Ege 11 physics. Preparation for the exam in physics: examples, solutions, explanations
Unified State Exam in Physics - an exam that is not included in the list of tests required for passing by all graduates. Physics is chosen by potential engineering students. Moreover, each university sets its own bar - in prestigious educational institutions it can be very high. This must be understood by the graduate when starting the preparation for the exam.Exam purpose - checking the level of knowledge and skills acquired during school education, for compliance with the norms and standards specified in the program.
- Almost 4 hours - 235 minutes are allotted for the exam, this time must be properly distributed between tasks in order to successfully cope with all without wasting a single minute.
- It is allowed to take a calculator with you, as many complex calculations are required to complete tasks. You can also take a ruler.
- The work consists of three parts, each has its own characteristics, consists of tasks of different difficulty levels.
Second part divided into 2 blocks. In the first block, it is necessary to apply formulas, laws and theories to solve tasks and get an answer. The candidate is offered options from which to choose the correct one.
In the second block - tasks, it is required to provide a detailed solution, a complete explanation of each action. Persons checking the assignment should also see here the formulas, laws that are used to solve - with them you need to start a detailed analysis of the assignment.
Physics is a difficult subject, approximately every 15-1 takes this exam annually to enter a technical university. It is assumed that a graduate with such goals will not study the subject "from scratch" in order to prepare for the USE.
To successfully pass the test, you must:
- Start repetition of the material in advance, approach the issue comprehensively;
- Actively apply theory in practice - solve many tasks of different difficulty levels;
- Self-education;
- Pass online testing on questions from past years.
Preparation for the exam and exam
The average general education
UMK line A. V. Grachev. Physics (10-11) (basic, advanced)
UMK line A. V. Grachev. Physics (7-9)
UMK line A.V. Peryshkin. Physics (7-9)
Preparation for the exam in physics: examples, solutions, explanations
We analyze the tasks of the exam in physics (Option C) with a teacher.Lebedeva Alevtina Sergeevna, physics teacher, work experience 27 years. Certificate of honor of the Ministry of Education of the Moscow Region (2013), Letter of Gratitude from the Head of the Resurrection Municipal District (2015), Certificate of honor of 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. Tasks increased level aimed at testing the ability to use the concepts and laws of physics to analyze various processes and phenomena, as well as the ability to solve problems on the application of one or two laws (formulas) on any of the topics school course physics. In work, 4 tasks of part 2 are tasks high level difficulties and test the ability to use the laws and theories of physics in a changed or new situation. The fulfillment of such tasks requires the application of knowledge from two three sections of physics at once, i.e. high level of training. This option is fully consistent with the demo version of the exam 2017, tasks are taken from open bank tasks of the exam.
The figure shows a graph of the dependence of the speed module on time t... Determine the distance covered by the car in the time interval from 0 to 30 s.
Decision. The distance traveled by a car in the time interval from 0 to 30 s is easiest to define as the area of \u200b\u200ba trapezoid, the bases of which are the time intervals (30 - 0) \u003d 30 s and (30 - 10) \u003d 20 s, and the height is the speed v \u003d 10 m / s, i.e.
| S = | (30 + 20) from | 10 m / s \u003d 250 m. |
| 2 |
Answer. 250 m.
A load weighing 100 kg is lifted vertically upward using a rope. The figure shows the dependence of the speed projection V load on the upward axle from time t... Determine the modulus of the cable tension during the ascent.
Decision. According to the graph of the dependence of the projection of speed v load on an axle directed vertically upward, from time t, you can determine the projection of the acceleration of the load
| a = | ∆v | = | (8 - 2) m / s | \u003d 2 m / s 2. |
| ∆t | 3 sec |
The load is influenced by: gravity directed vertically downward and cable tension force directed vertically upward along the cable, see fig. 2. Let us write down the basic equation of dynamics. Let's use Newton's second law. The geometric sum of the forces acting on the body is equal to the product of the body's mass by the acceleration imparted to it.
