Demo version of the Unified State Examination in Physics. Changes in the Unified State Examination in Physics

Secondary general education

Preparing for the Unified State Exam 2018: analysis of the demo version in physics

Unified State Exam 2018. Physics. Thematic training tasks

The publication contains:
assignments of different types on all topics of the Unified State Exam;
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 them to systematically prepare for the exam when passing each topic.

A point body at rest begins to move along the axis Ox. The figure shows the projection dependence graph ax acceleration of this body with time t.

Determine the distance the body traveled in the third second of movement.

Answer: _________ m.

Solution

Knowing how 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 projection of acceleration versus time, the path that the body has traveled in the third second of movement. 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 equal to zero. We determine the nature of the movement in this area: the body moved uniformly. The path is easy to determine if you know the speed and time of movement. However, in the interval from 0 to 2 s, the body moved uniformly accelerated. Using the definition of acceleration, we write the velocity projection equation V x = V 0x + a x t; since the body was initially at rest, the velocity projection at the end of the second second became

Then the distance traveled by the body in the third second

Answer: 8 m.

Rice. 1

Two bars connected by a light spring lie on a smooth horizontal surface. To a block of mass m= 2 kg apply a constant force equal in magnitude 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 block moves with an acceleration of 1 m/s 2.

Answer: _________ N.

Solution

Horizontally on a body of mass m= 2 kg two forces act, this is a force F= 10 N and the elastic force on the side of the spring. The resultant of these forces imparts acceleration to the body. Let's 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.

In projection onto axis 0 X: F – F control = ma (2)

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

Let's substitute the numerical values ​​into formula (3) and get, F control = 10 N – 2 kg · 1 m/s 2 = 8 N.

Answer: 8 N.

Task 3

A body with a mass of 4 kg located on a rough horizontal plane is given a speed of 10 m/s along it. Determine the modulus of work done 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 acted upon by the force of gravity, the reaction force of the support, the friction force, which creates a braking acceleration. The body was initially given a speed of 10 m/s. Let's write down Newton's second law for our case.

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

N – mg = 0; N = mg (2)

In projection onto the axis X: –F tr = – ma; F tr = ma; (3) We need to determine the modulus of the work of the friction force at the time when the speed becomes half as much, 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, we take the timeless formula:

Let's substitute (3) and (5) into (4)

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

Let's substitute numerical values

Answer: 150 J.

Unified State Exam 2018. Physics. 30 practice versions of exam papers

The publication contains:
30 training options for the Unified State Exam
instructions for implementation and evaluation criteria
answers to all tasks
Training options will help the teacher organize preparation for the Unified State Exam, and students will independently test their knowledge and readiness to take 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 inner pulley of the block if the system is in equilibrium?

Rice. 1

Answer: _________ see.

Solution

According to the conditions of the problem, the system is in equilibrium. On the image L 1, shoulder strength L 2nd arm of force Equilibrium condition: moments of forces rotating bodies clockwise must be equal to moments of forces rotating the body counterclockwise. Recall that the moment of force is the product of the modulus of force and the arm. The forces acting on the threads from the loads differ by a factor of 3. This means that the radius of the inner pulley of the block differs from the outer one by 3 times. Therefore, the shoulder L 2 will be equal to 8 cm.

Answer: 8 cm

Task 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 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 time 2.0 s is minimal.
4. The amplitude of the ball's oscillations is 30 mm.
5. The total mechanical energy of a pendulum consisting of a ball and a spring at time 3.0 s is minimal.

Solution

The table presents 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 the right two statements. The system is a spring 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 using the formula

Where k– spring stiffness coefficient, X– displacement of the body from the equilibrium position. If the displacement is maximum, then the speed at this point is zero, which means 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 goes through half of the oscillation in t= 2 s, complete oscillation takes twice as long T= 4 s. Therefore, statements 1 will be true; 2.

