Task 2 exam in physics. USE topics in physics, which will be in the examination paper. What can I take for the exam

The video course "Get an A" includes all the topics necessary for the successful passing of the exam in mathematics by 60-65 points. Completely all tasks 1-13 of the Profile USE in mathematics. Also suitable for passing the Basic USE in mathematics. If you want to pass the exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!

Preparation course for the exam for grades 10-11, as well as for teachers. Everything you need to solve part 1 of the exam in mathematics (the first 12 problems) and problem 13 (trigonometry). And this is more than 70 points on the Unified State Examination, and neither a hundred-point student nor a humanist can do without them.

All the necessary theory. Quick solutions, traps and secrets of the exam. All relevant tasks of part 1 from the Bank of FIPI tasks have been analyzed. The course fully complies with the requirements of the USE-2018.

The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.

Hundreds of exam tasks. Text problems and probability theory. Simple and easy to remember problem solving algorithms. Geometry. Theory, reference material, analysis of all types of USE tasks. Stereometry. Cunning tricks for solving, useful cheat sheets, development of spatial imagination. Trigonometry from scratch - to task 13. Understanding instead of cramming. Visual explanation of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. Base for solving complex problems of the 2nd part of the exam.

Changes in the tasks of the exam in physics for 2019 year no.

The structure of the tasks of the exam in physics-2019

The examination paper consists of two parts, including 32 tasks.

Part 1 contains 27 tasks.

• 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, 23 and 24 is a sequence of two numbers.
• The answer to tasks 19 and 22 are two numbers.

Part 2 contains 5 tasks. The answer to tasks 28–32 includes a detailed description of the entire progress of the task. The second part of the tasks (with a detailed answer) are evaluated by the expert commission on the basis of .

USE topics in physics, which will be in the examination paper

1. Mechanics(kinematics, dynamics, statics, conservation laws in mechanics, mechanical oscillations and waves).
2. Molecular physics(molecular-kinetic theory, thermodynamics).
3. Electrodynamics and fundamentals of SRT(electric field, direct current, magnetic field, electromagnetic induction, electromagnetic oscillations and waves, optics, fundamentals of SRT).
4. Quantum physics and elements of astrophysics(particle wave dualism, physics of the atom, physics of the atomic nucleus, elements of astrophysics).

The duration of the exam in physics

To complete the entire examination work is given 235 minutes.

Estimated time to complete the tasks of various parts of the work is:

1. for each task with a short answer - 3-5 minutes;
2. for each task with a detailed answer - 15–20 minutes.

What can I take for the exam:

• A non-programmable calculator is used (for each student) with the ability to calculate trigonometric functions (cos, sin, tg) and a ruler.
• The list of additional devices and, the use of which is allowed for the exam, is approved by Rosobrnadzor.

Important!!! do not rely on cheat sheets, tips and the use of technical means (phones, tablets) in the exam. Video surveillance at the Unified State Exam-2019 will be strengthened with additional cameras.

USE scores in physics

• 1 point - for 1-4, 8, 9, 10, 13, 14, 15, 19, 20, 22, 23, 25, 26, 27 tasks.
• 2 points - 5, 6, 7, 11, 12, 16, 17, 18, 21, 24.
• 3 points - 28, 29, 30, 31, 32.

Total: 52 points(maximum primary score).

What you need to know when preparing assignments for the exam:

• Know/understand the meaning of physical concepts, quantities, laws, principles, postulates.
• Be able to describe and explain the physical phenomena and properties of bodies (including space objects), the results of experiments ... give examples of the practical use of physical knowledge
• Distinguish hypotheses from scientific theory, draw conclusions based on experiment, etc.
• To be able to apply the acquired knowledge in solving physical problems.
• Use the acquired knowledge and skills in practical activities and everyday life.

How to start preparing for the exam in physics:

1. Learn the theory required for each assignment.
2. Train in test tasks in physics, developed on the basis of the Unified State Examination. On our website, tasks and options in physics will be replenished.
3. Allocate your time correctly.

We wish you success!

The second sampler in physics from the online school of Vadim Gabitov "USE for 5".

Assessment system for examination paper in physics

Tasks 1-26

For the correct answer to each of the tasks 1-4, 8-10, 13-15, 19, 20, 22-26, 1 point is given. These tasks are considered completed correctly if the required number, two numbers or a word are correctly indicated.