+ = (1)
Let us write the equation for the projection of vectors in the frame of reference connected with the earth, the OY axis is directed upwards. The projection of the tensile force is positive, since the direction of the force coincides with the direction of the OY axis, the projection of the gravity is negative, since the force vector is oppositely directed to the OY axis, the projection of the acceleration vector is also positive, so the body moves with acceleration upward. We have
T – mg = ma (2);
from formula (2) modulus of tensile force
T = m(g + a) \u003d 100 kg (10 + 2) m / s 2 \u003d 1200 N.
Answer... 1200 N.
The body is dragged along a rough horizontal surface at a constant speed, the modulus of which 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?
Decision. Imagine a physical process specified in the condition of the problem 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 + + \u003d (1)
Having chosen a frame of reference associated with a fixed surface, we write down the equations for the projection of vectors onto the selected coordinate axes. According to the condition of the problem, the body moves uniformly, since its speed is constant and equal to 1.5 m / s. This means that the acceleration of the body is zero. Horizontally, two forces act on the body: 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... Force projection 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. With this in mind, we have: F cosα - F tr \u003d 0; (1) express the projection of the force F, this is Fcosα \u003d F tr \u003d 16 N; (2) then the power developed by the force will be equal to N = Fcosα V (3) Let's make a substitution, taking into account equation (2), and substitute the corresponding data into equation (3):
N \u003d 16 N 1.5 m / s \u003d 24 W.
Answer. 24 watts
The load, fixed on a light spring with a stiffness of 200 N / m, makes vertical vibrations. The figure shows a plot of the dependence of the displacement x cargo from time t... Determine what the weight of the load is. Round your answer to the nearest whole number.
Decision. The spring loaded vibrates vertically. According to the graph of the dependence of the displacement of the load x from time t, we define the period of fluctuations of the load. The oscillation period is T \u003d 4 s; from the formula T \u003d 2π express the mass m cargo.
| = | T | ; | m | = | T 2 | ; m = k | T 2 | ; m \u003d 200 H / m | (4 s) 2 | \u003d 81.14 kg ≈ 81 kg. |
| 2π | k | 4π 2 | 4π 2 | 39,438 |
Answer: 81 kg.
The figure shows a system of two lightweight blocks and a weightless cable, with which you can balance or lift a load weighing 10 kg. Friction is negligible. Based on the analysis of the above figure, select twocorrect statements and indicate their numbers in the 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 a power gain.
- h, you need to stretch out a section of rope with a length of 3 h.
- In order to slowly raise the load to a height hh.
Decision. In this task, it is necessary to remember simple mechanisms, namely blocks: a movable and fixed block. The movable block gives a twofold gain in strength, with the rope section having to be pulled twice as long, and the stationary block is used to redirect the force. In operation, simple mechanisms of winning do not give. After analyzing the problem, we immediately select the necessary statements:
- In order to slowly raise the load to a height h, you need to pull out a section of rope with a length of 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 is completely immersed in a vessel with water, fixed on a weightless and inextensible thread. The cargo does not touch the walls and bottom of the vessel. Then an iron weight is immersed in the same vessel with water, the weight of which is equal to the weight of the aluminum weight. 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?
- Increases;
- Decreases;
- Doesn't change.
Decision. We analyze the condition of the problem and select those parameters that do not change in the course of the study: these are the body mass and the liquid into which the body is immersed on threads. After that, it is better to perform a schematic drawing and indicate the forces acting on the load: the tension force of the thread F upr directed upward along the thread; gravity directed vertically downward; Archimedean force a acting on the submerged body from the side of the liquid and directed upwards. By the condition of the problem, the mass of the loads is the same, therefore, the modulus of the gravity force acting on the load does not change. Since the density of cargo is different, the volume will also be different
| V = | m | . |
| p |
The density of iron is 7800 kg / m 3, and the density of aluminum is 2700 kg / m 3. Hence, V f< 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 up. The basic equation of dynamics taking into account the projection of forces is written in the form F control + F a – mg \u003d 0; (1) Express the pulling force F control \u003d mg – F a (2); Archimedean force depends on the density of the fluid and the volume of the submerged part of the body F a = ρ gVp.h.t. (3); The density of the liquid does not change, and the volume of the iron body is less V f< V a, therefore, the Archimedean force acting on the iron load will be less. We draw a conclusion about the modulus of the thread tension force, working with equation (2), it will increase.
Answer. 13.
