Physics, 11th Class 2 Draft codifier of elements of content and requirements for the level of training of educational organizations for a single state exam in physics Codifier of elements of content in physics and requirements for the level of training graduates of educational organizations for a single state exam is one of the documents, a united state exam In physics defining the structure and content of KIM EGE. It is based on the federal component of state standards of the main general and secondary (full) general education in physics (basic and profile levels) (order of the Ministry of Education of Russia of 05.03.2004 No. 1089). Codifier Section 1. The list of content elements tested on the single elements of the content and requirements for the level of preparation of the state exam in the physics of graduates of educational organizations for conducting a section code in the first column, which corresponds to a large unified state exam in physics blocks of content. In the second column, the content item code for which verification tasks are created. Large blocks of content are broken into smaller elements. The code was prepared by the Federal State Budget Code of the Research Institution Code LIRUE of various elements of the content, "Federal Institute of Pedagogical Measurement" of Cases of Elemes Checked Questions Kima 1 1.1 Kinematics 1.1.1 Mechanical Movement. The relativity of mechanical movement. Covering system 1.1.2 material point. z Trajectory Its radius-vector:  R (t) \u003d (x (t), y (t), z (t)),   Trajectory, R1 Δ r Movement:     R2 Δ r \u003d R (t 2) - R (T1) \u003d (Δ x, δ y, Δ z), o y path. Adjustment of movements: X    Δ R1 \u003d Δ R 2 + δ R0 © 2018 Federal Service for Supervision in Education and Science Russian Federation

Physics, grade 11 3 Physics, grade 11 4 1.1.3 Master point speed: 1.1.8 Movement of the point around the circle.   ΔR  2π υ \u003d R "T \u003d (υ x, υ y, υ z), angular and linear velocity point: υ \u003d ωr, ω \u003d 2πν. Δt Δt → 0 t Δx υ2 υx \u003d x" t, similar to υ y \u003d yt ", υ z \u003d zt". Centripetal acceleration point: ACS \u003d \u003d ω2 R ΔT Δt → 0 R    1.1.9 solid body. Protective and rotational motion Addition of speeds: υ1 \u003d υ 2 + υ0 solid body 1.1.4 Acceleration of the material point: 1.2 Dynamics   Δυ  a \u003d υt "\u003d (AX, AY, AZ), 1.2.1 inertial reference systems. First Newton's law. ΔT Δt → 0 The principle of the relativity of Galilee Δυ x 1.2.2 m ax \u003d (υ x) t ", similarly to Ay \u003d (υ y)", AZ \u003d (υ Z) T ". Body mass. The density of the substance: ρ \u003d Δt Δt → 0 T  V   1.1.5 Uniform straight movement: 1.2.3 force. Principle of Superposition of Force: Fravnother in \u003d F1 + F2 +  x (t) \u003d x0 + υ0 xt 1.2.4 Second  Newton's law: The material point in ISO    υ x (t) \u003d υ0 x \u003d const f \u003d MA; Δp \u003d FΔt at f \u003d const 1.1.6 Equal asked straight movement: 1.2.5 The third Newton law     A T2 of material points: F12 \u003d - F21 F12 F21 x (T) \u003d x0 + υ0 xt + x 2 υ x (t) \u003d υ0 x + AXT 1.2.6 The Law of the World Machine: The forces of attraction between MM AX \u003d const Spotties are equal to F \u003d G 1 2 2. R υ22x - υ12x \u003d 2AX (x2 - x1) The power of gravity. The dependence of gravity from the height H over 1.1.7 is free drop. Y  The surface of the planet with R0 radius: acceleration of free V0 GMM drop. Body movement, Mg \u003d (R0 + H) 2 abandoned at an angle α to Y0 α 1.2.7 The movement of the celestial bodies and their artificial satellites. Horizon: First Space Speed: GM O x0 x υ1k \u003d G 0 R0 \u003d R0  x (t) \u003d x0 + υ0 xt \u003d x0 + υ0 cosα ⋅ t Second Space Speed:   G YT 2 GT 2 2GM  Y (T ) \u003d Y0 + υ0 yt + \u003d y0 + υ0 sin α ⋅ t - υ 2 k \u003d 2υ1k \u003d  2 2 r0  x (t) \u003d υ0 x \u003d υ0 cosα 1.2.8 The force of elasticity. The law of the throat: f x \u003d - kx   y (t) \u003d υ0 y + g yt \u003d υ0 sin α - gt 1.2.9 friction force. Dry friction. Slip friction force: FTR \u003d μN gx \u003d 0  Rain friction force: ftr ≤ μn  gy \u003d - g \u003d const friction coefficient 1.2.10 f Pressure: P \u003d ⊥ Sl © 2018 Federal Service for Supervision in the field of education and science of the Russian Federation © 2018 Federal Service for Supervision in the field of education and science of the Russian Federation

Physics, 11th grade 5 Physics, grade 11 6 1.4.8 Law of changes and conservation of mechanical energy: 1.3 Static E Fur \u003d E KIN + E Potent, 1.3.1 Moment of force relative to the axis in ISO ΔE Fur \u003d AWEX Nepotenz. Forces, rotation:  L M \u003d FL, where L is the shoulder of the force F in ISO ΔE Fur \u003d 0, if the AVEX is notopenz. Forces \u003d 0 → O with respect to the axis passing through f 1.5 Mechanical oscillations and waves point O perpendicular to Figure 1.5.1 Harmonic oscillations. The amplitude and phase of oscillations. 1.3.2 Conditions for the equilibrium of a solid body in ISO: kinematic description: m 1 + m 2 +  \u003d 0 x (t) \u003d a sin (ωt + φ 0),   υ x (t) \u003d x "T, F1 + F2 +  \u003d 0 1.3.3 The law of Pascal AX (T) \u003d (υ X) "T \u003d -ω2 x (t). 1.3.4 Pressure in the liquid, resting in ISO: P \u003d P 0 + ρ GH Dynamic description:   1.3.5 Act Archimedes: FARK \u003d - PXEST. , Ma x \u003d - kx, where k \u003d mω. 2 If the body and liquid rest in ISO, then the FARK \u003d ρ GV extrusion. Energy Description (the law of maintaining a mechanical condition for swimming bodies MV 2 KX 2 MV MAX 2 Ka 2 of energy): + \u003d \u003d \u003d subst. 1.4 Laws of conservation in mechanics 2 2 2 2   Communication of the amplitude of the oscillation of the initial value from 1.4.1 pulse of the material point: p \u003d mυ     amplitudes of oscillations of its speed and acceleration: 1.4.2 Pulse system tel: p \u003d p1 + p2 + ... 2 V max \u003d ωa, a max \u003d ω A 1.4.3 The law of change and conservation  pulse:       δ p ≡ δ (p1 + p 2 + ...) \u003d f1 External Δ t + F2 External Δ T + ; 1.5.2 2π 1   Period and frequency of oscillations: T \u003d \u003d.    Ω ν in ISO Δp ≡ Δ (p1 + p2 + ...) \u003d 0, if F1 external + F2 external +  \u003d 0 period of small free oscillations of mathematical 1.4.4 work force: on a small movement    L A \u003d f ⋅ ΔR ⋅ cos α \u003d fx ⋅ Δx α  F pendulum: T \u003d 2π. ΔR G Period of free oscillations of spring pendulum: 1.4.5 Power power:  F M ΔA α t \u003d 2π p \u003d f ⋅ υ ⋅ cosα  k Δt Δt → 0 V 1.5.3 Forced oscillations. Resonance. Resonant curve 1.4.6 Kinetic energy material: 1.5.4 Transverse and longitudinal waves. Speed \u200b\u200bMυ 2 P 2 υ Ekin \u003d \u003d. distribution and wavelength: λ \u003d υt \u003d. 2 2M ν The law of changes in the kinetic energy system interference and diffraction of waves of material points: in ISO ΔEkin \u003d A1 + A2 +  1.5.5 sound. Sound speed 1.4.7 Potential energy: 2 Molecular physics. Thermodynamics for the potential forces A12 \u003d E 1 potset - E 2 potential \u003d - Δ E potent. 2.1 Molecular physics Potential body energy in a homogeneous field of gravity: 2.1.1 Models of the structure of gases, liquids and solids E potent \u003d Mgh. 2.1.2 Thermal motion of atoms and substance molecules The potential energy of the elastic deformed body: 2. 1.3 The interaction of particles of substance 2.1.4 diffusion. Brownian movement KX 2 E Potential \u003d 2.1.5 Model of the ideal gas in MTC: Gas particles move 2 chaotically and do not interact with each other © 2018 Federal Service for Supervision in Education and Science of the Russian Federation © 2018 Federal Service for Supervision in Education and Education and Science of the Russian Federation

