At meetings with classmates, friends still make fun of me, recalling the heart-rending cries of a physics teacher, “What acceleration can a horse have !!!”, followed by unprintable expressions that I will not give here. Physics was my favorite subject at school and only a couple of students in the class, including myself, were able to successfully solve problems in it. Now students come to me to learn how to solve problems in physics. The overwhelming majority formulate their problems in the following way: "In physics, I understand and know the whole theory, but I can't solve problems."

This is the first misconception that a student needs to get rid of. Only a deep understanding of the theory will give us the key to solving problems. The problem of solving problems is faced primarily by those who do not understand enough theoretical material. I noticed that schoolchildren simply do not open the theoretical part of the textbook, which is only 1-2 pages from the given task. The statement "I understand the theory part" is based on what he heard in the teacher's explanation class and had no questions. But the teacher's explanation does not exhaust the material necessary for solving problems! What I am trying to convey to schoolchildren is the need to read and look for answers to questions that will certainly arise in the process of reading. Long live progress, finding the answer to a question in physics is not difficult now - GOOGLE knows everything.

My main task, as a tutor in physics, is, first of all, to teach a child to formulate questions, and for this, first of all, he must learn to read thoughtfully. If the student does not have questions in the process of studying, this is a sure sign that he does not understand the material. Well, as a result - problems with solving problems.

Now I will explain in more detail what it means not to understand the theory. This is, first of all, not knowing the connections between the formulas that are given in the theoretical part of the textbook. To do this, you yourself need to carry out all the calculations and proofs. In the process of proof, several questions will arise, having dealt with which, the student will master the theoretical part of the material and, therefore, will make it easier for himself to solve problems on this topic.

Having calculated g in this way, it would not be out of place to note that the same constant can be calculated empirically by throwing a ball from a height and timing the fall, thereby recalling the formulas that describe free fall. In general, it is always a good idea to make comments based on the material covered as often as possible. Then the students will perceive each topic in relation to the previous ones, and the likelihood of hearing questions from him on the topic will be much higher. A well-formulated question is already half the answer.

Often problems arise in the process of calculating formulas. It would seem - which is easier - to substitute the numbers given in the condition of the problem into a ready-made formula and calculate the answer using a calculator. Yes, it was not there - the answer does not converge. What could be the problem? Most often, this is a mismatch in dimensions - for example, the length is given in meters, and the speed is in kilometers per second. So, the first question that a student should ask himself is whether everything is in order in his task with dimensions, and only after the dimensions have been reduced, you can start substituting data into formulas.

Well, the second problem, no less common, is elementary ignorance of mathematics and the inability to apply mathematical skills in life. 99.9% of students are trying to make their lives easier with enviable persistence by driving endless zeros into the calculator window. But this is the very case where laziness is the engine of progress. But no, in a physics class, all the knowledge acquired in a mathematics lesson evaporates without a trace. Here and now is the time to show the student why this knowledge may be needed.

Of course, the problems described are not the only ones when solving problems in physics, but by solving at least them, you will already feel an improvement in the situation and help your children get rid of the fear of problems, and possibly instill interest in solving unfamiliar problems.

What advice can I give parents? Before calling the tutor, have the child read the last paragraph in physics assigned to him, preceding the tasks with which he had problems. Ask him the questions that are at the end of each paragraph. Try to reason with your child when answering the question. You can even have a discussion. To do this, of course, you will also have to look through the textbook, in which there are "a lot of letters." Again, there is Google, which knows everything. It is a thorny path, but it can bring wonderful results. If the problem still remains, there are more than enough tutors. It is important to avoid a situation in which the tutor simply decides in class homework for your student. I believe that my task is to teach to solve independently, to find the necessary information for solving in a textbook and on the net, and for this ask and formulate questions correctly.

In the following Notes, I will tell you how to check the correctness of the solution of the problem, if there is no way to peep the answer. This can be useful on tests and, in addition, helps to memorize the necessary formulas.

With the kind permission of the administration, I add my contact details:
Skype: olga.kalyakina
email: [email protected]
Tel. 8-9649559520

Preparation for the OGE and the Unified State Examination

The average general education

UMK line A. V. Gracheva. Physics (10-11) (basic, advanced)

Line UMK A. V. Grachev. Physics (7-9)

Line UMK A. V. Peryshkin. Physics (7-9)

Preparation for the exam in physics: examples, solutions, explanations

Parsing USE assignments in physics (Option C) with a teacher.

Lebedeva Alevtina Sergeevna, teacher of physics, work experience 27 years. Certificate of honor Ministry of Education of the Moscow Region (2013), Gratitude of the Head of Voskresensky municipal district(2015), Diploma of the President of the Association of Teachers of Mathematics and Physics of the Moscow Region (2015).

The work presents tasks different levels difficulty: basic, advanced and high. Tasks basic level, This simple tasks checking the assimilation of the most important physical concepts, models, phenomena and laws. Tasks advanced level aimed at testing the ability to use the concepts and laws of physics for analysis various processes and phenomena, as well as the ability to solve problems for 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 complexity 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 compatible demo version USE 2017, assignments taken from open bank USE assignments.

