PHYSICS, grade 11 2 Draft Codifier of content elements and requirements for the level of training of graduates of educational institutions for the unified state examination in PHYSICS The codifier of content elements in physics and requirements for the level of training of graduates of educational organizations for the unified state examination is one of the documents, the Unified state examination in PHYSICS determining the structure and content of the CMM USE. It is drawn up on the basis of the Federal component of state standards for basic general and secondary (complete) general education in physics (basic and specialized levels) (order of the Ministry of Education of Russia dated 05.03.2004 No. 1089). Codifier Section 1. List of content elements tested on a single content elements and requirements for the level of preparation of the state exam in physics for graduates of educational organizations for conducting The first column contains the code of the section, which corresponds to large unified state exam in physics content blocks. The second column contains the code of the content item for which the test tasks are created. Large blocks of content are broken down into smaller pieces. The code was prepared by the Federal State Budgetary Control Scientific Institution. The code is different. Elements of content, "FEDERAL INSTITUTE OF PEDAGOGICAL MEASUREMENTS" The items are checked by tasks KIM and 1 MECHANICS 1.1 KINEMATICS 1.1.1 Mechanical movement. Relativity of mechanical movement. Reference system 1.1.2 Material point. z trajectory Its radius vector:  r (t) = (x (t), y (t), z (t)),   trajectory, r1 Δ r displacement:     r2 Δ r = r (t 2) - r (t1) = (Δ x, Δ y, Δ z), O y path. Addition of displacements: x    Δ r1 = Δ r 2 + Δ r0 © 2018 Federal Service for Supervision in Education and Science Russian Federation

PHYSICS, 11 grade 3 PHYSICS, 11 grade 4 1.1.3 Speed ​​of a material point: 1.1.8 Movement of a point along a circle.   Δr  2π υ = = r "t = (υ x, υ y, υ z), Angular and linear velocity of the point: υ = ωR, ω = 2πν. Δt Δt → 0 T Δx υ2 υx = = x" t, similarly to υ y = yt ", υ z = zt". Centripetal acceleration of a point: acs = = ω2 R Δt Δt → 0 R    1.1.9 Rigid body. Translational and rotational motion Addition of velocities: υ1 = υ 2 + υ0 of a rigid body 1.1.4 Acceleration of a material point: 1.2 DYNAMICS   Δυ  a = = υt "= (ax, ay, az), 1.2.1 Inertial reference systems. First Newton's law. Δt Δt → 0 Galileo's principle of relativity Δυ x 1.2.2 m ax = = (υ x) t ", similarly to ay = (υ y)", az = (υ z) t ". Body mass. Density of matter: ρ = Δt Δt → 0 t  V   1.1.5 Uniform rectilinear motion: 1.2.3 Force. The principle of superposition of forces: Equal effective in = F1 + F2 +  x (t) = x0 + υ0 xt 1.2.4 The second  Newton's law:  for a material point in IFR    υ x (t) = υ0 x = const F = ma; Δp = FΔt at F = const 1.1.6 Uniformly accelerated rectilinear motion: 1.2.5 Newton's third law   for   a t2 material points: F12 = - F21 F12 F21 x (t) = x0 + υ0 xt + x 2 υ x (t) = υ0 x + axt 1.2.6 The law of universal gravitation: the forces of attraction between mm ax = const point masses are equal to F = G 1 2 2. R υ22x - υ12x = 2ax (x2 - x1) Gravity. Dependence of the force of gravity on the height h above 1.1.7 Free fall. y  planet surface with radius R0: Acceleration of free fall v0 GMm. Body motion, mg = (R0 + h) 2 thrown at an angle α to y0 α 1.2.7 Movement of celestial bodies and their artificial satellites. horizon: First space speed: GM O x0 x υ1к = g 0 R0 = R0  x (t) = x0 + υ0 xt = x0 + υ0 cosα ⋅ t Second space speed:   g yt 2 gt 2 2GM  y (t ) = y0 + υ0 yt + = y0 + υ0 sin α ⋅ t - υ 2 к = 2υ1к =  2 2 R0 υ x ​​(t) = υ0 x = υ0 cosα 1.2.8 Elastic force. Hooke's law: F x = - kx  υ y (t) = υ0 y + g yt = υ0 sin α - gt 1.2.9 Friction force. Dry friction. Sliding friction force: Ftr = μN gx = 0  Static friction force: Ftr ≤ μN  gy = - g = const Friction coefficient 1.2.10 F Pressure: p = ⊥ S © 2018 Federal Service for Supervision of Education and Science of the Russian Federation © 2018 Federal Service for Supervision of Education and Science of the Russian Federation

PHYSICS, 11 class 5 PHYSICS, 11 class 6 1.4.8 The law of change and conservation of mechanical energy: 1.3 STATIC E mech = E kin + E potential, 1.3.1 Moment of force about the axis in ISO ΔE fur = A all nonpotent. forces, rotation:  l M = Fl, where l is the shoulder of the force F in IFR ΔE fur = 0, if A all nonpotent. forces = 0 → O relative to the axis passing through F 1.5 MECHANICAL VIBRATIONS AND WAVES point O perpendicular to Figure 1.5.1 Harmonic vibrations. Amplitude and phase of oscillations. 1.3.2 Equilibrium conditions of a rigid body in ISO: Kinematic description: M 1 + M 2 +  = 0 x (t) = A sin (ωt + φ 0),   υ x (t) = x "t, F1 + F2 +  = 0 1.3.3 Pascal's law ax (t) = (υ x) "t = −ω2 x (t). 1.3.4 Pressure in a liquid at rest in ISO: p = p 0 + ρ gh Dynamic description:   1.3.5 Archimedes' law: FArch = - Pdisplaced. , ma x = - kx, where k = mω. 2 if the body and the liquid are at rest in IFR, then FArkh = ρ gV displaced. Energy description (the law of conservation of mechanical conditions of floating bodies mv 2 kx 2 mv max 2 kA 2 energy): + = = = сonst. 1.4 CONSERVATION LAWS IN MECHANICS 2 2 2 2   Connection of the amplitude of fluctuations of the initial value with 1.4.1 The momentum of a material point: p = mυ    the amplitudes of fluctuations of its velocity and acceleration: 1.4.2 The momentum of a system of bodies: p = p1 + p2 + ... 2 v max = ωA, a max = ω A 1.4.3 Law of change and conservation of  momentum:     in ISO Δ p ≡ Δ (p1 + p 2 + ...) = F1 external Δ t + F2 external Δ t + ; 1.5.2 2π 1   Period and frequency of oscillations: T = =.    ω ν in ISO Δp ≡ Δ (p1 + p2 + ...) = 0, if F1 external + F2 external +  = 0 Period of small free oscillations of mathematical 1.4.4 Work of force: at small displacement    l A = F ⋅ Δr ⋅ cos α = Fx ⋅ Δx α  F of the pendulum: T = 2π. Δr g Period of free oscillations of a spring pendulum: 1.4.5 Force power:  F m ΔA α T = 2π P = = F ⋅ υ ⋅ cosα  k Δt Δt → 0 v 1.5.3 Forced oscillations. Resonance. Resonance curve 1.4.6 Kinetic energy of a material point: 1.5.4 Transverse and longitudinal waves. Speed ​​mυ 2 p 2 υ Ekin = =. propagation and wavelength: λ = υT =. 2 2m ν The law of change in the kinetic energy of the system Interference and diffraction of waves of material points: in IFR ΔEkin = A1 + A2 +  1.5.5 Sound. Speed ​​of sound 1.4.7 Potential energy: 2 MOLECULAR PHYSICS. THERMODYNAMICS for potential forces A12 = E 1 potential - E 2 potential = - Δ E potential 2.1 MOLECULAR PHYSICS Potential energy of a body in a uniform gravity field: 2.1.1 Models of the structure of gases, liquids and solids E potential = mgh. 2.1.2 Thermal motion of atoms and molecules of a substance Potential energy of an elastically deformed body: 2. 1.3 Interaction of particles of matter 2.1.4 Diffusion. Brownian motion kx 2 E potential = 2.1.5 Model of an ideal gas in the ICT: gas particles move 2 chaotically and do not interact with each other science of the Russian Federation