Block weight m slides off a fixed rough inclined plane with an angle α at the base. The block acceleration modulus is a, the speed modulus of the bar increases. Air resistance is negligible.
Establish a correspondence between physical quantities and formulas with which they can be calculated. For each position of 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 of the bar on an inclined plane
3) mg cosα
| 4) sinα - | a |
| gcosα |
Decision. This task requires the application of Newton's laws. We recommend making a schematic drawing; indicate all the kinematic characteristics of the movement. If possible, depict the acceleration vector and the vectors of all forces applied to the moving body; remember that the forces acting on the body are the result of interaction with other bodies. Then write down the basic equation of dynamics. Choose a reference system and write down the resulting equation for the projection of the vectors of forces and accelerations;
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 bar and the coordinate axes of the frame of reference associated with the surface of the inclined plane. Since all forces are constant, the movement of the bar will be equally variable with increasing speed, i.e. the acceleration vector is directed towards the movement. 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 + \u003d (1)
Let's write given equation (1) for force projection and acceleration.
On the OY axis: the projection of the support reaction force is positive, since the vector coincides with the direction of the OY axis N y = N; the projection of the friction force is zero because the vector is perpendicular to the axis; the projection of gravity will be negative and equal mg y= – mgcosα; acceleration vector projection a y \u003d 0, since the acceleration vector is perpendicular to the axis. We have N – mgcosα \u003d 0 (2) from the equation we express the force of the reaction acting on the bar, from the side of the inclined plane. N = mgcosα (3). Let's write projections on the OX axis.
OX axis: force projection N 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 = mgsinα (4) from right triangle... Acceleration projection positive a x = a; Then we write equation (1) taking into account the projection mgsinα - F tr \u003d ma (5); F tr \u003d m(gsinα - a) (6); Remember that the friction force is proportional to the normal pressure force N.
A-priory F tr \u003d μ N (7), we express the coefficient of friction of the bar on the inclined plane.
| μ = | F tr | = | m(gsinα - a) | \u003d tgα - | a | (8). |
| N | mgcosα | gcosα |
We select the appropriate positions for each letter.
Answer. A - 3; B - 2.
Task 8. Oxygen gas is in a 33.2 liter vessel. The gas pressure is 150 kPa, its temperature is 127 ° С. Determine the mass of gas in this vessel. Express your answer in grams and round to the nearest whole number.
Decision. It is important to pay attention to the conversion of units to the SI system. Converting temperature to Kelvin T = t° С + 273, volume V \u003d 33.2 l \u003d 33.2 · 10 –3 m 3; We translate the pressure P \u003d 150 kPa \u003d 150,000 Pa. Using the ideal gas equation of state
express the mass of the gas.
Be sure to pay attention to the unit in which you are asked to write down the answer. It is very important.
Answer. 48 g
Task 9. An ideal monatomic gas in the amount of 0.025 mol has expanded adiabatically. At the same time, its temperature dropped from + 103 ° С to + 23 ° С. What work did the gas do? Express your answer in Joules and round to the nearest whole number.
Decision. First, the gas is a monatomic number of degrees of freedom i \u003d 3, secondly, the gas expands adiabatically - this means without heat exchange Q \u003d 0. Gas does work by decreasing internal energy. Taking this into account, we write the first law of thermodynamics in the form 0 \u003d ∆ U + A g; (1) express the work of the gas A r \u003d –∆ U (2); We write the change in the 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 in order for its relative humidity to increase by 25% at a constant temperature?
Decision. Questions related to saturated steam and air humidity are most often difficult for schoolchildren. Let's use the formula to calculate the relative humidity
According to the condition of the problem, the temperature does not change, which means that the saturated vapor pressure remains the same. Let's write down the formula (1) for two states of air.
φ 1 \u003d 10%; φ 2 \u003d 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 substance in the liquid state was slowly cooled in a constant power melting furnace. The table shows the results of measurements of the temperature of a substance over time.
Choose from the list provided two statements that correspond to the results of the measurements 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 a liquid and solid state 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.
Decision. As the substance cooled down, its internal energy decreased. The temperature measurement results allow you to determine the temperature at which the substance begins to crystallize. As long as a substance passes from a liquid to a solid state, the temperature does not change. Knowing that the melting point 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 true statement is:
4. After 30 minutes. 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 are brought into thermal contact with each other. After some time, thermal equilibrium has come. How did the temperature of body B and the total internal energy of body A and B change as a result?
