Physics is a rather complicated subject, so the preparation for the exam in physics 2020 will take a sufficient amount of time. In addition to theoretical knowledge, the Commission will check the skill to read the charts of the scheme, solve the tasks.

Consider the structure of the examination work

It consists of 32 tasks distributed over two blocks. To understand, it is more convenient to arrange all the information in the table.

All the theory of exam in physics on sections

• Mechanics. This is a very large, but relatively simple section that studies the movement of bodies and occurring interaction between them, which includes the dynamics and kinematics, the laws of conservation in mechanics, statics, oscillations and waves of mechanical nature.
• Physics molecular. In this topic, special attention is paid to thermodynamics and molecular kinetic theory.
• Quantum physics and components of astrophysics. These are the most complex sections that cause difficulties in both studies and during testing. But, perhaps, one of the most interesting sections. Knowledge of such topics as physics of the atom and atomic nucleus, corpuscular-wave dualism, astrophysics are checked here.
• Electrodynamics and specialty of relativity. It is not necessary to do without studying the optics, the basics of a hundred, you need to know how the electrical and magnetic field acts, what is the constant current, what are the principles of electromagnetic induction, as electromagnetic oscillations and waves occur.

Yes, there are many information, the volume is very decent. In order to successfully pass the exam in physics, you need to speak very well to the whole school course on the subject, and it is studied for five years. Therefore, in a few weeks or even a month, it will not be possible to prepare for this exam. You need to start now, so that during testing feel calm.

Unfortunately, the subject of the physics causes difficulties in very many graduates, especially those who chose it as profile objects for admission to the university. Effective study of this discipline has nothing to do with the combaling rules, formulas and algorithms. In addition, to learn physical ideas and read as much theory as possible, it is not enough to own mathematical technique. Often, unimportant mathematical preparation does not give a schoolboy to pass well to physics.

How to prepare?

Everything is very simple: Choose theoretical section, carefully read it, study, trying to understand all physical concepts, principles, postulates. After that, reinforce the preparation by solving practical tasks on the selected topic. Use online tests to check your knowledge, this will allow you to immediately understand where you make mistakes and get used to the fact that a certain time is given to the solution of the problem. We wish you good luck!

Preparation for OGE and EGE

Secondary education

Line Ukk A. V. Gracheva. Physics (10-11) (bases., Condition)

Line Ukk A. V. Gracheva. Physics (7-9)

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

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

We disassemble the tasks of the exam in physics (Option C) with the teacher.

Lebedeva Alevtina Sergeevna, teacher of physics, work experience of 27 years. Honorary Mission of the Ministry of Education of the Moscow Region (2013), the gratitude of the head of the Voskresensky Municipal District (2015), the graduate of the President of the Mathematics and Physics Association of Mathematics and Physics (2015).

The paper presents the tasks of different levels of complexity: basic, elevated and high. Tasks of the baseline, these are simple tasks that check the assimilation of the most important physical concepts, models, phenomena and laws. The tasks of the elevated level are aimed at checking the ability to use the concepts and laws of physics for analyzing various processes and phenomena, as well as the ability to solve the tasks for the application of one or two laws (formulas) for any of the school courses of physics. In the work of 4 tasks of part 2 are the tasks of a high level of complexity and check the ability to use the laws and theory of physics in a modified or new situation. Performing such tasks requires the use of knowledge at once from two three sections of physics, i.e. High level training. This option is fully consistent with the demo version of the EGE 2017, the tasks are taken from the open bank of the tasks of the USE.

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

Decision. The path passed by the car in the time interval from 0 to 30 with the easiest way to determine as the area of \u200b\u200bthe trapez, the bases of which are the intervals of time (30 - 0) \u003d 30 C and (30 - 10) \u003d 20 s, and the speed is the height v. \u003d 10 m / s, i.e.

S. =	(30 + 20) from	10 m / s \u003d 250 m.
	2

Answer. 250 m.

A weight of 100 kg weighs up vertically up using a cable. The figure shows the dependence of the speed projection V. cargo on the axis directed upwards t.. Determine the cable tension force module during the lifting.

Decision. According to the chart of the drug projection v. cargo on the axis directed upright upwards t., you can define the projection of the acceleration of cargo

a. =	∆v.	=	(8 - 2) m / s	\u003d 2 m / s 2.
	∆t.		3 S.

The load is valid: the force of gravity, directed vertically down and the force of tensioning the cable, directed along the cable vertically, look up. 2. We write the main equation of speakers. We use the second law of Newton. The geometric sum of the forces acting on the body is equal to the product of the body mass on the acceleration reported to it.

