Early exam in physics profile level. Preparation for the exam in physics: examples, solutions, explanations

Many graduates will take physics in 2017 as well, as this exam is in great demand. Many universities need you to have the Unified State Exam in Physics, so that in 2017 they can accept, and you can enter certain specialties of the faculties of their institutes. And for this reason, a future graduate who studies in the 11th grade, not knowing that he will have to pass such a difficult exam, and not just like that, but with results that will allow him to actually enter a good specialty, which requires knowledge of physics as a subject and availability uSE results, as an indicator that this year you have the right to apply for admission to study, guided by the fact that you passed the USE in physics in 2017, you have good points, and you think that you will enter at least the commercial department, although I would like to go to the budgetary one.

And that's why we think that besides school textbooks, the knowledge available in the brain of the head, as well as those books that you have already bought, you will need at least two more files that we recommend you download for free.

Firstly, these are years, because this is the basis on which you will rely first. There will also be specifications and codifiers, according to which you will learn the topics that need to be repeated and, in general, the entire exam procedure and conditions for its conduct.

Secondly, this is the KIMs of the mock exam in physics conducted by FIPI in early spring, that is, in March-April.

Here they are just what we offer you to download here, and not only because it's all free, but to a greater extent for the reason that you need it, not us. These uSE assignments in physics taken from open bank data, in which FIPI places tens of thousands of problems and questions in all subjects. And you understand that it is simply unrealistic to solve them all, because it takes 10 or 20 years, and you don’t have that time, you need to act urgently in 2017, because you don’t want to lose one year, and besides there will arrive in time new graduates whose level of knowledge is unknown to us, and therefore it is not clear how it will be to compete with them, whether it is easy or difficult.

Taking into account the fact that knowledge fades over time, it is also necessary to study now, that is, while there is fresh knowledge in the head.

Based on these facts, we come to the conclusion that it is necessary to make maximum efforts in order to prepare ourselves in an original way for any exam, including exam exam physics 2017, trial early tasks of which we offer you right now and download here.

This is all and you need to understand thoroughly and to the end, because it will be difficult to digest everything the first time, and what you see in the tasks you downloaded will give you food for thought in order to be ready for all the troubles that await you on exam in the spring!

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 levels of difficulty: basic, advanced and high. Tasks basic level, these are simple tasks that test the assimilation of the most important physical concepts, models, phenomena and laws. Tasks increased level are 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) for any of the topics of the school physics course. 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 implementation of such tasks requires the application of knowledge from two three sections of physics at once, i.e. high level of training. This option fully complies demo version USE 2017, tasks are taken from the open bank of USE tasks.

The figure shows a graph of the dependence of the speed module on time t... Determine the path covered by the car in the time interval from 0 to 30 s.

Decision. The path 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 an 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: the force of gravity directed vertically downward and the cable tension force directed along the cable vertically upward, 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 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 is the weight of the load. 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, 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 being 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 yarn tension and the modulus of the 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. According to 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. Consequently, 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 liquid 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 reference frame 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 us 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 since 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 will express the force of the reaction acting on the bar, from the side of the inclined plane. N = mgcosα (3). Let's write projections onto the OX axis.

On the 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.

By definition 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 ° C. 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 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 us write down 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 performed 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 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 \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 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 propagation of a wave 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 propagation of light 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 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. From 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 the 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 electric 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 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.

Substituting numerical values \u200b\u200binto 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 coming from one medium to another; at the point of incidence of the ray 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 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).

Let's 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 the condition, it means that we can represent p 2 = p 1 + Δ p (1). The energy of a photon can be expressed in terms of the momentum of a photon 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 - 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 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?

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 into atomic nucleus occurs during the transformation of a proton into a neutron with the emission of a positron. As a result, 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 on the observation of diffraction using various diffraction gratings. Each of the gratings was illuminated with 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 were observed. First indicate 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 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 for light, 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 - an integer called the order of the diffraction maximum. Let us express from equation (1)

When choosing 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 long period was used is 2.

Answer. 42.

Current flows through the wirewound resistor. The resistor was replaced with another one, with a wire of the same metal and the same length, but with 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).

By the condition 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 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 period of oscillation of a mathematical pendulum on the Earth's surface is 1, 2 times longer than the period of its oscillation on a certain planet. What is the free fall acceleration modulus 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 size of the ball and the ball itself. The difficulty can arise if Thomson's formula for the period of oscillation 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 AT \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 must 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. (The diode only passes current in one direction, as shown at the top of the figure). Which of the lamps will light up if the north pole of the magnet is brought closer to the loop? Explain the answer by indicating what phenomena and patterns you used in the explanation.

Decision. The lines of magnetic induction come out of 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 (when viewed from the left). A diode in the circuit of the second lamp passes in this direction. This means that the second lamp will light up.

Answer. The second lamp comes on.