PHYSICS, 11 class 7 PHYSICS, 11 class 8 2.1.6 Relationship between pressure and average kinetic energy 2.1.15 Change in the state of aggregation of matter: evaporation and translational thermal motion of molecules of ideal condensation, boiling of liquid gas (basic equation of MKT): 2.1.16 Change aggregate states of matter: melting and 1 2 m v2  2 crystallization p = m0nv 2 = n ⋅  0  = n ⋅ ε post 3 3  2  3 2.1.17 Energy conversion in phase transitions 2.1.7 Absolute temperature : T = t ° + 273 K 2.2 THERMODYNAMICS 2.1.8 Relationship between gas temperature and average kinetic energy 2.2.1 Thermal equilibrium and temperature of translational thermal motion of its particles: 2.2.2 Internal energy 2.2.3 Heat transfer as a way of changing internal energy m v2  3 ε post =  0  = kT without doing work. Convection, thermal conductivity,  2  2 radiation 2.1.9 Equation p = nkT 2.2.4 Amount of heat. 2.1.10 Ideal gas model in thermodynamics: Specific heat of a substance with: Q = cmΔT.  Mendeleev - Clapeyron equation 2.2.5 Specific heat of vaporization r: Q = rm.  Specific heat of fusion λ: Q = λ m.  Expression for internal energy Mendeleev – Clapeyron equation (applicable forms Specific heat of combustion of fuel q: Q = qm records): 2.2.6 Elementary work in thermodynamics: A = pΔV. m ρRT Calculation of work according to the process schedule on the pV-diagram pV = RT = νRT = NkT, p =. μ μ 2.2.7 The first law of thermodynamics: Expression for the internal energy of a monatomic Q12 = ΔU 12 + A12 = (U 2 - U 1) + A12 of an ideal gas (applicable notation forms): Adiabat: 3 3 3m Q12 = 0  A12 = U1 - U 2 U = νRT = NkT = RT = νc νT 2 2 2μ 2.2.8 The second law of thermodynamics, irreversibility 2.1.11 Dalton's law for the pressure of a mixture of rarefied gases: 2.2.9 Principles of operation of heat engines. Efficiency: p = p1 + p 2 +  A Qheat - Qcold Q 2.1.12 Isotherm (T = const): pV = const, 2.2.10 Maximum efficiency value. Carnot cycle Tload - T cold T cold p max η = η Carnot = = 1− isochore (V = const): = const, Tload Tload TV 2.2.11 Heat balance equation: Q1 + Q 2 + Q 3 + ... = 0. isobar (p = const): = const. T 3 ELECTRODYNAMICS Graphical representation of isoprocesses on pV-, pT- and VT- 3.1 ELECTRIC FIELD diagrams 3.1.1 Electrification of bodies and its manifestations. Electric charge. 2.1.13 Saturated and unsaturated vapors. Qualitative Two types of charge. Elementary electric charge. The law of the dependence of the density and pressure of saturated vapor on the conservation of the electric charge of temperature, their independence from the volume of saturated vapor 3.1.2 Interaction of charges. Point charges. Coulomb's law: pair q ⋅q 1 q ⋅q 2.1.14 Air humidity. F = k 1 2 2 = ⋅ 1 2 2 r 4πε 0 r p steam (T) ρ steam (T) Relative humidity: ϕ = = 3.1.3 Electric field. Its effect on electric charges p is saturated. steam (T) ρ sat. pair (T) © 2018 Federal Service for Supervision of Education and Science of the Russian Federation © 2018 Federal Service for Supervision of Education and Science of the Russian Federation

PHYSICS, 11 class 9 PHYSICS, 11 class 10  3.1.4  F 3.2.4 Electrical resistance. Resistance dependence Electric field strength: E =. homogeneous conductor from its length and cross-section. Specific q test l q resistance of the substance. R = ρ Field of a point charge: E r = k 2, S  r 3.2.5 Current sources. EMF and internal resistance homogeneous field: E = const. A Pictures of the lines of these fields of the current source.  = external forces 3.1.5 Potential of the electrostatic field. q Potential difference and voltage. 3.2.6 Ohm's law for full (closed) A12 = q (ϕ1 - ϕ 2) = - q Δ ϕ = qU of the electric circuit:  = IR + Ir, whence ε, r R Potential charge energy in an electrostatic field:  I = W = qϕ. R + r W 3.2.7 Parallel connection of conductors: Electrostatic potential: ϕ =. q 1 1 1 I = I1 + I 2 + , U 1 = U 2 = , = + +  Relationship between field strength and potential difference for Rparallels R1 R 2 of a uniform electrostatic field: U = Ed. Series connection of conductors: 3.1.6 Principle   superposition  electric fields: U = U 1 + U 2 + , I 1 = I 2 = , Rpos = R1 + R2 +  E = E1 + E 2 + , ϕ = ϕ 1 + ϕ 2 +  3.2.8 Work of electric current: A = IUt 3.1.7 Conductors in an electrostatic  field. Condition Joule – Lenz law: Q = I 2 Rt of equilibrium of charges: inside the conductor E = 0, inside and on 3.2.9 ΔA of the surface of the conductor ϕ = const. Electric current power: P = = IU. Δt Δt → 0 3.1.8 Dielectrics in an electrostatic field. Dielectric Thermal power dissipated across the resistor: substance permeability ε 3.1.9 q U2 Capacitor. Capacitor capacity: C =. P = I 2R =. U R εε 0 S ΔA Electric capacity of a flat capacitor: C = = εC 0 Power of the current source: P = st. forces = I d Δ t Δt → 0 3.1.10 Parallel connection of capacitors: 3.2.10 Free carriers of electric charges in conductors. q = q1 + q 2 + , U 1 = U 2 = , C parallel = C1 + C 2 +  Mechanisms of conductivity of solid metals, solutions and Series connection of capacitors: molten electrolytes, gases. Semiconductors. 1 1 1 Semiconductor diode U = U 1 + U 2 + , q1 = q 2 = , = + +  3.3 MAGNETIC FIELD C after C1 C 2 3.3.1 Mechanical interaction of magnets. A magnetic field. 3.1.11 qU CU 2 q 2 Vector of magnetic induction. Superposition principle Energy of a charged capacitor: WC = = =    2 2 2C magnetic fields: B = B1 + B 2 + . Lines of magnetic 3.2 DIRECT CURRENT FIELDS LAWS. Pattern of the lines of the stripe and horseshoe-shaped field 3. 2.1 Δq of permanent magnets Amperage: I =. Direct current: I = const. Δ t Δt → 0 3.3.2 Oersted experiment. The magnetic field of a conductor with current. For direct current q = It Picture of the field lines of a long straight conductor and 3.2.2 Conditions for the existence of an electric current. closed ring conductor, coil with current. Voltage U and EMF ε 3.2.3 U Ohm's law for the circuit section: I = R © 2018 Federal Service for Supervision of Education and Science of the Russian Federation © 2018 Federal Service for Supervision of Education and Science of the Russian Federation

