Preparation for the OGE and the Unified State Exam
Secondary general education
Line UMK A.V. Grachev. Physics (10-11) (basic, advanced)
Line UMK A.V. Grachev. Physics (7-9)
Line UMK A.V. Peryshkin. Physics (7-9)
Preparing for the Unified State Exam in Physics: examples, solutions, explanations
We analyze the tasks of the Unified State Exam in physics (Option C) with the teacher.Lebedeva Alevtina Sergeevna, physics teacher, 27 years of work experience. Certificate of Honor from the Ministry of Education of the Moscow Region (2013), Gratitude from the Head of the Voskresensky Municipal District (2015), Certificate from the President of the Association of Teachers of Mathematics and Physics of the Moscow Region (2015).
The work presents tasks of different difficulty levels: basic, advanced and high. Basic level tasks are simple tasks that test the mastery of the most important physical concepts, models, phenomena and laws. Advanced level tasks are aimed at testing the ability to use concepts and laws of physics to analyze various processes and phenomena, as well as the ability to solve problems using one or two laws (formulas) on any of the topics of the school physics course. In work 4, the tasks of part 2 are tasks of a high level of complexity and test the ability to use the laws and theories of physics in a changed or new situation. Completing such tasks requires the application of knowledge from two or three sections of physics at once, i.e. high level of training. This option fully corresponds to the demo version of the Unified State Exam 2017; the tasks are taken from the open bank of Unified State Exam tasks.
The figure shows a graph of the speed modulus versus time t. Determine from the graph the distance traveled by the car in the time interval from 0 to 30 s.
Solution. The path traveled by a car in the time interval from 0 to 30 s can most easily be defined as the area of a trapezoid, the bases of which are the time intervals (30 – 0) = 30 s and (30 – 10) = 20 s, and the height is the speed v= 10 m/s, i.e.
| S = | (30 + 20) With | 10 m/s = 250 m. |
| 2 |
Answer. 250 m.
A load weighing 100 kg is lifted vertically upward using a cable. The figure shows the dependence of the velocity projection V load on the axis directed upwards, as a function of time t. Determine the modulus of the cable tension force during the lift.
Solution. According to the velocity projection dependence graph v load on an axis directed vertically upward, as a function of time t, we can determine the projection of the acceleration of the load
| a = | ∆v | = | (8 – 2) m/s | = 2 m/s 2. |
| ∆t | 3 s |
The load is acted upon by: the force of gravity directed vertically downward and the tension force of the cable directed vertically upward along the cable (see Fig. 2. Let's write down the basic equation of dynamics. Let's use Newton's second law. The geometric sum of the forces acting on a body is equal to the product of the mass of the body and the acceleration imparted to it.
+ = (1)
Let's write the equation for the projection of vectors in the reference system associated with the earth, directing the OY axis upward. The projection of the tension force is positive, since the direction of the force coincides with the direction of the OY axis, the projection of the gravity force is negative, since the force vector is opposite to the OY axis, the projection of the acceleration vector is also positive, so the body moves with upward acceleration. We have
T – mg = ma (2);
from formula (2) tensile force modulus
T = m(g + a) = 100 kg (10 + 2) m/s 2 = 1200 N.
Answer. 1200 N.
The body is dragged along a rough horizontal surface with a constant speed whose modulus is 1.5 m/s, applying a force to it as shown in Figure (1). In this case, the modulus of the sliding friction force acting on the body is 16 N. What is the power developed by the force? F?
Solution. Let's imagine the physical process specified in the problem statement and make a schematic drawing indicating all the forces acting on the body (Fig. 2). Let's write down the basic equation of dynamics.
Tr + + = (1)
Having chosen a reference system associated with a fixed surface, we write the equations for the projection of vectors onto the selected coordinate axes. According to the conditions of the problem, the body moves uniformly, since its speed is constant and equal to 1.5 m/s. This means the acceleration of the body is zero. Two forces act horizontally on the body: the sliding friction force tr. and the force with which the body is dragged. The projection of the friction force is negative, since the force vector does not coincide with the direction of the axis X. Projection of force F positive. We remind you that to find the projection, we lower the perpendicular from the beginning and end of the vector to the selected axis. Taking this into account we have: F cosα – F tr = 0; (1) let us express the projection of force F, This F cosα = F tr = 16 N; (2) then the power developed by the force will be equal to N = F cosα V(3) Let’s make a replacement, taking into account equation (2), and substitute the corresponding data into equation (3):
N= 16 N · 1.5 m/s = 24 W.
Answer. 24 W.
A load attached to a light spring with a stiffness of 200 N/m undergoes vertical oscillations. The figure shows a graph of the displacement dependence x load from time to time t. Determine what the mass of the load is. Round your answer to a whole number.
Solution. A mass on a spring undergoes vertical oscillations. According to the load displacement graph X from time t, we determine the period of oscillation of the load. The period of oscillation is equal to T= 4 s; from the formula T= 2π let's express the mass m cargo
| = | T | ; | m | = | T 2 | ; m = k | T 2 | ; m= 200 N/m | (4 s) 2 | = 81.14 kg ≈ 81 kg. |
| 2π | k | 4π 2 | 4π 2 | 39,438 |
Answer: 81 kg.
The figure shows a system of two light blocks and a weightless cable, with which you can keep in balance or lift a load weighing 10 kg. Friction is negligible. Based on the analysis of the above figure, select two true statements and indicate their numbers in your answer.
- In order to keep the load in balance, you need to act on the end of the rope with a force of 100 N.
- The block system shown in the figure does not give any gain in strength.
