The article presents an analysis of the tasks of the second part of the exam in physics under numbers 25-27. There is also a video tutorial from a physics tutor with detailed and understandable explanations for each of the tasks. If you have just started your preparation for the exam in physics, this article may be very, very useful for you.

Let's start by defining the acceleration at which the elevator moves. It moves from a state of rest, so the formula is valid:, where S- passed let, a- elevator acceleration, t- movement time. From here we get: m / s 2.

Let's depict the forces acting on this load. The force of gravity is directed vertically downward, and the force of elasticity of the spring (Hooke's force) is directed vertically upward, where k- spring stiffness, x- spring extension:

After the damping of the oscillations of the load on the spring, caused by the start of the movement of the elevator, the load will lower relative to the ground synchronously with the elevator with acceleration. For this situation, in projection onto the vertical axis OY co-directional with acceleration, from Newton's Second Law we obtain:

Calculations give kg.

Let's find first what is equal to p 2. For this we use the fact that the dependency p from V in the specified process is directly proportional to:, whence we get kPa.

It is known from the school course of thermodynamics that the work of a gas is numerically equal to the area under the graph of the gas process in coordinates ( p;V). This work is positive if the gas has expanded, and negative otherwise. Consequently, in this process, the work of the gas is positive and is numerically equal to the area of ​​the trapezoid 12 V 2 V 1 (in the figure it is highlighted in yellow):

The area of ​​the trapezoid is equal to the product of the half-sum of the bases and the height. That is, we get in this case:

Calculations give meaning:

In our calculations, we used that 1 liter is equal to 10 -3 m 3.

The photon energy is related to the wavelength by the known relation:, where h- Planck's constant, c- the speed of light in a vacuum, λ - the length of the light wave in a vacuum. This means that if the sought photon energy in the first case was equal to E, then when the wavelength of the incident radiation is halved, the photon energy becomes equal to 2 E... Let us write down the Einstein equations for the photoelectric effect in both cases:

Here E K1 and E K2 are the maximum kinetic energies of photoelectrons in the first and second cases, respectively, A- work function of electrons from metal. Then, subtracting the first equation from the second by term, we obtain eV.

Analysis of problems is presented by Sergey Valerievich

Preparation for the exam and exam

Secondary general education

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

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

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

Preparation for the exam in physics: examples, solutions, explanations

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

The work presents tasks of different levels of difficulty: basic, advanced and high. Basic level tasks are simple tasks that test the mastery of the most important physical concepts, models, phenomena and laws. Tasks of an advanced level are aimed at testing the ability to use the concepts and laws of physics to analyze various processes and phenomena, as well as the ability to solve problems on the application of one or two laws (formulas) for any of the topics of the school physics course. In work 4, the tasks of part 2 are tasks of a high level of complexity and test the ability to use the laws and theories of physics in a changed or new situation. The fulfillment of such tasks requires the application of knowledge from two three sections of physics at once, i.e. high level of training. This option is fully consistent with the demo version of the USE in 2017, the tasks are taken from the open bank of USE tasks.

The figure shows a graph of the dependence of the speed module on time t... Determine the path covered by the car in the time interval from 0 to 30 s.

Solution. The distance traveled by a car in the time interval from 0 to 30 s is easiest to define 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 rope. The figure shows the dependence of the speed projection V load on the upward axle from time t... Determine the modulus of the cable tension during the ascent.

Solution. According to the graph of the dependence of the projection of speed v load on an axle directed vertically upward, from time t, you can determine the projection of the acceleration of the load

The load is influenced by: the force of gravity directed vertically downward and the tension force of the rope directed vertically upward along the rope, 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 body's mass by the acceleration imparted to it.

+ = (1)

Let us write the equation for the projection of the vectors in the frame of reference connected with the earth, the OY axis is directed upwards. The projection of the tensile force is positive, since the direction of the force coincides with the direction of the OY axis, the projection of the gravity is negative, since the force vector is oppositely directed to the OY axis, the projection of the acceleration vector is also positive, so the body moves upward with acceleration. We have

T – mg = ma (2);

from formula (2) modulus of tensile force

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 at a constant speed, the modulus of which is 1.5 m / s, applying force to it, as shown in figure (1). In this case, the modulus of the sliding friction force acting on the body is 16 N. What is the power developed by the force F?