Task 6

A small piece of ice was lowered into a cylindrical glass of water to float. After some time, the ice completely melted. Determine how the pressure on the bottom of the glass and the level of water in the glass changed as a result of the melting of the ice.

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

Write to table

Solution

Rice. 1

Problems of this type are quite common in different versions of the Unified State Exam. And as practice shows, students often make mistakes. Let's try to analyze this task in detail. Let's denote m– mass of a piece of ice, ρ l – density of ice, ρ в – density of water, V pcht – the volume of the submerged part of the ice, equal to the volume of the displaced liquid (volume of the hole). Let's mentally remove the ice from the water. Then there will be a hole in the water whose volume is equal to V pcht, i.e. volume of water displaced by a piece of ice Fig. 1( b).

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

F a = mg (1)

ρ in V p.m. 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 water obtained from ice into a 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, will also not change. Therefore the answer will be

Unified State Exam 2018. Physics. Training tasks

The publication is addressed to high school students to prepare for the Unified State Exam in physics.
The benefit includes:
20 training options
answers to all tasks
Unified State Exam answer forms for each option.
The publication will assist teachers in preparing students for the Unified State Exam in Physics.

A 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 begins to deform by applying an external force to its free end A and uniformly moving point A.

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

Solution

From the figure for the problem it is clear that when the spring is not deformed, then its free end, and accordingly point A, are in a position with the coordinate X 0 . At some point in time, the spring begins to deform by applying an external force to its free end A. Point A moves uniformly. Depending on whether the spring is stretched or compressed, the direction and magnitude of the elastic force generated 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 letter B) shows the dependence of the projection of the external force on the magnitude of the deformation. Because with increasing external force, the magnitude of deformation and elastic force increase.

Answer: 24.

Task 8

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

Answer: ________ eV.

Solution

The problem is interesting because it is necessary to compare two temperature measurement scales. These are the Reaumur temperature scale and the Celsius scale. The melting points of ice are the same on the scales, but the boiling points are different; we can obtain a formula for converting from degrees Réaumur to degrees Celsius. This

Let's convert the temperature 29 (°R) to degrees Celsius

Let's convert the 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 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. From the formula it is clear that the dependence of the average kinetic energy on temperature is direct, that is, the number of times the temperature changes, the number of times the average kinetic energy of the thermal motion of molecules changes. Let's substitute the numerical values:

Let's convert the result into electronvolts and round to the nearest hundredth. Let's remember 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-diagram. For this process, determine 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 conditions of the problem in process 1–2, the graph of which is shown in VT-diagram, one mole of a monatomic ideal gas is involved. To answer the question of the problem, it is necessary to obtain expressions for the change in 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 the first law of thermodynamics:

Q 12 = A 12+Δ U 12 (5),

Where A 12 – gas work during expansion. By definition, work is equal to

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 reference book contains in full the theoretical material for the physics course required to pass the Unified State 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 compiled - test and measuring materials (CMM) of the Unified State Examination. Theoretical material is presented in a concise, accessible form. Each topic is accompanied by examples of exam tasks that correspond to the Unified State Exam format. This will help the teacher organize preparation for the unified state exam, and students will independently test their knowledge and readiness to take the final exam.

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

Answer: _________ g.

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 immersing the hot horseshoe, the energy received by the water will go directly to steam formation. Let us write the heat balance equation

With and · m P · ( 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 the units in which you want to leave the mass of water.

Answer: 90

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

Select two true statements based on the analysis of the presented graph.

1. The gas pressure in state 2 is greater than the 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 section 1–2 is less than the change in the internal energy of the gas in 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 what dependence graphs look like for isoprocesses in different axes, in particular R= const. In our example at TV The diagram shows two isobars. Let's see how pressure and volume 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 do any work; it is zero. The statement is incorrect. In section 1–2, the pressure increases, which is also incorrect. We just showed 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 a constant temperature as the gas is compressed.

Answer: 14.