- h, you need to pull out a section of rope length 3 h.
- To slowly lift a load to a height hh.
Solution. In this problem, it is necessary to remember simple mechanisms, namely blocks: a movable and a fixed block. The movable block gives a double gain in strength, while the section of the rope needs to be pulled twice as long, and the fixed block is used to redirect the force. In work, simple mechanisms of winning do not give. After analyzing the problem, we immediately select the necessary statements:
- To slowly lift a load to a height h, you need to pull out a section of rope length 2 h.
- In order to keep the load in balance, you need to act on the end of the rope with a force of 50 N.
Answer. 45.
An aluminum weight attached to a weightless and inextensible thread is completely immersed in a vessel with water. The load does not touch the walls and bottom of the vessel. Then an iron weight, the mass of which is equal to the mass of the aluminum weight, is immersed in the same vessel with water. How will the modulus of the tension force of the thread and the modulus of the force of gravity acting on the load change as a result of this?
- Increases;
- Decreases;
- Doesn't change.
Solution. We analyze the condition of the problem and highlight those parameters that do not change during the study: these are the mass of the body and the liquid into which the body is immersed on a thread. After this, it is better to make a schematic drawing and indicate the forces acting on the load: thread tension F control, directed upward along the thread; gravity directed vertically downward; Archimedean force a, acting from the side of the liquid on the immersed body and directed upward. According to the conditions of the problem, the mass of the loads is the same, therefore, the modulus of the force of gravity acting on the load does not change. Since the density of the cargo is different, the volume will also be different.
| V = | m | . |
| p |
The density of iron is 7800 kg/m3, and the density of aluminum cargo is 2700 kg/m3. Hence, V and< V a. The body is in equilibrium, the resultant of all forces acting on the body is zero. Let's direct the OY coordinate axis upward. We write the basic equation of dynamics, taking into account the projection of forces, in the form F control + F a – mg= 0; (1) Let us express the tension force F control = mg – F a(2); Archimedean force depends on the density of the liquid and the volume of the immersed part of the body F a = ρ gV p.h.t. (3); The density of the liquid does not change, and the volume of the iron body is smaller V and< V a, therefore the Archimedean force acting on the iron load will be less. We conclude about the modulus of the tension force of the thread, working with equation (2), it will increase.
Answer. 13.
A block of mass m slides off a fixed rough inclined plane with an angle α at the base. The acceleration modulus of the block is equal to a, the modulus of the block's velocity increases. Air resistance can be neglected.
Establish a correspondence between physical quantities and formulas with which they can be calculated. For each position in the first column, select the corresponding position from the second column and write down the selected numbers in the table under the corresponding letters.
B) Coefficient of friction between a block and an inclined plane
3) mg cosα
| 4) sinα – | a |
| g cosα |
Solution. This task requires the application of Newton's laws. We recommend making a schematic drawing; indicate all kinematic characteristics of movement. If possible, depict the acceleration vector and the vectors of all forces applied to the moving body; remember that the forces acting on a body are the result of interaction with other bodies. Then write down the basic equation of dynamics. Select a reference system and write down the resulting equation for the projection of force and acceleration vectors;
Following the proposed algorithm, we will make a schematic drawing (Fig. 1). The figure shows the forces applied to the center of gravity of the block and the coordinate axes of the reference system associated with the surface of the inclined plane. Since all forces are constant, the movement of the block will be uniformly variable with increasing speed, i.e. the acceleration vector is directed in the direction of motion. Let's choose the direction of the axes as shown in the figure. Let's write down the projections of forces on the selected axes.
Let's write down the basic equation of dynamics:
Tr + = (1)
Let us write this equation (1) for the projection of forces and acceleration.
On the OY axis: the projection of the ground reaction force is positive, since the vector coincides with the direction of the OY axis Ny = N; the projection of the friction force is zero since the vector is perpendicular to the axis; the projection of gravity will be negative and equal mg y= – mg cosα; acceleration vector projection a y= 0, since the acceleration vector is perpendicular to the axis. We have N – mg cosα = 0 (2) from the equation we express the reaction force acting on the block from the side of the inclined plane. N = mg cosα (3). Let's write down the projections on the OX axis.
On the OX axis: force projection N is equal to zero, since the vector is perpendicular to the OX axis; The projection of the friction force is negative (the vector is directed in the opposite direction relative to the selected axis); the projection of gravity is positive and equal to mg x = mg sinα (4) from a right triangle. Acceleration projection is positive a x = a; Then we write equation (1) taking into account the projection mg sinα – F tr = ma (5); F tr = m(g sinα – a) (6); Remember that the friction force is proportional to the force of normal pressure N.
A-priory F tr = μ N(7), we express the coefficient of friction of the block on the inclined plane.
| μ = | F tr | = | m(g sinα – a) | = tgα – | a | (8). |
| N | mg cosα | g cosα |
We select the appropriate positions for each letter.
Answer. A – 3; B – 2.
Task 8. Gaseous oxygen is in a vessel with a volume of 33.2 liters. The gas pressure is 150 kPa, its temperature is 127° C. Determine the mass of the gas in this vessel. Express your answer in grams and round to the nearest whole number.
Solution. It is important to pay attention to the conversion of units to the SI system. Convert temperature to Kelvin T = t°C + 273, volume V= 33.2 l = 33.2 · 10 –3 m 3 ; We convert the pressure P= 150 kPa = 150,000 Pa. Using the ideal gas equation of state
Let's express the mass of the gas.
Be sure to pay attention to which units are asked to write down the answer. It is very important.