Solution. Imagine a 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 frame of reference associated with a fixed surface, we write down the equations for the projection of vectors on the selected coordinate axes. According to the condition of the problem, the body moves uniformly, since its speed is constant and equal to 1.5 m / s. This means that the acceleration of the body is zero. Two forces act horizontally on the body: sliding friction force tr. and the force with which the body is dragged. The projection of the friction force is negative, since the force vector does not coincide with the direction of the axis NS... Force projection F positive. We remind you that to find the projection, we drop the perpendicular from the beginning and end of the vector to the selected axis. With this in mind, we have: F cosα - F tr = 0; (1) express the projection of the force F, this is 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 substitution, taking into account equation (2), and substitute the corresponding data into equation (3):

N= 16 N 1.5 m / s = 24 W.

Answer. 24 watts

The load, fixed on a light spring with a stiffness of 200 N / m, makes vertical vibrations. The figure shows a plot of the dependence of the displacement x cargo from time to time t... Determine what the weight of the load is. Round your answer to the nearest whole number.

Solution. A spring loaded weight vibrates vertically. According to the graph of the dependence of the displacement of the load NS from time t, we define the period of fluctuations of the load. The oscillation period is T= 4 s; from the formula T= 2π express the mass m cargo.

Answer: 81 kg.

The figure shows a system of two lightweight blocks and a weightless rope, with which you can balance or lift a load weighing 10 kg. Friction is negligible. Based on the analysis of the above figure, select two correct statements and indicate their numbers in the 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 a power gain.
3. h, you need to stretch out a section of rope with a length of 3 h.
4. In order to slowly raise the load to a height hh.

Solution. In this task, it is necessary to recall simple mechanisms, namely blocks: a movable and fixed block. The moveable block doubles in strength, with the rope stretching twice as long and the stationary block used to redirect the force. In operation, simple mechanisms of winning do not give. After analyzing the problem, we immediately select the necessary statements:

1. In order to slowly raise the load to a height h, you need to pull out a section of rope with a length of 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, fixed on a weightless and inextensible thread, is completely immersed in a vessel with water. The weight does not touch the walls and bottom of the vessel. Then an iron weight is immersed in the same vessel with water, the mass of which is equal to the mass of the aluminum weight. 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?

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

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

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

Answer. 13.

Block weight m slides off a fixed rough inclined plane with an angle α at the base. The block acceleration modulus is a, the speed modulus of the bar increases. Air resistance is negligible.

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

B) Coefficient of friction of the bar on an inclined plane

3) mg cosα

Solution. This task requires the application of Newton's laws. We recommend making a schematic drawing; indicate all the kinematic characteristics of the movement. If possible, depict the acceleration vector and vectors of all forces applied to the moving body; remember that the forces acting on the 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 the vectors of forces and accelerations;

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 bar and the coordinate axes of the frame of reference associated with the surface of the inclined plane. Since all forces are constant, the movement of the bar will be equally variable with increasing speed, i.e. the acceleration vector is directed towards the movement. Let's choose the direction of the axes as shown in the figure. Let's write down the projections of the 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 support reaction force is positive, since the vector coincides with the direction of the OY axis N y = N; the projection of the friction force is zero since the vector is perpendicular to the axis; the projection of gravity will be negative and equal mg y= – 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 force of the reaction acting on the bar, from the side of the inclined plane. N = mg cosα (3). Let's write projections onto the OX axis.

On the OX axis: force projection N 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 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 normal pressure force N.

A-priory F tr = μ N(7), we express the coefficient of friction of the bar 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. Gas pressure 150 kPa, its temperature 127 ° C. Determine the mass of 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. We convert the temperature to Kelvin T = t° С + 273, volume V= 33.2 l = 33.2 · 10 –3 m 3; We translate the pressure P= 150 kPa = 150,000 Pa. Using the ideal gas equation of state

express the mass of the gas.

Be sure to pay attention to the unit in which you are asked to write down the answer. It is very important.

