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

We disassemble USE assignments in physics (Option C) with a teacher.

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

The work presents tasks different levels difficulty: basic, advanced and high. Basic level tasks are simple tasks that test the assimilation of the most important physical concepts, models, phenomena and laws. Tasks increased 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 tasks of part 2 are tasks high level difficulties and tests 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 exam 2017, tasks are taken from open bank tasks of the exam.

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.

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

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, it is possible to determine the projection of the acceleration of the load

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

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 us 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 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 with acceleration upward. 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 lower 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 F cosα = F tr = 16 N; (2) then the power developed by the force will be equal to N = F cosα V(3) We 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 we express the mass m cargo.

=	T	;	m	=	T 2	; m = k	T 2	; m= 200 H / 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 lightweight blocks and a weightless cable, 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 movable block gives a twofold gain in strength, whereby the section of the rope must be pulled twice as long, and the stationary block is 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 stretch 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 cargo does not touch the walls and bottom of the vessel. Then an iron weight is immersed in the same vessel with water, the weight of which is equal to the weight 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 gravity force acting on the load does not change. Since the density of cargo is different, the volume will also be different.

V =	m	.
	p

The density of iron is 7800 kg / m 3, and the density of aluminum is 2700 kg / m 3. Consequently, 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 fluid 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 the 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

4) sinα -	a
	g 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. Choose a frame of reference 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 reference system 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's write down given equation(1) for force projection 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 the 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 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.

By definition F tr = μ N(7), we express the coefficient of friction of the bar 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. Oxygen gas is in a vessel with a volume of 33.2 liters. The gas pressure is 150 kPa, its temperature is 127 ° C. Determine the mass of 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 reducing 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); We write the change in the 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 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

According to the condition 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 substance in the liquid state was slowly cooled in a constant power melting furnace. 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, its internal energy decreased. The temperature measurement results allow you to determine the temperature at which the substance begins to crystallize. While the substance is passing from liquid state in solid, the temperature does not change. Knowing that the melting point and crystallization temperature are the same, we choose the statement:

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

The second true statement is:

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

For each value, determine the corresponding 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 there are no energy transformations 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.

∆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 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 induction vector magnetic field, as it shown on the picture. 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, thumb set back by 90 ° shows the direction of the Lorentz force acting on the particle. As a result, we have that the Lorentz force vector is directed away from the observer relative to the figure.

Answer. from the observer.

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 deals with 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

Pay 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 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 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

sinα	=	n 2	,
sinβ		n 1

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 the angles are measured 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 a 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 point in time t= 0.1 s, the change in magnetic flux through the circuit is 1 mVb.
2. Induction current in the jumper in the range from t= 0.1 s t= 0.3 s max.
3. Module EMF 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 lintel, 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 equal to 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 the law of refraction

sinβ =	sin50	= 0,4327 ≈ 0,433
	1,77

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 2–3 boundary 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 produced 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(one). The energy of a photon can be expressed in terms of the momentum of a photon using the following equations. This is 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:

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?

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 for light, 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 one, 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 a 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 gravitational acceleration 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 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 IN= 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 force of Ampere. 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 passed through it direct current, is equal to 120 J. How many times is it necessary to increase the strength of the current flowing through the coil winding in order for the stored energy of the magnetic field to increase by 5760 J.

Solution. The magnetic field energy 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 ratio of currents

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

Answer. The current strength must 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 when explaining.

Solution. The magnetic induction lines come out of north pole 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 (when 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 the vessel into which the 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;

- 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 - moment of 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

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 spoke presses on the bottom of the vessel, we write N = F e and from equation (7) we express this force:

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

Substitute 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 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). Final formula looks like:

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

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

C	U 2	+ L	I 2	= L	I m 2	(1)
	2		2		2

For the amplitude (maximum) values, we write

and from equation (2) we express

C	=	I m 2	(4).
L		U m 2

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 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

sinβ =	sinα	(3)
	n 2

Consider a rectangular ΔADB. In it AD = h, then DВ = АD

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

We get the following expression:

AC = 2 DB = 2 h	sinα	(5)

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.