+ = (1)

We write the equation for the projection of vectors in the land-related reference system, the Oy axis will send up. The projection of the tension force is positive, as the direction of force coincides with the axis direction of Oy, the projection of gravity is negative, as the vector of force is oppositely directed by the Oy axis, the projection of the acceleration vector is also positive, so the body moves with acceleration up. Have

T. – mG. = mA. (2);

from formula (2) module of tension force

T. = m.(g. + a.) \u003d 100 kg (10 + 2) m / s 2 \u003d 1200 N.

Answer. 1200 N.

The body drains on a rough horizontal surface with a constant speed of the module of which is 1, 5 m / s, applying force to it as shown in Figure (1). In this case, the module of the fiction force acting on the body is 16 N. What is equal to the power developed by force F.?

Decision. Imagine the physical process specified in the condition of the problem and make a schematic drawing with the indication of all forces acting on the body (Fig. 2). We write the main equation of speakers.

Tr + + \u003d (1)

By choosing a reference system associated with a fixed surface, write the equations for the projection of vectors on the selected coordinate axes. Under the condition of the problem, the body moves evenly, since its speed is constant and is equal to 1.5 m / s. This means, the acceleration of the body is zero. Horizontal on the body there are two forces: the force of friction slip TR. And the force with which the body is dragging. Projection of friction force negative, as the strength vector does not coincide with the direction of the axis H.. Projection of Power F. Positive. We remind you to find the projection by omit perpendicular from the beginning and end of the vector to the selected axis. With this, we have: F. COSα - F. Tr \u003d 0; (1) Express the projection of force F., this is F.cosα \u003d F. Tr \u003d 16 N; (2) Then the power developed by force will be equal to N. = F.cOSα. V. (3) We will make a replacement, considering equation (2), and substitut the relevant data to equation (3):

N. \u003d 16 N · 1.5 m / s \u003d 24 W.

Answer. 24 W.

Cargo fixed on a light spring with stiffness 200 n / m performs vertical oscillations. The figure shows a graph of displacement x. cargo from time t.. Determine what is equal to the mass of cargo. Answer round up to an integer.

Decision. The load on the spring performs vertical oscillations. On the schedule of the dependence of the shipment of cargo h. from time t., I will define the period of cargo oscillations. The period of oscillations is equal T. \u003d 4 s; from formula T. \u003d 2π express a lot m. cargo.

=	T.	;	m.	=	T. 2	; m. = k.	T. 2	; m. \u003d 200 h / m	(4 s) 2	\u003d 81.14 kg ≈ 81 kg.
	2π.		k.		4π 2.		4π 2.		39,438

Answer: 81 kg.

The figure shows a system of two light blocks and a weightless cable, with which you can hold in equilibrium or lift the load weighing 10 kg. The friction is negligible. Based on the analysis of the given pattern, select twofine allegations and indicate their numbers in response.

1. In order to keep the cargo in equilibrium, you need to act on the end of the rope with force of 100 N.
2. The blocks depicted in the figure does not give a winner.
3. h., you need to pull the rope length 3 h..
4. In order to slowly raise the load on the height h.h..

Decision. In this task, it is necessary to recall simple mechanisms, namely blocks: movable and stationary block. The movable block gives the winnings in force twice, while the area of \u200b\u200bthe rope should be pulled out twice as long, and the fixed block is used to redirect strength. In the work, simple winning mechanisms do not give. After analyzing the task, we immediately choose the necessary allegations:

1. In order to slowly raise the load on the height h., you need to pull the rope length 2 h..
2. In order to keep the cargo in equilibrium, you need to act on the end of the rope with force of 50 N.

Answer. 45.

In the vessel with water completely immersed aluminum cargo, fixed on the weightless and unpretentious thread. The cargo does not concern the walls and the bottom of the vessel. Then in the same vessel with water immerses the railway, the mass of which is equal to the mass of aluminum cargo. How as a result of this, the thread tension force module and the module of gravity acting on the load?

1. Increases;
2. Decreases;
3. Does not change.

Decision. We analyze the condition of the problem and allocate those parameters that do not change during the study: this is the mass of the body and the liquid into which the body is immersed on the thread. After that, it is better to perform a schematic drawing and indicate the force acting into the cargo: the thread of the thread F. UPR, directed along the thread up; gravity, directed vertically down; Archimedean Power a. , acting on the side of the liquid on the immersed body and directed upwards. By the condition of the problem, the mass of goods is the same, therefore, the module of the current force of gravity does not change. As the density of goods is different, the volume will also be different

V. =	m.	.
	p.

Iron density 7800 kg / m 3, and aluminum cargo 2700 kg / m 3. Hence, V. J.< V A.. The body in equilibrium, which is equal to all forces acting on the body is zero. Let's send a coordinate axis oy up. The main equation of dynamics, taking into account the projection of the forces we write down in the form F. UPR +. F A. – mG. \u003d 0; (1) Express the tension force F. UPR \u003d mG. – F A. (2); Archimedean force depends on the density of liquid and the volume of the immersed part of the body F A. = ρ gVp.Ch.T. (3); The density of the fluid does not change, and the volume of the body of iron is less V. J.< V A.So the Archimedean force acting on the railway will be less. We conclude a thread tension module, working with equation (2), it will increase.

