted on April 7, 2017.

Early versions of the Unified State Exam in Physics 2017

Demo version of the Unified State Exam 2017 in physics

Task option + answers variant + answer
Specification download
Codifier download

Demo versions of the Unified State Exam in Physics 2016-2015

Physics Download option
2016 version of the Unified State Exam 2016
2015 variant EGE fizika

Changes in the Unified State Exam KIM in 2017 compared to 2016

The structure of part 1 of the examination paper has been changed, part 2 has been left unchanged. Tasks with a choice of one correct answer have been excluded from the examination work and tasks with a short answer have been added.

When making changes to the structure of the examination work, the general conceptual approaches to assessing educational achievements were preserved. In particular, the maximum score for completing all tasks of the examination paper remained unchanged, the distribution of maximum points for tasks of different levels of complexity and the approximate distribution of the number of tasks by sections of the school physics course and methods of activity were preserved.

A complete list of questions that can be controlled at the 2017 Unified State Exam is given in the codifier of content elements and requirements for the level of training of graduates of educational organizations for the 2017 Unified State Exam in Physics.

The purpose of the demonstration version of the Unified State Examination in Physics is to enable any Unified State Examination participant and the general public to get an idea of ​​the structure of future CMMs, the number and form of tasks, and their level of complexity.

The given criteria for assessing the completion of tasks with a detailed answer, included in this option, give an idea of ​​the requirements for the completeness and correctness of recording a detailed answer. This information will allow graduates to develop a strategy for preparing and passing the Unified State Exam.

Approaches to selecting content and developing the structure of the KIM Unified State Examination in Physics

Each version of the examination paper includes tasks that test the mastery of controlled content elements from all sections of the school physics course, while tasks of all taxonomic levels are offered for each section. The most important content elements from the point of view of continuing education in higher educational institutions are controlled in the same version with tasks of different levels of complexity.

The number of tasks for a particular section is determined by its content and in proportion to the teaching time allocated for its study in accordance with the approximate physics program. The various plans by which examination options are constructed are built on the principle of content addition so that, in general, all series of options provide diagnostics for the development of all content elements included in the codifier.

Each option includes tasks in all sections of different levels of complexity, allowing you to test the ability to apply physical laws and formulas both in standard educational situations and in non-traditional situations that require the manifestation of a fairly high degree of independence when combining known action algorithms or creating your own plan for completing a task .

The objectivity of checking tasks with a detailed answer is ensured by uniform 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 Examination in Physics is an exam of choice for graduates and is intended for differentiation when entering higher educational institutions.

For these purposes, the work includes tasks of three difficulty levels. Completing tasks at a basic level of complexity allows you to assess the level of mastery of the most significant content elements of a high school physics course and mastery of the most important types of activities.

Among the tasks of the basic level, tasks are distinguished whose content corresponds to the standard of the basic level. The minimum number of Unified State Examination points in physics, confirming that a graduate has mastered a secondary (full) general education program in physics, is established based on the requirements for mastering the basic level standard. The use of tasks of increased and high levels of complexity in the examination work allows us to assess the degree of preparedness of a student to continue his education at a university.