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

Decision. The path traveled by the car in the time interval from 0 to 30 s is most simply defined as the area of ​​a trapezoid, the bases of which are the time intervals (30 - 0) = 30 s and (30 - 10) = 20 s, and the height is the speed v= 10 m/s, i.e.

S =	(30 + 20) with	10 m/s = 250 m.
	2

Answer. 250 m

A 100 kg mass is lifted vertically upwards with a rope. The figure shows the dependence of the velocity projection V load on the axis directed upwards, from time t. Determine the modulus of the cable tension during the lift.

Decision. According to the speed projection curve v load on an axis directed vertically upwards, 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 s

The load is acted upon by: gravity directed vertically downwards and cable tension force directed along the cable vertically upwards, see fig. 2. Let's write down the basic equation of dynamics. Let's use Newton's second law. geometric sum forces acting on the body is equal to the product of the mass of the body and the acceleration imparted to it.

+ = (1)

Let's write down the equation for the projection of vectors in the reference frame associated with the earth, the OY axis will be directed upwards. The projection of the tension force is positive, since the direction of the force coincides with the direction of the OY axis, the projection of the gravity force is negative, since the force vector is opposite to the OY axis, the projection of the acceleration vector is also positive, so the body moves with acceleration upwards. We have

T – mg = ma (2);

from formula (2) the modulus of the tension force

T = m(g + a) = 100 kg (10 + 2) m/s 2 = 1200 N.

Answer. 1200 N.

A body is dragged across a rough horizontal surface with constant speed the module of which is 1.5 m / s, applying a force to it as shown in figure (1). In this case, the module of the sliding friction force acting on the body is 16 N. What is the power developed by the force F?

Decision. Imagine 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 us write down the basic equation of dynamics.

Tr + + = (1)

Having chosen a reference system associated with a fixed surface, we write 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. Two forces act horizontally 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 cos- F tr = 0; (1) express the force projection F, This F cosα = F tr = 16 N; (2) then the power developed by the force will be equal to N = F cosα V(3) Let's make a replacement, taking into account equation (2), and substitute the corresponding data in equation (3):

N\u003d 16 N 1.5 m / s \u003d 24 W.

Answer. 24 W.

A load fixed on a light spring with a stiffness of 200 N/m oscillates vertically. The figure shows a plot of the offset x cargo from time t. Determine what the weight of the load is. Round your answer to the nearest whole number.

Decision. The weight on the spring oscillates vertically. According to the load displacement curve X from time t, determine the period of oscillation of the load. The oscillation period is T= 4 s; from the formula T= 2π we express the mass m cargo.

=	T	;	m	=	T 2	; m = k	T 2	; m= 200 H/m	(4 s) 2	= 81.14 kg ≈ 81 kg.
	2π		k		4π 2		4π 2		39,438

Answer: 81 kg.

The figure shows a system of two lightweight blocks and a weightless cable, with which you can balance or lift a load of 10 kg. Friction is negligible. Based on the analysis of the above figure, select twotrue statements and indicate their numbers in the answer.

1. In order to keep the load in balance, you need to act on the end of the rope with a force of 100 N.
2. The system of blocks shown in the figure does not give a gain in strength.
3. h, you need to pull out a section of rope with a length of 3 h.
4. To slowly lift a load to a height hh.

Decision. In this task, remember simple mechanisms, namely blocks: movable and fixed block. The movable block gives a gain in force twice, while the section of the rope must be pulled twice as long, and the fixed block is used to redirect the force. In work, simple mechanisms of winning do not give. After analyzing the problem, we immediately select the necessary statements:

1. To slowly lift a load to a height h, you need to pull out a section of rope with a length of 2 h.
2. 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, fixed on a weightless and inextensible thread, is completely immersed in a vessel with water. The load does not touch the walls and bottom of the vessel. Then, an iron load is immersed in the same vessel with water, the mass of which is equal to the mass of the aluminum load. How will the modulus of the tension force of the thread and the modulus of the force of gravity acting on the load change as a result of this?

1. increases;
2. Decreases;
3. Doesn't change.

Decision. We analyze the condition of the problem and select those parameters that do not change during the study: this is the mass of the body and the liquid into which the body is immersed on the threads. After that it is better to do schematic drawing and indicate the forces acting on the load: the tension force of the thread F control, directed along the thread up; gravity directed vertically downward; Archimedean force a, acting from the side of the liquid on the immersed body and directed upwards. According to the condition of the problem, the mass of the loads is the same, therefore, the modulus of the force of gravity acting on the load does not change. Since the density of goods is different, the volume will also be different.

V =	m	.
	p

The density of iron is 7800 kg / m 3, and the aluminum load is 2700 kg / m 3. Hence, V well< Va. The body is in equilibrium, the resultant of all forces acting on the body is zero. Let's direct the coordinate axis OY up. We write the basic equation of dynamics, taking into account the projection of forces, in the form F ex + Fa – mg= 0; (1) We express the tension force F extr = mg – Fa(2); Archimedean force depends on the density of the liquid and the volume of the submerged part of the body Fa = ρ gV p.h.t. (3); The density of the liquid does not change, and the volume of the iron body is less V well< Va, so 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.