PHYSICS, 11 class 11 PHYSICS, 11 class 12 3.3.3 Ampere force, its direction and magnitude: 3.5.2 The law of conservation of energy in the oscillatory circuit: FА = IBl sin α, where α is the angle between the direction CU 2 LI 2 CU max 2 LI 2  + = = max = const of the conductor and vector B 2 2 2 2 3.3.4 Lorentz force, its direction and magnitude:  3.5.3 Forced electromagnetic oscillations. Resonance  FLor = q vB sinα, where α is the angle between vectors v and B. 3.5.4 Alternating current. Production, transmission and consumption Movement of a charged particle in a uniform magnetic electric 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 Flux vector magnetic   3.5.6 Scale of electromagnetic waves. Application of n B induction: Ф = B n S = BS cos α of electromagnetic waves in technology and everyday life α 3.6 OPTICS S 3.6.1 Rectilinear propagation of light in a homogeneous medium. A ray of light 3.4.2 The phenomenon of electromagnetic induction. EMF of induction 3.6.2 Laws of light reflection. 3.4.3 Faraday's law of electromagnetic induction: 3.6.3 Construction of images in a flat mirror ΔΦ 3.6.4 Laws of refraction of light. i = - = −Φ "t Refraction of light: n1 sin α = n2 sin β. Δt Δt → 0 c 3.4.4 EMF of induction in a straight conductor of length l moving Absolute refractive index: n abs =.    v  () with velocity υ υ ⊥ l in a uniform magnetic Relative refractive index: n rel = n 2 v1 =. n1 v 2 field B:   i = Blυ sin α, where α is the angle between vectors B and υ; if Stroke rays in the prism. = n 2 λ 2 3.4.6 Ф 3.6.5 Total internal reflection. Inductance: L =, or Φ = LI. n2 I Limiting angle of total internal reflection ΔI: Self-induction. EMF of self-induction: si = - L = - LI "t 1 n n1 Δt Δt → 0 sin αpr = = 2 αpr 3.4.7 nrel n1 LI 2 Magnetic field energy of the coil with current: WL = 3.6.6 Collecting and scattering lenses. Thin lens. 2 Focal length and optical power of a thin lens: 3.5 ELECTROMAGNETIC VIBRATIONS AND WAVES 1 3.5.1 Oscillatory circuit. Free D = electromagnetic oscillations in an ideal C L F oscillatory circuit: 3.6.7 Formula of a thin lens: d 1 1 1 q (t) = q max sin (ωt + ϕ 0) + =. H  df FF  I (t) = qt ′ = ωq max cos (ωt + ϕ 0) = I max cos (ωt + ϕ 0) Increase given by 2π 1 F h Thomson's formula: T = 2π LC, whence ω = =. lens: Γ = h = f f T LC H d Connection of the amplitude of the capacitor charge with the amplitude of the current I in the oscillatory circuit: q max = 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, Grade 11 13 PHYSICS, Grade 11 14 3.6.8 The path of the beam passed through the lens at an arbitrary angle to its 5.1.4 Einstein's equation for the photoelectric effect: the main optical axis. Construction of images of a point and E of a photon = A output + E kin max, a straight line segment in collecting and scattering lenses and their hc hc systems where Ephoton = hν =, Aout = hν cr =, 3.6.9 Camera as an optical device. λ λ cr 2 Eye as an optical system mv max E kin max = = eU zap 3.6.10 Interference of light. Coherent sources. Conditions 2 observation of maxima and minima in 5.1.5 Wave properties of particles. De Broglie waves. interference pattern from two in-phase h h De Broglie wavelength of a moving particle: λ = =. coherent sources p mv λ Wave-corpuscle dualism. Electron diffraction maxima: Δ = 2m, m = 0, ± 1, ± 2, ± 3, ... on crystals 2 λ 5.1.6 Light pressure. Light pressure on a fully reflecting minima: Δ = (2m + 1), m = 0, ± 1, ± 2, ± 3, ... surface and on a completely absorbing surface 2 5.2 ATOM PHYSICS 3.6.11 Light diffraction. Diffraction grating. Condition 5.2.1 Planetary model of the atom of observation of the main maxima at normal incidence 5.2.2 Bohr's postulates. Emission and absorption of photons for monochromatic light with a wavelength λ onto a lattice with the transition of an atom from one energy level to another: period d: d sin ϕ m = m λ, m = 0, ± 1, ± 2, ± 3, ... hс 3.6.12 Dispersion of light hν mn = = En - Em λ mn 4 BASES OF THE SPECIAL THEORY OF RELATIVITY 4.1 Invariance of the modulus of the speed of light in vacuum. Principle 5.2.3 Linear spectra. Einstein's relativity Spectrum of energy levels of a hydrogen atom: 4.2 - 13.6 eV En =, n = 1, 2, 3, ... 2 Energy of a free particle: E = mc. v2 n2 1− 5.2.4 Laser c2  5.3 PHYSICS OF THE ATOMIC NUCLEUS Particle momentum: p = mv . v 2 5.3.1 Nucleon model of the Heisenberg – Ivanenko nucleus. Nuclear charge. 1 - The mass number of the kernel. Isotopes c2 4.3 Relation of mass and energy of a free particle: 5.3.2 Binding energy of nucleons in the nucleus. Nuclear forces E 2 - (pc) = (mc 2). 2 2 5.3.3 Nuclear mass defect AZ X: Δ m = Z ⋅ m p + (A - Z) ⋅ m n - m nuclei Rest energy of a free particle: E 0 = mc 2 5.3.4 Radioactivity. 5 QUANTUM PHYSICS AND ELEMENTS OF ASTROPHYSICS Alpha decay: AZ X → AZ −− 42Y + 42 He. 5.1 CORPUSCULAR-WAVE DUALISM A A 0 ~ Beta decay. Electronic β-decay: Z X → Z + 1Y + −1 e + ν e. 5.1.1 M. Planck's hypothesis about quanta. Planck's formula: E = hν Positron β-decay: AZ X → ZA − 1Y + + 10 ~ e + νe. 5.1.2 hc Gamma radiation Photons. Photon energy: E = hν = = pc. λ 5.3.5 - t E hν h The law of radioactive decay: N (t) = N 0 ⋅ 2 T Photon momentum: p = = = c c λ 5.3.6 Nuclear reactions. Fission and fusion of nuclei 5.1.3 Photoelectric effect. The experiments of A.G. Stoletov. The laws of the photoelectric effect 5.4 ELEMENTS OF ASTROPHYSICS 5.4.1 Solar system: terrestrial planets and giant planets, small bodies of the solar system © 2018 Federal Service for Supervision of Education and Science of the Russian Federation © 2018 Federal Service for Supervision of Education and Science of the Russian Federation

PHYSICS, 11 grade 15 PHYSICS, 11 grade 16 5.4.2 Stars: variety of stellar characteristics and their patterns. Sources of energy of stars 2.5.2 give examples of experiments illustrating that: 5.4.3 Modern ideas about the origin and evolution of observations and experiments serve as the basis for the advancement 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 findings; 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; physical theory makes it possible to predict yet unknown phenomena and their features; when explaining natural phenomena, Section 2 is used. The list of requirements for the level of training verified by physical models; one and the same natural object or on a single state exam in physics, a phenomenon can be investigated based on the use of different models; the laws of physics and physical theories have their own Code Requirements for the level of training of graduates, the development of certain limits of applicability the requirements of which are checked on the exam 2.5.3 measure physical quantities, present the results 1 Know / Understand: measurements taking into account their errors 1.1 the meaning of physical concepts 2.6 apply the acquired knowledge to solve physical 1.2 meaning physical quantities of tasks 1.3 the meaning of physical laws, principles, postulates 3 Use the acquired knowledge and skills in practical 2 Be able to: activity and Everyday life to: 2.1 describe and explain: 3.1 life safety during use Vehicle, household 2.1.1 physical phenomena, physical phenomena and properties of the bodies of electrical appliances, radio and telecommunications 2.1.2 results of communication experiments; assessing the impact on the human body and others 2.2 describe the fundamental experiments that have had organisms of environmental pollution; rational significant influence on the development of the physics of nature management and environmental protection; 2.3 give examples of the practical application of physical 3.2 determination of one's own position in relation to knowledge, laws of physics environmental issues and behavior in the natural environment 2.4 to determine the nature of the physical process according to the schedule, table, formula; products of nuclear reactions based on the laws of conservation of electric charge and mass number 2.5 2.5.1 to distinguish hypotheses from scientific theories; draw conclusions based on experimental data; give examples showing that: observations and experiment are the basis for the advancement of hypotheses and theories, allow you to check the truth of theoretical conclusions; physical theory makes it possible to explain well-known natural phenomena and scientific facts, to predict yet unknown phenomena; © 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

The average general education

UMK line G. Ya. Myakisheva, M.A. Petrova. Physics (10-11) (B)

Codifier of the Unified State Exam-2020 in Physics FIPI

The codifier of the content elements and requirements for the level of training of graduates of educational institutions for the USE in physics is one of the documents that determine the structure and content of the CMM of the unified state examination, the objects of the list of which have a specific code. A codifier has been compiled on the basis of the Federal component of state standards of basic general and secondary (complete) general education in physics (basic and specialized levels).

Major changes in the new demo

Most of the changes have become minor. So, in the tasks in physics there will be not five, but six questions that imply a detailed answer. The task number 24 on knowledge of the elements of astrophysics has become more complicated - now, instead of two obligatory correct answers, there can be either two or three correct options.

Soon we will talk about the upcoming USE on and on the air our YouTube channel.

Schedule of the exam in physics in 2020

On this moment it is known that the Ministry of Education and Rosobrnadzor have published the draft USE timetable for public discussion. The physics exams are expected to be held on June 4th.

The codifier is information divided into two parts:

part 1: "List of content elements tested in the unified state exam in physics";

part 2: "The list of requirements for the level of training of graduates, checked on the unified state exam in physics."

List of content elements tested in the unified state exam in physics

We present the original table with the list of content elements provided by FIPI. You can download the USE codifier in physics in full version at official website.