Each of tasks 5-7, 11, 12, 16-18 and 21 is worth 2 points if

both elements of the answer are correctly specified; 1 point if one mistake is made;

0 points if both items are incorrect. If more than two are specified

elements (including, possibly, correct ones) or the answer

missing - 0 points.

job number		job number

View document content
"Unified State Examination for 5". Training variant in physics No. 2 (with answers) "

Unified State Exam
in PHYSICS

Work instructions

To complete the examination paper in physics, 3 hours are allotted

55 minutes (235 minutes). The work consists of two parts, including

31 tasks.

In tasks 1-4, 8-10, 14, 15, 20, 24-26, the answer is an integer or a final decimal fraction. Write the number in the answer field in the text of the work, and then transfer according to the example below to the answer form No. 1. Units of measurement of physical quantities do not need to be written.

The answer to tasks 5-7, 11, 12, 16-18, 21 and 23 is

sequence of two digits. Write your answer in the answer field in the text

work, and then transfer according to the example below without spaces,

commas and other additional characters in the answer sheet No. 1.

The answer to task 13 is a word. Write your answer in the answer field

the text of the work, and then transfer according to the sample below into the form

answers number 1.

The answer to tasks 19 and 22 are two numbers. Write the answer in the answer field in the text of the work, and then transfer it according to the example below, without separating the numbers with a space, into the answer form No. 1.

The answer to tasks 27–31 includes a detailed description of the entire progress of the task. In the answer form No. 2, indicate the number of the task and

write down its complete solution.

When calculating, it is allowed to use a non-programmable

calculator.

All USE forms are filled in with bright black ink. It is allowed to use a gel, or capillary, or fountain pen.

When completing assignments, you can use a draft. Entries

in the draft are not taken into account when evaluating the work.

The points you get for completed tasks are summed up.

Try to complete as many tasks as possible and score the most

number of points.

We wish you success!

The following are reference data that you may need when doing your job.

Decimal Prefixes

Name	Designation	Factor	Name	Designation	Factor

Constants free fall acceleration on earth gravitational constant universal gas constant R = 8.31 J/(mol K) Boltzmann's constant Avogadro's constant speed of light in vacuum coefficient proportionality in Coulomb's law, the electron charge modulus (elementary electric charge) Planck's constant

Ratio between different units temperature 0 K = -273 °С atomic mass unit 1 atomic mass unit equivalent to 931 MeV 1 electron volt

Particle mass electron neutron

Specific heat water 4.2∙10³ J/(kg∙K) aluminum 900 J/(kg∙K) ice 2.1∙10³ J/(kg∙K) copper 380 J/(kg∙K) iron 460 J/(kg∙K) cast iron 800 J/(kg∙K) lead 130 J/(kg∙K) Specific heat water vaporization J/K melting lead J/K ice melt J/K

Normal conditions: pressure - Pa, temperature - 0 °С

Molar mass nitrogen 28∙ kg/mol helium 4∙ kg/mol argon 40∙ kg/mol oxygen 32∙ kg/mol hydrogen 2∙ kg/mol lithium 6∙ kg/mol air 29∙ kg/mol neon 20∙ kg/mol water 2.1∙10³ J/(kg∙K) carbon dioxide 44∙ kg/mol