Answer.'48
Task 9. An ideal monatomic gas in an amount of 0.025 mol expanded adiabatically. At the same time, its temperature dropped from +103°C to +23°C. How much work has been done by the gas? Express your answer in Joules and round to the nearest whole number.
Solution. Firstly, the gas is monatomic number of degrees of freedom i= 3, secondly, the gas expands adiabatically - this means without heat exchange Q= 0. The gas does work by decreasing internal energy. Taking this into account, we write the first law of thermodynamics in the form 0 = ∆ U + A G; (1) let us express the gas work A g = –∆ U(2); We write the change in internal energy for a monatomic gas as
Answer. 25 J.
The relative humidity of a portion of air at a certain temperature is 10%. How many times should the pressure of this portion of air be changed so that, at a constant temperature, its relative humidity increases by 25%?
Solution. Questions related to saturated steam and air humidity most often cause difficulties for schoolchildren. Let's use the formula to calculate relative air humidity
According to the conditions of the problem, the temperature does not change, which means that the saturated vapor pressure remains the same. Let us write down formula (1) for two states of air.
φ 1 = 10%; φ 2 = 35%
Let us express the air pressure from formulas (2), (3) and find the pressure ratio.
| P 2 | = | φ 2 | = | 35 | = 3,5 |
| P 1 | φ 1 | 10 |
Answer. The pressure should be increased by 3.5 times.
The hot liquid substance was slowly cooled in a melting furnace at constant power. The table shows the results of measurements of the temperature of a substance over time.
Select from the list provided two statements that correspond to the results of the measurements taken and indicate their numbers.
- The melting point of the substance under these conditions is 232°C.
- In 20 minutes. after the start of measurements, the substance was only in a solid state.
- The heat capacity of a substance in liquid and solid states is the same.
- After 30 min. after the start of measurements, the substance was only in a solid state.
- The crystallization process of the substance took more than 25 minutes.
Solution. As the substance cooled, its internal energy decreased. The results of temperature measurements allow us to determine the temperature at which a substance begins to crystallize. While a substance changes from liquid to solid, the temperature does not change. Knowing that the melting temperature and crystallization temperature are the same, we choose the statement:
1. The melting point of the substance under these conditions is 232°C.
The second correct statement is:
4. After 30 min. after the start of measurements, the substance was only in a solid state. Since the temperature at this point in time is already below the crystallization temperature.
Answer. 14.
In an isolated system, body A has a temperature of +40°C, and body B has a temperature of +65°C. These bodies were brought into thermal contact with each other. After some time, thermal equilibrium occurred. How did the temperature of body B and the total internal energy of bodies A and B change as a result?
For each quantity, determine the corresponding nature of the change:
- Increased;
- Decreased;
- Hasn't changed.
Write down the selected numbers for each physical quantity in the table. The numbers in the answer may be repeated.
Solution. If in an isolated system of bodies no energy transformations occur other than heat exchange, then the amount of heat given off by bodies whose internal energy decreases is equal to the amount of heat received by bodies whose internal energy increases. (According to the law of conservation of energy.) In this case, the total internal energy of the system does not change. Problems of this type are solved based on the heat balance equation.
| ∆U = ∑ | n | ∆U i = 0 (1); |
| i = 1 |
where ∆ U– change in internal energy.
In our case, as a result of heat exchange, the internal energy of body B decreases, which means the temperature of this body decreases. The internal energy of body A increases, since the body received an amount of heat from body B, its temperature will increase. The total internal energy of bodies A and B does not change.
Answer. 23.
Proton p, flying into the gap between the poles of the electromagnet, has a speed perpendicular to the magnetic field induction vector, as shown in the figure. Where is the Lorentz force acting on the proton directed relative to the drawing (up, towards the observer, away from the observer, down, left, right)
Solution. A magnetic field acts on a charged particle with the Lorentz force. In order to determine the direction of this force, it is important to remember the mnemonic rule of the left hand, do not forget to take into account the charge of the particle. We direct the four fingers of the left hand along the velocity vector, for a positively charged particle, the vector should enter perpendicularly into the palm, the thumb set at 90° shows the direction of the Lorentz force acting on the particle. As a result, we have that the Lorentz force vector is directed away from the observer relative to the figure.
Answer. from the observer.
The modulus of the electric field strength in a flat air capacitor with a capacity of 50 μF is equal to 200 V/m. The distance between the capacitor plates is 2 mm. What is the charge on the capacitor? Write your answer in µC.
Solution. Let's convert all units of measurement to the SI system. Capacitance C = 50 µF = 50 10 –6 F, distance between plates d= 2 · 10 –3 m. The problem talks about a flat air capacitor - a device for storing electric charge and electric field energy. From the formula of electrical capacitance
Where d– distance between the plates.
Let's express the voltage U=E d(4); Let's substitute (4) into (2) and calculate the charge of the capacitor.
q = C · Ed= 50 10 –6 200 0.002 = 20 µC
Please pay attention to the units in which you need to write the answer. We received it in coulombs, but present it in µC.
Answer. 20 µC.
The student conducted an experiment on the refraction of light, shown in the photograph. How does the angle of refraction of light propagating in glass and the refractive index of glass change with increasing angle of incidence?
- Increases
- Decreases
- Doesn't change
- Record the selected numbers for each answer in the table. The numbers in the answer may be repeated.