Section code	Controlled item code	Content Items Checked by CMM Jobs
1	Mechanics
1.1	Kinematics
1.2	Dynamics
1.3	Statics
1.4	Conservation laws in mechanics
1.5	Mechanical vibrations 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 vibrations and waves
3.6	Optics
4	Fundamentals of the special theory of relativity
5	Quantum physics and the elements of astrophysics
5.1	Wave-corpuscle dualism
5.2	Atomic physics
5.3	Atomic Nuclear Physics
5.4	Elements of astrophysics

The book contains materials for successfully passing the exam: brief theoretical information on all topics, assignments different types and levels of difficulty, solving problems of an increased level of complexity, answers and assessment criteria. Students do not have to search the Internet for additional information and buy other manuals. In this book, they will find everything they need to prepare independently and effectively for the exam.

Requirements for the level of training of graduates

KIM FIPI are developed based on specific requirements for the level of preparation of the examiners. Thus, in order to successfully cope with the physics exam, the graduate must:

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. Be able to:

2.1. describe and explain:

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

2.1.2. experimental results;

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

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

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

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 the advancement of hypotheses and theories and make it possible to check the truth of theoretical conclusions, physical theory makes it possible to explain well-known natural phenomena and scientific facts, to predict yet unknown phenomena;

2.5.2. give examples of experiments illustrating that: observations and experiment serve as the basis for hypothesis and the construction of scientific theories; the experiment allows you to check the truth of theoretical conclusions; physical theory makes it possible to explain natural phenomena and scientific facts; physical theory makes it possible to predict yet unknown phenomena and their features; when explaining natural phenomena, physical models are used; one and the same natural object or phenomenon can be investigated using different models; the laws of physics and physical theories have their own certain limits of applicability;

2.5.3. measure physical quantities, present the results of measurements, taking into account their errors;

2.6. apply the knowledge gained to solve physical problems.

3. Use the acquired knowledge and skills in practice and everyday life:

3.1. to ensure the safety of life in the process of using vehicles, household electrical appliances, radio and telecommunication means; assessing the impact on the human body and other organisms of environmental pollution; rational use of natural resources and environmental protection;

3.2. defining one's own position in relation to environmental problems and behavior in the natural environment.

Secondary general education

Preparing for the exam-2018: analysis of the demo in physics

We bring to your attention an analysis of the USE tasks in physics from the 2018 demo. The article contains explanations and detailed algorithms for solving tasks, as well as recommendations and links to useful materials that are relevant in preparation for the exam.

USE-2018. Physics. Thematic training tasks

The edition contains:
tasks of different types for all USE topics;
answers to all tasks.
The book will be useful both for teachers: it makes it possible to effectively organize the preparation of students for the Unified State Exam directly in the classroom, in the process of studying all topics, and for students: training tasks will allow you to systematically prepare for the exam when passing each topic.

A stationary point body begins to move along the axis Ox... The figure shows a graph of the dependence of the projection ax acceleration of this body from time to time t.

Determine which path the body has traveled in the third second of the movement.

Answer: _________ m.

Solution

Being able to read graphs is very important for every student. The question in the problem is that it is required to determine from the graph of the dependence of the projection of acceleration on time, the path that the body has traveled in the third second of motion. the graph shows that in the time interval from t 1 = 2 s to t 2 = 4 s, acceleration projection is zero. Consequently, the projection of the resultant force in this area, according to Newton's second law, is also zero. Determine the nature of the movement in this area: the body moved evenly. The path is easy to determine, knowing the speed and time of movement. However, in the interval from 0 to 2 s, the body moved uniformly. Using the definition of acceleration, we write the equation for the projection of velocity V x = V 0x + a x t; since the body was initially at rest, the projection of the velocity by the end of the second second became

Then the path traversed by the body in a third second

Answer: 8 m.

Rice. 1

On a smooth horizontal surface there are two bars connected by a light spring. To a bar with a mass m= 2 kg apply a constant force equal in modulus F= 10 N and directed horizontally along the axis of the spring (see figure). Determine the modulus of elasticity of the spring at the moment when this bar moves with an acceleration of 1 m / s 2.

Answer: _________ N.

Solution

Horizontally on a body with a mass m= 2 kg two forces act, this is the force F= 10 N and the elastic force from the side of the spring. The resultant of these forces imparts acceleration to the body. Choose a coordinate line and direct it along the action of the force F... Let's write down Newton's second law for this body.

Projected onto axis 0 NS: F – F control = ma (2)

Let us express from formula (2) the modulus of the elastic force F control = F – ma (3)

Substitute the numerical values ​​into the formula (3) and get, F control = 10 N - 2 kg 1 m / s 2 = 8 N.

Answer: 8 N.

Assignment 3

A body weighing 4 kg, located on a rough horizontal plane, was told along it a speed of 10 m / s. Determine the modulus of work performed by the friction force from the moment the body begins to move until the moment when the body's speed decreases by 2 times.

Answer: _________ J.

Solution

The body is affected by gravity, the reaction force of the support is the friction force that creates the braking acceleration. The body was initially told a speed of 10 m / s. Let's write Newton's second law for our case.

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

N – mg = 0; N = mg (2)

Projected onto the axis X: –F tr = - ma; F tr = ma; (3) We need to determine the modulus of work of the friction force by the time when the speed becomes two times less, i.e. 5 m / s. Let's write down the formula for calculating the work.

A · ( F tr) = - F tr S (4)

To determine the distance traveled, take the timeless formula:

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

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

Then the modulus of work of the friction force will be equal to:

Substitute numerical values

A(F tr) =		4 Kg	((	5 m	) 2 – (10	m	) 2)	= 150 J
		2		with		with

Answer: 150 J.

USE-2018. Physics. 30 training options for exam papers

The edition contains:
30 training options for the exam
instruction for implementation and evaluation criteria
answers to all tasks
Training options will help the teacher organize preparation for the exam, and the students will independently test their knowledge and readiness for the final exam.

The stepped block has an outer pulley with a radius of 24 cm. Weights are suspended from the threads wound on the outer and inner pulleys as shown in the figure. There is no friction in the block axis. What is the radius of the block's inner pulley if the system is in equilibrium?

Rice. 1

Answer: _________ see.

Solution

According to the condition of the problem, the system is in equilibrium. On the image L 1, shoulder strength L 2 shoulder of force Equilibrium condition: the moments of forces rotating bodies clockwise should be equal to the moments of forces rotating the body counterclockwise. Recall that the moment of force is the product of the modulus of force per shoulder. The forces acting on the thread from the side of the weights differ by a factor of 3. This means that the radius of the inner pulley of the block also differs from the outer one by 3 times. Hence the shoulder L 2 will be equal to 8 cm.

Answer: 8 cm.

Assignment 5

Oh, at different points in time.

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

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

Solution

The table shows data on the position of a ball attached to a spring and oscillating along a horizontal axis. Oh, at different points in time. We need to analyze this data and choose two statements correctly. The system is a spring-loaded pendulum. At a moment in time t= 1 s, the displacement of the body from the equilibrium position is maximum, which means this is the amplitude value. by definition, the potential energy of an elastically deformed body can be calculated by the formula

E p = k	x 2	,
	2

where k- coefficient of spring stiffness, NS- displacement of the body from the equilibrium position. If the displacement is maximum, then the velocity at this point is zero, which means that the kinetic energy will be zero. According to the law of conservation and transformation of energy, potential energy should be maximum. From the table we see that the body passes half of the vibration in t= 2 s, full oscillation in twice as long T= 4 s. Therefore, statements 1 will be true; 2.

Assignment 6

A small piece of ice was dropped into a cylindrical glass of water. After a while, the piece of ice completely melted. Determine how the pressure on the bottom of the glass and the level of water in the glass have changed as a result of the melting of the ice.

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

Write in table

Solution

Rice. 1

Problems of this type are quite common in different variants of the exam... And as practice shows, students often make mistakes. We will try to analyze this task in detail. We denote m Is the mass of a piece of ice, ρ l is the density of ice, ρ in is the density of water, V pcht - the volume of the submerged part of the ice, equal to the volume of the displaced liquid (the volume of the hole). Let's mentally remove the ice from the water. Then a hole will remain in the water, the volume of which is equal to V pht, i.e. the volume of water displaced by a piece of ice Fig. 1( b).

Let us write down the condition of ice floating in Fig. 1( a).

F a = mg (1)

ρ in V pht g = mg (2)

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

USE-2018. Physics. Training tasks

The publication is addressed to high school students to prepare for the exam in physics.
The manual includes:
20 training options
answers to all tasks
USE answer forms for each option.
The publication will assist teachers in preparing students for the exam in physics.

The weightless spring is on a smooth horizontal surface and is attached to the wall at one end (see figure). At some point in time, the spring begins to deform, applying an external force to its free end A and uniformly moving point A.