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

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

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

Write to 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 exam tasks. First of all, you need to remember the formula for calculating the efficiency factor. Be able to write it down using the temperature of the heater and the temperature of the refrigerator

in addition, be able to write down the efficiency through the useful work of gas A g and the amount of heat received from the heater Q n.

We carefully read the condition and determined which parameters were changed: in our case, we increased the temperature of the refrigerator, leaving the temperature of the heater 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), 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 picture). Where is it directed relative to the drawing ( right, left, up, down, towards the observer, away from the observer) charge acceleration – q in this moment in time, if only charges + act on it Q And –Q? Write the answer in word(s)

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 of where the charge acceleration is directed - q, at the moment of time when only charges +Q and – act on it 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 are the forces directed this way? Let us remember how similarly charged bodies interact, they repel, the force Coulomb force of interaction of charges is the central force. the force with which oppositely charged bodies attract. From the figure we see that the charge is q equidistant from stationary charges whose moduli are equal. Therefore, they will also be equal in modulus. The resulting force will be directed relative to the drawing down. The charge acceleration will also be directed - q, i.e. down.

Answer: Down.

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

Two series-connected resistors with a resistance of 4 Ohms and 8 Ohms are connected to a battery whose terminal voltage is 24 V. What thermal power is released in the lower value resistor?

Answer: _________ Tue.

Solution

To solve the problem, it is advisable to draw a diagram of the series connection of resistors. Then remember the laws of series 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 by a resistor of a lower value, we write:

P = I 2 R= (2 A) 2 · 4 Ohm = 16 W.

Answer: P= 16 W.

A wire frame with an area of ​​2·10–3 m2 rotates in a uniform magnetic field around an axis perpendicular to the magnetic induction vector. The magnetic flux penetrating the frame area varies according to the law

Ф = 4 10 –6 cos10π t,

where all quantities are expressed in SI. What is the magnetic induction module?

Answer: ________________ mT

Solution

The magnetic flux changes according to the law

Ф = 4 10 –6 cos10π t,

where all quantities are expressed in SI. You need to understand what magnetic flux is in general and how this quantity is related to the magnetic induction module B and frame area S. Let's write the equation in general form 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 module 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 = IN · S cosα, the flow is maximum at cosα = 1; let's express the induction modulus

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

Answer: 2.

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

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

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

Solution

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

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

where ρ is the resistivity of the conductor; l– length of the conductor; S– cross-sectional area. From the graph it can be seen that the length of the silver wire L c = 8 m; aluminum wire length L a = 14 m. Voltage on a section of silver wire U c = Δφ = 6 V – 2 V = 4 V. Voltage on a 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 will 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 the SI system for calculations.

Correct statement option 2.

Let's check the expressions for power.

P a = I 2 · R a(4);

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

The reference book contains in full the theoretical material for the physics course required to pass the Unified State 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 compiled - test and measuring materials (CMM) of the Unified State Examination. Theoretical material is presented in a concise, accessible form. Each topic is accompanied by examples of exam tasks that correspond to the Unified State Exam format. This will help the teacher organize preparation for the unified state exam, and students will independently test their knowledge and readiness to take the final exam. At the end of the manual, answers to self-test tasks are provided that will help schoolchildren and applicants to objectively assess the level of their knowledge and the degree of preparedness for the certification exam. The manual is addressed to high school students, applicants and teachers.

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

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

1. increases
2. decreases
3. does not change

Write to 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 the focal and double focal lengths 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.

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

Rice. 1

Rice. 2

We 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– resistor resistance).

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

Solution

Rice.1

According to the conditions of the problem, we neglect the internal resistance of the source. 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 strength through the source with the switch closed. If the key is closed, the two resistors will be connected in parallel. Ohm's law for the complete circuit in this case will look like:

Where I– current strength through the source with the switch closed;

Where N– the number of conductors connected in parallel with the same resistance.