Aluminum spoke length 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. Length of the submerged spoke l \u003d 10 cm. Find the force F, with which the needle presses on the bottom of the vessel if it is known that the thread is vertical. 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. Let's make an explanatory drawing.

- Thread tension;

- Force of reaction of the bottom of the vessel;

a - 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 is applied to the center of the entire spoke.

By definition, the weight 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 moment of the reaction force 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

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 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 ° С, having a fivefold safety factor? Molar mass of nitrogen M 1 \u003d 28 g / mol, hydrogen M 2 \u003d 2 g / mol.

Decision. Let us write the equation of state of the ideal gas of Mendeleev - Clapeyron for nitrogen

where V - the volume of the cylinder, T 1 = t 1 + 273 ° C. By condition, hydrogen can be stored at pressure p 2 \u003d p 1/5; (3) Taking into account 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 in the coil at this 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 the 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 equals

I \u003d 4.0 mA.

Answer. I \u003d 4.0 mA.

There is a mirror at the bottom of the 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 °

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 a working program in physics for grades 7–9 for the line of the UMK Peryshkin A. V. and working program of an in-depth level for grades 10-11 for the educational complex Myakisheva G.Ya. The programs are available for viewing and free download for all registered users.

Option No. 3109295

Early exam in physics 2017, option 101

When completing tasks with a short answer, write in the answer field the number that corresponds to the number of the correct answer, or a number, word, sequence of letters (words) or numbers. The answer should be recorded without spaces or any additional characters. Separate the fractional part from the whole decimal point. You do not need to write the measurement units. In tasks 1–4, 8–10, 14, 15, 20, 25–27, the answer is an integer or a final decimal fraction. The answer to tasks 5-7, 11, 12, 16-18, 21 and 23 is a sequence of two numbers. The answer to problem 13 is a word. The answer to tasks 19 and 22 is two numbers.

If the variant is set by the teacher, you can enter or upload answers to the tasks with a detailed answer into the system. The teacher will see the results of the assignments with a short answer and will be able to rate the uploaded answers to the assignments with a detailed answer. The points given by the teacher will appear in your statistics.

Version for printing and copying in MS Word

The figure shows a graph of the dependence of the projection of the body speed v x from time.

Determine the projection of the acceleration of this body a x in the time interval from 15 to 20 s. Express your answer in m / s 2.

Answer:

Cube mass M \u003d 1 kg, laterally compressed by springs (see figure), rests on a smooth horizontal table. The first spring is compressed by 4 cm, and the second is compressed by 3 cm. The stiffness of the first spring k 1 \u003d 600 N / m. What is the stiffness of the second spring k 2? Express your answer in N / m.

Answer:

Two bodies are moving at the same speed. The kinetic energy of the first body is 4 times less than the kinetic energy of the second body. Determine the mass ratio of the bodies.

Answer:

At a distance of 510 m from the observer, workers drive piles with a pile driver. How long will it take from the moment the observer sees the impact of the pile driver until the moment when he hears the sound of the impact? The speed of sound in air is 340 m / s. Express your answer in p.

Answer:

The figure shows graphs of pressure dependence p immersion depth h for two liquids at rest: water and heavy liquid diiodomethane, at a constant temperature.

Choose two correct statements that agree with the charts.

1) If the pressure inside the hollow ball is equal to atmospheric, then in water at a depth of 10 m the pressures on its surface from the outside and from the inside will be equal to each other.

2) The density of kerosene is 0.82 g / cm 3, a similar graph of the pressure versus depth for kerosene will be between the graphs for water and diiodomethane.

3) In water at a depth of 25 m pressure p 2.5 times more than atmospheric.

4) With increasing immersion depth, the pressure in diiodomethane increases faster than in water.

5) The density of olive oil is 0.92 g / cm 3, a similar pressure versus depth graph for oil will be between the graph for water and the abscissa axis (horizontal axis).

Answer:

A massive weight suspended from the ceiling on a weightless spring makes vertical free vibrations. The spring remains stretched all the time. How do the potential energy of the spring and the potential energy of the load behave in the gravity field when the load moves upward from the equilibrium position?

1) increases;

2) decreases;

3) does not change.

Answer:

Truck moving on a straight horizontal road at speed v, braked so that the wheels stopped turning. Truck weight m, coefficient of friction of wheels on the road μ ... Formulas A and B allow you to calculate the values \u200b\u200bof physical quantities that characterize the movement of the truck.

Establish a correspondence between formulas and physical quantities, the value of which can be calculated using these formulas.

AND	B

Answer:

As a result of cooling the rarefied argon, its absolute temperature decreased by 4 times. How many times has the average kinetic energy thermal motion of argon molecules?

Answer:

The working body of the heat engine per cycle receives from the heater an amount of heat equal to 100 J, and performs work of 60 J. What is the efficiency of the heat engine? Express your answer in%.