Establish correspondence between graphs of dependences of physical quantities on deformation x springs and these values. For each position of the first column, select the corresponding position from the second column and write in table

Solution

It can be seen from the figure to the problem that when the spring is not deformed, then its free end, and, accordingly, point A are in the position with the coordinate NS 0. At some point in time, the spring begins to deform, applying an external force to its free end A. At the same time, point A moves evenly. Depending on whether the spring is stretched or compressed, the direction and magnitude of the elastic force arising in the spring will change. Accordingly, under the letter A), the graph is the dependence of the modulus of the elastic force on the deformation of the spring.

The graph under the letter B) is the dependence of the projection of the external force on the amount of deformation. Because with an increase in the external force, the amount of deformation and the elastic force increase.

Answer: 24.

Assignment 8

When constructing the Reaumur temperature scale, it is assumed that at normal atmospheric pressure, ice melts at a temperature of 0 degrees Reaumur (° R), and water boils at a temperature of 80 ° R. Find what is the average kinetic energy of translational thermal motion of an ideal gas particle at a temperature of 29 ° R. Express your answer in eV and round to hundredths.

Answer: ________ eV.

Solution

The problem is interesting in that it is necessary to compare two scales for measuring temperature. These are the Reaumur temperature scale and the Celsius scale. The melting points of ice are the same on the scales, and the boiling points are different, we can get a formula for converting from Reaumur degrees to Celsius degrees. it

Convert temperature 29 (° R) to degrees Celsius

We convert the obtained result to Kelvin using the formula

T = t° C + 273 (2);

T= 36.25 + 273 = 309.25 (K)

To calculate the average kinetic energy of the translational thermal motion of ideal gas particles, we use the formula

where k- Boltzmann constant equal to 1.38 · 10 -23 J / K, T- absolute temperature on the Kelvin scale. It can be seen from the formula that the dependence of the average kinetic energy on temperature is straight, that is, how many times the temperature changes, how many times the average kinetic energy of the thermal motion of molecules changes. Substitute the numerical values:

The result is converted into electron volts and rounded to the nearest hundredth. Recall that

1 eV = 1.6 · 10 -19 J.

For this

Answer: 0.04 eV.

One mole of a monatomic ideal gas participates in process 1–2, the graph of which is shown in VT-chart. Determine for this process the ratio of the change in the internal energy of the gas to the amount of heat imparted to the gas.

Answer: ___________ .

Solution

According to the condition of the problem in the process 1–2, the graph of which is shown on VT-diagram, one mole of a monoatomic ideal gas is involved. To answer the question of the problem, it is necessary to obtain expressions for the change in the internal energy and the amount of heat imparted to the gas. The process is isobaric (Gay-Lussac's law). The change in internal energy can be written in two forms:

For the amount of heat imparted to the gas, we write down the first law of thermodynamics:

Q 12 = A 12 + Δ U 12 (5),

where A 12 - gas work during expansion. By definition, work is

A 12 = P 0 2 V 0 (6).

Then the amount of heat will be equal, taking into account (4) and (6).

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

Let's write the relation:

Answer: 0,6.

The handbook contains in full theoretical material in the course of physics, required for passing the exam. The structure of the book corresponds to the modern codifier of content elements in the subject, on the basis of which examination tasks are drawn up - control and measuring materials (CMM) of the exam. The theoretical material is presented in a concise, accessible form. Each topic is accompanied by examples of exam assignments that correspond to the USE format. This will help the teacher organize the preparation for the unified state exam, and the students independently test their knowledge and readiness for the final exam.

A blacksmith forges an iron horseshoe weighing 500 g at a temperature of 1000 ° C. When he finishes forging, he throws the horseshoe into a vessel of water. A hiss is heard and steam rises from the vessel. Find the mass of water that evaporates when a hot horseshoe is immersed in it. Assume that the water is already heated to the boiling point.

Answer: _________

Solution

To solve the problem, it is important to remember the heat balance equation. If there are no losses, then heat transfer of energy occurs in the system of bodies. As a result, the water evaporates. Initially, the water was at a temperature of 100 ° C, which means that after the immersion of the hot horseshoe, the energy received by the water will go directly to vaporization. Let us write down the heat balance equation

with f m NS · ( t n - 100) = Lm in 1),

where L- specific heat of vaporization, m c - the mass of water that has turned into steam, m n is the mass of the iron horseshoe, with g - specific heat capacity of iron. From formula (1), we express the mass of water

When writing down the answer, pay attention to what units you want to leave the mass of water.

Answer: 90 g

One mole of a monatomic ideal gas participates in a cyclic process, the graph of which is shown in TV- diagram.

Please select two correct statements based on the analysis of the presented schedule.

1. Gas pressure in state 2 is greater than gas pressure in state 4
2. Gas work in section 2-3 is positive.
3. In section 1–2, the gas pressure increases.
4. In section 4-1, a certain amount of heat is removed from the gas.
5. The change in the internal energy of the gas in the section 1–2 is less than the change in the internal energy of the gas in the section 2–3.

Solution

This type of task tests the ability to read graphs and explain the presented dependence of physical quantities. It is important to remember how the dependence plots for isoprocesses look in different axes, in particular R= const. In our example, on TV- the diagram shows two isobars. Let's see how the pressure and volume will change 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 means P 1 > P 4 . State 2 corresponds to pressure P 1 . Consequently, the gas pressure in state 2 is greater than the gas pressure in state 4. In section 2–3, the process is isochoric, the gas does not perform work, it is equal to zero. The statement is incorrect. In section 1-2, the pressure increases, also incorrect. We have just shown above that this is an isobaric transition. In section 4-1, a certain amount of heat is removed from the gas in order to maintain the temperature constant when the gas is compressed.

Answer: 14.

The heat engine operates according to the Carnot cycle. The temperature of the heat engine cooler was increased, leaving the heater temperature the same. The amount of heat received by the gas from the heater during the cycle did not change. How did the efficiency of the heat engine and the work of the gas change during the cycle?

For each value, determine the corresponding change pattern:

1. increased
2. decreased
3. hasn't changed

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

Solution

Heat engines operating according to the Carnot cycle are often found in the tasks on the exam. First of all, you need 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 to be able to write down the efficiency in terms of the useful work of the gas A g and the amount of heat received from the heater Q n.

We carefully read the condition and determined what parameters we changed: in our case, we increased the temperature of the refrigerator, leaving the heater temperature the same. Analyzing formula (1), we conclude that the numerator of the fraction decreases, the denominator does not change, therefore, the efficiency of the heat engine decreases. If we work with formula (2), then we will immediately answer the second question of the problem. The gas work per cycle will also decrease, with all current changes in the parameters of the heat engine.

Answer: 22.

Negative charge - qQ and negative - Q(see figure). Where is it directed relative to the figure ( to the right, left, up, down, towards the observer, from the observer) charge acceleration - q in this moment of time, if only charges act on it + Q and –Q? Write down the answer in a word (words)

Solution

Rice. 1

Negative charge - q is in the field of two stationary 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 moment of time when only charges + Q and - Q it is necessary to find the direction of the resulting force, as a geometric sum of forces according to Newton's second law, it is known that the direction of the acceleration vector coincides with the direction of the resulting force. The figure shows a geometric construction to determine the sum of two vectors. The question arises why the forces are directed in this way? Let's remember how similarly charged bodies interact, they repel, force The Coulomb force of interaction of charges is the central force. the force with which oppositely charged bodies are attracted. From the figure we see that the charge is q equidistant from stationary charges, the modules of which are equal. Therefore, the modulus will also be equal. The resulting force will be directed relative to the drawing way down. Charge acceleration will also be directed - q, i.e. way down.

Answer: Way down.

The book contains materials for successfully passing the exam in physics: brief theoretical information on all topics, assignments of different types and levels of difficulty, solving problems of an increased level of complexity, answers and assessment criteria. Students do not have to search the Internet for additional information and buy other manuals. In this book, they will find everything they need to prepare independently and effectively for the exam. The publication contains assignments of various types on all topics tested in the exam in physics, as well as solving problems of an increased level of complexity. The publication will provide invaluable assistance to students in preparation for the exam in physics, and can also be used by teachers in organizing the educational process.

Two series-connected resistors of 4 ohms and 8 ohms are connected to a battery with a voltage at the terminals of 24 V. What thermal power is released in a resistor of a lower rating?

Answer: _________ Tue.