Answer. 13.

Bar mass m. Slinds with a fixed rough rubber plane with an angle α at the base. The acceleration module of BROsa is equal a., Brawn speed module increases. Air resistance can be neglected.

Install the correspondence between physical quantities and formulas with which they can be calculated. To each position of the first column, select the appropriate position from the second column and write the selected numbers in the table under the appropriate letters.

B) friction coefficient Bruck about inclined plane

3) mG. COSα.

4) sinα -	a.
	g.cOSα.

Decision. This task requires the application of Newton's laws. We recommend to make a schematic drawing; Specify all the kinematic characteristics of the movement. If possible, portray the speed of acceleration and the vectors of all the forces applied to the moving body; Remember that the forces acting on the body are the result of interaction with other bodies. Then write the basic equation of speakers. Select the reference system and write the resulting equation for the projection of the forces and acceleration vectors;

Following the proposed algorithm, we will make a schematic drawing (Fig. 1). The figure shows the forces attached 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 the forces are constant, then the might of the bar will be equally referred with increasing speed, i.e. The speed of the acceleration is directed towards the movement. Choose axes directions as indicated in the figure. We write the projection forces on the selected axes.

We write the main dynamics equation:

Tr + \u003d (1)

We write this equation (1) for the projection of forces and acceleration.

On the OY axis: the projection of the reaction force is positive, as the vector coincides with the axis direction of Oy N y. = N.; The projection of the friction force is zero as the vector is perpendicular to the axis; The projection of gravity will be negative and equal mG Y.= – mG.cOSα; Projection of the acceleration vector a Y. \u003d 0, since the spelling vector is perpendicular to the axis. Have N. – mG.cOSα \u003d 0 (2) From the equation, we will express the reaction force of the reaction to the bar, from the side of the inclined plane. N. = mG.cOSα (3). We write projections on the OX axis.

On the OX axis: Projection of Power N. equal to zero, since the vector is perpendicular to the axis oh; The projection of the friction force is negative (vector is directed in the opposite direction relative to the selected axis); The projection of gravity is positive and equal mG X. = mG.sINα (4) from a rectangular triangle. Acceleration projection positive a X. = a.; Then equation (1) write down the projection mG.sINα - F. Tr \u003d. mA. (5); F. Tr \u003d. m.(g.sINα - a.) (6); Remember that the friction force is proportional to the strength of normal pressure N..

A-priory F. Tr \u003d μ. N. (7), we express the friction coefficient of Bruck about the inclined plane.

μ =	F. Tr.	=	m.(g.sINα - a.)	\u003d TGα -	a.	(8).
	N.		mG.cOSα.		g.cOSα.

Select the corresponding positions for each letter.

Answer. A - 3; B - 2.

Task 8. Gaseous oxygen is in a volume vessel with a volume of 33.2 liters. Gas pressure 150 kPa, its temperature is 127 ° C. Determine the gas mass in this vessel. Answer express in grams and round up to an integer.

Decision. It is important to pay attention to the translation of units into the SI system. Temperature translate to Kelvin T. = t.° C + 273, volume V. \u003d 33.2 l \u003d 33.2 · 10 -3 m 3; Pressure Translate P. \u003d 150 kPa \u003d 150 000 PA. Using the ideal gas equation

express gas mass.

We definitely pay attention to which unit is asked to write down the answer. It is very important.

Answer. 48

Task 9. The ideal single-variable gas in the amount of 0.025 mol adiabatically expanded. In this case, its temperature dropped from + 103 ° C to + 23 ° C. What kind of work made gas? Answer express in Joules and round up to an integer.

Decision. First, the gas is a single andomic number of degrees of freedom i. \u003d 3, secondly, gas expands adiabatically - it means without heat exchange Q. \u003d 0. Gas makes work by reducing internal energy. Taking into account this, the first law of thermodynamics will be recorded in the form 0 \u003d Δ U. + A. r; (1) Express the operation of the gas A. r \u003d -δ. U. (2); Changing the internal energy for single-variable gas Write as

Answer. 25 J.

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

Decision. Questions related to a saturated ferry and air humidity, most often cause difficulties from schoolchildren. We use the formula for calculating the relative humidity

Under the condition of the problem, the temperature does not change, it means that the pressure of the saturated steam remains the same. We write formula (1) for two air condition.

φ 1 \u003d 10%; φ 2 \u003d 35%

Express air pressure from formulas (2), (3) and find the reference ratio.

P. 2	=	φ 2.	=	35	= 3,5
P. 1		φ 1.		10

Answer. Pressure should be increased by 3.5 times.