Bar mass m slips off the fixed rough inclined plane with angle α at the base. The bar acceleration modulus is equal to a, the bar velocity modulus increases. Air resistance can be neglected.

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) The coefficient of friction of the bar on the inclined plane

3) mg cosα

4) sinα -	a
	g cosα

Decision. This task requires the application of Newton's laws. We recommend making a schematic drawing; specify all kinematic characteristics 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 force and acceleration vectors;

Following the proposed algorithm, we will make a schematic drawing (Fig. 1). The figure shows the forces applied to the center of gravity of the bar, and the coordinate axes of the reference system associated with the surface of the inclined plane. Since all forces are constant, the movement of the bar will be equally variable with increasing speed, i.e. the acceleration vector is directed in the direction of motion. Let's choose the direction of the axes as shown in the figure. Let's write down the projections of forces on the selected axes.

Let us write down the basic equation of dynamics:

Tr + = (1)

Let's write down given equation(1) for the projection of forces and acceleration.

On the OY axis: the projection of the reaction force of the support 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 to mgy= – mg cosα ; acceleration vector projection a y= 0, since the acceleration vector is perpendicular to the axis. We have N – mg cosα = 0 (2) from the equation we express the reaction force acting on the bar from the side of the inclined plane. N = mg cosα (3). Let's write down the projections on the OX axis.

On the OX axis: force projection N is equal to zero, since the vector is perpendicular to the OX axis; The projection of the friction force is negative (the vector is directed towards opposite side relative to the selected axis); the projection of gravity is positive and equal to mg x = mg sinα(4) of right triangle. Positive acceleration projection a x = a; Then we write equation (1) taking into account the projection mg sinα- F tr = ma (5); F tr = m(g sinα- a) (6); Remember that the force of friction is proportional to the force of normal pressure N.

A-priory F tr = μ N(7), we express the coefficient of friction of the bar on the inclined plane.

μ =	F tr	=	m(g sinα- a)	= tanα –	a	(8).
	N		mg cosα		g cosα

We select the appropriate positions for each letter.

Answer. A-3; B - 2.

Task 8. gaseous oxygen is in a vessel with a volume of 33.2 liters. The gas pressure is 150 kPa, its temperature is 127 ° C. Determine the mass of 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. Convert temperature to Kelvin T = t°С + 273, volume V\u003d 33.2 l \u003d 33.2 10 -3 m 3; We translate pressure P= 150 kPa = 150,000 Pa. Using the equation of state ideal gas

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

Task 9. An ideal monatomic gas in an amount of 0.025 mol expanded adiabatically. At the same time, its temperature dropped from +103°С to +23°С. What is the work done by the gas? Express your answer in Joules and round to the nearest whole number.

Decision. First, the gas is monatomic number of degrees of freedom i= 3, secondly, the gas expands adiabatically - this means no heat transfer Q= 0. The gas does work by reducing the internal energy. With this in mind, we write the first law of thermodynamics as 0 = ∆ U + A G; (1) we express the work of the gas A g = –∆ U(2); We write the change in internal energy for a monatomic gas as

Answer. 25 J.

The relative humidity of a portion of air at a certain temperature is 10%. How many times should the pressure of this portion of air be changed in order for its relative humidity to increase by 25% at a constant temperature?

Decision. Questions related to saturated steam and air humidity, most often cause difficulties for schoolchildren. Let's use the formula for calculating the relative humidity of the air

According to the condition of the problem, the temperature does not change, which means that the pressure saturated steam remains the same. Let's write formula (1) for two states of air.

φ 1 \u003d 10%; φ 2 = 35%

We express the air pressure from formulas (2), (3) and find the ratio of pressures.

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 melting furnace with a constant power. The table shows the results of measurements of the temperature of a substance over time.

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

1. The melting point of the substance under these conditions is 232°C.
2. In 20 minutes. after the start of measurements, the substance was only in the solid state.
3. The heat capacity of a substance in the liquid and solid state is the same.
4. After 30 min. after the start of measurements, the substance was only in the solid state.
5. The process of crystallization of the substance took more than 25 minutes.

Decision. Since the substance is cooled, it internal energy decreased. The results of temperature measurements allow to determine the temperature at which the substance begins to crystallize. While the substance is moving from liquid state into a solid, the temperature does not change. Knowing that the melting temperature and the crystallization temperature are the same, we choose the statement:

1. The melting point of a substance under these conditions is 232°C.

The second correct statement is:

4. After 30 min. after the start of measurements, the substance was only in the 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 is reached. 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 appropriate nature of the change:

1. Increased;
2. Decreased;
3. Hasn't changed.

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

Decision. If in an isolated system of bodies there are no energy transformations other than heat transfer, 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 on the basis of the heat balance equation.

∆U = ∑	n	∆U i = 0 (1);
	i = 1

where ∆ U- change in internal energy.