Answer. 48 g

Task 9. An ideal monatomic gas in the amount of 0.025 mol adiabatically expanded. At the same time, its temperature dropped from + 103 ° С to + 23 ° С. What kind of work did the gas do? Express your answer in Joules and round to the nearest whole number.

Solution. First, the gas is a monoatomic number of degrees of freedom i= 3, secondly, the gas expands adiabatically - this means without heat exchange Q= 0. 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) express the work of the gas A r = –∆ U(2); The change in the internal energy for a monatomic gas can be written as

Answer. 25 J.

The relative humidity of a portion of air at a certain temperature is 10%. How many times should the pressure of this portion of air be changed in order for its relative humidity to increase by 25% at a constant temperature?

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

According to the condition of the problem, the temperature does not change, which means that the saturated vapor pressure remains the same. Let's write down the 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 substance in liquid state 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.

Choose from the list provided two statements that correspond to the results of the measurements carried out 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 a liquid and solid state 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 down, its internal energy decreased. The temperature measurement results allow you to determine the temperature at which the substance begins to crystallize. As long as a substance passes from a liquid to a solid state, the temperature does not change. Knowing that the melting point and crystallization temperature are the same, we choose the statement:

1. The melting point of the substance under these conditions is 232 ° С.

The second true statement is:

4. After 30 minutes. after the start of measurements, the substance was only in a solid state. Since the temperature at this point in time 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 are brought into thermal contact with each other. After a while, thermal equilibrium has come. How did the body temperature B and the total internal energy of the body A and B change as a result?

For each value, determine the corresponding change pattern:

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 except for heat exchange, then the amount of heat given off by bodies, the internal energy of which decreases, is equal to the amount of heat received by bodies, the internal energy of which 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 that the temperature of this body decreases. The internal energy of body A increases, since the body has received the amount of heat from body B, then its temperature will increase. The total internal energy of bodies A and B does not change.

Answer. 23.

Proton p, flown into the gap between the poles of the electromagnet, has a velocity perpendicular to the magnetic induction vector, as shown in the figure. Where is the Lorentz force acting on the proton directed relative to the figure (up, towards the observer, from the observer, down, left, right)

Solution. The 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, not to forget to take into account the particle charge. We direct four fingers of the left hand along the velocity vector, for a positively charged particle, the vector should enter the palm perpendicularly, the thumb set back 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 50 μF flat air capacitor is 200 V / m. The distance between the capacitor plates is 2 mm. What is the charge of a capacitor? Write down the 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 speaks of a flat air capacitor - a device for accumulating electric charge and electric field energy. From the formula for electrical capacity

where d Is the distance between the plates.

Express the tension U= E d(4); Substitute (4) in (2) and calculate the capacitor charge.

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

We draw your attention to the units in which you need to write the answer. We got it in pendants, but we represent it in μC.

Answer. 20 μC.

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

1. Is increasing
2. Decreases
3. Does not change
4. Write down the selected numbers for each answer in the table. The numbers in the answer may be repeated.

Solution. In tasks of this kind, we recall 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 propagation of waves in these media are different. Having figured out from which medium to which light it propagates, we write the law of refraction in the form

where n 2 - the absolute refractive index of the glass, the medium where the light goes; n 1 is the absolute refractive index of the first medium from which the light is coming. For air n 1 = 1. α is the angle of incidence of the beam on the surface of the glass semi-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 propagation of light in glass is slower. Please note that we measure the angles from the perpendicular restored at the point of incidence of the ray. If you increase the angle of incidence, then the angle of refraction will also increase. The refractive index of the glass will not change from this.

Answer.

Copper jumper at a point in time t 0 = 0 begins to move at a speed of 2 m / s along parallel horizontal conductive 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 lintel and rails is negligible, the lintel is always perpendicular to the rails. The flux Ф of the magnetic induction vector through the circuit formed by a jumper, rails and a resistor changes over time t as shown in the graph.

Using the graph, select two correct statements and include their numbers in the answer.