Physics, 11th grade 7 Physics, grade 11 8 2.1.6 Communication between pressure and medium kinetic energy 2.1.15 Changes in aggregate states: evaporation and translational thermal motion of molecules of ideal condensation, boiling liquid gas (main equation MKT): 2.1.16 change aggregate states matter: melting and 1 2 m v2  2 Crystallization p \u003d m0nv 2 \u003d n ⋅  0  \u003d n ⋅ ε post 3 3  2  3 2.1.17 Energy conversion in phase transitions 2.1.7 Absolute temperature : T \u003d T ° + 273 K 2.2 Thermodynamics 2.1.8 Gas temperature connection with medium kinetic energy 2.2.1 Thermal equilibrium and temperature of translational heat movement of its particles: 2.2.2 Internal energy 2.2.3 Heat transfer as a method for changing internal energy m v2  3 ε post \u003d  0  \u003d KT without performing work. Convection, thermal conductivity,  2  2 Radiation 2.1.9 Equation P \u003d NKT 2.2.4 Number of heat. 2.1.10 The model of the ideal gas in thermodynamics: the specific heat capacity of the substance C: Q \u003d CmΔt.  Revision of Mendeleev - Klapaireron 2.2.5 Specific War Education R: Q \u003d Rm.  Specific melting heat λ: Q \u003d λ m.  Internal energy Energy equation Mendeleev-Klapairone equation (applicable shape specific heat combustion of fuel q: q \u003d qm records): 2.2.6 Elementary work in thermodynamics: a \u003d pΔV. M ρRT Calculation of work on the process graph on the PV diagram PV \u003d RT \u003d νRT \u003d nkt, p \u003d. μ μ 2.2.7 First law of thermodynamics: the expression for the internal energy of the same name Q12 \u003d ΔU 12 + a12 \u003d (U 2 - U 1) + A12 of the ideal gas (applicable forms of recording): adiabat: 3 3 3m q12 \u003d 0  a12 \u003d u1 - U 2 U \u003d νRT \u003d nkt \u003d rt \u003d νC νT 2 2 2μ 2.2.8 The second law of thermodynamics, irreversibility 2.1.11 Dalton law for pressure of a mixture of sparse gases: 2.2.9 Principles of thermal machinery. KPD: P \u003d P1 + P 2 +  A Q NGR - QHOL Q 2.1.12 Isoprocesses in a rarefied gas with a constant number η \u003d per cycle \u003d \u003d 1 - HOW QNAG Q Nagr Q Nagr particles N (with a constant amount of substance ν): isotherm (T \u003d Const): PV \u003d const, 2.2.10 The maximum efficiency of the efficiency. The cycle of carno TNAGR - T HOL T HOT P MAX η \u003d η carno \u003d \u003d 1 - heof (V \u003d const): \u003d const, TNAGR TNGR TV 2.2.11 The heat balance equation: Q1 + Q 2 + Q 3 + ... \u003d 0. Isobar (P \u003d const): \u003d const. T 3 Electrodynamics Graphic representation of isoproces on PV-, PT- and VT- 3.1 Electrical field of diagrams 3.1.1 Electrification of bodies and its manifestation. Electric charge. 2.1.13 Saturated and unsaturated pairs. High-quality two types of charge. Elementary electric charge. The law is the dependence of the density and pressure of a saturated steam on the conservation of the electrical charge of temperature, their independence from the volume of saturated 3.1.2 the interaction of charges. Point charges. Cool law: pair q ⋅q 1 Q ⋅Q 2.1.14 Air humidity. F \u003d k 1 2 2 \u003d ⋅ 1 2 2 r 4πε 0 R p pa steam (T) ρ pa steam (T) Relative humidity: φ \u003d \u003d 3.1.3 Electric field. His action on electrical charges P saturation. Couple (T) ρ sat. Couple (t) © 2018 Federal Service for Supervision in Education and Science of the Russian Federation © 2018 Federal Service for Supervision of Education and Science of the Russian Federation

Physics, grade 9 9 Physics, 11th grade 10  3.1.4  F 3.2.4 Electrical resistance. Resistance dependence of the electric field strength: E \u003d. Uniform conductor from its length and sections. Specific Q Trial L Q Resistance of the substance. R \u003d ρ Field of point charge: E R \u003d k 2, S  R 3.2.5 Current sources. EMF and internal resistance Uniform field: E \u003d const. A Pictures of the lines of these current source fields.  \u003d third-party forces 3.1.5 Potentialness of the electrostatic field. q Potential difference and voltage. 3.2.6 Ohma law for full (closed) A12 \u003d Q (φ1 - φ 2) \u003d - q Δ φ \u003d QU electrical circuit:  \u003d IR + IR, from where ε, R r Potential charge energy in the electrostatic field:  i \u003d W \u003d Qφ. R + R W 3.2.7 Parallel compound of conductors: The potential of the electrostatic field: φ \u003d. q 1 1 1 i \u003d i1 + i 2 + , u 1 \u003d u 2 \u003d , \u003d + +  Communication of the field of field and potential difference for RArall R1 R 2 of a homogeneous electrostatic field: U \u003d ED. Sequential compound of conductors: 3.1.6 Principle   Superposition  Electric fields: U \u003d u 1 + U 2 + , i 1 \u003d i 2 \u003d , Rapp \u003d R1 + R2 +  E \u003d E1 + E 2 + , φ \u003d φ 1 + φ 2 +  3.2.8 Electrical current operation: a \u003d iUT 3.1.7 Explorers in the electrostatic  field. Condition of the Joule-Lenza law: Q \u003d i 2 RT balance of charges: inside the conductor E \u003d 0, inside and 3.2.9 ΔA the surface of the conductor φ \u003d const. Power of electrical current: p \u003d \u003d IU. ΔT ΔT → 0 3.1.8 Dielectrics in the electrostatic field. The dielectric thermal power released on the resistor: the permeability of the substance ε 3.1.9 q U2 condenser. Electricity capacitor: C \u003d. P \u003d i 2r \u003d. U R εε 0 S ΔA Electricity of a flat capacitor: C \u003d \u003d εC 0 Current source power: P \u003d art. Forces \u003d i D Δ t Δt → 0 3.1.10 Parallel compound of capacitors: 3.2.10 Free media electrical charges in conductors. q \u003d Q1 + Q 2 + , u 1 \u003d U 2 \u003d , C param \u003d C1 + C 2 +  The mechanisms of conductivity of solid metals, solutions and a consecutive connection of capacitors: electrolyte melts, gases. Semiconductors. 1 1 1 semiconductor diode U \u003d U 1 + U 2 + , q1 \u003d q 2 \u003d , \u003d + +  3.3 Magnetic field C 1 C 2 3.3.1 Mechanical interaction of magnets. A magnetic field. 3.1.11 QU CU 2 Q 2 vector magnetic induction. The principle of superposition The energy of the charged condenser: wc \u003d \u003d    2 2 2C of magnetic fields: B \u003d B1 + B 2 + . Magnetic line 3.2 DC locations. Picture of the lines of the field of strip and horseshoe-shaped 3. 2.1 ΔQ permanent magnets current: i \u003d. Permanent current: i \u003d const. Δ T Δt → 0 3.3.2 Ersted Experience. Magnetic field of conductor with current. For DC C \u003d IT Pattern Line Lines Long Direct Explorer Fields and 3.2.2 Conditions of Electric Current. Closed ring conductor, coils with current. Voltage U and EDC ε 3.2.3 U Ohm law for a section of a chain: I \u003d r © 2018 Federal Service for Supervision in Education and Science of the Russian Federation © 2018 Federal Service for Supervision in the field of education and science of the Russian Federation

Physics, 11th grade 11 Physics, 11 class 12 3.3.3Sille of ampere, its direction and value: 3.5.2 The law of conservation of energy in the oscillatory circuit: FA \u003d IBL SIN α, where α is the angle between the direction Cu 2 Li 2 Cu Max 2 Li 2  + \u003d \u003d max \u003d Const conductor and vector B 2 2 2 2 3.3.4 Lorentz power, its direction and value:  3.5.3 Forced electromagnetic oscillations. Resonance  Flor \u003d Q VB Sinα, where α is the angle between vectors V and b. 3.5.4 Alternating current. Production, transmission and consumption Movement of the charged particle in a homogeneous magnetic electrical energy field 3.5.5 properties of electromagnetic waves. Mutual orientation   3.4 Electromagnetic induction of vectors in an electromagnetic wave in vacuum: E ⊥ B ⊥ C. 3.4.1 Magnetic vector stream   3.5.6 Scale of electromagnetic waves. Application N B induction: Φ \u003d b n s \u003d bs cos α electromagnetic waves in the technique and everyday life α 3.6 Optics S 3.6.1 The straight-line propagation of light in a homogeneous medium. Light Light 3.4.2 Phenomenon of electromagnetic induction. EMF Induction 3.6.2 Light reflection laws. 3.4.3 Faraday electromagnetic induction law: 3.6.3 Construction of images in a flat mirror Δφ 3.6.4 Laws of refraction of light. i \u003d - \u003d --φ "T refraction of light: n1 sin α \u003d n2 sin β. ΔT Δt → 0 C 3.4.4 EDC induction in a direct conductor length L, moving absolute refractive index: N ABS \u003d.    V  () With the speed υ υ ⊥ L in a homogeneous magnetic relative refractive index: N, n 1 V1 \u003d. N1 V 2 field B:   i \u003d BLυ SIN α, where α is the angle between the vectors B and υ; if the move Rays in the prism \u003d N 2 λ 2 3.4.6 f 3.6.5 Complete internal reflection. Inductance: L \u003d, or φ \u003d Li. N2 I The extreme angle of complete Δi internal reflection: self-induction. EMF self-induction: si \u003d - L \u003d - Li "T 1 N N1 ΔT ΔT → 0 SIN αPR \u003d \u003d 2 αPR 3.4.7 NOTE N1 Li 2 Magnetic field energy Coil: Wl \u003d 3.6.6 collecting and scattering lenses. Thin lens. 2 focal length and optical power of thin lens: 3.5 Electromagnetic oscillations and waves 1 3.5.1 oscillating circuit. Free d \u003d electromagnetic oscillations in the ideal C L F oscillatory circuit: 3.6.7 Formula of fine lens: D 1 1 1 Q (t) \u003d Q max sin (ωt + φ 0) + \u003d. H  df ff  i (t) \u003d qt '\u003d ωq max cos (ωt + φ 0) \u003d i max cos (ωt + φ 0) increase given by 2π 1 F H formula Tomson: T \u003d 2π Lc, from where Ω \u003d \u003d. lens: γ \u003d h \u003d f f t Lc H D Communication of the amplitude of the charge of the capacitor with the amplitude of the current I current in the oscillatory circuit: q Max \u003d max. ω © 2018 Federal Service for Supervision in Education and Science of the Russian Federation © 2018 Federal Service for Supervision in Education and Science of the Russian Federation