In our case, as a result of heat transfer, 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 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, flying into the gap between the poles of the electromagnet, has a speed , perpendicular to the vector induction magnetic field, as it shown on the picture. Where is the Lorentz force acting on the proton directed relative to the figure (up, towards the observer, away from the observer, down, left, right)

Decision. A magnetic field acts on a charged particle with the Lorentz force. In order to determine the direction of this force, it is important to remember the mnemonic rule of the left hand, not to forget to take into account the charge of the particle. We direct the four fingers of the left hand along the velocity vector, for a positively charged particle, the vector must enter the palm perpendicularly, thumb set aside 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.

Tension modulus electric field in a flat air capacitor with a capacity of 50 microfarads is 200 V / m. The distance between the capacitor plates is 2 mm. What is the charge on the capacitor? Write your 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= 2 10 -3 m. The problem deals with a flat air capacitor - a device for accumulating electric charge and electric field energy. From the electric capacitance formula

where d is the distance between the plates.

Let's Express the Tension U= E d(4); Substitute (4) in (2) and calculate the charge of the capacitor.

q = C · Ed\u003d 50 10 -6 200 0.002 \u003d 20 μC

Pay attention to the units in which you need to write the answer. We received it in pendants, but we present it in μC.

Answer. 20 µC.

The student conducted the 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?

1. is increasing
2. Decreases
3. Doesn't change
4. Record the selected numbers for each answer in the table. Numbers in the answer may be repeated.

Decision. In tasks of such a plan, 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 speeds of wave propagation in these media are different. Having figured out from which medium into which light propagates, we write the law of refraction in the form

sinα	=	n 2	,
sinβ		n 1

where n 2 – absolute indicator glass refraction, medium where is going light; n 1 is the absolute refractive index of the first medium, from which the light is coming. For air n 1 = 1. α is the angle of incidence of the beam on the surface of the glass half-cylinder, β is the angle of refraction of the beam in the glass. Moreover, the angle of refraction will be less than the angle of incidence, since glass is an optically denser medium - a medium with a high refractive index. The speed of light propagation in glass is slower. Please note that the angles are measured from the perpendicular restored at the point of incidence of the beam. If you increase the angle of incidence, then the angle of refraction will also increase. The refractive index of glass will not change from this.

Answer.

Copper jumper at time t 0 = 0 starts moving 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 jumper and the rails is negligible, the jumper is always perpendicular to the rails. The flux Ф of the magnetic induction vector through the circuit formed by the jumper, rails and resistor changes over time t as shown in the chart.

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

1. By the time t\u003d 0.1 s, the change in the magnetic flux through the circuit is 1 mWb.
2. Induction current in the jumper in the range from t= 0.1 s t= 0.3 s max.
3. Module EMF induction, arising in the circuit, is equal to 10 mV.
4. The strength of the inductive current flowing in the jumper is 64 mA.
5. To maintain the movement of the jumper, a force is applied to it, the projection of which on the direction of the rails is 0.2 N.

Decision. According to the graph of the dependence of the flow of the magnetic induction vector through the circuit on time, we determine the sections where the flow Ф changes, and where the change in the flow is zero. This will allow us to determine the time intervals in which the inductive current will occur in the circuit. Correct statement:

1) By the time t= 0.1 s the change in the magnetic flux through the circuit is 1 mWb ∆F = (1 - 0) 10 -3 Wb; The EMF module of induction that occurs in the circuit is determined using the EMP law

Answer. 13.

According to the graph of the dependence of current strength on time in electrical circuit, whose inductance is 1 mH, determine the module EMF self-induction in the time interval from 5 to 10 s. Write your answer in microvolts.

Decision. Let's convert all quantities to the SI system, i.e. we translate the inductance of 1 mH into H, we get 10 -3 H. The current strength shown in the figure in mA will also be converted to A by multiplying by 10 -3.

The self-induction EMF formula has the form

in this case, the time interval is given according to the condition of the problem

∆t= 10 s – 5 s = 5 s

seconds and according to the schedule we determine the interval of current change during this time:

∆I= 30 10 –3 – 20 10 –3 = 10 10 –3 = 10 –2 A.

Substitute numerical values into 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 beam of light falls from the air onto the surface of the first plate (see figure). It is known that the refractive index of the upper plate is equal to n 2 = 1.77. Establish a correspondence between physical quantities and their 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 passage 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 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 beam and the surface, and 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° = 50°, the refractive index n 2 = 1,77; n 1 = 1 (air).

Let's write the law of refraction

sinβ =	sin50	= 0,4327 ≈ 0,433
	1,77

Let's build an approximate path of the beam through the plates. We use formula (1) for the 2–3 and 3–1 boundaries. In response we get

A) The sine of the angle of incidence of the beam on the boundary 2–3 between the plates is 2) ≈ 0.433;

B) The angle of refraction of the beam when crossing the boundary 3–1 (in radians) is 4) ≈ 0.873.

Answer. 24.