Solution. In problems of this kind, we remember what refraction is. This is a change in the direction of propagation of a wave when passing from one medium to another. It is caused by the fact that the speeds of wave propagation in these media are different. Having figured out which medium the light is propagating to which, let us write the law of refraction in the form
| sinα | = | n 2 | , |
| sinβ | n 1 |
Where n 2 – absolute refractive index of glass, the medium where the light goes; n 1 is the absolute refractive index of the first medium from which the light comes. For air n 1 = 1. α is the angle of incidence of the beam on the surface of the glass half-cylinder, β is the angle of refraction of the beam in the glass. Moreover, the angle of refraction will be less than the angle of incidence, since glass is an optically denser medium - a medium with a high refractive index. The speed of light propagation in glass is slower. Please note that we measure angles from the perpendicular restored at the point of incidence of the beam. If you increase the angle of incidence, then the angle of refraction will increase. This will not change the refractive index of glass.
Answer.
Copper jumper at a point in time t 0 = 0 begins to move at a speed of 2 m/s along parallel horizontal conducting rails, to the ends of which a 10 Ohm resistor is connected. The entire system is in a vertical uniform magnetic field. The resistance of the jumper and the rails is negligible; the jumper is always located perpendicular to the rails. The flux Ф of the magnetic induction vector through the circuit formed by the jumper, rails and resistor changes over time t as shown in the graph.
Using the graph, select two correct statements and indicate their numbers in your answer.
- By the time t= 0.1 s change in magnetic flux through the circuit is 1 mWb.
- Induction current in the jumper in the range from t= 0.1 s t= 0.3 s max.
- The module of the inductive emf arising in the circuit is 10 mV.
- The strength of the induction current flowing in the jumper is 64 mA.
- To maintain the movement of the jumper, a force is applied to it, the projection of which on the direction of the rails is 0.2 N.
Solution. Using a graph of the dependence of the flux of the magnetic induction vector through the circuit on time, we will determine the areas where the flux F changes and where the change in flux is zero. This will allow us to determine the time intervals during which an induced current will appear in the circuit. True statement:
1) By the time t= 0.1 s change in magnetic flux through the circuit is equal to 1 mWb ∆Ф = (1 – 0) 10 –3 Wb; The module of the inductive emf arising in the circuit is determined using the EMR law
Answer. 13.
Using the graph of current versus time in an electrical circuit whose inductance is 1 mH, determine the self-inductive emf module in the time interval from 5 to 10 s. Write your answer in µV.
Solution. Let's convert all quantities to the SI system, i.e. we convert the inductance of 1 mH into H, we get 10 –3 H. We will also convert the current shown in the figure in mA to A by multiplying by 10 –3.
The formula for self-induction emf has the form
in this case, the time interval is given according to the conditions of the problem
∆t= 10 s – 5 s = 5 s
seconds and using the graph we determine the interval of current change during this time:
∆I= 30 10 –3 – 20 10 –3 = 10 10 –3 = 10 –2 A.
We substitute numerical values into formula (2), we get
| Ɛ | = 2 ·10 –6 V, or 2 µV.
Answer. 2.
Two transparent plane-parallel plates are pressed tightly against each other. A ray of light falls from the air onto the surface of the first plate (see figure). It is known that the refractive index of the upper plate is equal to n 2 = 1.77. Establish a correspondence between physical quantities and their meanings. For each position in the first column, select the corresponding position from the second column and write down the selected numbers in the table under the corresponding letters.
Solution. To solve problems on the refraction of light at the interface between two media, in particular problems on the passage of light through plane-parallel plates, the following solution procedure can be recommended: make a drawing indicating the path of rays coming from one medium to another; At the point of incidence of the beam at the interface between the two media, draw a normal to the surface, mark the angles of incidence and refraction. Pay special attention to the optical density of the media under consideration and remember that when a light beam passes from an optically less dense medium to an optically denser medium, the angle of refraction will be less than the angle of incidence. The figure shows the angle between the incident ray and the surface, but we need the angle of incidence. Remember that angles are determined from the perpendicular restored at the point of impact. We determine that the angle of incidence of the beam on the surface is 90° – 40° = 50°, refractive index n 2 = 1,77; n 1 = 1 (air).
Let's write down the law of refraction
| sinβ = | sin50 | = 0,4327 ≈ 0,433 |
| 1,77 |
Let's plot the approximate path of the beam through the plates. We use formula (1) for the boundaries 2–3 and 3–1. In response we get
A) The sine of the angle of incidence of the beam on the boundary 2–3 between the plates is 2) ≈ 0.433;
B) The angle of refraction of the beam when crossing the boundary 3–1 (in radians) is 4) ≈ 0.873.
Answer. 24.
Determine how many α - particles and how many protons are produced as a result of the thermonuclear fusion reaction
+ → x+ y;
Solution. In all nuclear reactions, the laws of conservation of electric charge and number of nucleons are observed. Let us denote by x the number of alpha particles, y the number of protons. Let's make up equations
+ → x + y;
solving the system we have that x = 1; y = 2
Answer. 1 – α-particle; 2 – protons.
The momentum modulus of the first photon is 1.32 · 10 –28 kg m/s, which is 9.48 · 10 –28 kg m/s less than the momentum modulus of the second photon. Find the energy ratio E 2 /E 1 of the second and first photons. Round your answer to the nearest tenth.
Solution. The momentum of the second photon is greater than the momentum of the first photon according to the condition, which means it can be represented p 2 = p 1 + Δ p(1). The energy of a photon can be expressed in terms of the momentum of the photon using the following equations. This E = mc 2 (1) and p = mc(2), then
E = pc (3),
Where E– photon energy, p– photon momentum, m – photon mass, c= 3 · 10 8 m/s – speed of light. Taking into account formula (3) we have:
| E 2 | = | p 2 | = 8,18; |
| E 1 | p 1 |
We round the answer to tenths and get 8.2.