Physics, 11th grade 13 Physics, 11th grade 14 3.6.8 The stroke of the ray that has passed the lens at an arbitrary angle to its 5.1.4 Einstein equation for a photo effect: the main optical axis. Building images of the point and E photon \u003d A exit + E CIN MAX, the segment is straight into collecting and scattering lenses and their HC HCs of systems where EFhoton \u003d Hν \u003d, AVOD \u003d Hν kr \u003d, 3.6.9 camera as an optical device. λ λ Cr 2 Eye as an optical system MV MAX E KIN MAX \u003d EU ZAP 3.6.10 Light Interference. Coherent sources. Conditions 2 of observation of highs and minima in 5.1.5 wave properties of particles. Waves de Broglie. interference pattern from two syphase H h wavelength de Broglya moving particles: λ \u003d \u003d. coherent sources P MV λ corpuscular wave dualism. Maximum electron diffraction: δ \u003d 2m, m \u003d 0, ± 1, ± 2, ± 3, ... on crystals 2 λ 5.1.6 Light pressure. The pressure of light on the fully reflective minimum: δ \u003d (2m + 1), m \u003d 0, ± 1, ± 2, ± 3, ... surface and completely absorbing surface 2 5.2 atomic physics 3.6.11 Light diffraction. Diffraction grating. Condition 5.2.1 The planetary model of the supervision of the main maxima at normal fall 5.2.2 Bohr's postulates. Radiation and absorption of photons with monochromatic light with a wavelength λ on the grille with the transition of an atom from one level of energy to another: the period d: d sin φ m \u003d m λ, m \u003d 0, ± 1, ± 2, ± 3, ... HC 3.6.12 Light dispersion Hν Mn \u003d EN - EM λ Mn 4 Basics of special relativity Theory 4.1 Invariance of the light velocity module in vacuum. Principle 5.2.3Line spectra. Relativity Einstein Spectrum of the energy levels of the hydrogen atom: 4.2 - 13.6 eV en \u003d, n \u003d 1, 2, 3, ... 2 The energy of the free particle: E \u003d Mc. V2 N2 1- 5.2.4 Laser C2  5.3 Physics of the atomic nucleus Pulse particles: P \u003d MV . V 2 5.3.1 Nucleon model of Heisenberg-Ivanhenko kernel. Shoulder kernel. 1- Mass number of kernel. Isotopes C2 4.3 Communication of the mass and energy of the free particle: 5.3.2 Energy of the bonding of nucleons in the kernel. Nuclear power E 2 - (PC) \u003d (MC 2). 2 2 5.3.3 Defect for the mass of the kernel AZ X: Δ m \u003d z ⋅ m P + (a - z) ⋅ M n - M cores of rest energy of a free particle: E 0 \u003d Mc 2 5.3.4 Radioactivity. 5 Quantum physics and elements of astrophysics Alpha decay: AZ X → AZ - 42Y + 42 HE. 5.1 Vaccular wave dualism A A 0 ~ beta decay. Electronic β-decay: z x → z + 1y + -1 e + ν e. 5.1.1 Hypothesis M. Planck about quanta. Plank formula: E \u003d Hν positron β-decay: AZ X → ZA-1Y + +10 ~ E + νE. 5.1.2 HC Gamma radiation photons. Photon energy: E \u003d hν \u003d \u003d PC. λ 5.3.5 - T e hν h The law of radioactive decay: n (t) \u003d n 0 ⋅ 2 T pulse photon: p \u003d \u003d c c λ 5.3.6 nuclear reactions. Division and synthesis of nuclei 5.1.3 photoeff. Experiments A.G. Stoletova. CEO of the photo effect 5.4 Elements of astrophysics 5.4.1 Solar system: planets of the earth group and planets giants, Small Body Solar System © 2018 Federal Service for Supervision in Education and Science of the Russian Federation © 2018 Federal Service for Supervision in the field of education and science of the Russian Federation

Physics, 11th grade 15 Physics, 11th grade 16 5.4.2 Stars: a variety of star characteristics and their patterns. Sources of energy of stars 2.5.2 Conduct examples of experiments that illustrate that: 5.4.3 Modern ideas about the origin and evolution of observation and the experiment serve as the basis for the extension of the Sun and Stars. hypotheses and construction of scientific theories; Experiment 5.4.4 Our Galaxy. Other galaxies. Spatial allows you to check the truth of theoretical conclusions; The scale of the observed universe physical theory makes it possible to explain the phenomena 5.4.5. Modern views on the structure and evolution of the Universe of Nature and scientific facts; The physical theory allows you to predict not yet known phenomena and their features; With the explanation of natural phenomena, section 2. The list of requirements for the level of preparation verifiable physical models; The same natural object or on a single state examination in physics phenomenon can be explored based on the use of different models; Laws of Physics I. physical theories Requirements have their own code for the level of graduate training, the development of certain boundaries of the applicability of which is checked for EGE 2.5.3 Measure physical quantities, submit results 1 to know / understand: measurements, taking into account their errors 1.1 The meaning of physical concepts 2.6 Apply the knowledge to solve the physical 1.2 meaning The physical quantities of tasks 1.3 The meaning of physical laws, principles, postulates 3 use acquired knowledge and skills in practical 2 to be able to: Activity and everyday life For: 2.1, describe and explain: 3.1 Ensuring the safety of life in the process of use vehicle, household 2.1.1 Physical phenomena, physical phenomena and properties of electrical appliances, radio and telecommunication equipment 2.1.2 Results of communication experiments; Estimates of the effect on the human body and others 2.2 to describe the fundamental experiments that have provided environmental pollution organisms; rational significant impact on the development of environmental management and environmental physics; 2.3 Give examples of practical application of physical 3.2 definitions of your own position in relation to knowledge, laws of physics environmental issues and behavior in natural environment 2.4 to determine the nature of the physical process on graphics, table, formula; nuclear reaction products based on the laws of maintenance of electric charge and mass number 2.5 2.5.1 distinguish hypotheses from scientific theories; draw conclusions based on experimental data; Give examples showing that: observations and experiment are the basis for hypotheses and theories, allow you to check the truth of theoretical conclusions; The physical theory makes it possible to explain the well-known phenomena of nature and scientific facts, to predict not yet known phenomena; © 2018 Federal Service for Supervision in the Field of Education and Science of the Russian Federation © 2018 Federal Service for Supervision of Education and Science of the Russian Federation

Average general education

Line Ukk G. Ya. Myakisheva, M.A. Petrova. Physics (10-11) (b)

EGE-2020 Codifier in Phi Phi Phi

The codifier of the elements of the content and requirements for the level of training of graduates of educational organizations for the exam in physics is one of the documents defining the structure and content of the Kim Unified State Exam, the objects of which have a specific code. A codifier was drawn up on the basis of the federal component of state standards of the main general and secondary (full) general education in physics (basic and profile levels).

Main changes in new demo

Mostly change became minor. So, in the tasks in physics there will be not five, but six questions that imply a detailed answer. The task number 24 was complicated on the knowledge of the elements of astrophysics - now instead of two mandatory correct answers, either two or three correct options can be.

Soon we will talk about the coming ege on and on the air of our channel on YouTube.

Schedule Ege Physics in 2020

On the this moment It is known that the mines and Rosobrnadzor published for public discussion projects of the schedule of the exam. Exams in physics are supposed to be held on June 4.

The codifier is information separated into two parts:

part 1: "The list of elements of the contents verifiable at a single state examination in physics";

part 2: "The list of requirements for the preparation of graduates verifiable at a single state exam in physics."

The list of elements of the contents checked in a single state examination in physics

We present the original table with the list of content elements represented by the FII. Download the Codifier of the exam in physics in the full version you can Official website.

Section code	Code of controlled element	Content elements checked by kim jobs
1	Mechanics
1.1	Kinematics
1.2	Dynamics
1.3	Statics
1.4	Conservation laws in mechanics
1.5	Mechanical oscillations and waves
2	Molecular physics. Thermodynamics
2.1	Molecular physics
2.2	Thermodynamics
3	Electrodynamics
3.1	Electric field
3.2	DC laws
3.3	A magnetic field
3.4	Electromagnetic induction
3.5	Electromagnetic oscillations and waves
3.6	Optics
4	Basics of special theory of relativity
5	Quantum physics and elements of astrophysics
5.1	Corpuscular wave dualism
5.2	Atomic physics
5.3	Physics of the atomic nucleus
5.4	Elements of astrophysics

The book contains materials for the successful passing of EGE: short theoretical information on all topics, tasks different types and levels of complexity, solving problems of an increased level of complexity, answers and evaluation criteria. Students do not have to look for additional information on the Internet and buy other benefits. In this book, they will find everything necessary for independent and effective preparation for the exam.