Part 1

			The answers to tasks 1–23 are a word, a number, or a sequence of digits or numbers. Write your answer in the answer field in the text of the work, and then transfer it to the ANSWER FORM No. 1 to the right of the number of the corresponding task, starting from the first cell. Write each character in a separate box in accordance with the samples given in the form. Units of measurement of physical quantities do not need to be written.
			The bar lies on a rough inclined support (see figure). There are three forces acting on it: gravity mg =30 N , support reaction force N =15 N and friction force F Tp \u003d 15 N. The angle alpha is 60 0 . What is the modulus of the resultant forces N and Ftr , if the block is at rest? Answer: _________________________ N.
			What value did the student get for the acceleration of free fall when doing laboratory work, if a pendulum 80 cm long made 100 oscillations in 3 minutes? Round your answer to tenths. Answer: ___________________________m/s 2
			The block moves uniformly up the inclined plane. Pick two true statements: 1) The modulus of the traction force is equal to the force of sliding friction 2) The modulus of the friction force vector is proportional to the force of normal pressure 3) The resultant of all forces depends on the angle of the inclined plane 4) The modulus of the friction force vector does not depend on the surface area of ​​the bar 5) The modulus of the friction force vector is inversely proportional to the surface area of ​​the bar
			A ball of mass m, thrown horizontally from a height H with an initial velocity V0, flew a distance S in the horizontal direction during the flight. In another experiment, a ball of mass 2m is already thrown horizontally from a height H with an initial velocity V0/2. What will happen to the range and acceleration of the balloon? will increase decrease Will not change Answer: ____________
			A body is thrown at an angle of 30 0 to the horizon with an initial velocity V 0 . PHYSICAL VALUES FORMULA A) the speed V of the body in projection onto the Y axis 1) (V 0y) 2 / 2g when moving up 2) (V 0 *cos30 0) 2 /2g B) maximum lifting height 3) V 0y - gt
			To heat 96 g of molybdenum by 1 K, it is necessary to transfer to it an amount of heat equal to 24 J. What is the specific heat capacity of this substance? Answer: ________ J/(kg*K)
			An ideal gas is isobarically compressed at a pressure of 300 kPa from a volume of 3 liters to a volume of 1 liter . What is the work done by the gas in this process? Answer: _________ kJ
			The pressure of an ideal gas at a constant concentration of its molecules decreased by 2 times. Choose two true statements. 1) The temperature of the gas has doubled. 2) Gas volume remains unchanged 3) The temperature of the gas has decreased by 2 times. 4) The volume of gas has doubled. 5) The number of gas molecules has doubled
	1 2		The temperature of the heater of the heat engine was lowered, leaving the temperature of the refrigerator the same. The amount of heat given off by the gas to the refrigerator per cycle has not changed. How did the efficiency of the heat engine and the amount of heat received by the gas per cycle from the heater change? For each value, determine the appropriate nature of the change: 1) increased 2) decreased 3) has not changed Write in the table the selected numbers for each physical quantity. Numbers in the answer may be repeated.
			What is the voltage that an ideal voltmeter will show connected to the resistor R 2, if it is known that between the points A and B voltage is 8V? Answer: ______________ B
			The metal surface is illuminated with light of frequency ν. In this case, a photoelectric effect is observed. With an increase in the frequency of the incident light by 2 times: photoelectric effect will not occur the number of photoelectrons will increase by 2 times the wavelength of the light will decrease by 2 times the maximum kinetic energy of the photoelectron will increase by more than 2 times the maximum kinetic energy of a photoelectron will increase by 2 times Choose two true statements.
			Current flows through the wire resistor. How will the thermal power released by the resistor and its electrical resistance change when the wire length is reduced by 4 times and the current doubled? increases decreases does not change Write in the table the selected numbers for each physical quantity. Numbers in the answer may be repeated.
1 8		Establish a correspondence between physical quantities and formulas by which they can be calculated. PHYSICAL VALUES FORMULA A) the radius of the circle during the movement of a charged 1) mV / qB particles in a perpendicular magnetic field 2) 2πm/qB B) the period of circulation around the charged circle 3) qB / mV particles in a perpendicular magnetic field 4) 2πR/qB Write in the table the selected numbers under the corresponding letters.

For some atoms, a characteristic feature is the possibility of capture by the atomic nucleus of one of the electrons closest to it. How do the characteristics of the atomic nucleus listed below behave when an electron is captured by the nucleus: the number of neutrons in the nucleus, the charge of the nucleus?

increases

decreases

does not change

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

The ball rolls down the chute. The change in the coordinates of the ball over time in the inertial frame of reference is shown in the graph. Based on this graph, it can be confidently stated that

the speed of the ball is constantly increasing

the first 2 s the speed of the ball increased, and then remained constant

the first 2 s the ball moved with decreasing speed, and then rested

an ever-increasing force acting on the ball

Answer: ___________

Masses of water and water evaporation time

Water volume and water evaporation time

Masses of water, water evaporation time and humidity in the room

Masses of water, water evaporation time and room volume

The volume of oxygen weighing 160 g, the temperature of which is 27 0 C, doubled during isobaric heating. Find the amount of heat that went into heating oxygen?

Answer: ______________ kJ

Bar mass t is placed on a plane inclined at an angle α to the horizon and released with an initial velocity equal to zero. The coefficient of friction between the bar and the plane is μ. At what α will the bar move down the plane? What is the force of friction of the bar on the plane?

There is air in a vessel with a small crack. Air can slowly seep through the crack. During the experiment, the volume of the vessel was reduced by 8 times, the air pressure in the vessel increased by 2 times, and its absolute temperature increased by 1.5 times. What is the change in the internal energy of the air in the vessel? (Air is assumed to be an ideal gas.)