Answer. 8,2.
The nucleus of the atom has undergone radioactive positron β - decay. How did the electric charge of the nucleus and the number of neutrons in it change as a result of this?
For each quantity, determine the corresponding nature of the change:
- Increased;
- Decreased;
- Hasn't changed.
Write down the selected numbers for each physical quantity in the table. The numbers in the answer may be repeated.
Solution. Positron β - decay in the atomic nucleus occurs when a proton transforms into a neutron with the emission of a positron. As a result of this, the number of neutrons in the nucleus increases by one, the electric charge decreases by one, and the mass number of the nucleus remains unchanged. Thus, the transformation reaction of the element is as follows:
Answer. 21.
Five experiments were carried out in the laboratory to observe diffraction using various diffraction gratings. Each of the gratings was illuminated by parallel beams of monochromatic light with a specific wavelength. In all cases, the light fell perpendicular to the grating. In two of these experiments, the same number of main diffraction maxima was observed. Indicate first the number of the experiment in which a diffraction grating with a shorter period was used, and then the number of the experiment in which a diffraction grating with a larger period was used.
Solution. Diffraction of light is the phenomenon of a light beam into a region of geometric shadow. Diffraction can be observed when, on the path of a light wave, there are opaque areas or holes in large obstacles that are opaque to light, and the sizes of these areas or holes are commensurate with the wavelength. One of the most important diffraction devices is the diffraction grating. 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)
Selecting pairs according to the experimental conditions, we first select 4 where a diffraction grating with a shorter period was used, and then the number of the experiment in which a diffraction grating with a larger period was used - this is 2.
Answer. 42.
Current flows through a wirewound resistor. The resistor was replaced with another, with a wire of the same metal and the same length, but having half the cross-sectional area, and half the current was passed through it. How will the voltage across the resistor and its resistance change?
For each quantity, determine the corresponding nature of the change:
- Will increase;
- Will decrease;
- Will not change.
Write down the selected numbers for each physical quantity in the table. The numbers in the answer may be repeated.
Solution. It is important to remember on what values the conductor resistance depends. The formula for calculating resistance is
Ohm's law for a section of the circuit, from formula (2), we express the voltage
U = I R (3).
According to the conditions of the problem, the second resistor is made of wire of the same material, the same length, but different cross-sectional area. The area is twice as small. Substituting into (1) we find that the resistance increases by 2 times, and the current decreases by 2 times, therefore, the voltage does not change.
Answer. 13.
The period of oscillation of a mathematical pendulum on the surface of the Earth is 1.2 times greater than the period of its oscillation on a certain planet. What is the magnitude of the acceleration due to gravity on this planet? The influence of the atmosphere in both cases is negligible.
Solution. A mathematical pendulum is a system consisting of a thread whose dimensions are much larger than the dimensions of the ball and the ball itself. Difficulty may arise if Thomson's formula for the period of oscillation of a mathematical pendulum is forgotten.
T= 2π (1);
l– length of the mathematical pendulum; g- acceleration of gravity.
By condition
Let us express from (3) g n = 14.4 m/s 2. It should be noted that the acceleration of gravity depends on the mass of the planet and the radius
Answer. 14.4 m/s 2.
A straight conductor 1 m long carrying a current of 3 A is located in a uniform magnetic field with induction IN= 0.4 Tesla at an angle of 30° to the vector. What is the magnitude of the force acting on the conductor from the magnetic field?
Solution. If you place a current-carrying conductor in a magnetic field, the field on the current-carrying conductor will act with an Ampere force. Let's write down the formula for the Ampere force modulus
F A = I LB sinα ;
F A = 0.6 N
Answer. F A = 0.6 N.
The magnetic field energy stored in the coil when a direct current is passed through it is equal to 120 J. How many times must the strength of the current flowing through the coil winding be increased in order for the magnetic field energy stored in it to increase by 5760 J.
Solution. The energy of the magnetic field of the coil is calculated by the formula
| W m = | LI 2 | (1); |
| 2 |
By condition W 1 = 120 J, then W 2 = 120 + 5760 = 5880 J.
| I 1 2 = | 2W 1 | ; I 2 2 = | 2W 2 | ; |
| L | L |
Then the current ratio
| I 2 2 | = 49; | I 2 | = 7 |
| I 1 2 | I 1 |
Answer. The current strength must be increased 7 times. You enter only the number 7 on the answer form.
An electrical circuit consists of two light bulbs, two diodes and a turn of wire connected as shown in the figure. (A diode only allows current to flow in one direction, as shown at the top of the picture.) Which of the bulbs will light up if the north pole of the magnet is brought closer to the coil? Explain your answer by indicating what phenomena and patterns you used in your explanation.
Solution. Magnetic induction lines emerge from the north pole of the magnet and diverge. As the magnet approaches, the magnetic flux through the coil of wire increases. In accordance with Lenz's rule, the magnetic field created by the inductive current of the coil must be directed to the right. According to the gimlet rule, the current should flow clockwise (as viewed from the left). The diode in the second lamp circuit passes in this direction. This means the second lamp will light up.
Answer. The second lamp will light up.