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

Choose from the proposed list two Approvals that meet the results of the measurements and specify their numbers.

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

Decision. Since the substance was cooled, its internal energy decreased. The results of temperature measurement, allow to determine the temperature at which the substance begins to crystallize. So far, the substance moves from a liquid state into solid, the temperature does not change. Knowing that the melting point and crystallization temperature are the same, choose the assertion:

1. The temperature of the melting of the substance under these conditions is equal to 232 ° C.

The second right statement is:

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

Answer.14.

In an isolated system, the body A has a temperature of + 40 ° C, and the body B is a temperature of + 65 ° C. These bodies led to a thermal contact with each other. After a while there was a thermal equilibrium. As a result, the body temperature used changed and the total internal energy of the body A and B?

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

1. Increased;
2. Decreased;
3. Not changed.

Record the selected numbers in the table for each physical value. Figures in response can be repeated.

Decision. If no energy transformations occurs in an isolated system of bodies, except heat exchange, the amount of heat, given by bodies, the internal energy of which decreases, is equal to the amount of heat obtained by the 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. The tasks of this type are solved on the basis of the thermal balance equation.

∆U \u003d Σ.	N.	∆U i \u003d.0 (1);
	i. = 1

where Δ. U. - Change in internal energy.

In our case, as a result of heat exchange, the internal energy of the body B decreases, which means that the temperature of this body decreases. The internal energy of the body is increasing, as the body received the amount of heat from the body B, then the temperature will increase it. The total internal energy of bodies A and B does not change.

Answer. 23.

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

Decision. On the charged particle, the magnetic field acts with the force of Lorentz. In order to determine the direction of this force, it is important to remember the mnemonic rule of the left hand, do not forget to consider the particle charge. Four fingers of the left hand we guide the velocity vector, for a positively charged particle, the vector must perpendicular to the palm, the thumb replied 90 ° shows the direction of Lorentz acting on a particle. As a result, we have that Lorentz's strength vector is directed from the observer regarding the picture.

Answer. from the observer.

The electrical field strength module in a flat air capacitor with a capacity of 50 μF is 200 V / m. The distance between the plates of the condenser is 2 mm. What is the charge of the condenser? Record write to the ICR.

Decision. We translate all units of measurement to the SI system. Capacity C \u003d 50 μF \u003d 50 · 10 -6 F, distance between plates d. \u003d 2 · 10 -3 m. The problem refers to a flat air capacitor - a device for the accumulation of electrical charge and electric field energy. Formula of electrical capacity

where d. - Distance between the plates.

Express tension U. \u003d E · d.(four); Substitute (4) in (2) and calculate the charge of the condenser.

q. = C. · ED\u003d 50 · 10 -6 · 200 · 0.002 \u003d 20 μKl

We pay attention to which units you need to record the answer. Received in the coulons, but we present to the ICR.

Answer. 20 μKl.

The student spent experience in the refraction of the light, presented in the photo. How does it change when increasing the angle of incidence of the refractive area spreading in glass and the refractive index of glass?

1. Increases
2. Decreases
3. Does not change
4. Write down the selected numbers for each response into the table. Figures in response can be repeated.

Decision. In the tasks of such a plan, you remember what refraction. This is a change in the direction of wave propagation when passing from one environment to another. It is caused by the fact that the velocities of the propagation of waves in these environments are different. Having understood from which environment to what light it applies to, write down the law of refraction in the form of

sINα.	=	n. 2	,
sINβ.		n. 1

where n. 2 - an absolute refractive index of glass, Wednesday where there is light; n. 1 - Absolute refractive index of the first environment, where the light comes from. For air n. 1 \u003d 1. α is an angle of falling the beam on the surface of a glass half-cylinder, β is the beam refractive angle in glass. Moreover, the refractive angle will be less than the angle of fall, as the glass is optically more dense medium with a large refractive index. The speed of propagation of light in the glass is smaller. We draw attention to that the angles measure from the perpendicular restored at the point of the fall of the beam. If you increase the angle of falling, then the refractive angle will grow. The refractive index of glass will not change from this.

Answer.

Copper jumper at time t. 0 \u003d 0 Begins to move with a speed of 2 m / s along the parallel horizontal conductive rails, to the ends of which the resistor resistance is connected to 10 ohms. The whole system is in a vertical homogeneous magnetic field. The resistance of the jumper and rails are negligible, the jumper all the time is perpendicular to the rails. The flow of the magnetic induction vector through the circuit formed by the jumper, rails and the resistor, changes over time t. So, as shown in the graph.

Using a schedule, select two true statements and indicate in response their numbers.