Solution

To solve the problem, it is advisable to draw a diagram of a series connection of resistors. Then remember the laws of the serial connection of conductors.

The scheme will be as follows:

Where R 1 = 4 Ohm, R 2 = 8 ohms. The voltage at the battery terminals is 24 V. When the conductors are connected in series at each section of the circuit, the current will be the same. The total resistance is defined as the sum of the resistances of all resistors. According to Ohm's law for a section of the circuit, we have:

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

P = I 2 R= (2 A) 2 4 ohms = 16 W.

Answer: P= 16 W.

A wire frame with an area of ​​2 · 10 –3 m 2 rotates in a uniform magnetic field around an axis perpendicular to the magnetic induction vector. The magnetic flux penetrating the frame area changes according to the law

Ф = 4 · 10 –6 cos10π t,

where all quantities are expressed in SI units. What is the modulus of magnetic induction?

Answer: ________________ mTl.

Solution

The magnetic flux changes according to the law

Ф = 4 · 10 –6 cos10π t,

where all quantities are expressed in SI units. You need to understand what a magnetic flux is in general and how this value is related to the modulus of magnetic induction. B and frame area S... Let's write the equation in general form in order to understand what quantities are included in it.

Φ = Φ m cosω t(1)

We remember that before the cos or sin sign there is an amplitude value of a changing value, which means Φ max = 4 · 10 -6 Wb, on the other hand, the magnetic flux is equal to the product of the magnetic induction modulus by the area of ​​the circuit and the cosine of the angle between the normal to the circuit and the magnetic induction vector Φ m = V · S cosα, the flux is maximum at cosα = 1; express the modulus of induction

The answer must be recorded in mT. Our result is 2 mT.

Answer: 2.

The section of the electrical circuit consists of silver and aluminum wires connected in series. A constant electric current of 2 A flows through them.The graph shows how the potential φ changes in this section of the circuit when it is displaced along the wires by a distance x

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

1. The cross-sectional areas of the wires are the same.
2. Cross-sectional area of ​​a silver wire 6.4 · 10 –2 mm 2
3. Cross-sectional area of ​​a silver wire 4.27 · 10 –2 mm 2
4. A thermal power of 2 W is generated in the aluminum wire.
5. Silver wire produces less heat output than aluminum wire.

Solution

The answer to the question in the problem will be two correct statements. To do this, let's try to solve a few simple problems using a graph and some data. The section of the electrical circuit consists of silver and aluminum wires connected in series. A constant electric current of 2 A flows through them.The graph shows how the potential φ changes in this section of the circuit when it is displaced along the wires by a distance x... The specific resistances of silver and aluminum are equal to 0.016 μOhm · m and 0.028 μOhm · m, respectively.

The connection of the wires is serial, therefore, the current strength in each section of the circuit will be the same. The electrical resistance of a conductor depends on the material of which the conductor is made, the length of the conductor, the cross-sectional area of ​​the wire

R =	ρ l	(1),
	S

where ρ is the resistivity of the conductor; l- conductor length; S- cross-sectional area. The graph shows that the length of the silver wire L s = 8 m; length of aluminum wire L a = 14 m. Voltage on a section of silver wire U s = Δφ = 6 V - 2 V = 4 V. Voltage in the section of aluminum wire U a = Δφ = 2 V - 1 V = 1 V. According to the condition, it is known that a constant electric current of 2 A flows through the wires, knowing the voltage and current strength, we determine the electrical resistance according to Ohm's law for a section of the circuit.

It is important to note that the numerical values ​​must be in SI for calculations.

Correct statement option 2.

Let's check the expressions for the cardinality.

P a = I 2 R a (4);

P a = (2 A) 2 0.5 Ohm = 2 W.

Answer:

The handbook contains in full the theoretical material on the physics course required for passing the exam. The structure of the book corresponds to the modern codifier of content elements in the subject, on the basis of which examination tasks are drawn up - control and measuring materials (CMM) of the exam. The theoretical material is presented in a concise, accessible form. Each topic is accompanied by examples of exam assignments that correspond to the USE format. This will help the teacher organize the preparation for the unified state exam, and the students independently test their knowledge and readiness for the final exam. At the end of the manual, answers to tasks for self-examination are given, which will help students and applicants to objectively assess their level of knowledge and the degree of preparedness for the certification exam. The manual is addressed to senior students, applicants and teachers.

A small object is located on the main optical axis of a thin converging lens between focal length and double focal length from it. The subject begins to move closer to the focus of the lens. How do the size of the image and the optical power of the lens change?

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

1. increases
2. decreases
3. does not change

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

Solution

The object is located on the main optical axis of a thin converging lens between focal length and double focal length from it. The object begins to be brought closer to the focus of the lens, while the optical power of the lens does not change, since we do not change the lens.

D =	1	(1),
	F

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

Rice. 1

Rice. 2

Constructed two images for two positions of the object. Obviously, the size of the second image has increased.

Answer: 13.

The figure shows a DC circuit. The internal resistance of the current source can be neglected. Establish a correspondence between physical quantities and formulas by which they can be calculated (- EMF of the current source; R Is the resistance of the resistor).

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

Solution

Rice.1

By the condition of the problem, the internal resistance of the source is neglected. The circuit contains a constant current source, two resistors, resistance R, each and the key. The first condition of the problem requires determining the current through the source with a closed switch. If the key is closed, then the two resistors will be connected in parallel. Ohm's law for a complete circuit in this case will look like:

where I- current through the source when the key is closed;

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

- EMF of the current source.

Substituting (2) in (1) we have: this is the formula under the number 2).

According to the second condition of the problem, the key must be opened, then the current will flow only through one resistor. Ohm's law for a complete circuit in this case will be:

Solution

Let's write down the nuclear reaction for our case:

As a result of this reaction, the law of conservation of charge and mass numbers is fulfilled.

Z = 92 – 56 = 36;

M = 236 – 3 – 139 = 94.

Therefore, the charge of the nucleus is 36, and the mass number of the nucleus is 94.

The new handbook contains all the theoretical material on the physics course required to pass the unified state exam. It includes all the elements of the content checked by control and measuring materials, and helps to generalize and systematize the knowledge and skills of the school physics course. The theoretical material is presented in a concise and accessible form. Each topic is accompanied by examples of test items. Practical assignments correspond to the format of the Unified State Exam. At the end of the manual you will find the answers to the tests. The manual is addressed to schoolchildren, applicants and teachers.

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

Answer: _________ mg.

Solution

The problem of using the law of radioactive decay. It can be written as

where m 0 - the initial mass of the substance, t- the time during which the substance disintegrates, T- half life. Substitute numerical values

Answer: 0.3 mg.

A beam of monochromatic light falls on the metal plate. In this case, the phenomenon of the photoelectric effect is observed. The graphs in the first column show the dependence of the energy on the wavelength λ and the frequency of light ν. Establish a correspondence between the graph and the energy for which it can determine the presented dependence.

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

Solution

It is useful to remember the definition of photoeffect. This is the phenomenon of the interaction of light with matter, as a result of which the energy of the photons is transferred to the electrons of the matter. Distinguish between external and internal photoelectric effect. In our case, we are talking about an external photoelectric effect. When, under the influence of light, electrons are pulled out of the substance. The work function depends on the material from which the photocell photocathode is made, and does not depend on the light frequency. The energy of the incident photons is proportional to the frequency of the light.

E= hν (1)

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

Substitute (3) into (1) We get

We analyze the resulting formula. Obviously, with increasing wavelength, the energy of incident photons decreases. This type of dependence corresponds to the graph under the letter A)

Let's write down the Einstein equation for the photoelectric effect:

hν = A out + E to (5),

where hν is the energy of the photon incident on the photocathode, A out - work function, E k is the maximum kinetic energy of photoelectrons emitted from the photocathode under the action of light.

From formula (5), we express E k = hν – A out (6), therefore, with an increase in the frequency of the incident light the maximum kinetic energy of photoelectrons increases.

Red border

ν cr =	A out	(7),
	h

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

Answer:

Determine the ammeter readings (see figure) if the error in direct current measurement is equal to the division value of the ammeter.

Answer: (___________ ± ___________) A.

Solution

The task tests the ability to record the readings of the measuring device, taking into account the specified measurement error. Determine the scale division value with= (0.4 A - 0.2 A) / 10 = 0.02 A. According to the condition, the measurement error is equal to the division price, i.e. Δ I = c= 0.02 A. The final result is written in the form:

I= (0.20 ± 0.02) A

It is necessary to assemble an experimental setup with which it is possible to determine the coefficient of sliding friction of steel on wood. for this, the student took a steel bar with a hook. Which two items from the list of equipment below should be used additionally for this experiment?