Requirements for graduate training

Kim FIPI is developed with support for specific requirements for the training level. Thus, to successfully cope with the physics exam, the graduate is necessary:

1. Know / understand:

1.1. the meaning of physical concepts;

1.2. the meaning of physical quantities;

1.3. The meaning of physical laws, principles, postulates.

2. To be able to:

2.1. Describe and explain:

2.1.1. physical phenomena, physical phenomena and properties of bodies;

2.1.2. the results of experiments;

2.2. describe the fundamental experiments that provided a significant impact on the development of physics;

2.3. give examples of practical use of physical knowledge, physics laws;

2.4. determine the nature of the physical process on schedule, table, formula; nuclear reaction products based on the laws of conservation of an electric charge and mass number;

2.5.1. distinguishing hypotheses from scientific theories; draw conclusions based on experimental data; Conduct examples showing that: observations and experiment are the basis for hypotheses and theories and make it possible to check the truth of theoretical conclusions, the physical theory makes it possible to explain the well-known phenomena of nature and scientific facts, to predict still unknown phenomena;

2.5.2. Conduct examples of experiments illustrating that: observations and experiment serve as the basis for nomination of the hypotheses and the construction of scientific theories; The experiment allows you to check the truth of theoretical conclusions; Physical theory makes it possible to explain the phenomena of nature and scientific facts; Physical theory allows you to predict unknown phenomena and their features; With the explanation of natural phenomena, physical models are used; The same natural object or phenomenon can be explored based on the use of different models; The laws of physics and physical theories have their certain borders of applicability;

2.5.3. measure physical quantities, represent the measurement results, taking into account their errors;

2.6. Apply the knowledge gained to solve physical problems.

3. Use acquired knowledge and skills in practical activity and everyday life:

3.1. to ensure the safety of life in the process of using vehicles, household electrical appliances, radio and telecommunication communications; Estimates of the effect on the human body and other environmental pollution organisms; rational environmental management and environmental protection;

3.2. definitions of their own position in relation to environmental issues and behavior in the natural environment.

Secondary education

Preparing for the EEG-2018: Demals demo physics

We bring to your attention the analysis of the tasks of the exam in physics from the demolism of 2018. The article contains explanations and detailed assurance algorithms, as well as recommendations and references to useful materials, topical when preparing for the USE.

EGE-2018. Physics. Thematic training tasks

The publication contains:
tasks of different types in all themes of the ESE;
Answers to all tasks.
The book will be useful as teachers: it makes it possible to effectively organize the training of students to the EE directly in the lessons, in the process of studying all topics and students: training tasks will allow systematically when passing each topic, prepare for the exam.

Rose pointing point begins to move along the axis O.x.. Figure shows a graph of projection a. X.accelerate this body from time t..

Determine what path the body passed over the third second of the movement.

Answer: _________ m.

Decision

To be able to read graphics is very important for each student. The question is that it is required to determine the chart of the dependence of the projection of acceleration from time, the path that the body passed over the third second of the movement. The graph shows that in the time interval from t. 1 \u003d 2 s to t. 2 \u003d 4 s, the acceleration projection is zero. Consequently, the projection of the resultant force on this site, according to the second law of Newton, is also zero. Determine the nature of the movement on this site: the body moved evenly. The way is easy to determine, knowing the speed and time of movement. However, in the range from 0 to 2 s, the body moved equally. Using the definition of acceleration, write the speed projection equation V X. = V. 0x. + a X T.; Since the body initially rested, the projection of speed by the end of the second second was

Then the way passed by the body for the third second

Answer: 8 m.

Fig. 1

On a smooth horizontal surface, two bars are connected by a light spring. To the bar mass m.\u003d 2 kg apply constant force equal to module F.\u003d 10 H and directional horizontally along the axis of the spring (see Figure). Determine the modulus of the elasticity of the spring at the moment when this bar is moving with an acceleration of 1 m / s 2.

Answer: _________ N.

Decision

Horizontally on the body mass m. \u003d 2 kg two forces act, it's power F.\u003d 10 H and the strength of elasticity, from the side of the spring. The equitive of these forces informs the body acceleration. We choose the coordinate direct and send it along the action of power F.. We write Newton's second law for this body.

In the projection on the axis 0 H.: F. – F. UPR \u003d mA. (2)

Express the elasticity force module from formula (2) F. UPR \u003d F. – mA. (3)

Substitute numeric values \u200b\u200bin formula (3) and get F. Ex \u003d 10 H - 2 kg · 1 m / s 2 \u003d 8 N.

Answer: 8 N.

Task 3.

The body weighing 4 kg, located on a rough horizontal plane, told the speed of 10 m / s along her. Determine the module of work performed by the friction force, since the start of the body moves to the moment when the body speed decreases by 2 times.

Answer: _________ J.

Decision

The body has the power of gravity, the force of the support reaction force of friction that creates the braking acceleration of the body initially reported the speed of 10 m / s. We write Newton's second law for our case.

Equation (1), taking into account the projection on the selected axis Y. will look at:

N. – mG. = 0; N. = mG. (2)

In the projection on the axis X.: –F. Tr \u003d - mA.; F. Tr \u003d. mA.; (3) We need to define the module of the friction force by the time when the speed becomes twice as fewer, i.e. 5 m / s. We write the formula to calculate the work.

A. · ( F. Tr) \u003d - F. Tr · S. (4)

To determine the distance traveled, take a missing formula:

S. =	v 2 - v. 0 2	(5)
	2a.

Substitute (3) and (5) in (4)

Then the module of the friction force will be equal to:

Substitute numeric values

A.(F. Tr) \u003d.		4 kg	((	5 M.	) 2 – (10	m.	) 2)	\u003d 150 j.
		2		from		from

Answer: 150 J.

EGE-2018. Physics. 30 training options for examination

The publication contains:
30 Training Options EGE
Instructions for the implementation and criteria for estimation
Answers to all tasks
Training options will help the teacher to organize preparations for the exam, and students - independently test their knowledge and readiness for the delivery of the final exam.

The step block has an external pulley with a radius of 24 cm. To threads wound on external and internal pulleys, loads are suspended as shown in the figure. The friction in the axis of the block is absent. What is the radius of the internal pulley of the block, if the system is in equilibrium?

Fig. one

Answer: _________ cm.

Decision

Under the condition of the task, the system is in equilibrium. On the image L. 1, shoulder strength L. 2 shoulder strength Condition of equilibrium: moments of forces rotating tel clockwise should be equal to the moments of forces that rotate the body counterclockwise. Recall that the moment of power is the work of the force module on the shoulder. Forces acting on the threads from the side of goods, differ 3 times. So, the radius of the inner pulley of the block differs from the external one, too 3 times. Consequently, shoulder L. 2 will be 8 cm.

Answer:8 cm.

Task 5.

Ohat various points in time.

From the list below, select twoproper statements and specify their numbers.

1. The potential energy of the spring at the time of 1.0 from the maximum.
2. The oscillation period of the ball is 4.0 s.
3. The kinetic energy of the ball at the time of time 2.0 with minimal.
4. The amplitude of the ball oscillation is 30 mm.
5. The total mechanical energy of the pendulum, consisting of a ball and spring, at the time of 3.0 s min is minimal.

Decision

The table presents the data on the position of the ball attached to the spring and fluctuating along the horizontal axis Ohat various points in time. We need to analyze this data and choose the correct two statements. The system is a spring pendulum. At the time of time t. \u003d 1 C, the bodies offset from the equilibrium position is maximally, then it is an amplitude value. By definition, the potential energy of the elastic deformed body can be calculated by the formula

E P. = k.	x. 2	,
	2

where k. - Spring rigidity coefficient, h. - Body displacement from equilibrium position. If the offset is maximum, then the speed at this point is zero, it means that the kinetic energy will be zero. According to the law of conservation and turning energy, the potential energy must be maximal. From the table we see that half fluctuations the body passes for t. \u003d 2 C, full fluctuations in time twice T. \u003d 4 C. Therefore, the claims will be correct; 2.

Task 6.

The cylindrical glass with water was lowered to float a small iceclock. After a while, the ice was completely melted. Determine how as a result of melting of the ice, the pressure has changed on the bottom of the glass and the water level in the glass.

1. increased;
2. decreased;
3. not changed.

Write down B. table

Decision

Fig. one

The tasks of this type are quite common in different embassions. And as practice shows, students often allow mistakes. We will try to disassemble this task in detail. Denote m. - mass of a piece of ice, ρ l - the density of ice, ρ in - water density, V. PCT is the volume of the immersed part of the ice equal to the volume of the displaced fluid (the volume of the well). Mentally remove ice from the water. Then the hole remains in the water, the volume of which is equal V. PCT, i.e. The volume of water displaced with a piece of ice rice. one( b.).

Write ice floating condition Fig. one( but).

F A. = mG. (1)

ρ B. V. PCT g. = mG. (2)

Comparing formulas (3) and (4) we see that the volume of the well is exactly equal to the volume of water obtained from melting our piece of ice. Therefore, if we are now (mentally), the water obtained from ice in the hole, the well, the whole will be fully filled with water, and the water level in the vessel will not change. If the water level does not change, the hydrostatic pressure (5), which in this case depends only on the height of the liquid, will not change. Consequently, the answer will

EGE-2018. Physics. Training tasks

The publication is addressed to students of high schools to prepare for the exam in physics.
The manual includes:
20 training options
Answers to all tasks
Forms of EGE responses for each option.
The publication will help teachers in the preparation of students to the exam in physics.

The weightless spring is located on a smooth horizontal surface and one end is attached to the wall (see Figure). At some point in time, the spring is beginning to deform, applying to its free end and external force and evenly moving point A.