A flat frame of 5 ohm wire is placed in a uniform magnetic field. The projection of the magnetic field induction on the Ox axis, perpendicular to the plane of the frame, varies from AT 1x = 3 T to AT 2x = -1 T During the change in the field, a charge of 1.6 C flows through the frame. Determine the area of ​​the frame?

http://vk.com/physic_100/

MINISTRY OF EDUCATION AND SCIENCE OF THE RUSSIAN FEDERATION

FEDERAL STATE BUDGET EDUCATIONAL INSTITUTION OF HIGHER PROFESSIONAL EDUCATION

"DON STATE TECHNICAL UNIVERSITY"

Tasks for classroom practical classes and independent work of students

Electricity and Magnetism Study Guide

Rostov-on-Don 2012

S.I. Egorova, V.S. Kovaleva, V.S. Kunakov, G.F. Lemeshko, Yu.M. heirs

Physics: Assignments for classroom practical

F 48 classes and independent work of students. Part 2. Electricity and magnetism: textbook. allowance / S.I. Egorova and others - Rostov n / D: DSTU Publishing Center, 2012. - 52 p.

The purpose of the manual is to provide a student-centered approach to practical classes in the general course of physics, taking into account the degree of preparation of students and the number of classroom hours allocated for lectures and practical classes.

The manual is intended for teaching and monitoring the work of students in practical classes in the section "electricity and magnetism" during the first and second semesters of study.

Published by decision of the editorial and publishing council of the Don State Technical University

Scientific editor

Dr. tech. sciences, prof. V.S. Kunakov

© DSTU Publishing Center, 2012

General guidelines

When solving and designing tasks, the following requirements must be observed:

1. Write down a brief condition of the problem, express all known quantities in the same system of units (usually in SI). If necessary, introduce additional constant physical quantities.

2. Problem solving should be accompanied by brief but comprehensive explanations. If necessary, provide a drawing or diagram.

3. It is necessary to solve the problem in a general way, i.e. express the desired value in the letter notation of the quantities given

in task condition. Perform calculations according to the calculation formula in compliance with the rules of approximate calculations.

The results of the control of classroom and independent work of students in practical classes are taken into account by the lecturer when taking exams and differentiated tests.

1. ELECTROSTATICS Basic formulas and laws

Coulomb's law

where F is the modulus of the force of interaction of two point charges q 1

and q

; r is the distance between charges;

8.85 10 12 f/m -

electric

constant,

The dielectric constant

the medium in which the charges are located (for vacuum 1 )

Strength and potential of the electrostatic field

W p,

E F ;

F - strength,

acting on a point positive charge

q0,

placed in this

field point;

W p–

potential

charge energy q 0 ;A is the work to move the charge q 0 from a given point of the field to infinity.

The intensity and potential of the electrostatic field created by a point charge q at a distance r from it

		4r2

The flow of the intensity vector through the area dS

E EdS En dS,

							dS ,
where dS dS	n - vector, module			whom
direction	matches		normal		to the site;
component			towards
site.
		tension				arbitrary

surface S

E EdS En dS.

The intensity and potential of the field created by a system of point charges (the principle of superposition (superposition) of electrostatic fields)

Ei;

I are the field strength and potential, respectively,

where E i

generated by the charge q i

n is the number of charges that create the field.

Connection between

tension

potential

electrostatic field

E grad , or

where i ,j ,k are the unit vectors of the coordinate axes.

In the case of a field with central or axial symmetry,

E d dr.

For a uniform field (plate capacitor fields)

E 1 2 , d

where (1 2 ) is the potential difference between the capacitor plates, d is the distance between them.

Electric moment of dipole (dipole moment)

pql ,

where l is the dipole arm (a vector quantity directed from

negative charge to positive).

Linear, surface and volume charge density, i.e. charge per unit length, area and volume, respectively:

Gauss' theorem for an electrostatic field in vacuum

E E ndS

dV ,

where q i is the algebraic sum of the charges enclosed inside

closed surface

N is the number of charges;

– volumetric

charge density.

tension

created

evenly

charged infinite plane,

tension

potential

created

conducting charged sphere with radius								R with charge q on
distance r	from the center of the sphere
								(inside the sphere);

								for r R (outside the sphere).