1. By the time t. \u003d 0.1 C change of magnetic flux through the contour is 1 MVB.
2. Induction current in the jumper in the interval from t. \u003d 0.1 C. t. \u003d 0.3 s Maximum.
3. The emf induction module arising in the circuit is 10 mV.
4. The power of the induction current flowing in the jumper is 64 mA.
5. To maintain the movement of the jumper to it apply force, the projection of which on the direction of rails is 0.2 N.

Decision. According to a graph of the magnetic induction vector dependence through the contour, we define the sections where the flow F is changing, and where the flow change is zero. This will allow us to determine the time intervals in which the induction current will occur in the circuit. True statement:

1) by the time of time t. \u003d 0.1 C change of magnetic flux through the circuit is 1 MVB ΔF \u003d (1 - 0) · 10 -3 WB; Module EMF induction arising in the circuit Determine using the AM law

Answer. 13.

According to the flow rate of the current from time to time in the electrical circuit, the inductance of which is 1 mpn, define the self-induction EMF module in the time range from 5 to 10 s. Record write to the MKV.

Decision. We translate all the values \u200b\u200binto the SI system, i.e. The inductance of 1 MGN translates into GNs, we obtain 10 -3 Gn. The current strength shown in the figure in Ma will also be translated into A by multiplying the value of 10 -3.

Formula EMF self-induction has the form

at the same time, the time interval is given by the condition of the problem

∆t.\u003d 10 C - 5 C \u003d 5 C

seconds and on schedule determine the current change interval during this time:

∆I.\u003d 30 · 10 -3 - 20 · 10 -3 \u003d 10 · 10 -3 \u003d 10 -2 A.

We substitute numeric values \u200b\u200bin formula (2), we get

| Ɛ | \u003d 2 · 10 -6 V, or 2 μV.

Answer. 2.

Two transparent plane-parallel plates are tightly pressed to each other. From the air to the surface of the first plate there is a beam of light (see Figure). It is known that the refractive index of the upper plate is equal n. 2 \u003d 1.77. Set the correspondence between physical values \u200b\u200band their values. To each position of the first column, select the appropriate position from the second column and write the selected numbers in the table under the appropriate letters.

Decision. To solve problems about the refractiveness of light on the border of the section of two media, in particular tasks for the passage of light through the plane-parallel plates, you can recommend the following procedure for the solution: make a drawing with the progress of the rays that run out of one environment to another; At the fall point of the beam on the border of the section of two environments, it is normal to the surface, mark the angles of drop and refraction. Especially pay attention to the optical density of the media under consideration and remember that when moving the beam of light from an optically less dense medium in an optically more dense medium, the refractive angle will be less than the angle of the fall. The figure is given an angle between the incident beam and the surface, and we need an angle of falling. Remember that the angles are determined from the perpendicular restored at the fall point. We define that the angle of falling the beam to the surface 90 ° - 40 ° \u003d 50 °, the refractive index n. 2 = 1,77; n. 1 \u003d 1 (air).

We write the law of refraction

sinβ \u003d	sIN50.	= 0,4327 ≈ 0,433
	1,77

We construct an approximate course of the beam through the plates. Use formula (1) for border 2-3 and 3-1. In response, get

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

B) the refractive angle of the beam in the transition of the boundary 3-1 (in radians) is 4) ≈ 0.873.

Answer. 24.

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

+ → x.+ y.;

Decision. With all nuclear reactions, the laws of conservation of the electrical charge and the number of nucleons are observed. Denote by x - the amount of alpha particles, y- the number of protons. Make an equation

+ → x + y;

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

Answer. 1 - α-partition; 2 - proton.

The first photon impulse module is 1.32 · 10 -28 kg · m / s, which is 9.48 · 10 -28 kg · m / s less than the pulse module of the second photon. Find the energy ratio of E 2 / E 1 second and first photons. Answer round up to tenths.

Decision. The pulse of the second photon is greater than the impulse of the first photon by condition means you can imagine p. 2 = p. 1 + Δ. p. (one). The photon energy can be expressed through the photon pulse using the following equations. it E. = mC. 2 (1) and p. = mC. (2), then

E. = pC. (3),

where E. - photon energy, p. - Photon pulse, m - photon mass, c. \u003d 3 · 10 8 m / s - speed of light. With the formula (3) we have:

E. 2	=	p. 2	= 8,18;
E. 1		p. 1

The answer is round to the tenth and get 8.2.

Answer. 8,2.

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

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

1. Increased;
2. Decreased;
3. Not changed.

Record the selected numbers in the table for each physical value. Figures in response can be repeated.

Decision. Positron β - the decay in the atomic core occurs when the proton transforms into the neutron with the emission of the positron. As a result, the number of neutrons in the nucleus increases by one, the electrical charge decreases by one, and the mass number of the kernel remains unchanged. Thus, the transformation reaction of the element is as follows:

Answer. 21.