For each value, determine the corresponding change pattern:
- 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.
Decision. If in an isolated system of bodies there are no energy transformations except for heat exchange, then the amount of heat given off by bodies, the internal energy of which decreases, is equal to the amount of heat received by bodies, the internal energy of which increases. (According to the law of conservation of energy.) In this case, the total internal energy of the system does not change. Problems of this type are solved based on the heat balance equation.
| ∆U \u003d ∑ | n | ∆U i \u003d0 (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 that the temperature of this body decreases. The internal energy of body A increases, since the body has received the amount of heat from body B, then its temperature will increase. The total internal energy of bodies A and B does not change.
Answer. 23.
Proton p, flown into the gap between the poles of the electromagnet, has a velocity perpendicular to the magnetic induction vector, as shown in the figure. Where is the Lorentz force acting on the proton directed relative to the figure (up, towards the observer, from the observer, down, left, right)
Decision. The 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 left-hand mnemonic rule, not to forget to take into account the particle charge. We direct four fingers of the left hand along the velocity vector, for a positively charged particle, the vector should enter the palm perpendicularly, thumb set back by 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 50 μF flat air capacitor is 200 V / m. The distance between the capacitor plates is 2 mm. What is the charge of a capacitor? Write down the answer in μC.
Decision. Let's convert all units of measurement to the SI system. Capacitance C \u003d 50 μF \u003d 50 · 10 -6 F, distance between plates d \u003d 2 · 10 –3 m. The problem deals with a flat air capacitor - a device for accumulating electric charge and electric field energy. From the formula for electrical capacity
where d Is the distance between the plates.
Express the tension U \u003d E d(4); Substitute (4) in (2) and calculate the capacitor charge.
q = C · Ed\u003d 50 · 10 –6 · 200 · 0.002 \u003d 20 μC
Please note in which units you need to write the answer. We got it in pendants, but we represent it in μC.
Answer. 20 μC.
The student conducted an experiment on the refraction of light, presented 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
- Does not change
- Write down the selected numbers for each answer in the table. The numbers in the answer may be repeated.
Decision. In tasks of this kind, we recall what refraction is. This is a change in the direction of wave propagation when passing from one medium to another. It is caused by the fact that the wave propagation speeds in these media are different. Having figured out from which medium to which light it propagates, we write the law of refraction in the form
| sinα | = | n 2 | , |
| sinβ | n 1 |
where n 2 - the absolute refractive index of the 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 \u003d 1. α is the angle of incidence of the beam on the surface of the glass semi-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 the angles are measured from the perpendicular restored at the point of incidence of the ray. If you increase the angle of incidence, then the angle of refraction will also increase. The refractive index of the glass will not change from this.
Answer.
Copper jumper at time t 0 \u003d 0 starts to move at a speed of 2 m / s along parallel horizontal conductive 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 lintel and the rails is negligible, the lintel is always perpendicular to the rails. The flux Ф of the magnetic induction vector through the circuit formed by a jumper, rails and a resistor changes over time t as shown in the graph.
Using the graph, select two correct statements and include their numbers in the answer.
- By the time t \u003d 0.1 s, the change in magnetic flux through the circuit is equal to 1 mVb.
- Induction current in the jumper in the range from t \u003d 0.1 s t \u003d 0.3 s max.
- The EMF modulus of the induction 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 lintel, a force is applied to it, the projection of which on the direction of the rails is 0.2 N.
Decision. According to the graph of the dependence of the flux of the magnetic induction vector through the circuit on time, we determine the sections where the flux Ф changes, and where the flux change is zero. This will allow us to determine the time intervals in which the induction current will occur in the circuit. Correct statement:
1) By the time t \u003d 0.1 s change in magnetic flux through the circuit is equal to 1 mWb ∆F \u003d (1 - 0) · 10 –3 Wb; The EMF modulus of induction arising in the circuit is determined using the EMR law
Answer. 13.
According to the graph of the dependence of the current strength on time in an electrical circuit, the inductance of which is 1 mH, determine the EMF modulus of self-induction in the time interval from 5 to 10 s. Write down the answer in μV.