– EMF of the current source.

Substituting (2) into (1) we have: this is the formula numbered 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 the complete circuit in this case will be:

Solution

Let's write the nuclear reaction for our case:

As a result of this reaction, the law of conservation of charge and mass number is satisfied.

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 reference book contains all the theoretical material for the physics course required to pass the unified state exam. It includes all content elements tested by test 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 tasks. Practical tasks correspond to the Unified State Exam format. Answers to the tests are provided at the end of the manual. 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 task is to use the law of radioactive decay. It can be written in the form

Where m 0 – initial mass of the substance, t- the time it takes for a substance to decay, T- half life. Let's substitute numerical values

Answer: 0.3 mg.

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

For each position in 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 recall the definition of the photoelectric effect. This is the phenomenon of interaction of light with matter, as a result of which the energy of photons is transferred to the electrons of the substance. There are external and internal photoeffects. In our case we are talking about the external photoelectric effect. When, under the influence of light, electrons are ejected from a substance. The work function depends on the material from which the photocathode of the photocell is made, and does not depend on the frequency of the light. The energy of incident photons is proportional to the frequency of light.

E= h v(1)

where λ is the wavelength of light; With– speed of light,

Let's substitute (3) into (1) We get

Let's analyze the resulting formula. It is obvious that as the wavelength increases, the energy of the incident photons decreases. This type of dependence corresponds to the graph under the letter A)

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

hν = A out + E to (5),

Where hν is the energy of a photon incident on the photocathode, A out – work function, E k is the maximum kinetic energy of photoelectrons emitted from the photocathode under the influence of light.

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

red border

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).

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

Answer: (___________±___________) A.

Solution

The task tests the ability to record the readings of a measuring device, taking into account a given measurement error. Let's determine the price of the scale division With= (0.4 A – 0.2 A)/10 = 0.02 A. The measurement error according to the condition is equal to the division price, i.e. Δ I = c= 0.02 A. We write the final result in the form:

I= (0.20 ± 0.02) A

It is necessary to assemble an experimental setup that can be used to determine the coefficient of sliding friction between steel and wood. To do this, the student took a steel bar with a hook. Which two additional items from the list of equipment below must be used to conduct this experiment?

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

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

Solution

The task requires determining the sliding friction coefficient of steel on wood, so to conduct the experiment it is necessary to take a wooden ruler and a dynamometer from the proposed list of equipment to measure force. It is useful to recall the formula for calculating the modulus of sliding friction force

Fck = μ · N (1),

where μ is the sliding friction coefficient, N– ground reaction force equal in modulus to body weight.

The reference book contains detailed theoretical material on all topics tested by the Unified State Exam in physics. After each section, multi-level tasks are given in the form of the Unified State Exam. For the final control of knowledge, training options corresponding to the Unified State Exam are given at the end of the reference book. Students will not have to search for additional information on the Internet and buy other textbooks. In this guide, they will find everything they need to independently and effectively prepare for the exam. The reference book is addressed to high school students to prepare for the Unified State Exam in physics. The manual contains detailed theoretical material on all topics tested by the exam. After each section, examples of Unified State Examination tasks and a practice test are given. Answers are provided for all tasks. The publication will be useful to physics teachers and parents for effectively preparing students for the Unified State Exam.

Consider the table containing information about bright stars.

Select two statements that correspond to the characteristics of stars.

1. The surface temperature and radius of Betelgeuse indicate that this star is a red supergiant.
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 the Earth and, therefore, belong to the same constellation.
4. The star Vega belongs to the white stars of spectral class A.
5. Since the masses of the Vega and Capella stars are the same, they belong to the same spectral class.

Solution

In the task you need to choose two correct statements that correspond to the characteristics of the stars. The table shows that Betelgeuse has the lowest temperature and largest radius, which means that this star belongs to the red giants. Therefore, the correct answer is (1). To correctly choose the second statement, you need to know the distribution of stars by spectral types. We need to know the temperature range and the color of the star corresponding to this temperature. Analyzing the table data, we conclude that the correct statement is (4). The star Vega belongs to the white stars of spectral class A.