1. By the time t= 0.1 s, the change in magnetic flux through the circuit is equal to 1 mVb.
2. Induction current in the jumper in the range from t= 0.1 s t= 0.3 s max.
3. The EMF modulus of the induction 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 bulkhead, a force is applied to it, the projection of which on the direction of the rails is 0.2 N.

Solution. According to the graph of the dependence of the flux of the magnetic induction vector through the circuit on time, we determine the sections where the flux Ф changes, and where the flux change is zero. This will allow us to determine the time intervals in which the induction current will occur in the circuit. Correct statement:

1) By the time t= 0.1 s the change in magnetic flux through the circuit is equal to 1 mWb ∆F = (1 - 0) · 10 –3 Wb; The EMF modulus of induction arising in the circuit is determined using the EMR law

Answer. 13.

According to the graph of the dependence of the current strength on time in an electric circuit, the inductance of which is 1 mH, determine the EMF modulus of self-induction in the time interval from 5 to 10 s. Write down the answer in μV.

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

The EMF formula of self-induction has the form

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

∆t= 10 s - 5 s = 5 s

seconds and according to 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.

Substituting numerical values ​​into formula (2), we obtain

| Ɛ | = 2 · 10 –6 V, or 2 µV.

Answer. 2.

Two transparent plane-parallel plates are tightly pressed against each other. A ray of light falls from the air onto the surface of the first plate (see figure). It is known that the refractive index of the upper plate is n 2 = 1.77. Establish a correspondence between physical quantities and their values. For each position of the first column, select the corresponding position from the second column and write down the selected numbers in the table under the corresponding letters.

Solution. To solve problems on the refraction of light at the interface between two media, in particular, problems on the transmission of light through plane-parallel plates, the following order of solution can be recommended: make a drawing indicating the path of rays going from one medium to another; at the point of incidence of the ray 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 the angles are determined from the perpendicular restored at the point of incidence. We determine that the angle of incidence of the beam on the surface is 90 ° - 40 ° = 50 °, the refractive index n 2 = 1,77; n 1 = 1 (air).

Let's write down the law of refraction

Let's construct an approximate path of the ray through the plates. We use formula (1) for the boundaries 2–3 and 3–1. In the answer we get

A) The sine of the angle of incidence of the beam on the boundary 2–3 between the plates is 2) ≈ 0.433;

B) The angle of refraction of the ray when crossing the boundary 3–1 (in radians) is 4) ≈ 0.873.

Answer. 24.

Determine how many α - particles and how many protons are obtained as a result of a thermonuclear fusion reaction

+ → x+ y;

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

+ → x + y;

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

Answer. 1 - α -particle; 2 - proton.

The modulus of the momentum of the first photon is 1.32 · 10 –28 kg · m / s, which is 9.48 · 10 –28 kg · m / s less than the modulus of the momentum of the second photon. Find the energy ratio E 2 / E 1 of the second and first photons. Round your answer to tenths.

Solution. The momentum of the second photon is greater than the momentum of the first photon by the condition, it means that we can represent p 2 = p 1 + Δ p(1). The energy of a photon can be expressed in terms of the momentum of a photon using the following equations. it E = mc 2 (1) and p = mc(2) then

E = pc (3),

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

Round the answer to tenths and get 8.2.

Answer. 8,2.

The atomic nucleus 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?

For each value, determine the corresponding change pattern:

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 an atomic nucleus occurs during the transformation of a proton into a neutron with the emission of a positron. As a result, 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.

In the laboratory, five experiments were carried out to observe diffraction using various diffraction gratings. Each of the gratings was illuminated with parallel beams of monochromatic light with a specific wavelength. In all cases, the light was incident perpendicular to the grating. In two of these experiments, the same number of main diffraction maxima were observed. First indicate 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 longer period was used.

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

d sinφ = kλ (1),

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

When choosing 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 long period was used is 2.

Answer. 42.

Current flows through the 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 value, determine the corresponding change pattern:

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 resistance of the conductor depends. The formula for calculating the resistance is

Ohm's law for a section of the circuit, from formula (2), we express the voltage

U = I R (3).

According to the condition 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 half the size. Substituting in (1) we get that the resistance increases by 2 times, and the current decreases by 2 times, therefore, the voltage does not change.