Install the correspondence between the dependences of the physical quantities from the deformation x.springs and these values. To each position of the first column, select the appropriate position from the second column and write down table

Decision

From the drawing to the task it is clear that when the spring is not deformed, its free end, and, accordingly, T. A are in the position with the coordinate h. 0. At some point in time, the spring begins to deform, applying to its free end and external power. Point and at the same time moves evenly. Depending on whether the spring is stretched or compressed, the direction and the amount of elasticity arising in the spring will change. Accordingly, under the letter a) the graph is the dependence of the modulus of the force of the elasticity of the springs deformation.

The schedule under the letter b) is the dependence of the projection of the external force on the magnitude of the deformation. Because With an increase in external force, the deformation value and the force of elasticity increases.

Answer: 24.

Task 8.

When constructing a temperature scale, the reomuyur is assumed that at normal atmospheric pressure, the ice melt at a temperature of 0 degree re-results (° R), and water boils at 80 ° R. Find what is equal to the average kinetic energy of the progressive thermal motion of the particle of the ideal gas at a temperature of 29 ° R. Answer express in EV and round up to hundredths.

Answer: ________ EV.

Decision

The task is interesting in that it is necessary to compare the two temperature measurement scales. This is the temperature range of the Reomyur and Celsius scale. The ice melting point coincides on the scales, and the boiling point is different we can get a formula to transfer from degrees to the reality in degrees Celsius. it

Translate temperature 29 (° R) to degrees Celsius

The resulting result will be translated into Kelvin using the formula

T. = t.° C + 273 (2);

T. \u003d 36.25 + 273 \u003d 309.25 (k)

To calculate the average kinetic energy of the progressive heat movement of the particles of the ideal gas, we use the formula

where k. - Boltzmann's constant 1.38 · 10 -23 J / k, T. - Absolute temperature on the Kelvin scale. From the formula, it can be seen that the dependence of the average kinetic energy from the temperature is straight, that is, how many times the temperature changes, the average kinetic energy of the thermal motion of molecules changes in so many times. Substitute numeric values:

The result will be transferred to electronic content and rounded to hundredths. Recall that

1 eV \u003d 1.6 · 10 -19 J.

For this

Answer: 0.04 eV.

One mole of monoatomic ideal gas participates in the process of 1-2, whose graph is depicted on Vt.-Diagram. Determine for this process the ratio of changes in the internal gas energy to the magnitude of the gas reported by the amount of heat.

Answer: ___________.

Decision

Under the condition of the task in process 1-2, the graph of which is depicted on Vt.-Diagram, one mole of one-nominal ideal gas participates. To answer the question of the task, it is necessary to obtain expressions to change the internal energy and the amount of heat, reported by gas. The process isobaric (the law of gay-lousak). Changing internal energy can be written in two types:

For the amount of heat, reported by gas, we write the first law of thermodynamics:

Q. 12 = A. 12 + Δ. U. 12 (5),

where A. 12 - Gas operation when expanding. By definition, the work is equal

A. 12 = P. 0 · 2. V. 0 (6).

Then the amount of heat will be equal to (4) and (6).

Q. 12 = P. 0 · 2. V. 0 + 3P. 0 · V. 0 = 5P. 0 · V. 0 (7)

We write the ratio:

Answer: 0,6.

The directory contains in full theoretical Material At the rate of physics needed to pass the exam. The structure of the book corresponds to the modern codifier of the elements of the content on the subject, on the basis of which examination tasks are compiled - test measurement materials (KIM) of the EGE. Theoretical material is set out in a brief, accessible form. Each topic is accompanied by examples of examination tasks corresponding to the format of the USE. This will help the teacher to organize preparations for a single state exam, and students - independently test their knowledge and readiness for the promotion of the final exam.

Kuznets Kuzn iron horseshoe weighing 500 g at a temperature of 1000 ° C. Having finished the forging, he throws a horseshoe into a water vessel. There is a hiss, and steam rises above the vessel. Find a mass of water evaporated when immersed in it hot horseshoe. Consider the water is already heated to boiling point.

Answer: _________

Decision

To solve the problem, it is important to recall the thermal balance equation. If there are no losses, there is heat transfer of energy in the body. As a result, water evaporates. Initially, the water was at a temperature of 100 ° C, this means that after immersing the hot horseshoe, the energy obtained by water will go immediately on the vaporization. We write the thermal balance equation

from Well · m. P · ( t. n - 100) \u003d LM. in 1),

where L. - the specific heat of the vaporization, m. in - the mass of water that turned into steam, m. P - Mass of iron horseshoe, from F - specific iron heat capacity. From formula (1) we will express a lot of water

When writing a response, pay attention to which units you need to leave a lot of water.

Answer: 90

One mole of monoatomic ideal gas participates in the cyclic process, whose graph is depicted on TV.- diagram.

Choose twothe right statements on the basis of the analysis of the schedule presented.

1. Gas pressure in a state 2 More gas pressure is able to 4
2. The operation of gas on a plot 2-3 is positive.
3. In section 1-2 gas pressure increases.
4. On the area 4-1 from gas, a certain amount of heat is given.
5. The change in the internal gas energy in the section 1-2 is less than the change in the internal energy of the gas on the plot 2-3.

Decision

This type of task checks the ability to read graphs and explain the dependence of physical quantities. It is important to remember how the graphics of dependence looks for isoprocesses in different axes, in particular r \u003d const. In our example on TV.-Diagram is presented two isobars. Let's see how pressure and volume are changed at a fixed temperature. For example, for points 1 and 4 lying on two isobars. P. 1 . V. 1 = P. 4 . V. 4, we see that V. 4 > V. 1, meaning P. 1 > P. four . Condition 2 corresponds to pressure P. one . Consequently, the gas pressure is in a state 2 more gas pressure in the state 4. On the section 2-3, the process isochorn, the gas does not make it equal to zero. The approval is incorrect. In the section 1-2 pressure increases, also incorrectly. Just above we have shown that this isolation. In a section 4-1 of gas, a certain amount of heat is given in order to maintain the temperature constant, with a gas compression.

Answer: 14.

The heat machine works along the carno cycle. The temperature of the cooler of the heat machine was raised, leaving the temperature of the heater former. The amount of heat obtained by gas from the heater for the cycle has not changed. How have changed the efficiency of the heat machine and the gas operation of the cycle?

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

1. increased
2. decreased
3. not changed

Write down B. table Selected numbers for each physical size. Figures in response can be repeated.

Decision

The heat machines operating on the Carnot cycle are often found in the tasks on the exam. First of all, it is necessary to remember the formula for calculating the efficiency. Be able to record it through the temperature of the heater and the temperature of the refrigerator

in addition, be able to record efficiency through the useful operation of the gas A. r and the amount of heat obtained from the heater Q. n.

Carefully read the condition and determined which parameters were changed: in our case, the temperature of the refrigerator increased, leaving the temperature of the heater for the same. Analyzing formula (1), we conclude that the fraction numerator decreases, the denominator does not change, therefore, the efficiency of the thermal machine is reduced. If we work with formula (2), you will immediately answer the second question of the problem. The operation of gas per cycle will also decrease, with all current changes in the parameters of the thermal machine.

Answer: 22.

Negative charge - q.Q.and negative - Q.(See Figure). Where is directed relative to the picture ( right, left, up, down, to the observer, from the observer) Charge acceleration - q B.this point in time, if only charges + Q. and –Q.? Answer write down the word (words)

Decision

Fig. one

Negative charge - q. Located in the field of two fixed charges: Positive + Q. and negative - Q., as shown in the figure. In order to answer the question where the charge acceleration is directed - q., at the time when only charges + Q and - - Q. It is necessary to find the direction of the resulting force as the geometric amount of forces According to the second Law of Newton, it is known that the direction of the acceleration vector coincides with the direction of the resulting force. The figure performed a geometric construction, to determine the sum of two vectors. The question arises why the forces are sent? Recall how the charged bodies interact are interacting, they are repelled, the strength of the Coulomb force of the interaction of charges is central. The force with which oppositely charged bodies are attracted. From the drawing we see that the charge is q. Equifold from fixed charges, the modules of which are equal. Therefore, in the module will also be equal. The resulting force will be directed relative to the drawing down.Charge acceleration will also be directed - q.. down.

Answer: Down.

The book contains materials for the successful examination of the exam in physics: short theoretical information on all topics, tasks of different types and levels of complexity, solving problems of an increased level of complexity, answers and evaluation criteria. Students do not have to look for additional information on the Internet and buy other benefits. In this book, they will find everything necessary for independent and effective preparation for the exam. The publication contains the tasks of different types according to all themes verifiable for the exam in physics, as well as solving problems of an increased level of complexity. The publication will have invaluable assistance to students when preparing for the exam in physics, and can also be used by teachers when organizing the educational process.

Two successively connected resistors with a resistance of 4 ohms and 8 ohms are connected to the battery, voltage on the terminals of which is 24 V. What kind of thermal power is highlighted in a resistor of a smaller nominal?

Answer: _________ W.

Decision

To solve the problem, it is advisable to draw a sequential connection scheme of resistors. After that, remember the laws of the sequential connection of the conductors.

The scheme will be as follows:

Where R. 1 \u003d 4 ohm, R. 2 \u003d 8 ohms. The voltage on the terminals of the battery is 24 V. With a sequential connection of the conductors on each plot of the circuit, the current will be the same. General resistance is defined as the sum of the resistance of all resistors. According to the Ohm law, we have:

To determine the thermal power released on a resistor of a smaller nominal, write:

P. = I. 2 R. \u003d (2 a) 2 · 4 Ohm \u003d 16 W.

Answer: P. \u003d 16 W.