Answer:

The relative humidity of the air in a closed vessel with a piston is 50%. What will be the relative humidity of the air in the vessel if the volume of the vessel at a constant temperature is reduced by 2 times? Express your answer in%.

Answer:

The hot substance, initially in a liquid state, was slowly cooled. The heat sink capacity is constant. The table shows the results of measurements of the temperature of a substance over time.

Select from the proposed list two statements that correspond to the results of the measurements, and indicate their numbers.

1) The crystallization process of the substance took more than 25 minutes.

2) The specific heat capacity of a substance in liquid and solid states is the same.

3) The melting point of the substance under these conditions is 232 ° C.

4) After 30 min. after the start of measurements, the substance was only in a solid state.

5) After 20 minutes. after the start of measurements, the substance was only in a solid state.

Answer:

Diagrams A and B show diagrams p − T and p − V for processes 1–2 and 3–4 (hyperbola) carried out with 1 mol of helium. Diagrams p - pressure, V - volume and T Is the absolute gas temperature. Establish a correspondence between charts and statements characterizing the processes depicted in the charts. For each position of the first column, select the corresponding position of the second column and write down the selected numbers in the table under the corresponding letters.

AND	B

Answer:

How is the Ampere force acting on conductor 1 from the side of conductor 2 directed relative to the drawing (to the right, left, up, down, towards the observer, from the observer) if the conductors are thin, long, straight, parallel to each other? ( I - current strength.) Write down the answer in a word (words).

Answer:

A direct current flows through a section of the circuit (see figure) I \u003d 4 A. What current will be shown by an ideal ammeter included in this circuit, if the resistance of each resistor r \u003d 1 ohm? Express your answer in amperes.

Answer:

In the experiment on observing electromagnetic induction, a square frame of one turn of a thin wire is in a uniform magnetic field perpendicular to the plane of the frame. Magnetic field induction increases uniformly from 0 to maximum value AT max for time T... In this case, an EMF of induction is excited in the frame, equal to 6 mV. What EMF of induction will arise in the frame if T reduce by 3 times, and AT reduce max by 2 times? Express your answer in mV.

Answer:

A uniform electrostatic field is created by a uniformly charged extended horizontal plate. The field strength lines are directed vertically upward (see figure).

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

1) If to point AND place a test point negative charge, then a force directed vertically downward will act on it from the side of the plate.

2) The plate has a negative charge.

3) Potential of the electrostatic field at the point AT lower than at the point FROM.

5) The work of the electrostatic field by moving the test point negative charge from the point AND and to the point AT is zero.

Answer:

An electron moves in a circle in a uniform magnetic field. How will the Lorentz force acting on the electron and the period of its revolution change if its kinetic energy is increased?

For each value, determine the corresponding change pattern:

1) will increase;

2) 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.

Answer:

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

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

PHYSICAL VALUES

FORMULAS

A) the current through the source with the open key K

B) the current through the source with the closed key K

Answer:

In a vacuum, two monochromatic electromagnetic waves... The photon energy of the first wave is 2 times higher than the photon energy of the second wave. Determine the ratio of the lengths of these electromagnetic waves.

Answer:

How will they change with β - decay mass number of the nucleus and its charge?

For each value, determine the corresponding change pattern:

1) will increase

2) 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.

Answer:

Determine the voltmeter readings (see figure) if the error of direct voltage measurement is equal to the division value of the voltmeter. Indicate the answer in volts. In the answer, write down the value and the error together without a space.

Answer:

For laboratory work upon detecting the dependence of the resistance of the conductor on its length, the student was given five conductors, the characteristics of which are indicated in the table. Which two of the following guides should a student take to conduct this research?

As in the previous year, in 2017 there are two "streams" of holding a single state examination - an early period (it takes place in the middle of spring) and the main one, traditionally starting at the end school year, last days of May. The official draft timetable for the USE "spelled out" all the dates for passing exams in all subjects in both of these periods - including additional reserve days provided for those who, for a good reason (illness, coincidence of exam dates, etc.), could not pass the USE within the specified time frame.

Schedule of the early period for the USE - 2017

In 2017, the early "wave" of the unified state examination starts earlier than usual. If last year the peak of the spring exam period fell on the last week of March, then this season the spring break period will be free from the Unified State Exam.

The main dates of the early period are from 14 to 24 March... Thus, by the beginning of spring school holidays many "early adopters" will have time to pass the tests. And this may turn out to be convenient: among the graduates who have the right to take the Unified State Exam in an early wave are the guys who will participate in Russian or international competitions and contests in May, and during spring break they often leave for sports camps, profile shifts to camps, etc. etc. Shifting exams to an earlier date will allow them to use the latter to its fullest.