Aluminum spoke length L= 25 cm and cross-sectional area S= 0.1 cm 2 suspended on a thread by the upper end. The lower end rests on the horizontal bottom of the vessel into which water is poured. Length of the submerged part of the spoke l= 10 cm. Find the force F, with which the knitting needle presses on the bottom of the vessel, if it is known that the thread is located vertically. Density of aluminum ρ a = 2.7 g/cm 3, density of water ρ b = 1.0 g/cm 3. Acceleration of gravity g= 10 m/s 2
Solution. Let's make an explanatory drawing.
– Thread tension force;
– Reaction force of the bottom of the vessel;
a is the Archimedean force acting only on the immersed part of the body, and applied to the center of the immersed part of the spoke;
– the force of gravity acting on the spoke from the Earth and applied to the center of the entire spoke.
By definition, the mass of the spoke m and the Archimedean force modulus are expressed as follows: m = SLρ a (1);
F a = Slρ in g (2)
Let's consider the moments of forces relative to the point of suspension of the spoke.
M(T) = 0 – moment of tension force; (3)
M(N)= NL cosα is the moment of the support reaction force; (4)
Taking into account the signs of the moments, we write the equation
| NL cosα + Slρ in g (L – | l | )cosα = SLρ a g | L | cosα (7) |
| 2 | 2 |
considering that according to Newton's third law, the reaction force of the bottom of the vessel is equal to the force F d with which the knitting needle presses on the bottom of the vessel we write N = F d and from equation (7) we express this force:
| F d = [ | 1 | Lρ a– (1 – | l | )lρ in ] Sg (8). |
| 2 | 2L |
Let's substitute the numerical data and get that
F d = 0.025 N.
Answer. F d = 0.025 N.
Cylinder containing m 1 = 1 kg nitrogen, during strength testing exploded at temperature t 1 = 327°C. What mass of hydrogen m 2 could be stored in such a cylinder at a temperature t 2 = 27°C, having a fivefold safety margin? Molar mass of nitrogen M 1 = 28 g/mol, hydrogen M 2 = 2 g/mol.
Solution. Let us write the Mendeleev–Clapeyron ideal gas equation of state for nitrogen
Where V– volume of the cylinder, T 1 = t 1 + 273°C. According to the condition, hydrogen can be stored at pressure p 2 = p 1 /5; (3) Considering that
we can express the mass of hydrogen by working directly with equations (2), (3), (4). The final formula looks like:
| m 2 = | m 1 | M 2 | T 1 | (5). | ||
| 5 | M 1 | T 2 |
After substituting numeric data m 2 = 28 g.
Answer. m 2 = 28 g.
In an ideal oscillatory circuit, the amplitude of current fluctuations in the inductor is I m= 5 mA, and the voltage amplitude on the capacitor Um= 2.0 V. At time t the voltage across the capacitor is 1.2 V. Find the current in the coil at this moment.
Solution. In an ideal oscillatory circuit, the oscillatory energy is conserved. For a moment of time t, the law of conservation of energy has the form
| C | U 2 | + L | I 2 | = L | I m 2 | (1) |
| 2 | 2 | 2 |
For amplitude (maximum) values we write
and from equation (2) we express
| C | = | I m 2 | (4). |
| L | Um 2 |
Let's substitute (4) into (3). As a result we get:
I = I m (5)
Thus, the current in the coil at the moment of time t equal to
I= 4.0 mA.
Answer. I= 4.0 mA.
There is a mirror at the bottom of a reservoir 2 m deep. A ray of light, passing through the water, is reflected from the mirror and comes out of the water. The refractive index of water is 1.33. Find the distance between the point of entry of the beam into the water and the point of exit of the beam from the water if the angle of incidence of the beam is 30°
Solution. Let's make an explanatory drawing
α is the angle of incidence of the beam;
β is the angle of refraction of the beam in water;
AC is the distance between the point of entry of the beam into the water and the point of exit of the beam from the water.
According to the law of refraction of light
| sinβ = | sinα | (3) |
| n 2 |
Consider the rectangular ΔADB. In it AD = h, then DB = AD
| tgβ = h tgβ = h | sinα | = h | sinβ | = h | sinα | (4) |
| cosβ |
We get the following expression:
| AC = 2 DB = 2 h | sinα | (5) |
Let's substitute the numerical values into the resulting formula (5)
Answer. 1.63 m.
In preparation for the Unified State Exam, we invite you to familiarize yourself with work program in physics for grades 7–9 to the UMK line of Peryshkina A.V. And advanced level work program for grades 10-11 for teaching materials Myakisheva G.Ya. The programs are available for viewing and free downloading to all registered users.
This article presents an analysis of tasks in mechanics (dynamics and kinematics) from the first part of the Unified State Exam in Physics with detailed explanations from a physics tutor. There is a video analysis of all tasks.
Let us select a section on the graph corresponding to the time interval from 8 to 10 s:
The body moved over this time interval with the same acceleration, since the graph here is a section of a straight line. During these s, the speed of the body changed by m/s. Consequently, the acceleration of the body during this period of time was equal to 
m/s 2 . Graph number 3 is suitable (at any time the acceleration is -5 m/s 2).
 2. Two forces act on the body: and . By force and resultant of two forces  find the modulus of the second force (see figure). |
The vector of the second force is equal to 
. Or, which is similar, 
. Then we add the last two vectors according to the parallelogram rule:
The length of the total vector can be found from a right triangle ABC, whose legs AB= 3 N and B.C.= 4 N. According to the Pythagorean theorem, we find that the length of the desired vector is equal to 
N.
Let us introduce a coordinate system with a center coinciding with the center of mass of the block and an axis OX, directed along an inclined plane. Let us depict the forces acting on the block: gravity, support reaction force and static friction force. The result will be the following picture:
The body is at rest, therefore the vector sum of all forces acting on it is equal to zero. Including zero and the sum of the projections of forces on the axis OX.