The wire frame 2 · 10 -3 m 2 rotates in a homogeneous magnetic field around the axis perpendicular to the magnetic induction vector. Magnetic flow, permeating the frame area, varies by law

F \u003d 4 · 10 -6 COS10π t.,

where all the values \u200b\u200bare expressed in si. What is the magnetic induction module?

Answer: ________________ MTL.

Decision

Magnetic flow varies by law

F \u003d 4 · 10 -6 COS10π t.,

where all the values \u200b\u200bare expressed in si. It should be understood that this is generally a magnetic flux and how this value is connected with the magnetic induction module B. and framework S.. We write the equation in general form to understand what values \u200b\u200bare included in it.

Φ \u003d φ m cosω t.(1)

Remember that before the COS or SIN sign there is an amplitude value that changes the value, which means φ max \u003d 4 · 10 -6 WB on the other side, the magnetic flux is equal to the product of the magnetic induction module on the contour area and the causina of the angle between the normal to the contour and the magnetic induction vector M \u003d. IN · S.cOSα, the flow is maximum at COSα \u003d 1; Express the induction module

The answer is required to write to MTL. Our result is 2 mT.

Answer: 2.

The plot of the electrical circuit is a successively connected silver and aluminum wire. Through them proceed with a constant electric current by force 2 A. The graph shows how the potential φ changes on this section of the chain when the wire is displaced along x.

Using a schedule, select twofine allegations and indicate their numbers in response.

1. Cross sections area Wires are the same.
2. Cross section of silver wire 6.4 · 10 -2 mm 2
3. Cross-section area of \u200b\u200bsilver wire 4.27 · 10 -2 mm 2
4. In the aluminum wire there is a thermal power of 2 watts.
5. In silver wire, less thermal power is highlighted than aluminum

Decision

The answer to the question of the task will be two right statements. To do this, let's try to solve a few simple tasks using a schedule and some data. The plot of the electrical circuit is a successively connected silver and aluminum wire. Through them proceed with a constant electric current by force 2 A. The graph shows how the potential φ changes on this section of the chain when the wire is displaced along x.. Specific resistances of silver and aluminum are 0.016 μm · m and 0.028 μC · m, respectively.

The connection of the wire serial, therefore, the current of the current on each section of the chain will be the same. The electrical resistance of the conductor depends on the material from which the conductor is made, the length of the conductor, the cross-sectional area of \u200b\u200bthe wire

R. =	ρ l.	(1),
	S.

where ρ is the specific resistance of the conductor; l. - the length of the conductor; S. - Cross-section area. From the graph, it can be seen that the length of the silver wire L. C \u003d 8 m; Length Aluminum Wire L. A \u003d 14 m. Voltage on a plot of silver wire U. C \u003d Δφ \u003d 6 V - 2 V \u003d 4 V. Voltage on a plot of aluminum wire U. A \u003d Δφ \u003d 2 B - 1 B \u003d 1 V. under the condition it is known that a constant electric current 2 A occurs through the wire, knowing the voltage and current strength, we will determine the electrical resistance according to the Ohm's law for the chain section.

It is important to notice that numeric values \u200b\u200bshould be in the SI system for calculations.

Option of proper statement 2.

Check the expressions for power.

P. a \u003d. I. 2 · R. A (4);

P. a \u003d (2 a) 2 · 0.5 Ohm \u003d 2 W.

Answer:

The directory contains in full theoretical material at the rate of physics necessary for the exam. The structure of the book corresponds to the modern codifier of the elements of the content on the subject, on the basis of which examination tasks are compiled - test measurement materials (KIM) of the EGE. Theoretical material is set out in a brief, accessible form. Each topic is accompanied by examples of examination tasks corresponding to the format of the USE. This will help the teacher to organize preparations for a single state exam, and students - independently test their knowledge and readiness for the promotion of the final exam. At the end of the manual, there are answers to the tasks for self-test, which will help to schoolchildren and applicants objectively assess the level of their knowledge and the degree of preparedness for the attestation exam. The manual is addressed to senior schoolchildren, applicants and teachers.

A small object is located on the main optical axis of a thin collecting lens between the focal and double focal length from her. The object is beginning to bring the focus of the lenses. How do the size of the image and the optical power of the lenses change?

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

1. increases
2. decreases
3. does not change

Write down B. table Selected numbers for each physical size. Figures in response can be repeated.

Decision

The subject is located on the main optical axis of the thin collecting lens between the focal and double focal length from it. The object is beginning to bring the focus of the lens, while the optical power of the lens does not change, as we do not change the lens.

D. =	1	(1),
	F.

where F. - focal length of lenses; D. - optical power of lenses. To answer the question of how the image size changes, it is necessary to build an image for each position.

Fig. 1

Fig. 2.

Built two images for two positions of the subject. Obviously, the size of the second image has increased.

Answer:13.

The figure shows the DC chain. Internal resistance of the current source can be neglected. Install the correspondence between physical quantities and formulas by which they can be calculated (- EMF of the current source; R. - resistance of the resistor).

To each position of the first column, select the appropriate position of the second and write down table Selected numbers under the appropriate letters.

Decision

Fig.1

Under the condition of the task in the internal resistance of the source neglege. The scheme contains a source of direct current, two resistors, resistance R., Each and key. The first task condition requires to determine the current strength through the source when a closed key. If the key is closed, the two resistors will connect in parallel. Ohm's law for the full chain in this case will be:

where I. - Current power through a source when a closed key;

where N. - The number of conductors connected in parallel with the same resistance.

- EMF of the current source.

We substitute (2) in (1) we have: This is a formula for a number 2).

According to the second condition of the task, the key must be opened, then the current will only go through one resistor. Ohm's law for the total chain in this case will be the form:

Decision

We write a nuclear reaction for our case:

As a result of this reaction, the law of preserving the charge and mass number is performed.

Z. = 92 – 56 = 36;

M. = 236 – 3 – 139 = 94.

Consequently, the charge of the nucleus 36, and the mass number of the nucleus 94.

The new reference book contains all theoretical material at the rate of physics necessary for the commissioning of a single state exam. It includes all elements of the content being checked by control and measuring materials, and helps to summarize and systematize the knowledge and skills of the school courses of physics. Theoretical material is set forth in a brief and accessible form. Each topic is accompanied by examples of test tasks. Practical tasks comply with the format of the USE. At the end of the manual, answers are answered to tests. The manual is addressed to schoolchildren, applicants and teachers.

Period T.the half-life of potassium isotope is 7.6 min. Initially, the sample contained 2.4 mg of this isotope. How much of this isotope will remain in the sample in 22.8 min.?

Answer: _________ mg.

Decision

The task of using the law of radioactive decay. It can be written in the form

where m. 0 - the initial mass of the substance, t. - the time for which the substance disintegates, T. - half life. Substitute numeric values

Answer: 0.3 mg.

A bunch of monochromatic light falls on the metal plate. In this case, the phenomenon of the photo effect is observed. On the charts in the first column, the dependences of the energy from the wavelength λ and the frequency of the light ν are presented. Set the match between the schedule and the energy for which it can determine the presented dependency.

To each position of the first column, select the appropriate position from the second column and write down table Selected numbers under the appropriate letters.

Decision

It is useful to remember the identification of the photo effect. This is the phenomenon of the interaction of light with the substance, as a result of which the photon energy is transmitted to the electrons of the substance. There are external and internal photoeffs. In our case, we are talking about an external photo effect. When the electrons from the substance occurs under the action of light. The operation of exit depends on the material from which the photocell photocathode is made, and does not depend on the frequency of light. The energy of falling photons is proportional to the frequency of light.

E.= h.ν (1)

where λ is the wavelength of light; from - light speed,

Substitute (3) in (1) we get

We analyze the resulting formula. Obviously, with an increase in wavelength, the energy of falling photons decreases. This type of dependence corresponds to the schedule under the letter a)

We write Einstein equation for photo effect:

h.ν = A. out +. E. K (5),

where h.ν - the photon energy falling on the photocathode, A. exit work, E. K is the maximum kinetic energy of photoelectrons departing from a photocathode under the action of light.

From formula (5) express E. K \u003d. h.ν – A. out (6), therefore, with increasing frequency of falling light the maximum kinetic energy of photoelectrons increases.

Red border

ν kr \u003d	A. out	(7),
	h.

this is the minimum frequency at which the photo effect is still possible. The dependence of the maximum kinetic energy of photoelectrons on the frequency of the incident light is reflected by the schedule under the letter B).

Answer:

Determine the readings of the ammeter (see Figure) if the error of direct measurement of current force is equal to the ammeter division price.

Answer: (___________ ± ___________) A.

Decision

The task checks the skill to record the testing of the measuring instrument, taking into account the specified measurement error. Determine the price of dividing the scale from \u003d (0.4 A - 0.2 A) / 10 \u003d 0.02 A. Measurement error by condition is equal to the price of division, i.e. Δ. I. = c. \u003d 0.02 A. The final result is recorded in the form:

I. \u003d (0.20 ± 0.02) a

It is necessary to collect an experimental setup, with which you can determine the coefficient of friction of the sliding steel on the tree. For this, the schoolboy took a steel bar with a crochet. What two objects from the list below should be additionally used to carry out this experiment?

1. wooden rail
2. dynamometer
3. beaker
4. plastic Reika
5. stopwatch

In response, write down the numbers of selected items.