Decision. Let's translate all the quantities into the SI system, i.e. the inductance of 1 mH is converted into H, we get 10 –3 H. The strength of the current shown in the figure in mA will also be converted into A by multiplying by 10 –3.
The self-induction EMF formula has the form
the time interval is given according to the problem statement
∆t\u003d 10 s - 5 s \u003d 5 s
seconds and according to the graph we determine the interval of current change during this time:
∆I\u003d 30 · 10 –3 - 20 · 10 –3 \u003d 10 · 10 –3 \u003d 10 –2 A.
Substitute the numerical values \u200b\u200bin the formula (2), we obtain
| Ɛ | \u003d 2 · 10 –6 V, or 2 μV.
Answer. 2.
Two transparent plane-parallel plates are tightly pressed 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 n 2 \u003d 1.77. Establish a correspondence between physical quantities and their values. For each position of the first column, select the corresponding position from the second column and write down the selected numbers in the table under the corresponding letters.
Decision. To solve problems on the refraction of light at the interface between two media, in particular, problems on the transmission of light through plane-parallel plates, the following order of solution can be recommended: make a drawing indicating the path of rays going 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, note the angles of incidence and refraction. Pay particular 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 the angles are determined from the perpendicular restored at the point of incidence. We determine that the angle of incidence of the beam on the surface is 90 ° - 40 ° \u003d 50 °, the refractive index n 2 = 1,77; n 1 \u003d 1 (air).
We write the law of refraction
| sinβ \u003d | sin50 | = 0,4327 ≈ 0,433 |
| 1,77 |
Let's construct an approximate path of the ray through the plates. We use formula (1) for the boundaries 2–3 and 3–1. In the answer 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 ray 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 by a thermonuclear fusion reaction
+ → x+ y;
Decision. In all nuclear reactions, the laws of conservation of electric charge and the number of nucleons are observed. Let us denote by x - the number of alpha particles, y - the number of protons. Let's make the equations
+ → x + y;
solving the system, we have that x = 1; y = 2
Answer. 1 - α-particle; 2 - proton.
The module of the momentum of the first photon is 1.32 · 10 –28 kg · m / s, which is 9.48 · 10 –28 kg · m / s less than the module of the momentum of the second photon. Find the energy ratio E 2 / E 1 of the second and first photons. Round your answer to tenths.
Decision. The momentum of the second photon is greater than the momentum of the first photon by condition means it can be imagined p 2 = p 1 + Δ p (1). The photon energy can be expressed in terms of the photon momentum using the following equations. it E = mc 2 (1) and p = mc (2) then
E = pc (3),
where E - photon energy, p - photon momentum, m - photon mass, c \u003d 3 · 10 8 m / s is the speed of light. Taking into account formula (3), we have:
| E 2 | = | p 2 | = 8,18; |
| E 1 | p 1 |
Round the answer to tenths and get 8.2.
Answer. 8,2.
The atomic nucleus 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?
For each value, determine the corresponding change pattern:
- 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.
Decision. Positron β - decay in atomic nucleus occurs during the conversion of a proton 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.
In the laboratory, five experiments were carried out to observe diffraction using various diffraction gratings. Each of the gratings was illuminated by parallel beams of monochromatic light with a specific wavelength. The light in all cases fell perpendicular to the grating. In two of these experiments, the same number of main diffraction maxima was observed. First indicate the number of the experiment in which the diffraction grating with a shorter period was used, and then the number of the experiment in which the diffraction grating with a longer period was used.
Decision. Diffraction of light is the phenomenon of a light beam in the area of \u200b\u200ba geometric shadow. Diffraction can be observed when on the path of the light wave there are opaque areas or holes in large and opaque obstacles, and the sizes of these areas or holes are commensurate with the wavelength. One of the most important diffraction devices is a diffraction grating. The angular directions to the maxima of the diffraction pattern are determined by the equation
dsinφ \u003d k λ (1),
where d Is the period of the diffraction grating, φ is the angle between the normal to the grating and the direction to one of the maxima of the diffraction pattern, λ is the light wavelength, k Is an integer called the order of the diffraction maximum. Express from equation (1)
Selecting pairs according to the conditions of the experiment, we first select 4 where the diffraction grating with a shorter period was used, and then the number of the experiment in which the diffraction grating with a longer period was used - this is 2.