Projection of gravity onto the axis OX equal to leg AB corresponding right triangle (see figure). Moreover, from geometric considerations, this leg lies opposite the angle in . That is, the projection of gravity onto the axis OX equal to .
The static friction force is directed along the axis OX, therefore the projection of this force onto the axis OX equal to simply the length of this vector, but with the opposite sign, since the vector is directed against the axis OX. As a result we get:
We use the formula known from the school physics course:
Let us determine from the figure the amplitudes of steady-state forced oscillations at driving force frequencies of 0.5 Hz and 1 Hz:
The figure shows that at a driving force frequency of 0.5 Hz, the amplitude of steady-state forced oscillations was 2 cm, and at a driving force frequency of 1 Hz, the amplitude of steady-state forced oscillations was 10 cm. Consequently, the amplitude of steady-state forced oscillations increased 5 times.
 6. A ball thrown horizontally from a height H with initial speed, during flight t flew horizontal distance L(see picture). What will happen to the flight time and acceleration of the ball if, at the same installation, with a constant initial speed of the ball, we increase the height H? (Neglect air resistance.) For each value, determine the corresponding nature of its change: 1) will increase 2) will decrease 3) will not change Write down the selected numbers for each physical quantity in the table. The numbers in the answer may be repeated. |
In both cases, the ball will move with the acceleration of gravity, so the acceleration will not change. In this case, the flight time does not depend on the initial speed, since the latter is directed horizontally. The flight time depends on the height from which the body falls, and the higher the height, the longer the flight time (it takes longer for the body to fall). Consequently, the flight time will increase. Correct answer: 13.
In the second task of the Unified State Exam in physics, it is necessary to solve a problem on Newton’s laws or related to the action of forces. Below we present the theory with formulas that are necessary to successfully solve problems on this topic.
Theory for task No. 2 of the Unified State Exam in Physics
Newton's second law
Newton's second law formula F =ma . Here F And a – vector quantities. Magnitude a – This is the acceleration of a body's motion under the influence of a specified force. It is directly proportional to the force acting on a given body and is directed in the direction of the force.
Resultant
Resultant force is a force whose action replaces the action of all forces applied to the body. Or, in other words, the resultant of all forces applied to the body is equal to the vector sum of these forces.
Friction force
F tr =μN , Where μ μ, which is a constant value for a given case. Knowing the friction force and the normal pressure force (this force is also called the support reaction force), you can calculate the friction coefficient.
Gravity
The vertical component of movement depends on the forces acting on the body. Knowledge of the gravity formula is required F=mg, since, as a rule, only it acts on a body thrown at an angle to the horizontal.
Elastic force
Elastic force is a force that arises in a body as a result of its deformation and tends to return it to its original (initial) state. For elastic force, Hooke's law is used: F = kδl, Where k— elasticity coefficient (body stiffness), δl— magnitude of deformation.
Law of Gravity
The force F of gravitational attraction between two material points of mass m1 and m2, separated by a distance r, is proportional to both masses and inversely proportional to the square of the distance between them:
Analysis of typical options for tasks No. 2 of the Unified State Exam in Physics
Demo version 2018
The graph shows the dependence of the sliding friction force modulus on the normal pressure force modulus. What is the coefficient of friction?
Solution algorithm:
- Let us write down a formula connecting these forces. Express the coefficient of friction.
- We examine the graph and set a pair of corresponding values of the forces of normal pressure N and friction.
- We calculate the coefficient based on the force values taken from the graph.
- We write down the answer.
Solution:
- The friction force is related to the normal pressure force by the formula F tr=μ N, Where μ – friction coefficient. From here, knowing the magnitude of the friction force and pressure normal to the surface, we can determine μ, which is a constant value for a given case. Knowing the friction force and the normal pressure force (this force is also called the support reaction force), you can calculate the friction coefficient. From the above formula it follows that: μ = F tr: N
- Let's look at the dependence graph. Let's take any point on the graph, for example, when N = 12 (N), and F tr = 1.5 (N).
- Let's take the selected force values and calculate the coefficient value μ : μ= 1,5/12 = 0,125
Answer: 0.125
First version of the task (Demidova, No. 3)
Force F imparts an acceleration a to a body of mass m in the inertial frame of reference. Determine the acceleration of a body of mass 2m under the influence of a force of 0.5F in this frame of reference.
1) ; 2) ; 3) ; 4)
Solution algorithm:
- Let's write down Newton's second law. We express the acceleration from the formula.
- We substitute the changed values of mass and force into the resulting expression and find the new value of acceleration, expressed through its original value.
- Choose the correct answer.
Solution:
1. According to Newton's second law F=m a, force F, which acts on a body of mass m, imparts acceleration to the body A. We have:
2. By condition m 2 = 2m, F 2 =0,5F.
Then the changed acceleration will be equal to:
In vector form the notation is similar.
Second version of the task (Demidova, No. 9)
A stone weighing 200 g is thrown at an angle of 60° to the horizontal with an initial speed v = 20 m/s. Determine the modulus of gravity acting on the stone at the top point of the trajectory.
If a body is thrown at an angle to the horizontal and the drag force can be neglected, the resultant of all forces is constant. The vertical component of movement depends on the forces acting on the body. It is necessary to know the formula of gravity F=mg, since, as a rule, only it acts on a body thrown at an angle to the horizontal.
Solution algorithm:
- Convert the mass value to SI.
- We determine what forces act on the stone.