Decision

The task requires to determine the friction coefficient of steel coefficient by wood, so it is necessary to take a wooden ruler and a dynamometer from the proposed equipment list of equipment, to measure force. It is useful to remember the formula for calculating the sliding friction force module

F ck. = μ · N. (1),

where μ is the spray friction coefficient, N. - Support reaction force equal to the body weight module.

Answer:

The directory contains detailed theoretical material on all topics verified by the physics. After each section, there are multi-level tasks in uniform. For the final knowledge control at the end of the reference, there are training options that correspond to the exam. Students do not have to look for additional information on the Internet and buy other benefits. In this directory, they will find everything necessary for independent and effective preparation for the exam. The directory is addressed to students of high schools to prepare for the exam in physics. The manual contains detailed theoretical material on all the topics verified by the exam. After each section, there are examples of the assignments of the USE and the training test. All tasks are answers. The publication will be useful to teachers of physics, parents for effective training of students to the exam.

Consider a table containing information about bright stars.

Name Star	Temperature, TO	Weight (in the masses of the sun)	Radius (in the radius of the Sun)	Distance to Star (St. Year)
Aldebaran.		5
Bethelgeuse

Choose twoapprovals that correspond to the characteristics of stars.

1. The surface temperature and radius of Bethelgeuse say that this star belongs to red superdigants.
2. The temperature on the survey surface is 2 times lower than on the surface of the Sun.
3. Stars Castor and Capella are at the same distance from the Earth and, therefore, belong to one constellation.
4. Star Vega relates to white stars of spectral class A.
5. Since the masses of the stars of Vega and Chapel are the same, they relate to the same spectral class.

Decision

Name Star	Temperature, TO	Weight (in the masses of the sun)	Radius (in the radius of the Sun)	Distance to Star (St. Year)
Aldebaran.
Bethelgeuse
			2,5

In the task, you need to choose two true statements that correspond to the characteristics of stars. From the table it is clear that the most low temperature And the big radius of Betelgeuse, it means that this star belongs to the red giants. Consequently, the right answer (1). To correctly choose the second approval, you need to know the distribution of stars by spectral classes. We need to know the temperature range and corresponding to this temperature color of the star. Analyzing the data of the table, we conclude that the faithful statement will (4). Star Vega relates to white stars of spectral class A.

The shell weighing 2 kg, flying at a speed of 200 m / s, breaks into two fragments. The first fragment of a mass of 1 kg flies at an angle of 90 ° to the initial direction at a speed of 300 m / s. Find the speed of the second fragment.

Answer: _______ m / s.

Decision

At the moment of breaking the projectile (Δ t. → 0) The action of gravity can be neglected and consider the projectile as a closed system. According to the law of preservation of the impulse: the vector sum of the pulses of bodies, which are included in the closed system, remains constant for any interactions of the body of this system among themselves. For our case, we write:

- the speed of the projectile; m. - the mass of the projectile until the gap; - the speed of the first fragment; m. 1 - the mass of the first fragment; m. 2 - the mass of the second fragment; - The speed of the second fragment.

Choose a positive axis direction H.coinciding with the direction of the speed of the projectile, then in the projection on this axis, the equation (1) write:

mV X. = m. 1 v. 1x. + m. 2 v. 2x. (2)

Under the condition, the first shard flies at an angle of 90 ° to the initial direction. The length of the desired pulse vector is determined by the Pythagora theorem for a rectangular triangle.

p. 2 = √p. 2 + p. 1 2 (3)

p. 2 \u003d √400 2 + 300 2 \u003d 500 (kg · m / s)

Answer: 500 m / s.

In compressing the perfect single-nuclear gas at constant pressure, the external forces made a job of 2000 J. What amount of warmth was transmitted with the gas surrounding bodies?

Answer: _____ J.

Decision

Task for the first law of thermodynamics.

Δ U. = Q. + A. Sun, (1)

Where Δ. U. – change in the internal energy of gas, Q. - the amount of heat transmitted by gas surrounding bodies, A. Sun is the work of external forces. By condition, the gas is single andomic and compress it at constant pressure.

A. Sun \u003d - A. g (2),

Q. = Δ U.– A. Sun \u003d Δ. U.+ A. r \u003d.	3	p.Δ V. + p.Δ V. =	5	p.Δ V.,
	2		2

where p.Δ V. = A. G.

Answer: 5000 J.

Flat monochromatic light wave with a frequency of 8.0 · 10 14 Hz drops on the normal on the diffraction grid. In parallel with the grid behind it placed the collecting lens with a focal length of 21 cm. The diffraction pattern is observed on the screen in the rear focal plane of the lens. The distance between its main maxima of the 1st and 2nd orders is 18 mm. Find the lattice period. Answer express in micrometers (μm), rounded to the tenths. Read for small angles (φ ≈ 1 in radians) TGα ≈ sinφ ≈ φ.

Decision

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 per one of the maxima of the diffraction pattern λ - the length of the light wave, k. - an integer called a diffraction maximum. Express the diffraction lattice period from equation (1)

Fig. one

By the condition of the problem, we know the distance between its main maxima of the 1st and 2nd, we denote it as Δ x. \u003d 18 mm \u003d 1.8 · 10 -2 m, frequency of light wave ν \u003d 8.0 · 10 14 Hz, focal length lenses F. \u003d 21 cm \u003d 2.1 · 10 -1 m. We need to determine the period of the diffraction lattice. In fig. 1 shows the scheme of the ray through the grille and the lens standing behind it. On the screen in the focal plane of the collecting lens, a diffraction pattern is observed, as the result of the interference of the waves coming from all the cracks. We use the formula one for two maxima of the 1st and 2nd order.

d.sinφ 1 \u003d. k.λ (2),

if a k. \u003d 1, then d.sinφ 1 \u003d λ (3),

similarly write to for k. = 2,

Since the angle φ is small, tgφ ≈ sinφ. Then from fig. 1 see that

where x. 1 is the distance from the zero maximum, to the first order maximum. Similar to distance x. 2 .

Then have

Diffraction lattice period

as by definition

where from \u003d 3 · 10 8 m / s - speed of light, then substituting the numeric values

The answer was presented in micrometers, rounded to the tenths, as required in the condition of the task.

Answer: 4.4 microns.

Based on the laws of physics, find the ideal voltmeter reading in the diagram shown in the figure, before closing the key to and describe changes its testimony after closing the K. K. Initially, the capacitor is not charged.

Decision

Fig. one

The tasks of part C require a student full and unfolded response. Based on the laws of physics, it is necessary to determine the testimony of the voltmeter before closing the key to and after closing the key K. We take into account that the condenser in the chain is not charged. Consider two states. When the key is open, only a resistor is connected to the power source. The voltmeter readings are zero, as it is connected parallel to the condenser, and the condenser is originally charged, then q. 1 \u003d 0. The second state when the key is closed. Then the voltmeter readings will increase until the maximum value is achieved, which will not change with time,

where r. - Internal source resistance. Voltage on the condenser and resistor, according to the Ohma law for the chain section U. = I. · R.over time, it will not change, and the testimony of the voltmeter will no longer change.

Wooden ball tied the thread to the bottom of the cylindrical vessel with the bottom area S.\u003d 100 cm 2. In the vessel pour water so that the ball is completely immersed in the liquid, while the thread is stretched and acts on the ball with force T.. If you cut the thread, the ball pops up, and the water level will change to h. \u003d 5 cm. Find the strength of the thread tension T..

Decision

Fig. one		Fig. 2.

Initially, a wooden ball is tied to the thread of the bottom of the cylindrical vessel bottom area S. \u003d 100 cm 2 \u003d 0.01 m 2 and completely immersed in water. Three strengths act on the ball: the strength of gravity from the ground, - the force of Archimedes from the liquid, is the force of tension of the thread, the result of the interaction of the ball and the thread. By the condition of the ball equilibrium in the first case, the geometric sum of all the forces acting on the ball must be zero:

Choose a coordinate axis Oy. And send it up. Then, taking into account the projection, the equation (1) write:

F A. 1 = T. + mG. (2).

Archimedes with force:

F A. 1 \u003d ρ · V. 1 g. (3),

where V. 1 - the volume of the part of the ball immersed in the water, in the first it is the volume of the whole ball, m. - The mass of the ball, ρ is the density of water. Equilibrium condition in the second case

F A. 2 \u003d Mg (4)

Archimedized in this case, with the power of Archimedes:

F A. 2 \u003d ρ · V. 2 g. (5),

where V. 2 - the volume of the ball, immersed in the fluid in the second case.

We work with equations (2) and (4). You can use the substitution method or subtract from (2) - (4), then F A. 1 – F A. 2 = T.using formulas (3) and (5) get ρ · V. 1 g.– ρ · V. 2 g.= T.;

ρg ( V. 1 – V. 2) = T. (6)

Considering that

V. 1 – V. 2 = S. · h. (7),

where h. \u003d H 1 - H. 2; Receive

T. \u003d ρ · G · S. · h. (8)

Substitute numeric values

Answer: 5 N.

All the information necessary for the examinations in physics is presented in visual and affordable tables, after each topic - training tasks for knowledge control. With this book, students will be able to increase their knowledge in the shortest possible time, in a matter of days before the exam recall all the most important topics, practice in fulfilling the tasks in the format of the EGE and become more confident in their forces. After repeating all those presented in the manual, the long-awaited 100 points will be much closer! The manual contains theoretical information on all topics verifiable for the exam in physics. After each section, the training tasks of different types with answers are given. A visual and affordable presentation of the material will quickly find the necessary information, eliminate gaps in knowledge and in the shortest possible time to repeat the large amount of information. The publication will assist high school students when preparing for lessons, various forms of current and intermediate control, as well as to prepare for exams.