Answer. 42.
Current flows through the 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 value, determine the corresponding change pattern:
- 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.
Decision. It is important to remember on what values \u200b\u200bthe resistance of the conductor depends. The formula for calculating the 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 a wire of the same material, the same length, but different cross-sectional area. The area is half the size. Substituting in (1) we get that the resistance increases by 2 times, and the current decreases by 2 times, therefore, the voltage does not change.
Answer. 13.
The oscillation period of a mathematical pendulum on the surface of the Earth is 1, 2 times the period of its oscillations on a planet. What is the modulus of free fall acceleration on this planet? The influence of the atmosphere in both cases is negligible.
Decision. A mathematical pendulum is a system consisting of a thread whose dimensions are much larger than the sizes of the ball and the ball itself. Difficulty may arise if Thomson's formula for the oscillation period of a mathematical pendulum is forgotten.
T \u003d 2π (1);
l - the length of the mathematical pendulum; g - acceleration of gravity.
By condition
Let us express from (3) g n \u003d 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, through which a current of 3 A flows, is located in a uniform magnetic field with induction IN \u003d 0.4 T at an angle of 30 ° to the vector. What is the modulus of the force acting on the conductor from the side of the magnetic field?
Decision. If you place a conductor with current in a magnetic field, then the field on the conductor with current will act with the Ampere force. We write the formula for the modulus of the Ampere force
F A \u003d I LBsinα;
F A \u003d 0.6 N
Answer. F A \u003d 0.6 N.
The energy of the magnetic field stored in the coil when a direct current is passed through it is equal to 120 J. How many times should the current flowing through the coil winding be increased in order for the stored magnetic field energy to increase by 5760 J.
Decision. The magnetic field energy of the coil is calculated by the formula
| W m \u003d | LI 2 | (1); |
| 2 |
By condition W 1 \u003d 120 J, then W 2 \u003d 120 + 5760 \u003d 5880 J.
| I 1 2 = | 2W 1 | ; I 2 2 = | 2W 2 | ; |
| L | L |
Then the ratio of currents
| I 2 2 | = 49; | I 2 | = 7 |
| I 1 2 | I 1 |
Answer. The current strength needs to be increased by 7 times. In the answer form you enter only the number 7.
The electrical circuit consists of two light bulbs, two diodes and a coil of wire, connected as shown in the figure. (The diode only passes current in one direction, as shown at the top of the figure). Which of the bulbs will light up if the north pole of the magnet is brought closer to the loop? Explain the answer by indicating which phenomena and patterns you used in the explanation.
Decision. The lines of magnetic induction exit the north pole of the magnet and diverge. As the magnet approaches, the magnetic flux through the coil of wire increases. According to Lenz's rule, the magnetic field created by the induction current of the loop must be directed to the right. According to the rule of the gimbal, the current should flow clockwise (if viewed from the left). In this direction, passes a diode standing in the circuit of the second lamp. This means that the second lamp will light up.
Answer. The second lamp comes on.
Long spoke aluminum L \u003d 25 cm and cross-sectional area S \u003d 0.1 cm 2 suspended on a thread at the upper end. The lower end rests on the horizontal bottom of a vessel into which water is poured. The length of the submerged part of the spoke l \u003d 10 cm. Find strength F, with which the spoke presses on the bottom of the vessel, if it is known that the thread is located vertically. The density of aluminum ρ a \u003d 2.7 g / cm 3 the density of water ρ b \u003d 1.0 g / cm 3. Acceleration of gravity g \u003d 10 m / s 2
Decision. Perform an explanatory drawing.
- Thread tension;
- The reaction force of the bottom of the vessel;
a - Archimedean force acting only on the submerged part of the body, and applied to the center of the submerged part of the spoke;
- the force of gravity acting on the spoke from the side of the Earth and applied to the center of the entire spoke.
By definition, the mass of the spoke m and module archimedean force expressed as follows: m = SLρ a (1);
F a \u003d Slρ in g (2)
Consider the moments of forces relative to the suspension point of the spoke.