- We write down the formula for gravity. We calculate the magnitude of the force.
- We write down the answer.
Solution:
- Stone mass m=200 g=0.2 kg.
- A thrown stone is affected by gravity F T = mg. Since the condition does not stipulate otherwise, air resistance can be neglected.
- The force of gravity is the same at any point in the trajectory of the stone. This means the data in the condition (initial speed v and the angle to the horizon at which the body is thrown) are redundant. From here we get: F T = 0.2∙10 =2 N.
Answer : 2
Third version of the task (Demidova, No. 27)
A constant horizontal force of F = 9 N is applied to a system of a cube weighing 1 kg and two springs (see figure). The system is at rest. There is no friction between the cube and the support. The left edge of the first spring is attached to the wall. The stiffness of the first spring k1 = 300 N/m. The stiffness of the second spring is k2 = 600 N/m. What is the elongation of the second spring?
Solution algorithm:
- We write down Hooke's law for the 2nd spring. We find its connection with the force F given in the condition.
- From the resulting equation we express the elongation and calculate it.
- We write down the answer.
Solution:
- According to Hooke's law, the elongation of a spring is related to the spring stiffness k and the force applied to it F expression F= k∆ l. The second spring is subject to a tensile force F 2 = k2∆ l. 1st spring is stretched by force F. By condition F=9 H. Since the springs form a single system, the force F also stretches the 2nd spring, i.e. F 2 =F.
- Elongation Δ l is defined like this:
The video course “Get an A” includes all the topics necessary to successfully pass the Unified State Exam in mathematics with 60-65 points. Completely all tasks 1-13 of the Profile Unified State Exam in mathematics. Also suitable for passing the Basic Unified State Examination in mathematics. If you want to pass the Unified State Exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!
Preparation course for the Unified State Exam for grades 10-11, as well as for teachers. Everything you need to solve Part 1 of the Unified State Exam in mathematics (the first 12 problems) and Problem 13 (trigonometry). And this is more than 70 points on the Unified State Exam, and neither a 100-point student nor a humanities student can do without them.
All the necessary theory. Quick solutions, pitfalls and secrets of the Unified State Exam. All current tasks of part 1 from the FIPI Task Bank have been analyzed. The course fully complies with the requirements of the Unified State Exam 2018.
The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.
Hundreds of Unified State Exam tasks. Word problems and probability theory. Simple and easy to remember algorithms for solving problems. Geometry. Theory, reference material, analysis of all types of Unified State Examination tasks. Stereometry. Tricky solutions, useful cheat sheets, development of spatial imagination. Trigonometry from scratch to problem 13. Understanding instead of cramming. Clear explanations of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. A basis for solving complex problems of Part 2 of the Unified State Exam.
Changes in Unified State Examination tasks in physics for 2019 no year.
Structure of Unified State Examination tasks in physics-2019
The examination paper consists of two parts, including 32 tasks.
Part 1 contains 27 tasks.
- 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, 23 and 24 is a sequence of two numbers.
- The answer to tasks 19 and 22 are two numbers.
Part 2 contains 5 tasks. The answer to tasks 28–32 includes a detailed description of the entire progress of the task. The second part of the tasks (with a detailed answer) is assessed by an expert commission on the basis of.
Unified State Exam topics in physics that will be included in the exam paper
- Mechanics(kinematics, dynamics, statics, conservation laws in mechanics, mechanical vibrations and waves).
- Molecular physics(molecular kinetic theory, thermodynamics).
- Electrodynamics and fundamentals of SRT(electric field, direct current, magnetic field, electromagnetic induction, electromagnetic oscillations and waves, optics, fundamentals of SRT).
- Quantum physics and elements of astrophysics(wave-corpuscular dualism, atomic physics, physics of the atomic nucleus, elements of astrophysics).
Duration of the Unified State Exam in Physics
The entire examination work will be completed 235 minutes.
The approximate time to complete tasks of various parts of the work is:
- for each task with a short answer – 3–5 minutes;
- for each task with a detailed answer – 15–20 minutes.
What you can take for the exam:
- A non-programmable calculator is used (for each student) with the ability to calculate trigonometric functions (cos, sin, tg) and a ruler.
- The list of additional devices and devices, the use of which is permitted for the Unified State Examination, is approved by Rosobrnadzor.
Important!!! You should not rely on cheat sheets, tips or the use of technical means (phones, tablets) during the exam. Video surveillance at the Unified State Exam 2019 will be strengthened with additional cameras.
Unified State Exam scores in physics
- 1 point - for 1-4, 8, 9, 10, 13, 14, 15, 19, 20, 22, 23, 25, 26, 27 tasks.
- 2 points - 5, 6, 7, 11, 12, 16, 17, 18, 21, 24.
- 3 points - 28, 29, 30, 31, 32.
Total: 52 points(maximum primary score).
What you need to know when preparing tasks for the Unified State Exam:
- Know/understand the meaning of physical concepts, quantities, laws, principles, postulates.
- Be able to describe and explain physical phenomena and properties of bodies (including space objects), the results of experiments... give examples of the practical use of physical knowledge
- Distinguish hypotheses from scientific theory, draw conclusions based on experiment, etc.
- Be able to apply acquired knowledge when solving physical problems.
- Use acquired knowledge and skills in practical activities and everyday life.
Where to start preparing for the Unified State Exam in Physics:
- Study the theory required for each task.
- Practice test tasks in physics, developed on the basis of the Unified State Exam. On our website, tasks and options in physics will be updated.
- Manage your time correctly.
We wish you success!