Task 30.

In the room with dimensions of 4 × 5 × 3 M, in which the air has a temperature of 10 ° C and the relative humidity of 30%, included the air humidifier with a capacity of 0.2 l / h. What will the relative humidity in the room be equal to 1.5 hours? The pressure of a saturated water vapor at a temperature of 10 ° C is 1.23 kPa. The room is considered a hermetic vessel.

Decision

Getting Started to solve problems on pairs and humidity, it is always useful to keep in mind the following: if the temperature and pressure (density) of the saturating steam is set, its density (pressure) is determined from the Mendeleev equation - Klapairone. Write the Mendeleev equation - Klapaireron and the relative humidity formula for each state.

For the first case at φ 1 \u003d 30%. Partial water vapor pressure Express the formula:

where T. = t. + 273 (K), R. - Universal gas constant. Express the initial weight of the steam contained in the room using equation (2) and (3):

During the work of the humidifier, the mass of water will increase on

Δ m. = τ · ρ · I., (6)

where I.– Productivity of the humidifier under the condition it is equal to 0.2 l / h \u003d 0.2 · 10 -3 m 3 / h, ρ \u003d 1000 kg / m 3 - water density. Let the formula (4) and (5) in (6)

We transform expression and express

This is the desired formula for relative humidity, which will be in the room after the operation of the air humidifier.

Substitute numeric values \u200b\u200band get the following result

Answer:83 %.

By horizontally arranged rough rails with negligible low resistance, two identical rods mass can slide m. \u003d 100 g and resistance R. \u003d 0.1 Ohm each. The distance between the rails L \u003d 10 cm, and the friction coefficient between the rods and the rails μ \u003d 0.1. Rails with rods are in a homogeneous vertical magnetic field with induction B \u003d 1 TL (see Figure). Under the action of horizontal force acting on the first rod along the rail, both rods move forward evenly with different speeds. What is the speed of movement of the first rod relative to the second? Contour self-induction neglected.

Decision

Fig. one

The task is complicated by the fact that two rods move and need to determine the speed of the first relative to the second. Otherwise, the approach to solving the tasks of this type remains the same. The change in the magnetic flux of the flowing circuit leads to the occurrence of EDC induction. In our case, when the rods move with different speeds, the change in the flow of the magnetic induction vector, permeating the contour, during the period of time Δ t.determined by the formula

ΔΦ = B. · l. · ( v. 1 – v. 2) · Δ t. (1)

This leads to the emergence of EMF induction. According to Faraday law

By the condition of the problem by self-induction contour neglect. According to the law of Ohm for a closed chain for the current force arising in the chain, write down the expression:

The amperes force and modules of which are equal to each other, are equal to the conductors with a current in a magnetic field, and are equal to the product of the current, the module of the magnetic induction vector and the length of the conductor. Since the strength vector is perpendicular to the current direction, then sinα \u003d 1, then

F. 1 = F. 2 = I. · B. · l. (4)

On the rods still acts inhibiting the thoring force of friction,

F. Tr \u003d μ · m. · g. (5)

by condition, it is said that the rods are moving evenly, which means the geometric sum of the forces applied to each rod is zero. Only ampere power and the strength of the friction apply to the second rod F. Tr \u003d. F. 2, taking into account (3), (4), (5)

Express the relative speed from here

Substitute numeric values:

Answer: 2 m / s.

In the experience of studying the photoeffect, the light of ν \u003d 6.1 · 10 14 Hz falls on the surface of the cathode resulting in the circuit there is a current. Current Dependency Dependency I. from voltage U. Between the anode and the cathode is shown in the figure. What is the power of the falling light Rif on average one of the 20 photons falling on the cathode knocks out an electron?

Decision

By definition, the current is a physical value numerically equal to charge q.passing through the cross section of the conductor per unit time t.:

I. =	q.	(1).
	t.

If all photoelectrons, knocked out of the cathode, reach the anode, then the current in the chain reaches saturation. The full charge of the conductor passed through the cross section can be calculated

q. = N E. · e. · t. (2),

where e. - Electron Charge Module, N E.– The number of photoelectrons embroidered from the cathode for 1 s. By a condition, one of the 20 photons falling on the cathode knocks out an electron. Then

where N. F - the number of photons falling on the cathode for 1 s. The maximum current in this case will be

Our task to find the number of photons falling on the cathode. It is known that the energy of one photon is equal E. F \u003d. h. · v., then the power of the falling light

After the substitution of the corresponding values, we obtain the final formula

P. = N. f · h. · v. =

twenty · I. Max · h. EGE-2018. Physics (60x84 / 8) 10 training options for examination work for preparation for the Unified State Exam The attention of schoolchildren and applicants is offered a new physics allowance for preparation of the EGEwhich contains 10 options for training examinations. Each option is compiled in full compliance with the requirements of a unified state exam in physics, includes the tasks of different types and the level of complexity. At the end of the book there are answers for self-test for all tasks. The proposed training options will help the teacher to organize preparations for a single state exam, and students - independently test their knowledge and readiness for the delivery of the final exam. The manual is addressed to schoolchildren, applicants and teachers.

On the eve of the school year on the official website of the FIII published demo version Kim Ege 2018 in all subjects (including physics).

This section presents documents defining the structure and content of KIM EGE 2018:

Demonstration options for control measuring materials of a single state exam.
- codifiers of elements of content and requirements for the level of training of graduates of general education institutions for a single state exam;
- specifications of control measuring materials for a single state exam;

SEVERSION EGE 2018 on the physics of the task with the answers

Physics demoversion EGE 2018	variant + Otvet.
Specification	download
Codifier	download

Changes in Kim Ege in 2018 in physics compared with 2017

In the codifier of the content of the contents checked for the exam in physics, the subsection 5.4 "Elements of astrophysics" is included.

In part 1. examination Added one task with multiple selection, checking elements of astrophysics. The content of the plots 4, 10, 13, 14 and 18 is expanded. Part 2 is left unchanged. Maximum score For the implementation of all tasks of the examination work increased from 50 to 52 points.

The duration of the pro 2018 in physics

235 minutes is given to the fulfillment of the entire examination. An approximate time for performing the tasks of various parts of the work is:

1) for each task with a brief response - 3-5 minutes;

2) For each task with a detailed response - 15-20 minutes.

Structure Kim Eger

Each version of the examination work consists of two parts and includes 32 tasks that differ in the form and level of complexity.

Part 1 contains 24 tasks with a brief answer. Of these, 13 tasks with a response entry in the form of a number, words or two numbers, 11 tasks for conformity and multiple choices in which the answers must be written as a sequence of numbers.

Part 2 contains 8 tasks united by a common activity - solving problems. Of these, 3 tasks with a brief answer (25-27) and 5 tasks (28-32) for which the detailed response must be brought.

In 2018, graduates of graduates of class 11 and agencies vocational education Will pass the exam 2018 in physics. Latest newsRegarding the exam in physics in 2018 are based on the fact that some changes will be made in it, both large and insignificant.

What is the meaning of changes and how many of them

The main change relating to the exam in physics, compared with previous years, is the lack of a test part with a response choice. This means that the preparation for the USE must be accompanied by the skill of the student to give brief or deployed answers. Consequently, guess the option and score some points will not succeed and have to work seriously.

In basic part of the ege Physics added a new task 24, which requires the ability to solve the tasks of astrophysics. By adding №24, the maximum primary score rose to 52. The exam is divided into two parts by difficulty levels: the basic out of 27 tasks involving a brief or complete answer. In the second part there are 5 tasks of an elevated level, where you need to give a detailed answer and explain the course of your decision. One an important nuance: Many students skip this part, but even for trying to perform these tasks, you can get from one to two points.

All changes in the exam in physics are made with the goal to deepen the preparation and improve the learning of knowledge on the subject. In addition, the elimination of the test part motivates future applicants to accumulate the volume of knowledge more intense and arguing logically.

Exam Structure

Compared with the previous year, the structure of the USE did not undergo significant changes. 235 minutes are allocated for all work. Each task of the base part should be solved from 1 to 5 minutes. The tasks of increased complexity are solved about 5-10 minutes.

All kima are stored at the Exam's venue, the autopsy is made during the test. The structure is as follows: 27 basic tasks check the existence of knowledge from all sections of physics, from mechanics to quantum and nuclear physics. In 5 assignments of the high level of complexity, the student shows the skills in the logical substantiation of its solution and the correctness of the thoughts. The number of primary points can reach maximum 52. Then they are recalculated within the 100-point scale. Due to the change in the primary score, the minimum passing score can be changed.

Demo version

The demonstration version of the exam in physics is already lying on the official portal of FIPI, which is developing a single state exam. According to the structure and complexity, the demo version is similar to the one that will appear on the exam. Each task is described in detail, at the end there is a list of answers to questions by which the student is moresed with their decisions. Also at the end, a detailed layout for each of the five tasks is given indicating the number of points for true or partially performed actions. For each task of high complexity, you can get from 2 to 4 points, depending on the requirements and exploration of the solution. Tasks may contain a sequence of numbers that need to be correctly recorded, setting the conformity between the elements, as well as small tasks in one or two actions.

• Download demo: EGE-2018-FIZ-Demo.pdf
• Download archive with specification and codifier: EGE-2018-FIZ-DEMO.zip

We wish you successfully pass physics and go to the desired university, everything is in your hands!