Additional (reserve) days the early period of the USE-2017 will be held from 3 to 7 April... At the same time, many will probably have to write exams on reserve dates: if no more than two subjects were submitted on the same day in the schedule of last year, then in 2017 most elective exams are grouped "in triplets".

Separate days are allocated for only three subjects: the compulsory exam in the Russian language for graduates and all future applicants, as well as mathematics and the oral part of the exam in foreign languages... At the same time, this year, "early adopters" will hand over the "speaking" before the written part.

It is planned to distribute the March exams by dates as follows:

March 14th (Tuesday) - mathematics exam (both basic and profile level);

March 16 (Thursday) - chemistry, history, computer science;

18th of March (Saturday) - USE in foreign languages \u200b\u200b( oral part exam);

20th of March (Monday) - Russian language exam;

March 22 (Wednesday) - biology, physics, foreign languages \u200b\u200b(written exam);

March 24 (Friday) - USE, literature and social studies.

There is a nine-day pause between the main and reserve days of the early period. All additional tests for "reservists" will take place in three days:

April 3 (Monday) - chemistry, literature, computer science, foreign language (speaking);

5th of April (Wednesday) - foreign (in writing), geography, physics, biology, social studies;

7 april (Friday) - Russian, basic and.

As a rule, the bulk of those taking the USE ahead of schedule are graduates of previous years, as well as graduates of secondary special educational institutions (in colleges and vocational schools the program high school usually "pass" in the first year of study). In addition, school graduates who, during the main period of the USE, will be absent for valid reasons (for example, to participate in Russian or international competitions or to be treated in a sanatorium), or intend to continue their education outside of Russia, may “shoot out” with exams early.

2017 graduates can also on their own choose the deadline for passing exams in those subjects for which the program has been completed in full. This is true primarily for those who plan to - school course in this subject is read until grade 10, and early passing of one of the exams can reduce tension during the main period of the exam.

Schedule of the main period for passing the exam - 2017

The main period for passing the exam in 2017 starts on May 26, and by June 16, most of the graduates will have completed the exam epic. For those who could not pass the exam on time for a good reason or chose subjects that coincide in terms of delivery, there are reserve exam days from June 19... Like last year, the last day of the exam period will become a "single reserve" - \u200b\u200bon June 30 it will be possible to pass an exam in any subject.

At the same time, the schedule of exams for the main period of the USE-2017 is much less dense in comparison with the pre-term ones, and, most likely, most graduates will be able to avoid "overlapping" exams.

Separate examination days are allocated for the delivery of compulsory subjects: the Russian language, mathematics of the basic and specialized level (students have the right to take one of these exams, or both at once, therefore, in the schedule of the main period, they are traditionally spread over several days).

As in the previous year, a separate day has been set aside for the most demanded elective exam - social studies. And for passing the oral part of the exam in foreign languages, two separate days are allocated at once. In addition, a separate day is allocated for the less demanded for USE subject - geography. Perhaps this was done in order to distribute all subjects of the natural science profile in the schedule, reducing the number of matches.

Thus, in exam timetable there are two pairs and one "three" subjects, exams for which will be taken simultaneously:

chemistry, history and informatics;

foreign languages \u200b\u200band biology,

literature and physics.

The exams must be passed on the following dates:

26 of May(Friday) - geography,

May 29 (Monday) - Russian language,

May 31 (Wednesday) - history, chemistry, computer science and ICT,

2 June (Friday) - profile mathematics,

June 5th (Monday) - social studies;

June 7 (Wednesday) -,

the 9th of June (Friday) - written foreign, biology,

June 13 (Tuesday) - literature, physics,

June 15th (Thursday) and June 16 (Friday) - foreign oral.

Thus, the majority of schoolchildren will prepare for the graduation evenings "with a clear conscience", having already passed all the planned exams and received results in most subjects. Those who were ill during the main exam period, chose subjects that were matched in terms of time, received a "bad" in Russian or mathematics, were removed from the exam, or faced technical or organizational difficulties during the exam (for example, lack of additional forms or power outages), will take exams on reserve dates.

Reserve days will be distributed as follows:

June 19 (Monday) - informatics, history, chemistry and geography,

June 20 (Tuesday) - physics, literature, biology, social studies, written foreign,

21st of June (Wednesday) - Russian,

June, 22 (Thursday) - basic math,

June 28(Wednesday) - mathematics at the profile level,

June 29 (Thursday) - oral foreign,

30 June (Friday) - all items.

Can there be changes in the timetable for passing the exam

The draft official timetable for the exam is usually published at the beginning of the academic year, is discussed, and the final approval of the timetable for the exam takes place in the spring. Therefore, changes are possible in the USE schedule for 2017.

However, for example, in 2016 the project was approved without any changes and the actual dates of the exams completely coincided with those announced in advance - both in the early and in the main wave. So chances are high that the 2017 schedule will also be adopted unchanged.