On the official website of FIPI in the section "Analytical and methodological materials"published" Guidelines for teachers based on analysis typical mistakes participants of the USE 2017 ", it is here that you can find information about which average score The exam in physics was in 2017.

Download the document.

Table 1

Average USE score 2017 in physics

The USE in physics in 2017 was attended by 155,281 people, including 98.9% of the graduates of the current year. In percentage terms, the number of participants in the USE in physics has not changed and is about 24% of the total number of graduates of the current year.

The largest number of USE participants in physics is noted in Moscow (9943), the Moscow region (6745), St. Petersburg (5775), the Republic of Bashkortostan (5689) and the Krasnodar Territory (4869).

The average USE score in physics in 2017 was 53.16, which is higher than last year (50.02 test points).

The minimum USE score in physics in 2017, as in 2016, was 36 tb, which corresponded to 9 primary points. Percentage of exam participants who did not pass minimum score in 2017, amounted to 3.78%, which is significantly less than the share of participants who did not reach the minimum limit in 2016 (6.11%).

In comparison with the two previous years, in 2017, the share of unprepared and poorly trained participants (who scored up to 40 tb) decreased significantly.

The share of graduates showing average results (41–60 teb.) Remained practically unchanged, and the share of high-score students (81–100 te. B.) Increased, reaching the maximum values ​​in three years - 4.94%.

The maximum test score was scored by 278 exam participants, which is higher than in the previous two years.

The maximum primary score for a job is 50.

For the USE in physics, the range from 61 to 100 test points is also significant, which demonstrates the readiness of graduates to successfully continue their education in organizations. higher education... In 2017, this group of graduates increased significantly compared to the previous two years and amounted to 21.44%. These results indicate an improvement in the quality of teaching physics in specialized classes.

The Federal Service for Supervision in Education and Science has summed up the preliminary results of the 2017 USE in social science, literature and physics.

In the main period, the USE in social studies was taken by about 318 thousand participants, the USE in physics - more than 155 thousand participants, the USE in literature - more than 41 thousand participants. The average scores in all three subjects in 2017 are comparable to last year's results.

The number of USE participants who failed to overcome the established minimum threshold in subjects decreased: in social studies to 13.8% from 17.5% last year, in physics - to 3.8% from 6.1%, in literature - to 2 , 9% from 4.4% a year earlier.

“The average scores are comparable to the results of the previous year, which speaks about the stability of the exam and the objectivity of the assessment. It is important that the number of those who have not overcome the minimum thresholds is reduced. This is largely due to competent work with USE results when they are analyzed and used in the work of teacher training institutions. In a number of regions, the project "I will pass the exam" gave very serious results, - said the head of Rosobrnadzor Sergey Kravtsov.

Thanks to the use of scanning technology for the participants' work at the examination points, the USE results in social studies, literature and physics were processed ahead of the deadlines set by the schedule for issuing the results. Graduates will be able to find out their result a day earlier.

Analysis of the results of state (final) certification

in the form of a unified state examination (USE)

graduates of MBOU "Secondary School No. 6" NMR RT

in physics in 2017

The Unified State Exam (hereinafter - the Unified State Exam) is a form of objective assessment of the quality of training of persons who have mastered educational programs middle general education, using tasks of a standardized form (control measuring materials). The Unified State Exam is conducted in accordance with the Federal Law of December 29, 2012 No. 273-FZ "On Education in Russian Federation". Control measuring materials allow to establish the level of development by graduates of the Federal component of the state educational standard secondary (complete) general education in physics, basic and specialized levels.

The results of the unified state exam in physics are recognized educational organizations higher vocational education as results entrance examinations in physics.