Determine how many α - particles and how many protons are obtained as a result of a thermonuclear fusion reaction

+ → x+ y;

Decision. For all nuclear reactions the laws of conservation of electric charge and the number of nucleons are observed. Denote by x the number of alpha particles, y the number of protons. Let's make equations

+ → x + y;

solving the system we have that x = 1; y = 2

Answer. 1 – α-particle; 2 - protons.

The momentum modulus of the first photon is 1.32 · 10 -28 kg m/s, which is 9.48 · 10 -28 kg m/s less than the momentum module 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, so we can imagine p 2 = p 1 + ∆ p(one). The photon energy can be expressed in terms of the photon momentum using the following equations. This is E = mc 2(1) and p = mc(2), then

E = pc (3),

where E is the photon energy, p is the momentum of the photon, m is the mass of the photon, c= 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

We round the answer to tenths and get 8.2.

Answer. 8,2.

The nucleus of an atom has undergone radioactive positron β-decay. How did this change electric charge nucleus and the number of neutrons in it?

For each value, determine the appropriate nature of the change:

1. Increased;
2. Decreased;
3. Hasn't changed.

Write in the table the selected numbers for each physical quantity. 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 mass number core remains unchanged. Thus, the transformation reaction of an element is as follows:

Answer. 21.

Five experiments were carried out in the laboratory to observe diffraction using various diffraction gratings. Each of the gratings was illuminated by parallel beams of monochromatic light with certain length waves. The light in all cases was incident perpendicular to the grating. In two of these experiments, the same number of principal diffraction maxima were observed. First, indicate the number of the experiment in which the diffraction grating with a shorter period, 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 into the region of a geometric shadow. Diffraction can be observed when opaque areas or holes are encountered in the path of a light wave in large and opaque barriers for light, and the dimensions 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

d sinφ = kλ(1),

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

Selecting pairs according to the experimental conditions, we first choose 4 where a diffraction grating with a smaller period was used, and then the number of the experiment in which a diffraction grating with a large period was used is 2.

Answer. 42.

Current flows through the wire resistor. The resistor was replaced with another, with a wire of the same metal and the same length, but having half the area cross section, and passed through it half the current. How will the voltage across the resistor and its resistance change?

For each value, determine the appropriate nature of the change:

1. will increase;
2. will decrease;
3. Will not change.

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

Decision. It is important to remember on what quantities the resistance of the conductor depends. The formula for calculating the resistance is

Ohm's law for the circuit section, from formula (2), we express the voltage

U = I R (3).

According to the condition of the problem, the second resistor is made of wire of the same material, the same length, but different area cross section. The area is twice as small. 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.

Oscillation period mathematical pendulum on the surface of the Earth is 1.2 times the period of its oscillations on some planet. What equals module acceleration free fall on this planet? The effect of the atmosphere in both cases is negligible.

Decision. A mathematical pendulum is a system consisting of a thread, the dimensions of which are much larger than the dimensions of the ball and the ball itself. Difficulty may arise if the Thomson formula for the period of oscillation of a mathematical pendulum is forgotten.

T= 2π (1);

l is the length of the mathematical pendulum; g- acceleration of gravity.

By condition

Express from (3) g n \u003d 14.4 m / s 2. It should be noted that the acceleration of free fall depends on the mass of the planet and the radius

Answer. 14.4 m / s 2.

straight conductor 1 m long, through which a current of 3 A flows, is located in a uniform magnetic field with induction AT= 0.4 T at an angle of 30° to the vector . What is the modulus of the force acting on the conductor from the magnetic field?

Decision. If a current-carrying conductor is placed in a magnetic field, then the field on the current-carrying conductor will act with the Ampere force. We write the formula for the Ampère force modulus

F A = I LB sinα;

F A = 0.6 N

Answer. F A = 0.6 N.

The energy of the magnetic field stored in the coil when passed through it direct current, is equal to 120 J. How many times it is necessary to increase the strength of the current flowing through the coil winding in order for the energy of the magnetic field stored in it to increase by 5760 J.

Decision. The energy of the magnetic field of the coil is calculated by the formula

W m =	LI 2	(1);
	2

By condition W 1 = 120 J, then W 2 \u003d 120 + 5760 \u003d 5880 J.

I 1 2 =	2W 1	; I 2 2 =	2W 2	;
	L		L

Then the current ratio

I 2 2	= 49;	I 2	= 7
I 1 2		I 1

Answer. The current strength must be increased by 7 times. In the answer sheet, you enter only the number 7.

An electrical circuit consists of two light bulbs, two diodes, and a coil of wire connected as shown in the figure. (A diode only allows current to flow in one direction, as shown at the top of the figure.) Which of the bulbs will light up if the north pole of the magnet is brought closer to the coil? Explain your answer by indicating what phenomena and patterns you used in the explanation.

Decision. Lines of magnetic induction come out of north pole magnet and diverge. When a magnet approaches magnetic flux through a coil of wire increases. According to Lenz's rule, the magnetic field created by by induction current coil, should be directed to the right. According to the gimlet's rule, the current should flow clockwise (when viewed from the left). In this direction, the diode in the circuit of the second lamp passes. So, the second lamp will light up.