Answer. 13.

The period of oscillation of a mathematical pendulum on the surface of the Earth is 1, 2 times longer than the period of its oscillation on a certain planet. What is the modulus of acceleration of 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, the dimensions of which are much larger than the dimensions of the ball and the ball itself. Difficulty can arise if Thomson's formula for the period of oscillation of a mathematical pendulum is forgotten.

T= 2π (1);

l- the 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, through which a current of 3 A flows, is located in a uniform magnetic field with induction V= 0.4 T at an angle of 30 ° to the vector. What is the modulus of the force acting on the conductor from the side of the magnetic field?

Solution. If you place a conductor with current in a magnetic field, then the field on the conductor with current will act with the Ampere force. We write the formula for the modulus of the Ampere force

F A = I LB sinα;

F A = 0.6 N

Answer. F A = 0.6 N.

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

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

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

Then the ratio of currents

Answer. The current strength needs to be increased by 7 times. In the answer form, you enter only the number 7.

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

Solution. The magnetic induction lines leave the north pole of the magnet and diverge. As the magnet approaches, the magnetic flux through the coil of wire increases. According to Lenz's rule, the magnetic field created by the induction current of the loop must be directed to the right. According to the rule of the gimbal, the current should flow clockwise (if viewed from the left). A diode in the circuit of the second lamp passes in this direction. This means that the second lamp will light up.

Answer. The second lamp comes on.

Aluminum spoke length L= 25 cm and cross-sectional area S= 0.1 cm 2 suspended on a thread at the upper end. The lower end rests on the horizontal bottom of a vessel into which water is poured. Length of the submerged spoke l= 10 cm. Find the force F, with which the needle presses on the bottom of the vessel, if it is known that the thread is vertical. The density of aluminum ρ a = 2.7 g / cm 3, the 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 of reaction 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 is applied to the center of the entire spoke.

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

F a = Slρ in g (2)

Consider the moments of forces relative to the suspension point of the spoke.

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

M(N) = NL cosα is the moment of the reaction force of the support; (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 spoke presses on the bottom of the vessel, we write N = F e and from equation (7) we express this force:

Substitute the numerical data and get that

F d = 0.025 N.

Answer. F d = 0.025 N.

A container containing m 1 = 1 kg nitrogen, exploded in strength test at temperature t 1 = 327 ° C. What is the mass of hydrogen m 2 could be stored in such a container at a temperature t 2 = 27 ° C, having a fivefold safety factor? Molar mass of nitrogen M 1 = 28 g / mol, hydrogen M 2 = 2 g / mol.

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

where V- the volume of the cylinder, T 1 = t 1 + 273 ° C. By condition, hydrogen can be stored at pressure p 2 = p 1/5; (3) Taking into account that

we can express the mass of hydrogen by working directly with equations (2), (3), (4). The final formula is:

After substitution of numeric data m 2 = 28 g.

Answer. m 2 = 28 g.

In an ideal oscillatory circuit, the amplitude of the current fluctuations in the inductor I m= 5 mA, and the amplitude of the voltage across the capacitor U m= 2.0 V. At the 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 vibration energy is stored. For the moment of time t, the energy conservation law has the form

For the amplitude (maximum) values, we write

and from equation (2) we express

Substitute (4) in (3). As a result, we get:

I = I m (5)

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

I= 4.0 mA.

Answer. I= 4.0 mA.

There is a mirror at the bottom of the 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 ray 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 a rectangular ΔADB. In it AD = h, then DВ = АD

We get the following expression:

Substitute the numerical values ​​into the resulting formula (5)

Answer. 1.63 m.

In preparation for the exam, we suggest that you familiarize yourself with a working program in physics for grades 7–9 for the line of the UMK Peryshkina A.V. and working program of an in-depth level for grades 10-11 for the teaching materials Myakisheva G.Ya. The programs are available for viewing and free download for all registered users.

USE 2017. Physics. Typical test tasks. 25 options for tasks. Lukasheva E.V., Chistyakova N.I.

Moscow: 2017 - 280 p.