When preparing for the exam, all work was aimed at organizing group work with students, with the aim of orienting the preparation of "weak" students to overcome the required minimum, as well as with the aim of orienting the preparation of "strong" students to work out complex topics, analysis of the criteria for checking tasks of an advanced and high level. To increase the efficiency of mastering the course of physics in the classroom, basic notes were used, containing an obligatory minimum of knowledge on a particular topic; used in her work demo versions, tasks of the open segment of the federal bank test items posted on the website "FIPI" regularly used the site "Reshu Unified State Examination". Also, in preparation for the exam, it was planned to repeat the knowledge and skills formed when studying the material in the main and high school... The main direction of work was the organization of independent learning activities on the implementation of specific tasks with a written fixation of the results, their further analysis. When solving KIM tasks, the students independently processed the information provided in the tasks, made inferences and argued them.

Each version of the examination work consists of two parts and includes 31 tasks, differing in form and level of difficulty (Table 1).

Part 1 contains 23 tasks with a short answer. Of these, 13 tasks with recording the answer in the form of a number, word or two numbers, 10 tasks for establishing correspondence and multiple choice, in which the answers must be written in the form of a sequence of numbers.

Part 2 contains 8 tasks combined general view activity - solving problems. Of these, 3 tasks with a short answer (24-26) and 5 tasks (27-31), for which it is necessary to give a detailed answer.

Table 1. Distribution of tasks of examination work by parts of work

In total, several plans are used to form the KIM USE 2017.

In part 1, to ensure a more accessible perception of information, tasks 1–21 are grouped based on the thematic assignment of tasks: mechanics, molecular physics, electrodynamics, the quantum physics... In part 2, tasks are grouped depending on the form of assignments and in accordance with thematic affiliation.

In the examin physics took part4 (22.2%) graduates.

Overcame the "threshold" in physics (the minimum number of points is 36) 4 out of 4 graduates (100% of the total number of those who passed the exam in physics).

The maximum USE score was - 62 (Nikolaeva Anastasia).

Unified State Exam in Physics iselective exam and is intended for differentiation upon admission to higher schools... For these purposes, the work includes tasks of three levels of complexity. Among the tasks of the basic level of complexity, tasks are distinguished, the content of which corresponds to the standard of the basic level. The minimum number of USE scores in physics (36 points), confirming the graduate's mastering of the secondary general education program in physics, is established based on the requirements for mastering the basic level standard.

Table 2 - Sections and topics of the examination work of the exam in physics

The result of the completed Unified State Exam assignments in physics by the graduates of MBOU "Secondary School No. 6" NMR RT in 2017

Analyzing the completed tasks of part 1 (1-24) KIM USE in PHYSICS of various levels of complexity, it can be noted that more than half of the graduates successfully cope with the taskswith a choice of answer bymechanics.

3 people out of 4 gave correct answers to tasks with a short answer (1).

The analysis data allow us to conclude that graduates are most successful in completing tasks of 2-4 basic levels of complexity, for the implementation of which it is necessary to know / understand the lawuniversal gravitation, Hooke's law, as well as the formula for calculating the friction force.

There was also a high percentage of completion of task 5 of the basic level of complexity (3 people out of 4), in which the assimilation of basic concepts was checked on the topics “Condition of balance solid"," Force of Archimedes "," Pressure "," Mathematical and spring pendulums "," Mechanical waves and sound ".

Task 7 was of an increased difficulty level, in which in different options it was required to establish a correspondence between graphs and physical quantities, between physical quantities and formulas, units of measurement. Nevertheless, more than half of the graduates successfully completed this task: 25% of the graduates scored 1 point, making one mistake, and 50% scored the primary 2 points, completing this task completely correctly.

Practically the same result was demonstrated by the graduates when performing task 6 of the basic level of complexity.