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

In response, write down the numbers of the selected items.

Solution

In the task, it is required to determine the coefficient of sliding friction of steel on wood, therefore, for the experiment, it is necessary to take a wooden ruler and a dynamometer from the proposed list of equipment to measure the force. It is useful to recall the formula for calculating the modulus of the sliding friction force

F ck = μ · N (1),

where μ is the coefficient of sliding friction, N- the force of reaction of the support, equal in absolute value to the weight of the body.

Answer:

The handbook contains detailed theoretical material on all topics tested by the exam in physics. After each section, there are different levels of tasks in USE form... For the final control of knowledge at the end of the handbook, training options are given that correspond to the exam. Students do not have to search the Internet for additional information and buy other manuals. In this guide, they will find everything they need to prepare independently and effectively for the exam. The reference book is addressed to high school students to prepare for the exam in physics. The manual contains detailed theoretical material on all topics covered by the exam. After each section, examples of USE tasks and a practice test are given. All tasks are answered. The publication will be useful for physics teachers, parents for effective preparation of students for the Unified State Exam.

Consider the table for information on bright stars.

Star name	Temperature, TO	Weight (in the masses of the Sun)	Radius (in the radii of the Sun)	Distance to the star (holy year)
Aldebaran		5
Betelgeuse

Please select two statements that match the characteristics of the stars.

1. The surface temperature and radius of Betelgeuse indicate that this star belongs to red supergiants.
2. The temperature on the surface of Procyon is 2 times lower than on the surface of the Sun.
3. The stars Castor and Capella are at the same distance from Earth and therefore belong to the same constellation.
4. The star Vega belongs to white stars of spectral class A.
5. Since the masses of the stars Vega and Capella are the same, they belong to the same spectral type.

Solution

Star name	Temperature, TO	Weight (in the masses of the Sun)	Radius (in the radii of the Sun)	Distance to the star (holy year)
Aldebaran
Betelgeuse
			2,5

In the assignment, you need to choose two correct statements that correspond to the characteristics of the stars. The table shows that the most low temperature and Betelgeuse has a large radius, which means that this star belongs to the red giants. Therefore, the correct answer is (1). To choose the second statement correctly, it is necessary to know the distribution of stars by spectral type. We need to know the temperature range and the color of the star corresponding to this temperature. Analyzing the data in the table, we conclude that the correct statement will be (4). The star Vega belongs to white stars of spectral class A.

A projectile weighing 2 kg, flying at a speed of 200 m / s, explodes into two fragments. The first fragment weighing 1 kg flies at an angle of 90 ° to the original direction at a speed of 300 m / s. Find the speed of the second shard.

Answer: _______ m / s.

Solution

At the moment of projectile rupture (Δ t→ 0) the action of gravity can be neglected and the projectile can be considered as a closed system. According to the law of conservation of momentum: the vector sum of the momenta of the bodies included in a closed system remains constant for any interactions of the bodies of this system with each other. for our case, we will write:

- projectile speed; m- the mass of the projectile to rupture; - speed of the first fragment; m 1 - mass of the first fragment; m 2 - the mass of the second fragment; Is the speed of the second fragment.

Let's choose the positive direction of the axis NS coinciding with the direction of the velocity of the projectile, then in the projection onto this axis we write equation (1):

mv x = m 1 v 1x + m 2 v 2x (2)

According to the condition, the first fragment flies at an angle of 90 ° to the original direction. The length of the required impulse vector is determined by the Pythagorean theorem for a right-angled triangle.

p 2 = √p 2 + p 1 2 (3)

p 2 = √400 2 + 300 2 = 500 (kg m / s)

Answer: 500 m / s.

When an ideal monatomic gas was compressed at constant pressure, external forces performed work of 2000 J. What amount of heat was transferred by the gas to the surrounding bodies?

Answer: _____ J.

Solution

The problem for the first law of thermodynamics.

Δ U = Q + A sun, (1)

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

A sun = - A r (2),

Q = Δ U– A sun = Δ U+ A r =	3	pΔ V + pΔ V =	5	pΔ V,
	2		2

where pΔ V = A G

Answer: 5000 J.

A plane monochromatic light wave with a frequency of 8.0 · 10 14 Hz is incident along the normal onto a diffraction grating. A collecting lens with a focal length of 21 cm is placed in parallel to the grating behind it. 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. Express your answer in micrometers (μm), rounded to the nearest tenth. Calculate for small angles (φ ≈ 1 in radians) tanα ≈ sinφ ≈ φ.

Solution

The angular directions to the maxima of the diffraction pattern are determined by the equation

d Sinφ = kΛ (1),

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

Rice. 1

By the condition of the problem, we know the distance between its main maxima of the 1st and 2nd order, we denote it as Δ x= 18 mm = 1.8 · 10 –2 m, the frequency of the light wave ν = 8.0 · 10 14 Hz, the focal length of the lens F= 21 cm = 2.1 · 10 –1 m. We need to determine the period of the diffraction grating. In fig. 1 shows a diagram of the path of rays through the grating and the lens behind it. On the screen located in the focal plane of the collecting lens, a diffraction pattern is observed as a result of the interference of waves coming from all the slits. Let's use formula one for two maxima of the 1st and 2nd order.

d sinφ 1 = kλ (2),

if k = 1, then d sinφ 1 = λ (3),

similarly write for k = 2,

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

where x 1 is the distance from the zero maximum to the first order maximum. Likewise for distance x 2 .

Then we have

Diffraction grating period,

since by definition

where with= 3 10 8 m / s - the speed of light, then substituting the numerical values ​​we get

The answer was presented in micrometers, rounded to tenths, as required in the problem statement.

Answer: 4.4 microns.

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

Solution

Rice. 1

Part C assignments require a complete and detailed answer from the student. Based on the laws of physics, it is necessary to determine the voltmeter readings before the K key is closed and after the K key is closed. Let's take into account that the capacitor in the circuit is initially not charged. Consider two states. When the switch is open, only a resistor is connected to the power supply. The voltmeter readings are zero, since it is connected in parallel with the capacitor, and the capacitor is not initially charged, then q 1 = 0. Second state when the key is closed. Then the voltmeter readings will increase until they reach the maximum value, which will not change over time,

where r Is the internal resistance of the source. The voltage across the capacitor and resistor, according to Ohm's law for a section of the circuit U = I · R will not change over time, and the voltmeter readings will stop changing.

A wooden ball is tied with a thread to the bottom of a cylindrical vessel with a bottom area S= 100 cm 2. Water is poured into the vessel so that the ball is completely immersed in the liquid, while the thread is pulled and acts on the ball with force T... If the thread is cut, the ball will float, and the water level will change by h = 5 cm. Find the thread tension T.

Solution

Rice. 1		Rice. 2

Initially, a wooden ball is tied with a thread to the bottom of a cylindrical vessel with an area of ​​the bottom S= 100 cm 2 = 0.01 m 2 and is completely submerged in water. Three forces act on the ball: the force of gravity from the side of the Earth, - the force of Archimedes from the side of the liquid, - the tension force of the thread, the result of the interaction of the ball and the thread. According to the condition of the equilibrium of the ball in the first case, the geometric sum of all forces acting on the ball must be equal to zero:

Let's choose the coordinate axis OY and send it up. Then, taking into account the projection, equation (1) is written:

F a 1 = T + mg (2).

Let's write down the strength of Archimedes:

F a 1 = ρ V 1 g (3),

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

F a 2 = mg (4)

Let's write down the strength of Archimedes in this case:

F a 2 = ρ V 2 g (5),

where V 2 - the volume of the part of the ball immersed in the liquid in the second case.

Let's 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), we obtain 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= H 1 - H 2; get

T= ρ g S · h (8)

Substitute numerical values

Answer: 5 N.

All the information necessary for passing the exam in physics is presented in clear and accessible tables, after each topic there are training tasks to control knowledge. With the help of this book, students will be able to improve their knowledge in the shortest possible time, recall all the most important topics a few days before the exam, practice completing assignments in the USE format and become more confident in their abilities. After repeating all the topics presented in the manual, the long-awaited 100 points will become much closer! The manual contains theoretical information on all topics tested on the exam in physics. Each section is followed by training tasks of different types with answers. A clear and accessible presentation of the material will allow you to quickly find the information you need, eliminate knowledge gaps and quickly repeat a large amount of information. The publication will assist high school students in preparing for lessons, various forms of current and intermediate control, as well as to prepare for exams.