M(T) \u003d 0 - moment of tension force; (3)
M(N) \u003d Nlcosα is the reaction force moment of the support; (4)
Taking into account the signs of the moments, we write the equation
| Nlcosα + Slρ in g (L – | l | ) cosα \u003d SLρ a g | L | cosα (7) |
| 2 | 2 |
given 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 spoke presses on the bottom of the vessel, we write N = F e and from equation (7) we express this force:
| F d \u003d [ | 1 | Lρ a– (1 – | l | )lρ in] Sg (8). |
| 2 | 2L |
Substitute the numerical data and get that
F d \u003d 0.025 N.
Answer. Fd \u003d 0.025 N.
A container containing m 1 \u003d 1 kg nitrogen, exploded in strength test at temperature t 1 \u003d 327 ° C. What mass of hydrogen m 2 could be stored in such a container at a temperature t 2 \u003d 27 ° C, having a fivefold safety factor? Molar mass of nitrogen M 1 \u003d 28 g / mol, hydrogen M 2 \u003d 2 g / mol.
Decision. We write the equation of state of an ideal Mendeleev-Clapeyron gas for nitrogen
where V - cylinder volume T 1 = t 1 + 273 ° C. By condition, hydrogen can be stored at pressure p 2 \u003d p 1/5; (3) Given that
we can express the mass of hydrogen by working directly with equations (2), (3), (4). The final formula is:
| m 2 = | m 1 | M 2 | T 1 | (5). | ||
| 5 | M 1 | T 2 |
After substitution of numeric data m 2 \u003d 28 g.
Answer. m 2 \u003d 28 g.
In an ideal oscillatory circuit, the amplitude of the current fluctuations in the inductor I m \u003d 5 mA, and the amplitude of the voltage across the capacitor U m \u003d 2.0 V. At the time t the voltage across the capacitor is 1.2 V. Find the current strength in the coil at that moment.
Decision. In an ideal oscillatory circuit, the vibration energy is stored. For the moment of time t, the energy conservation law 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 | U m 2 |
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 is equal to
I \u003d 4.0 mA.
Answer. I \u003d 4.0 mA.
At the bottom of a reservoir 2 m deep lies a mirror. 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 where the beam enters the water and the point where the beam exits the water if the angle of incidence of the beam is 30 °
Decision. Let's make an explanatory drawing
α is the angle of incidence of the beam;
β is the angle of refraction of the ray 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 \u003d | sinα | (3) |
| n 2 |
Consider a rectangular ΔADB. In it AD \u003d h, then DВ \u003d АD
| tgβ \u003d htgβ \u003d h | sinα | = h | sinβ | = h | sinα | (4) |
| cosβ |
We get the following expression:
| AC \u003d 2 DB \u003d 2 h | sinα | (5) |
Substitute the numerical values \u200b\u200binto the resulting formula (5)
Answer. 1.63 m.
In preparation for the exam, we suggest that you familiarize yourself with work program in physics for grades 7–9 to the line of the CMC A. Peryshkina and advanced level work program for grades 10-11 to the teaching materials department Myakisheva G.Ya. The programs are available for viewing and free download for all registered users.
Physics exam duration - 3 hours 55 minutes
The work consists of two parts, including 31 tasks.
Part 1: tasks 1 - 23
Part 2: tasks 24 - 31.
In tasks 1–4, 8–10, 14, 15, 20, 24–26, the answer is
integer or trailing decimal.
The answer to tasks 5-7, 11, 12, 16-18, 21 and 23
is a sequence of two digits.
The answer to problem 13 is a word.
The answer to tasks 19 and 22 are two numbers.
The answer to tasks 27–31 includes
detailed description the entire course of the assignment.
Minimum test score (on a 100-point scale) - 36
Demo USE 2020 in Physics (PDF):
Unified State Exam
Demonstration option ege tasks is to enable any USE participant to get an idea of \u200b\u200bthe structure of the CMM, the number and form of tasks, and the level of their complexity.The above criteria for assessing the performance of tasks with a detailed answer, included in this var-t, give an idea of \u200b\u200bthe requirements for the completeness and correctness of recording a detailed answer.
For successful preparation for passing the exam I propose to analyze prototype solutions real assignments from var-tov exam.