Bymolecular physics in part 1 of the KIM USE, 3 tasks were presented with the choice and recording of the number of the correct answer (8-10), for the correct execution of which 1 point was given. All students coped with task 8, in task 9 I made a mistake 1 out of 4. In addition, 2 tasks with a short answer (11-12) are presented, these are tasks for establishing correspondence and multiple choice, in which the answers must be written in the form sequences of numbers. The students showed the most successful performance when completing 11 assignments. In general, with tasks formolecular physics graduates did well.

Byelectrodynamics in part 1 of the KIM USE, 4 tasks were presented with the choice and recording of the number of the correct answer (13-16), for the correct execution of which 1 point was given. In addition, there are 2 tasks with a short answer (17-18), these are tasks for establishing correspondence and multiple choice, in which the answers must be written in the form of a sequence of numbers.

The analysis data allow us to conclude that, in general, the graduates performed the tasks in electrodynamics much worse than similar tasks in mechanics and molecular physics.

The most difficult task for the graduates turned out to be task 13 of the basic level of complexity, in which their ideas aboutelectrization of bodies, behavior of conductors and dielectrics in an electric field, the phenomenon of electromagnetic induction, light interference, diffraction and dispersion of light.

The most successful graduates completed task 16 of the basic level of complexity, for the implementation of which it is necessary to have an idea of ​​Faraday's law of electromagnetic induction, the oscillatory circuit, the laws of reflection and refraction of light, the course of rays in the lens (75 %).

Task 18 of an increased level of complexity, in which in different versions it was required to establish a correspondence between graphs and physical quantities, between physical quantities and formulas, units of measurement, the graduates performed no worse than a similar task in mechanics and molecular physics.

Byquantum physics in part 1 of the KIM USE, 3 tasks were presented with the choice and recording of the number of the correct answer (19-21), for the correct execution of which 1 point was given. In addition, 1 task with a short answer is presented (22). The highest percentage of completion (2 people out of 2) was in the case of task 20 of the basic level of complexity, which tested the knowledge of graduates on the topics "Radioactivity", " Nuclear reactions"And" Fission and fusion of nuclei ".

Most of the students (3 out of 4) did not start and did not get primary points when completing tasks with a detailed answer (part C).

However, it should be noted that there were no students who would have coped successfully (by 3 maximum points) with at least one task. This is explained by the fact that physics is studied at school on basic level, and these assignments presuppose mainly specialized training in the subject.

Students showed average level preparation for the exam in physics. The presented data indicate that in part 1 of the KIM USE in physics, graduates performed much better tasks in mechanics and molecular physics than in electrodynamics and quantum physics.

Many students did not realize that the new assessment criteria for assignments require explanations for each formula for solving these problems.

Use the results of the analysis to prepare for the Unified State Exam - 2018.

To form in students the skills specified in the educational standard as the main goals in teaching physics:

Explain correctly physical phenomena;

Establish connections between physical quantities;

Give examples of confirmation of fundamental laws and their consequences.

4. Use the laws of physics to analyze phenomena at the qualitative and calculated levels.

5. Make calculations based on the data presented in graphical or tabular forms.

Physics teacher __________________ / Mochenova O.V. /

Year. The officials did not pass by and Unified State Exam in Physics... 2017 will bring several innovations to this exam that can affect the overall performance of students and reveal the true picture of their knowledge.

The main amendment is the exclusion of the test part. It should be noted that this innovation will occur not only in the physics exam, but also in many others (history, literature, chemistry).

The main changes in the exam-2017

Several months ago it became known that the deputies are seriously thinking about adding another discipline to the list of compulsory subjects submitted for the Unified State Exam. In total, their total number will increase to three.

Until 2017, students at the end passed the Russian language and mathematics, as well as additional subjects necessary for admission to a university for a specific specialty. Starting next year, it claims to be a compulsory subject in the first place.