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

Answer: _______ m/s.

Solution

At the moment the projectile bursts (Δ t→ 0) the effect 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 momentum 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 write:

– projectile speed; m– mass of the projectile before bursting; – speed of the first fragment; m 1 – mass of the first fragment; m 2 – mass of the second fragment; – speed of the second fragment.

Let's choose the positive direction of the axis X, coinciding with the direction of the projectile velocity, 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. We determine the length of the desired impulse vector using the Pythagorean theorem for a right 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 2000 J of work. How much heat was transferred by the gas to the surrounding bodies?

Answer: _____ J.

Solution

Problem on the first law of thermodynamics.

Δ U = Q + A sun, (1)

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

A sun = – A g (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 normally on a diffraction grating. A collecting lens with a focal length of 21 cm is placed 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) tgα ≈ sinφ ≈ φ.

Solution

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

d· sinφ = kλ (1),

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

Rice. 1

According to the conditions of the problem, we know the distance between its main maxima of the 1st and 2nd order, let us denote it as Δ x= 18 mm = 1.8 10 –2 m, light wave frequency ν = 8.0 10 14 Hz, lens focal length F= 21 cm = 2.1 · 10 –1 m. We need to determine the period of the diffraction grating. In Fig. Figure 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 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),

we write similarly for k = 2,

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

Where x 1 – distance from the zero maximum to the first order maximum. Same for distance x 2 .

Then we have

Diffraction grating period,

because 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 circuit shown in the figure before closing key K and describe the changes in its readings after closing key K. Initially, the capacitor is not charged.

Solution

Rice. 1

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

Where r– internal resistance of the source. 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 stretched and acts on the ball with force T. If the thread is cut, the ball will float and the water level will change to h = 5 cm. Find the tension in the thread T.

Solution

Initially, a wooden ball is tied with a thread to the bottom of a cylindrical vessel with the area of ​​the bottom S= 100 cm 2 = 0.01 m 2 and is completely immersed in water. Three forces act on the ball: the force of gravity from the Earth, – the Archimedes force from the liquid, – the tension force of the thread, the result of the interaction of the ball and the thread. According to the condition of 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 a coordinate axis OY and point it up. Then, taking into account the projection, we write equation (1):

F a 1 = T + mg (2).

Let us describe the Archimedes force:

F a 1 = ρ V 1 g (3),

Where V 1 – the volume of part of the ball immersed in water, in the first it is the volume of the entire ball, m is the mass of the ball, ρ is the density of water. Equilibrium condition in the second case

F a 2 = mg (4)

Let us describe the Archimedes force in this case:

F a 2 = ρ V 2 g (5),

Where V 2 is the volume of the part of the ball immersed in 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 ; we get

T= ρ g S · h (8)

Let's substitute numerical values

Answer: 5 N.

All information necessary to pass the Unified State 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 increase the level of their knowledge in the shortest possible time, remember all the most important topics a few days before the exam, practice completing tasks in the Unified State Exam 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 Unified State Exam in physics. After each section there are training tasks of different types with answers. A clear and accessible presentation of the material will allow you to quickly find the necessary information, eliminate gaps in knowledge and repeat a large amount of information in the shortest possible time. The publication will help high school students prepare for lessons, various forms of current and intermediate control, as well as prepare for exams.

Task 30

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

Solution

When starting to solve problems on steam and humidity, it is always useful to keep in mind the following: if the temperature and pressure (density) of the saturating steam are given, 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 (K), R– universal gas constant. Let us express the initial mass of steam contained in the room using equations (2) and (3):

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

Δ m = τ · ρ · I, (6)

Where I– According to the condition, the performance of the humidifier is equal to 0.2 l/h = 0.2 10 –3 m3/h, ρ = 1000 kg/m3 – water density. Let’s substitute formulas (4) and (5) into (6)

Let's transform the expression and express

This is the desired formula for the relative humidity that will be in the room after the humidifier is running.