In the laboratory, five experiments were carried out on the observation of diffraction with various diffraction gratings. Each of the lattices was illuminated by parallel bunches of monochromatic light with a certain wavelength. Light in all cases fell perpendicular to the grid. In two of these experiments, the same number of major diffraction maxima was observed. Specify the first number of the experiment in which the diffraction grille with a smaller period was used, and then the experiment number in which the diffraction lattice was used with a large period.

Decision. The diffraction of light is called the phenomenon of the light beam to the area of \u200b\u200bthe geometric shadow. The diffraction can be observed in the case when opaque areas or holes in large in size and opaque obstacles are found on the path of the light wave, and the size of these sections or holes is commensurate with a wavelength. One of the most important diffraction devices is a diffraction grating. The angular directions on the maxima of the diffraction pattern are determined by the equation

d.sinφ \u003d. k. λ (1),

where d. - the period of the diffraction lattice, φ is the angle between the normal to the lattice and the direction on one of the maxima of the diffraction pattern, λ is the length of the light wave, k. - an integer called a diffraction maximum. Express from equation (1)

Selecting the pairs according to the experimental condition, select first 4 where the diffraction grid was used with a smaller period, and then the experiment number in which the diffraction lattice was used with a large period is 2.

Answer. 42.

For wire resistor flows current. The resistor was replaced on another, with a wire from the same metal and the same length, but having a smaller cross-sectional area, and they missed a smaller current through it. How do the voltage on the resistor and its resistance change?

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

1. Will increase;
2. Will decrease;
3. Will not change.

Record the selected numbers in the table for each physical value. Figures in response can be repeated.

Decision. It is important to remember which values \u200b\u200bdepends on the resistance of the conductor. The formula for calculating the resistance is

ohm's law for the chain section, from formula (2), we will express the tension

U. = I R. (3).

By the condition of the problem, the second resistor is made of wire of the same material, the same length, but of different cross-sectional areas. The area is twice as smaller. Substituting in (1) we obtain that the resistance increases by 2 times, and the current power decreases by 2 times, therefore, the voltage does not change.

Answer. 13.

The period of oscillations of the mathematical pendulum on the surface of the Earth in 1, 2 times the period of its oscillations on some planet. What is the fluency acceleration module on this planet? The effect of the atmosphere in both cases is negligible.

Decision. The mathematical pendulum is a system consisting of a thread, the size of which is much more than the size of the ball and the ball itself. The difficulty may arise if the Thomson formula is forgotten for the oscillation period of the mathematical pendulum.

T. \u003d 2π (1);

l. - the length of the mathematical pendulum; g. - acceleration of gravity.

By condition

Express from (3) g. n \u003d 14.4 m / s 2. It should be noted that the acceleration of free fall depends on the mass of the planet and radius

Answer. 14.4 m / s 2.

Straight conductor with a length of 1 m, according to which the current flow 3 A is located in a homogeneous magnetic field with induction IN \u003d 0.4 TL at an angle of 30 ° to the vector. What is the module of force acting on the conductor from the magnetic field?

Decision. If in the magnetic field, place the conductor with a current, then the field on the conductor with the current will act with the force of the ampere. We write ampere power module formula

F. A \u003d. I LB.sINα;

F. A \u003d 0.6 n

Answer. F. A \u003d 0.6 N.

The energy of the magnetic field, stored in the coil when the DC passes through it is 120 J. Which time you need to increase the strength of the current flowing through the coil winding, in order to store the magnetic field energy in it increased by 5760 J.

Decision. The magnetic field of the coil is calculated by the formula

W. M \u003d	LI 2	(1);
	2

By condition W. 1 \u003d 120 J, then W. 2 \u003d 120 + 5760 \u003d 5880 J.

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

Then the attitude of currents

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

Answer. Current strength should be increased 7 times. In the answer blank, you only make a digit 7.

The electrical circuit consists of two light bulbs, two diodes and a wire of the wire connected, as shown in the figure. (Diode passes the current only in one direction, as shown on the top of the figure). Which of the lights will light up if the north pole of the magnet is brought to the turn? The answer explain, indicating which phenomena and patterns you used with the explanation.

Decision. The magnetic induction lines leave the north pole of the magnet and diverge. When the magnet approaches the magnetic flow through the coil of the wire increases. In accordance with the Lenza rule, the magnetic field created by the induction current of the cooler must be directed to the right. According to the rule of the reel, the current should go clockwise (if you look at the left). In this direction, the diode passes in the chain of the second lamp. So, the second lamp will light up.

Answer. The second lamp will light up.

Aluminum needker length L. \u003d 25 cm and cross-sectional area S. \u003d 0.1 cm 2 is suspended on the thread at the top end. The lower end relies on the horizontal bottom of the vessel in which water is poured. Length submerged parts of the knitting needles l. \u003d 10 cm. Find strength F.With which the needker presses the bottom of the vessel, if it is known that the thread is located vertically. Aluminum density ρ a \u003d 2.7 g / cm 3, water density ρ B \u003d 1.0 g / cm 3. Acceleration of gravity g. \u003d 10 m / s 2

Decision. Perform an explanatory drawing.