Typical test tasks in physics contain 25 options for sets of tasks, compiled taking into account all the features and requirements of the Unified State Exam in 2017. The purpose of the manual is to provide readers with information about the structure and content of 2017 control measuring materials in physics, as well as the degree of difficulty of the tasks. The collection provides answers to all test variants, as well as solutions to the most difficult problems in all 25 variants. In addition, there are samples of forms used on the exam. The team of authors is members of the Federal Subject Commission of the Unified State Examination in Physics. The manual is addressed to teachers to prepare students for the physics exam, and senior students for self-study and self-control.

Format: pdf

The size: 9.5 MB

Watch, download: drive.google

CONTENT
Work instructions 5
OPTION 1 10
Part 1 10
Part 2 16
OPTION 2 18
Part 1 18
Part 2 24
OPTION 3 26
Part 1 26
Part 2 32
OPTION 4 34
Part 1 34
Part 2 40
OPTION 5 42
Part 1 42
Part 2 48
OPTION 6 51
Part 1 51
Part 2 58
OPTION 7 60
Part 1 60
Part 2 66
OPTION 8 68
Part 1 68
Part 2 74
OPTION 9 76
Part 1 76
Part 2 82
OPTION 10 85
Part 1 85
Part 2 91
OPTION 11 93
Part 1 93
Part 2 99
OPTION 12 102
Part 1 102
Part 2 108
OPTION 13 111
Part 1 111
Part 2 118
OPTION 14 120
Part 1 120
Part 2 126
OPTION 15 128
Part 1 128
Part 2 134
OPTION 16 137
Part 1 137
Part 2 143
OPTION 17 .146
Part 1 146
Part 2 151
OPTION 18 154
Part 1 154
Part 2 159
OPTION 19162
Part 1 162
Part 2 168
OPTION 20 170
Part 1 170
Part 2 176
OPTION 21 178
Part 1 178
Part 2 185
OPTION 22 187
Part 1 187
Part 2 193
OPTION 23 196
Part 1 196
Part 2 203
OPTION 24 205
Part 1 205
Part 2 212
OPTION 25 214
Part 1 214
Part 2 220
ANSWERS. SCORING SYSTEM FOR EXAMINATION IN PHYSICS 223

• Problem 25, which was previously presented in part 2 as a task with a short answer, is now proposed for a detailed solution and is estimated at a maximum of 2 points. Thus, the number of tasks with a detailed answer increased from 5 to 6.
• For task 24, which checks the mastery of the elements of astrophysics, instead of choosing two obligatory correct answers, it is proposed to choose all correct answers, the number of which can be either 2 or 3.

The structure of the USE assignments in physics-2020

The examination paper consists of two parts, which include 32 tasks.

Part 1 contains 26 tasks.

• In tasks 1–4, 8–10, 14, 15, 20, 25–26, the answer is an integer or a final decimal fraction.
• The answer to tasks 5-7, 11, 12, 16-18, 21, 23 and 24 is a sequence of two numbers.
• The answer to problem 13 is a word.
• The answer to tasks 19 and 22 is two numbers.

Part 2 contains 6 tasks. The answer to tasks 27–32 includes a detailed description of the entire progress of the task. The second part of the tasks (with a detailed answer) are evaluated by an expert commission on the basis of.

Themes of the exam in physics, which will be 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 basics of SRT(electric field, direct current, magnetic field, electromagnetic induction, electromagnetic oscillations and waves, optics, basics of SRT).
4. Quantum physics and the elements of astrophysics(particle-wave dualism, physics of the atom, physics of the atomic nucleus, elements of astrophysics).

Duration of the exam in physics

All examination work is assigned 235 minutes.

The approximate time for completing tasks for various parts of the work is:

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

What can be taken for the exam:

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

Important!!! do not rely on cheat sheets, tips and the use of technical means (phones, tablets) during the exam. Video surveillance at the exam-2020 will be enhanced with additional cameras.

USE scores in physics

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

Total: 53 points(maximum primary score).

What you need to know when preparing assignments for the exam:

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

Where to start preparing for the exam in physics:

1. Learn the theory required for each assignment.
2. Train in physics test items designed based on