Answer. The second lamp will light up.

Aluminum spoke length L= 25 cm and cross-sectional area S\u003d 0.1 cm 2 is suspended on a thread by the upper end. The lower end rests on the horizontal bottom of the vessel in which water is poured. The length of the submerged part of the spoke l= 10 cm Find strength F, with which the needle presses on the bottom of the vessel, if it is known that the thread is located vertically. The density of aluminum ρ a = 2.7 g / cm 3, the density of water ρ in = 1.0 g / cm 3. Acceleration of gravity g= 10 m/s 2

Decision. Let's make an explanatory drawing.

– Thread tension force;

– Reaction force of the bottom of the vessel;

a is the Archimedean force acting only on the immersed part of the body and applied to the center of the immersed part of the spoke;

- the force of gravity acting on the spoke from the side of the Earth and is applied to the center of the entire spoke.

By definition, the mass of the spoke m and module Archimedean strength are expressed as follows: m = SLρ a (1);

F a = Slρ in g (2)

Consider the moments of forces relative to the suspension point of the spoke.

M(T) = 0 is the moment of tension force; (3)

M(N) = NL cosα is the moment of the reaction force of the support; (4)

Taking into account the signs of the moments, we write the equation

NL cos + Slρ in g (L –	l	) cosα = SLρ a g	L	cos(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 needle presses on the bottom of the vessel we write N = F e and from equation (7) we express this force:

F d = [	1	Lρ a– (1 –	l	)lρ in] Sg (8).
	2		2L

Plugging in the numbers, we get that

F d = 0.025 N.

Answer. F d = 0.025 N.

A bottle containing m 1 = 1 kg of nitrogen, when tested for strength exploded at a temperature t 1 = 327°C. What mass of hydrogen m 2 could be stored in such a cylinder at a temperature t 2 \u003d 27 ° C, with a fivefold margin of safety? Molar mass nitrogen M 1 \u003d 28 g / mol, hydrogen M 2 = 2 g/mol.

Decision. We write the equation of state of an ideal gas Mendeleev - Clapeyron for nitrogen

where V- the volume of the balloon, T 1 = t 1 + 273°C. According to the condition, hydrogen can be stored at a pressure p 2 = p 1 /5; (3) Given that

we can express the mass of hydrogen by working immediately with equations (2), (3), (4). Final Formula looks like:

m 2 =	m 1		M 2		T 1	(5).
	5		M 1		T 2

After substituting numerical data m 2 = 28

Answer. m 2 = 28

In the ideal oscillatory circuit amplitude of current fluctuations in the inductor I m= 5 mA, and the amplitude of the voltage across the capacitor Um= 2.0 V. At time t the voltage across the capacitor is 1.2 V. Find the current in the coil at this moment.

Decision. In an ideal oscillatory circuit, the energy of vibrations is conserved. 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		Um 2

Let us substitute (4) into (3). As a result, we get:

I = I m (5)

Thus, the current in the coil at the time t is equal to

I= 4.0 mA.

Answer. I= 4.0 mA.

There is a mirror at the bottom of a reservoir 2 m deep. A beam of light, passing through the water, is reflected from the mirror and exits 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 beam incidence angle;

β is the angle of refraction of the beam in water;

AC is the distance between the beam entry point into the water and the beam exit point from the water.

According to the law of refraction of light

sinβ =	sinα	(3)
	n 2

Consider a rectangular ΔADB. In it AD = h, then DВ = AD

tgβ = h tgβ = h	sinα	= h	sinβ	= h	sinα	(4)
	cosβ

We get following expression:

AC = 2 DB = 2 h	sinα	(5)

Substitute the numerical values ​​in the resulting formula (5)

Answer. 1.63 m

In preparation for the exam, we invite you to familiarize yourself with work program in physics for grades 7–9 to the line of teaching materials Peryshkina A.V. and the working program of the in-depth level for grades 10-11 to the TMC Myakisheva G.Ya. Programs are available for viewing and free download to all registered users.

If you want to learn how to solve problems in physics on your own, first of all you must study the necessary theoretical material. Those. know the laws, formulas, definitions, understand why they are written in this way, in which cases they can be applied, and in which not. However, when solving all problems, you have to perform a standard set of actions, which is even more related to mathematics. Let's start with him.

Brief statement of the condition.