Let's substitute the numerical values ​​and get the following result

Answer: 83 %.

Two identical rods of mass 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. The rails with rods are in a uniform vertical magnetic field with induction B = 1 T (see figure). Under the influence of a horizontal force acting on the first rod along the rails, both rods move forward uniformly at different speeds. What is the speed of the first rod relative to the second? Neglect self-induction of the circuit.

Solution

Rice. 1

The task is complicated by the fact that two rods are moving and you need 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 penetrating the circuit leads to the appearance of an induced emf. In our case, when the rods move at different speeds, the change in the flux of the magnetic induction vector penetrating the circuit over a period of time Δ t determined by the formula

ΔΦ = B · l · ( v 1 – v 2) Δ t (1)

This leads to the occurrence of induced emf. According to Faraday's law

According to the conditions of the problem, we neglect the self-inductance of the circuit. According to Ohm's law for a closed circuit, we write the expression for the current strength arising in the circuit:

Conductors carrying current in a magnetic field are acted upon by the Ampere force and their modules are equal to each other and are equal to the product of the current strength, the module 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 braking force of friction still acts on the rods,

F tr = μ · m · g (5)

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

Let us express from here the relative speed

Let's substitute the numerical values:

Answer: 2 m/s.

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

Solution

By definition, current strength is a physical quantity numerically equal to charge q passing through the cross section of the conductor per unit time 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 knocked out of 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 is the number of photons incident on the cathode in 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 equal to E f = h · v, then the power of the incident light

After substituting the corresponding values, we obtain the final formula

Specification
control measuring materials
for holding the unified state exam in 2018
in PHYSICS

1. Purpose of KIM Unified State Exam

The Unified State Exam (hereinafter referred to as the Unified State Exam) is a form of objective assessment of the quality of training of persons who have mastered educational programs of secondary general education, using tasks of a standardized form (control measurement materials).

The Unified State Exam is conducted in accordance with Federal Law No. 273-FZ dated December 29, 2012 “On Education in the Russian Federation.”

Control measuring materials make it possible to establish the level of mastery by graduates of the Federal component of the state educational standard of secondary (complete) general education in physics, basic and specialized levels.

The results of the unified state exam in physics are recognized by educational organizations of secondary vocational education and educational organizations of higher professional education as the results of entrance tests in physics.

2. Documents defining the content of the Unified State Exam KIM

3. Approaches to selecting content and developing the structure of the Unified State Exam KIM

Each version of the examination paper includes controlled content elements from all sections of the school physics course, while tasks of all taxonomic levels are offered for each section. The most important content elements from the point of view of continuing education in higher educational institutions are controlled in the same version with tasks of different levels of complexity. The number of tasks for a particular section is determined by its content and in proportion to the teaching time allocated for its study in accordance with the approximate physics program. The various plans by which examination options are constructed are built on the principle of content addition so that, in general, all series of options provide diagnostics for the development of all content elements included in the codifier.

The priority when designing a CMM is the need to test the types of activities provided for by the standard (taking into account the limitations in the conditions of mass written testing of students’ knowledge and skills): mastering the conceptual apparatus of a physics course, mastering methodological knowledge, applying knowledge in explaining physical phenomena and solving problems. Mastery of skills in working with information of physical content is tested indirectly by using various methods of presenting information in texts (graphs, tables, diagrams and schematic drawings).

The most important type of activity from the point of view of successful continuation of education at a university is problem solving. Each option includes tasks in all sections of different levels of complexity, allowing you to test the ability to apply physical laws and formulas both in standard educational situations and in non-traditional situations that require the manifestation of a fairly high degree of independence when combining known action algorithms or creating your own plan for completing a task .