Civil servants, from whose submission the above-mentioned innovations were made, justify their action by the fact that at present too few students have decent knowledge in the field of national and world history. Few of them are interested in the past and do not know what their ancestors lived and how they “built” the state. According to them, such a trend cannot be called positive, and if appropriate measures are not taken, very few worthy educated citizens will remain in the country soon.

What will change in the physics exam?

Let's take a look at the exam in physics. This item will not receive any special amendments. The only thing to pay attention to is the exclusion of the test block. It is planned to replace it with an oral and written response. It's too early to talk about any specific details on this issue, exactly the same as what can be included in the tasks submitted for the exam.

As for the cancellation of the test part, it is worth noting that the officials came to this decision not overnight. Over the course of many months, Rosobrnadzor held heated negotiations on the feasibility of this amendment. All pros and cons have been weighed and carefully negotiated.

Ultimately, as we can see, it was decided to implement the oral part in many final tests. The most important advantage of this approach to checking knowledge is the exclusion of guessing, or, as the people say, “the poke method”. Simply put, now you won't be able to count on “maybe you're lucky” and put the answer at random. In turn, written and oral answers of the student will be able to show the examiner his level of education, as well as the ability to learn.

Date of the exam

There is not much time left before the start of the tests, so you can already familiarize yourself with the official schedule. So, the USE in physics in 2017 will be held on the following dates:

• The early period is March 22 (Wednesday). Reserve day - April 5.
• The main period is June 13 (Tuesday). Reserve day - June 20.

Importance of the exam in Russia in the future

Note that in the next few years, the procedure for conducting the Unified State Exam in Russia will change radically. The test part will be removed in all subjects and this is not the limit.

Closer to 2022, Rosobrnadzor intends to expand the list of compulsory disciplines to four. Most likely, it will become a foreign language, because in our time knowledge, for example, of English language incredibly appreciated and gives the chance to apply for a prestigious high-paying position.

In addition to English, it will be possible to take German, French and Spanish.

You can already guess what education in the Russian Federation will be like in the future. IN currently even an ordinary person can see that the world and trends in it are changing every day. What was previously insignificant comes to the fore. IN modern society networking skills and diplomacy are incredibly prized.

To maintain business relationships with people of another nation, fluency in several languages ​​is required. Only by communicating with a person in his native dialect will it be possible to establish a close, trusting relationship. Actually, for this, already now in Russian schools, great attention is paid to foreign languages and their study among students.

How to prepare for the exam

Given the fact that physics is a complex subject and cannot be on a par with the Russian language or literature, eleventh graders should devote a little more time to it than to the rest of the subject. This is due to the fact that understanding a particular topic can drag on for a long time, and without understanding about good result you can forget on the exam. In addition, if you want to enroll in prestigious university, knowledge in the field of physics is extremely important.

It is worth noting that there is a category of people who claim that the USE will be canceled in 2017. No need to mislead yourself and others - there will be no cancellation! And the next 5-6 years, you can only dream of something like that. Besides, what can you exchange such an exam for? Despite all its rigor, the Unified State Exam is able to show the real level of knowledge and preparation of the student for adult student life.

Where to get knowledge from?

You can prepare for the exam in physics according to the same principle by which you plan to prepare for other subjects. First of all, of course, you should pay attention to educational materials: books and reference books. While studying at school, the teacher is obliged to give a huge amount of knowledge, which can later be used on. The main thing is to listen carefully to the teacher, ask again and understand the essence of the material presented.

Stock up on a collection of basic physical formulas so that this part of the exam does not become intimidating for you. Another tool for preparing for the exam in physics can be a collection of problems. Various problems with solutions are printed in it, which can be used as training. Of course, there will be completely different tasks on the test, but having gotten a hand in solving physical problems, examination paper will not seem so difficult to you.

You can start going to a tutor, as well as study on the Internet on your own. Now there are a lot of online resources with the help of which, you can understand how the physics exam will actually be held.

Any of your efforts will once again prove that at this stage of your life the main thing for you is study and you will do everything to make it successful!

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