A short entry begins with the word "Given:". Below you write the letter designations of those physical quantities that are given in the problem and what they are equal to. For example, such a task.
The proton, having passed the accelerating potential difference U=800 V, flies into uniform, crossed at a right angle magnetic (V=50 mT) and electric fields. Determine the strength E of the electric field if the proton moves in crossed fields in a straight line.
Here the authors of the task have done almost everything for you. You will only have to write out that U = 800 V, V = 50 mT, and you need to find E.
Look at another issue.
Magnetic field strength at the center circular coil with current is 30 A/m. The radius of the turn is 8 cm. Determine the field strength on the axis of the turn at a point located at a distance of 6 cm from the center of the turn.
It no longer indicates which letter to designate a given physical quantity. Therefore, we will have to remember that the magnetic field strength is H, the radius is R, the distance from the center can be denoted h. But note that one tension is given to us, and another must be found. Those. the same physical quantity is present twice in the condition. Therefore, they need to be designated by different indices (the symbol is at the bottom right of the letter). We get H1 and H2. As a result, we write "given" as follows:
Given:
H1 = 30 A/m
R=8cm=0.08m
h = 6 cm = 0.06 m
H2 - ?
I strongly recommend that you remember what letters denote physical quantities. Then the task will no longer seem overly difficult. You can already make a short note to it, and this is already part of the solution. For those who still do not remember all the designations, I made a cheat sheet. Use it until you remember. Trust me, it's not that hard to remember.
This stage of the solution is the simplest and usually does not cause any particular difficulties. True, there are tasks where the condition is a little confusing.
The current strength in a horizontally located conductor with a length of 20 cm and a mass of 4 g is 10 A. Find the minimum induction of the magnetic field in which gravity can be balanced by the Ampère force.
After the words "current strength" a number is given that indicates the length of the conductor, then the mass is given. Only at the end of the sentence is written "10 A", this is the value of the current strength. Such a record often confuses those who do not carefully read the condition. Do not rush, follow the logic of presentation, look at the units of measurement. Current cannot be measured in centimeters or grams. All this will help you write down the condition correctly and move on to the next step. Just in case, here's an example. abbreviation conditions of the last task.
Given:
I = 10 A
l= 20 cm = 0.2 m
m = 4 g = 0.004 kg
B-?

Converting units of measurement to the SI system.

You have probably already noticed that in the brief notation of the condition, some numerical values ​​are written as in the text of the problem, while others are translated into new units of measurement. For example, h \u003d 6 cm \u003d 0.06 m. This is done because each physical quantity has a basic unit of measurement. These units are listed in the cheat sheet. In order for the numerical answer in the problem to turn out to be correct, it is necessary to convert all non-basic units into basic ones. This is where the first difficulties usually begin. In fact, everything is quite simple. You just need to understand and remember the procedure. Almost all non-basic units of measurement are obtained by adding a prefix to the main ones. For example:
kN (kilonewton) - Newton is preceded by the prefix "kilo";
km (kilometer) - before the meter is the prefix "kilo";
cm (centimeter) - before the meter is the prefix "santi";
mm (millimeter) - before the meter is the prefix "milli";
MJ (megajoule) - before the Joule is the prefix "mega";
I think these examples are enough to understand how non-basic units are formed. Now we will learn how to translate them into the main ones. For this we need such a table.

Exponent	Name	Designation		Exponent	Name	Designation
18	exo	E		-1	deci	d
15	peta	P		-2	centi	with
12	tera	T		-3	Milli	m
9	giga	G		-6	micro	mk
6	mega	M		-9	nano	n
3	kilo	to		-12	pico	P
2	hecto	G		-15	femto	f
1	soundboard	Yes		-18	atto	a

Further, everything is simple. Consider how to convert to basic units with examples.
F = 3 kN. We look at the table, the prefix "k" corresponds to the number 3. So you need to move the comma three digits to the right. If there is no comma, then just add three zeros. Then we get F = 3 kN = 3000 N. Please note that we do not write the prefix "k" for the second time, because zeros appeared instead.
F = 3.2 kN. We move the comma to three decimal places. F = 3.2 kN = 3200 N
F = 3 mN. We look at the table, the prefix "m" corresponds to the number -3. So you need to move the comma three characters to the left. If there is no comma, put it after the three. Then we get F = 3 mN = 0.003 N. Please note that we do not write the prefix "m" again for the second time, because zeros appeared instead.
720 nm. The prefix "n" corresponds to -9. Then we get 0.000000720 m or 0.00000072 m. We moved the comma nine digits to the left.
5 MV (megavolt). Mega means 6. Move the comma six units to the right. 6000000 V.
The last two examples have a lot of zeros. This is not very convenient. But you can greatly simplify everything if you use a more convenient method of translation. See how it's done.

We write the number given to us, then we attribute "multiply by 10", and put the exponent corresponding to the prefix. Everything is simple. A little more complicated with volume and area, density and some other units. Read more about them on this page.

Derivation of formulas.

In almost all problems, you have to express an unknown quantity from the main formula. For example, you are solving a problem on the Joule-Lenz law.

All the quantities included in the formula are known, it is necessary to express the time from it and find it. To do this, you can use a simple rule. If there is no addition and subtraction in the formula, then letters can be transferred from the left side of the formula to the right and vice versa. Let me explain that the left side is what is written to the left of equals, the right - to the right of equals. When transferring, what was written in the numerator (above the fraction line) falls into the denominator (below the fraction line) and vice versa, from the denominator falls into the numerator.
Now let's learn how to apply this rule. We need to find time. See how it's done.

Everything is very simple. You don't even have to think! Transfer all unnecessary and the new formula is ready. See another example with the Mendeleev-Clapeyron equation.

The situation is a little more complicated when the unknown quantity is in the denominator.