Task 30

In a room measuring 4 × 5 × 3 m, in which the air has a temperature of 10 ° C and a relative humidity of 30%, a humidifier with a capacity of 0.2 l / h was turned on. What will be the relative humidity in the room after 1.5 hours? The pressure of saturated water vapor at a temperature of 10 ° C is 1.23 kPa. Consider the room as an airtight vessel.

Solution

When starting to solve problems for vapors and humidity, it is always useful to keep in mind the following: if the temperature and pressure (density) of the saturating vapor are set, then its density (pressure) is determined from the Mendeleev - Clapeyron equation. Write down the Mendeleev-Clapeyron equation and the relative humidity formula for each state.

For the first case at φ 1 = 30%. We express the partial pressure of water vapor from the formula:

where T = t+ 273 (C), R Is a universal gas constant. Let us express the initial mass of steam contained in the room using equations (2) and (3):

During the time τ of operation of the humidifier, the mass of water will increase by

Δ m = τ · ρ · I, (6)

where I– the productivity of the humidifier, according to the condition, is equal to 0.2 l / h = 0.2 · 10 –3 m 3 / h, ρ = 1000 kg / m 3 is the density of water. Let us substitute formulas (4) and (5) in (6)

We transform the expression and express

This is the desired formula for the relative humidity in the room after the humidifier is in operation.

Substitute the numerical values ​​and get the following result

Answer: 83 %.

On horizontally located rough rails with negligible resistance, two identical rods with a mass of m= 100 g and resistance R= 0.1 ohm each. The distance between the rails is l = 10 cm, and the coefficient of friction between the rods and the rails is μ = 0.1. Rails with rods are in a uniform vertical magnetic field with induction B = 1 T (see figure). Under the action of a horizontal force acting on the first rod along the rail, both rods move translationally uniformly at different speeds. What is the speed of the first rod relative to the second? Disregard the self-induction of the circuit.

Solution

Rice. 1

The task is complicated by the fact that two rods are moving and it is necessary to determine the speed of the first relative to the second. Otherwise, the approach to solving problems of this type remains the same. A change in the magnetic flux of the penetrating circuit leads to the emergence of an EMF of induction. In our case, when the rods move at different speeds, the change in the flux of the magnetic induction vector penetrating the contour over a time interval Δ t is 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's law

By the condition of the problem, we neglect the self-induction of the circuit. According to Ohm's law for a closed circuit for the current that occurs in the circuit, we write the expression:

On conductors with a current in a magnetic field, the Ampere force acts and the modules of which are equal to each other, and are equal to the product of the current strength, the modulus of the magnetic induction vector and the length of the conductor. Since the force vector is perpendicular to the direction of the current, then sinα = 1, then

F 1 = F 2 = I · B · l (4)

The rods are still affected by the braking force of friction,

F tr = μ m · g (5)

according to the condition, it is said that the rods move uniformly, which means that the geometric sum of the forces applied to each rod is equal to zero. Only the Ampere force and the friction force act on the second rod.Therefore F tr = F 2, taking into account (3), (4), (5)

Let us express from this the relative speed

Substitute the numerical values:

Answer: 2 m / s.

In an experiment on the study of the photoelectric effect, light with a frequency of ν = 6.1 · 10 14 Hz falls on the cathode surface, as a result of which a current arises in the circuit. Current dependence graph I from stresses U between the anode and cathode is shown in the figure. What is the power of the incident light R, if on average one of 20 photons incident on the cathode knocks out an electron?

Solution

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

I =	q	(1).
	t

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

q = N e · e · t (2),

where e- electron charge modulus, N e– the number of photoelectrons ejected from the cathode in 1 s. According to the condition, one of 20 photons incident on the cathode knocks out an electron. Then

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

Our task is to find the number of photons incident on the cathode. It is known that the energy of one photon is E f = h · v, then the power of the incident light

After substituting the corresponding values, we obtain the final formula

P = N f · h · v =

twenty · I max h USE-2018. Physics (60x84 / 8) 10 training options for examination papers to prepare for the unified state exam The attention of schoolchildren and applicants is offered a new textbook on physics for preparation of the exam, which contains 10 options for training exam papers. Each option is compiled in full accordance with the requirements of the unified state exam in physics, includes tasks of different types and levels of difficulty. At the end of the book, self-test answers are given for all tasks. The proposed training options will help the teacher organize preparation for the unified state exam, and the students - independently test their knowledge and readiness for the final exam. The manual is addressed to schoolchildren, applicants and teachers.

On the eve of the academic year, demos of the KIM USE 2018 in all subjects (including physics) have been published on the FIPI official website.

This section contains documents defining the structure and content of the KIM USE 2018:

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

Demo version of the Unified State Exam 2018 in physics tasks with answers

Physics demo version of the exam 2018	variant + otvet
Specification	download
Codifier	download

Changes in the KIM USE in 2018 in physics compared to 2017

The codifier of the content elements checked on the exam in physics includes subsection 5.4 "Elements of astrophysics".

To part 1 examination work added one multiple choice quest that tests astrophysics elements. The content of the quest lines 4, 10, 13, 14 and 18 has been expanded. Part 2 is left unchanged. Maximum score for the completion of all tasks of the examination work increased from 50 to 52 points.

Duration of the exam 2018 in physics

The entire examination work takes 235 minutes. The approximate time for completing tasks for various parts of the work is:

1) for each task with a short answer - 3-5 minutes;

2) for each task with a detailed answer - 15–20 minutes.

The structure of the KIM USE

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

Part 1 contains 24 tasks with a short answer. Of these, 13 tasks with recording the answer in the form of a number, word or two numbers, 11 tasks for establishing correspondence and multiple choice, in which the answers must be written in the form of a sequence of numbers.

Part 2 contains 8 tasks, united by a common type of activity - problem solving. Of these, 3 tasks with a short answer (25-27) and 5 tasks (28-32), for which it is necessary to give a detailed answer.

In 2018, graduates of grade 11 and secondary institutions vocational education will take the Unified State Exam 2018 in physics. Latest news concerning the exam in physics in 2018 are based on the fact that some changes, both large and insignificant, will be introduced into it.

What is the meaning of the changes and how many there are

The main change related to the USE in physics compared to previous years is the absence of a test part with a choice of answers. This means that preparation for the exam should be accompanied by the student's ability to give short or detailed answers. Therefore, it will not be possible to guess the option and score a certain number of points, and you will have to work hard.

To basic part of the exam in physics, a new task 24 has been added, which requires the ability to solve problems in astrophysics. Due to the addition of No. 24, the maximum primary score increased to 52. The exam is divided into two parts according to difficulty levels: the basic one of 27 tasks, which involves a short or complete answer. In the second part, there are 5 advanced level problems, where it is necessary to give a detailed answer and explain the course of your solution. One important nuance: Many students skip this part, but even trying these assignments can get you one to two points.

All changes in the exam in physics are made in order to deepen the preparation and improve the assimilation of knowledge in the subject. In addition, the elimination of the test part motivates future applicants to accumulate knowledge more intensively and reason logically.

Exam structure

Compared to the previous year, the structure of the USE has not undergone significant changes. All work is given 235 minutes. Each task of the basic part should be solved from 1 to 5 minutes. Problems of increased complexity are solved in about 5-10 minutes.

All CMMs are stored at the exam site, and autopsy is performed during the test. The structure is as follows: 27 basic tasks check whether the examinee has knowledge in all areas of physics, from mechanics to quantum and nuclear physics. In 5 tasks of a high level of difficulty, the student demonstrates skills in the rationale of his decision and the correctness of his train of thought. The number of primary points can reach a maximum of 52. Then they are recalculated within a 100-point scale. Due to the change in the primary score, the minimum passing score may also change.

Demo version

The demo version of the Unified State Exam in Physics is already on the official portal of FIPI, which is developing a unified state exam. The structure and complexity of the demo version is similar to the one that will appear on the exam. Each assignment is detailed, and at the end there is a list of answers to questions for which the student checks against his decisions. Also at the end there is a detailed layout for each of the five tasks, indicating the number of points for correctly or partially performed actions. For each task of high complexity, you can get from 2 to 4 points, depending on the requirements and deployment of the solution. Assignments can contain a sequence of numbers that you need to correctly write down, establishing correspondence between elements, as well as small tasks in one or two steps.

• Download demo: ege-2018-fiz-demo.pdf
• Download the archive with the specification and codifier: ege-2018-fiz-demo.zip

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