The objectivity of checking tasks with a detailed answer is ensured by uniform assessment criteria, the participation of two independent experts evaluating one work, the possibility of appointing a third expert and the presence of an appeal procedure.

The Unified State Examination in Physics is an exam of choice for graduates and is intended for differentiation when entering higher educational institutions. For these purposes, the work includes tasks of three difficulty levels. Completing tasks at a basic level of complexity allows you to assess the level of mastery of the most significant content elements of a high school physics course and mastery of the most important types of activities.

Among the tasks of the basic level, tasks are distinguished whose content corresponds to the standard of the basic level. The minimum number of Unified State Examination points in physics, confirming that a graduate has mastered a secondary (full) general education program in physics, is established based on the requirements for mastering the basic level standard. The use of tasks of increased and high levels of complexity in the examination work allows us to assess the degree of preparedness of a student to continue his education at a university.

4. Structure of KIM Unified State Exam

Each version of the examination paper consists of two parts and includes 32 tasks, differing in form and level of complexity (Table 1).

Part 1 contains 24 short answer questions. Of these, 13 are tasks with the answer written in the form of a number, a word or two numbers. 11 matching and multiple choice tasks that require you to write your answers as a sequence of numbers.

Part 2 contains 8 tasks united by a common activity - problem solving. Of these, 3 tasks with a short answer (25-27) and 5 tasks (28-32), for which you need to provide a detailed answer.

Option No. 3304330

Demo version of the Unified State Exam 2018 in physics.

When completing tasks with a short answer, enter in the answer field the number that corresponds to the number of the correct answer, or a number, a word, a sequence of letters (words) or numbers. The answer should be written without spaces or any additional characters. Separate the fractional part from the whole decimal point. There is no need to write units of measurement. In problems 1–4, 8–10, 14, 15, 20, 25–27, the answer is a whole number or a finite decimal fraction. The answer to tasks 5–7, 11, 12, 16–18, 21 and 23 is a sequence of two numbers. The answer to task 13 is a word. The answer to tasks 19 and 22 are two numbers.

If the option is specified by the teacher, you can enter or upload answers to tasks with a detailed answer into the system. The teacher will see the results of completing tasks with a short answer and will be able to evaluate the downloaded answers to tasks with a long answer. The scores assigned by the teacher will appear in your statistics.

On the eve of the academic year, demo versions of the KIM Unified State Exam 2018 in all subjects (including physics) have been published on the official website of the FIPI.

This section presents documents defining the structure and content of the KIM Unified State Exam 2018:

Demonstration versions of control measurement materials of the Unified State Exam.
- codifiers of content elements and requirements for the level of training of graduates of general education institutions for conducting a unified state exam;
- specifications of control measuring materials for the Unified State Exam;

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

Changes in the Unified State Exam KIM in 2018 in physics compared to 2017

The codifier of content elements tested on the Unified State Exam in Physics includes subsection 5.4 “Elements of Astrophysics”.

One multiple choice question testing elements of astrophysics has been added to Part 1 of the exam paper. The content of task lines 4, 10, 13, 14 and 18 has been expanded. Part 2 has been left unchanged. Maximum score for completing all tasks of the examination work increased from 50 to 52 points.

Duration of the Unified State Exam 2018 in physics

235 minutes are allotted to complete the entire examination work. The approximate time to complete tasks of 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.

Structure of KIM Unified State Examination

Each version of the examination paper consists of two parts and includes 32 tasks, differing in form and level of difficulty.

Part 1 contains 24 short answer questions. Of these, 13 tasks require the answer to be written in the form of a number, a word or two numbers, 11 tasks require matching and multiple choice, in which the answers must be written as a sequence of numbers.

Part 2 contains 8 tasks united by a common activity - problem solving. Of these, 3 tasks with a short answer (25–27) and 5 tasks (28–32), for which you need to provide a detailed answer.