In this article, we will tell you the basic scheme solving problems in physics.

By following this scheme, you will be less likely to get confused in own decision, and the person checking your work will have nothing to complain about. (Of course, if everything is done correctly)

1) First you need read the problem(thanks captain), but not just to read, but to try to understand its essence, to understand: what do they want from us? While re-reading, try to figure out in your mind the course of your future decision.

2) The first thing to do when starting to write down the solution is to write down "Given". All data for solving the problem is usually contained in the condition, but in some cases, the problems use constants whose values ​​are set in a separate table. When writing these values, you should pay attention to the dimensions in which they are presented, and if necessary, convert all values ​​to the SI system! Under "Given" you should write the question of the task.

Example 1: The problem is given a car speed of 72 km/h and a travel time of 10 seconds. You need to find the path that the car traveled during this time.

To find the path, you need to convert 72 km / h to m / s or 10 sec. at hours. It would not be rational to convert 10 seconds into hours, so we will convert 72 km/h to m/s and get 20 m/s.

It looks something like this:

3) For most problems in physics visual drawing required, representing the essence of the phenomenon described in the problem. The figure should show all the physical quantities necessary for the solution. A correctly drawn drawing will help you not only better understand a physical phenomenon, but also quickly come to a solution to this problem.

Example 2: The problem reads as follows: A bar, under the influence of a horizontal force, moves uniformly across the table. What forces are acting on it?

The question of the problem can be answered without a picture, but with a picture it is less likely that we will forget something.

Drawing all the forces in vector form, we get the following:

4) The next point is the most important: the decision. At first all formulas are written, which we will use in solving. From these formulas a system of equations is compiled(or one equation) in general view. Next comes mathematical transformation this system of equations (or one equation). When the value of the required quantity is obtained in general terms, a check of the dimensions should be carried out.

We look at the dimension of the desired value and do a check on the obtained value of the variable (in general terms).

Let's take the simplest example: find the path of a uniformly moving body.

After checking the dimension, we calmly calculate the value of the desired value, substituting the values ​​known to us.

5) The answer should be written down in general and numerically.

That's actually all. Our article "How to solve problems in physics" has come to an end. If you find any mistake, typo or have questions, be sure to write about it in the comments! Good luck with your decisions! © site

This article presents an analysis of tasks in mechanics (dynamics and kinematics) from the first part of the exam in physics with detailed explanations from a tutor in physics. There is a video analysis of all tasks.

Let's select a section on the graph corresponding to a time interval from 8 to 10 s:

The body moved on this time interval with the same acceleration, since the graph here is a section of a straight line. During these s, the speed of the body changed by m/s. Therefore, the acceleration of the body in this period of time was equal to m/s 2 . The schedule number 3 is suitable (at any time, the acceleration is -5 m / s 2).

2. Two forces act on the body: and. According to the force and the resultant of two forces find the modulus of the second force (see figure).

The vector of the second force is . Or, similarly, . Then we add the last two vectors according to the parallelogram rule:

The length of the sum vector can be found from the right triangle ABC, whose legs AB= 3 N and BC= 4 N. By the Pythagorean theorem, we obtain that the length of the desired vector is equal to N.

Let us introduce a coordinate system with the center coinciding with the center of mass of the bar and the axis OX directed along an inclined plane. Let's depict the forces acting on the bar: gravity, support reaction force and static friction force. The result is the following figure:

The body is at rest, so the vector sum of all forces acting on it is zero. Including zero and the sum of the projections of forces on the axis OX.

The projection of gravity on the axis OX equal to the leg AB the corresponding right triangle (see figure). At the same time, from geometric considerations this leg lies opposite the angle at. That is, the projection of gravity on the axis OX is equal to .

The static friction force is directed along the axis OX, so the projection of this force on the axis OX is simply equal to the length of this vector, but with opposite sign, since the vector is directed against the axis OX. As a result, we get:

We use the formula known from the school physics course:

Let us determine from the figure the amplitudes of steady forced oscillations at frequencies of the driving force of 0.5 Hz and 1 Hz:

It can be seen from the figure that at a driving force frequency of 0.5 Hz, the amplitude of steady-state forced oscillations was 2 cm, and at a driving force frequency of 1 Hz, the amplitude of steady-state forced oscillations was 10 cm. Therefore, the amplitude of steady-state forced hesitation increased 5 times.

6. A ball thrown horizontally from a height H with the initial speed , during the flight t distance flown horizontally L(see picture). What will happen to the flight time and acceleration of the ball, if on the same setting with the same initial speed ball increase height H? (Ignore air resistance.) For each value, determine the appropriate nature of its change: 1) increase 2) decrease 3) will not change Write in the table the selected numbers for each physical quantity. Numbers in the answer may be repeated.

In both cases, the ball will move with free fall acceleration, so the acceleration will not change. AT this case the flight time does not depend on the initial speed, since the latter is directed horizontally. The flight time depends on the height from which the body falls, and what more height, topics more time flight (the body takes longer to fall). Therefore, the flight time will increase. Correct answer: 13.