- thread tension force;

- the reaction force of the bottom of the vessel;

a - Archimedean force acting only on the immersed part of the body, and attached to the center of the immersed part of the knitting needles;

- The strength of gravity acting on the needle from the ground and is attached to the value of the whole needle.

By definition, the mass of the needles m. And the Archimedean Module is expressed as follows: m. = SL.ρ a (1);

F. a \u003d. SL.ρ B. g. (2)

Consider the moments of forces regarding the spokes suspension.

M.(T.) \u003d 0 - the moment of the tension force; (3)

M.(N) \u003d NLcosα - the moment of the reaction force of the support; (four)

Taking into account the signs of moments we write the equation

NLcOSα +. SL.ρ B. g. (L. –	l.	) COSα \u003d. SL.ρ A. g.	L.	cOSα (7)
	2		2

considering that according to the third law of Newton, the reaction force of the vessel bottom is equal to force F. d with which the needker presses the bottom of the vessel we write N. = F. D and from equation (7) Express this power:

F d \u003d [	1	L.ρ A.– (1 –	l.	)l.ρ in] SG. (8).
	2		2L.

Substitute numeric data and get that

F. d \u003d 0.025 N.

Answer. F.d \u003d 0.025 N.

Balon containing m. 1 \u003d 1 kg of nitrogen, when tested for strength exploded at temperatures t. 1 \u003d 327 ° C. What a mass of hydrogen m. 2 could be stored in such a cylinder at temperatures t. 2 \u003d 27 ° С, having a fivefold margin of safety? Nitrogen molar mass M. 1 \u003d 28 g / mol, hydrogen M. 2 \u003d 2 g / mol.

Decision. We write the equation of the status of the ideal gas of Mendeleev - Klapairone 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 \u003d p 1/5; (3) considering that

we can express the mass of hydrogen working immediately with equations (2), (3), (4). The final formula has the form:

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

After substitution of numeric data m. 2 \u003d 28 g

Answer. m. 2 \u003d 28 g

In the perfect oscillatory loaf of the amplitude of the fluctuations in the current strength in the inductance coil I M. \u003d 5 mA, and voltage amplitude on the condenser U M. \u003d 2.0 V. at the time of time t. The voltage on the condenser is 1.2 V. Find the strength of the current in the coil at that moment.

Decision. In the ideal oscillatory circuit, the energy of oscillations is preserved. For the moment T, the law of energy conservation has the form

C.	U. 2	+ L.	I. 2	= L.	I M. 2	(1)
	2		2		2

For amplitude (maximum) values \u200b\u200bwrite

and from equation (2) express

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

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

I. = I M. (5)

Thus, the power of the current in the coil at the time of time t. equal

I. \u003d 4.0 mA.

Answer. I. \u003d 4.0 mA.

At the bottom of the reservoir, a depth of 2 m is a mirror. The beam of light, passing through the water, reflected from the mirror and comes out of the water. The refractive index of water is 1.33. Find the distance between the entrance point of the beam to the water and the beam outlet point from the water if the beam drop angle is 30 °

Decision. Let's make an explanatory figure

α - the angle of falling the beam;

β is the beam refraction angle in water;

AC is the distance between the entrance point of the beam to the water and the beam outlet point from the water.

By the law of refraction of light

sinβ \u003d	sINα.	(3)
	n. 2

Consider rectangular ΔAdv. In it asd \u003d h., then DB \u003d AD

tGβ \u003d. h.tGβ \u003d. h.	sINα.	= h.	sINβ.	= h.	sINα.	(4)
	cosβ

We get the following expression:

AC \u003d 2 DB \u003d 2 h.	sINα.	(5)

Substitute numerical values \u200b\u200bin the resulting formula (5)

Answer. 1.63 m.

As part of preparation for the exam, we suggest you familiarize yourself with working program in physics for 7-9 class to line UMK Pryricina A. V. and the working program of the in-depth level for the 10-11 classes to the UMC Mikishheva G.Ya. Programs are available for viewing and free download to all registered users.

Is it possible to prepare for the exam in physics yourself, having only access to the Internet? There is always a chance. About what to do and in what order, the author of the textbook "Physics says. A full course of preparation for EGE "I. V. Yakovlev.

Independent preparation for the exam in physics begins with the study of the theory. Without this, it is impossible to learn to solve problems. We must first, taking any topic, thoroughly deal with the theory, read the corresponding material.

Take the topic "Newton Law." It is necessary to read about the inertial reference systems, find out that the forces are formed vector, about how vectors are designed to the axis, as it can work in a simple situation - for example, on an inclined plane. It is necessary to learn what kind of friction force is the difference in the power of slip friction from the strength of rest friction. If you do not distinguish them, then, most likely, be mistaken in the appropriate task. After all, the tasks are often given in order to understand those or other theoretical moments, therefore, with the theory, it is necessary to figure out the most clearly possible.

For the full development of the course of physics, we recommend you the textbook I. V. Yakovlev "Physics. Full preparation course for EGE. " You can purchase it or read the materials online on our website. The book is written in a simple and understandable language. It is also good because the theory in it is grouped precisely on the items of the Codifier of the EGE.

And then you need to take the tasks.
First stage. To begin with - take the easiest task, and this is the problem of Rymkiewicz. You need to break up 10-15 tasks on the chosen topic. In this collection, the task is quite simple, in one or two actions. You will understand how to solve problems on this topic, and at the same time they will remember all the formulas that are needed.

When you are preparing for the exam in physics yourself - do not specifically flaw the formula and write crib. Effectively all this is perceived only when it came through the solution of tasks. The problem of Rymkiewicz, like no other, meets this primary goal: learn to solve simple tasks and at the same time learn all formulas.

Second phase.It is time to move to training precisely on the tasks of the exam. It is best to prepare for wonderful benefits edited by Demidova (on the cover of the Russian tricolor). These collections are two species, namely - collections of typical options and collections of thematic options. It is recommended to start with thematic options. These collections are built as follows: first go options only on mechanics. They are connounced in accordance with the structure of the USE, but the tasks in them only on mechanics. Then - the mechanics is fixed, thermodynamics connect. Then - mechanics + thermodynamics + electrodynamics. The topics are then added, quantum physics, after which there are 10 full-fledged EGE options in this manual - on all topics.
Such a manual, which includes about 20 thematic options, is recommended as a second stage after Rymkiewicz's task for those who are independently prepared for the exam in physics.

For example, it can be a collection
"EGE physics. Thematic examination options. " M.Yu. Demidova, I.I. Nurminsky, V.A. Mushrooms.

Similarly, we use collections in which sample examinations are selected.

Third stage.If time allows, it is extremely desirable to reach the third stage. This is a preparation for the tasks of Fiztech, a higher level. For example, the collector of the baucanina, Belonochkin, the goat (publishing house "Enlightenment"). The tasks of such collections seriously exceed the level of the EGE. But in order to successfully pass the exam, you need to be ready for a couple of steps above - for a variety of reasons, up to a banal self-confidence.

It is not necessary to be limited only to the benefits of the USE. After all, it is not a fact that the tasks will repeat. There may be tasks that were not met before in the collections of the exam.

How to distribute time when independent preparation for the exam in physics?
What to do when you have one year and 5 large topics: mechanics, thermodynamics, electricity, optics, quantum and nuclear physics?

The maximum number is half of all the time of preparation - you need to take two topics: mechanics and electricity. These are dominant themes, the most complex. The mechanics are studied in the 9th grade, and it is believed that schoolchildren know her best. But actually it is not. Tasks for mechanics are as complex as possible. And electricity - the topic is difficult in itself.
Thermodynamics and molecular physics are a pretty simple topic. Of course, there are your underwater stones. For example, schoolchildren do not understand what saturated pairs are. But in general, experience shows that there are no such problems as in mechanics and electricity. Thermodynamics and molecular physics at the school level are a simpler section. And most importantly - this is a partition offline. It can be studied without mechanics, without electricity, he is in itself.

The same can be said about optics. Geometric optics is simple - it comes down to geometry. We must learn the basic things associated with thin lenses, the law of refraction - and that's it. Wave optics (interference, diffraction of light) is present in the EGE in minimal quantities. The compilers of options do not give any complex tasks in the exam on this topic.

And there remains quantum and nuclear physics. Schoolchildren are traditionally afraid of this section, and in vain, because it is the easiest of all. The last task of the final part of the EGE - on the photoeff, the pressure of light, nuclear physics is easier than others. It is necessary to know the Einstein equation for the photo effect and the law of radioactive decay.

In the version of the exam in physics there are 5 tasks where you need to write a detailed solution. The peculiarity of the physics is that the complexity of the task is not growing with the growth of the room. You never know what task will be in the exam in physics complex. Sometimes complicated mechanics, sometimes thermodynamics. But traditionally the task of quantum and nuclear physics is the simplest.

To prepare for the exam in physics on their own - it is possible.But if there is at least the smallest opportunity to turn to a qualified specialist, then it is better to do this. Schoolchildren, preparing for the exam in physics on their own, are very risking to lose many points on the exam, simply because they do not understand the strategy and tactics of preparation. The specialist knows what way to go, and the schoolboy may not know it.

We invite you to our training courses for the exam in physics. The year of classes is to master the course of physics at the level of 80-100 points. Success to you in preparation for the exam!

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