The intensity of the field created by a uniformly charged infinite cylindrical surface with a radius

R at a distance r from the axis of the cylinder,
							(inside the cylinder);
								(outside the cylinder).
	Work done by the forces of the electrostatic field
when moving charge q						from point 1(potential 1) to point 2
(potential 2 ),
	A 12q (1 2) , or A 12								q El dl,
where E l
	is the vector projection					E on elementary direction

displacements dl .

Dielectric polarization vector

where V is	dielectric volume;	pi–		dipole moment i - th
molecules,	N is the number of molecules.
Relationship between polarization vector and intensity
electrostatic field at the same point inside the dielectric
		æ 0 E ,

where æ is the dielectric susceptibility of the substance.

Relation between permittivity and dielectric susceptibility æ

1 + æ.

Relationship between the field strength E in a dielectric and

strength E 0 of the external field

E E 0.

Relationship between electric displacement vectors and

electrostatic field strength

Communication between vectors

E and P

0 E P.

Gauss's theorem for the electrostatic field in

dielectric

Dn dS qi

where q i

algebraic

prisoners inside

closed surface S of free electric charges;

component

direction

normals n

site

dS; dS

dS n is a vector whose modulus is equal to dS ,

direction

matches

normal

to the site.

Integration is carried out over the entire surface.

Capacitance of a solitary conductor and a capacitor

where q is the charge,	reported				conductor;					potential
conductor; U	- difference			potentials						plates
capacitor.
Electric capacitance of a flat capacitor

where S is the area of ​​the capacitor plate; d is the distance between

plates.

Capacitance of a capacitor bank: in series (a) and parallel (b) connections

b) C C i ,

where C i

– electric capacity of the i-th

condenser; n - number

capacitors.

The energy of a solitary charged conductor

Potential energy of a system of point charges

qi i,

2 i 1

where i

- the potential created at the point where the charge is located

q i , with all charges except the i-th, n is the number of charges. The energy of a charged capacitor

CU2

q - charge

condenser;

C-

electrical capacity; U-

potential difference between the plates.

attraction

differently

charged plates of a flat capacitor

The energy of the electrostatic field of a flat capacitor

S.U.2

S is the area of ​​one plate;

U - potential difference

between plates;

V Sd -

areas between

capacitor plates.

Volumetric energy density of the electrostatic field

where E is the field strength,

D is the electrical displacement.

1.1. The force of gravitational attraction of two identically charged water drops with radii of 0.1 mm is balanced by the Coulomb repulsive force. Determine the charge of the drops. The density of water is 1 g/cm 3 . .

1.2. How many times is the force of gravitational interaction between two protons less than the force of their Coulomb repulsion? The charge of a proton is numerically equal to the charge of an electron.

[by 1.25∙1038 times].

1.3. Three identical point charges q 1 \u003d q 2 \u003d q 3 \u003d 2 nC are located at the vertices of an equilateral triangle with sides of 10 cm. Determine the module and direction of the force acting on one of the charges from the other two.

1.4. The vertices of an equilateral triangle have equal positive charges. q = 2 nC. What negative charge q 1 must be placed in the center of the triangle so that the attractive force from the charge q 1 balances the repulsive forces of positive charges?

1.5. Four identical point charges q 1 \u003d q 2 \u003d q 3 \u003d q 4 \u003d 2 nC are located at the vertices of a square with a side of 10 cm. Determine the force acting on one of the charges from the other three. .

1.6. Two balls of the same radius and mass are suspended on two threads so that their surfaces are in contact. After informing the balls of charge 4. 10-7 C they pushed off each other and parted at an angle of 60˚. Find the mass of each ball if the length of the thread is 20 cm.

1.7. Two balls of mass 1 kg each are suspended on threads, the upper ends of which are connected together. The length of each thread is 10 cm. What identical charges must be imparted to the balls so that the threads diverge at an angle of 60˚? .

1.8. To an infinitely charged plane with a surface charge density of 8.85 nC/cm 2 a similarly charged ball with a mass of 1 g and a charge of 2 nC is attached to the thread. What angle does the thread on which the ball hangs form with the plane?

1.9. With what force per unit area do two equally charged infinitely extended planes repel each other? Surface charge density on each plane 2 µC/m 2 ? .

1.10. With what force per unit length do two similarly charged infinitely long filaments with the same linear charge density of 2 μC/m, located at a distance of 2 cm from each other, repel each other? .

1.11. With what force does the electric field of a charged infinite plane act on each meter of a charged infinitely long filament placed in this field? The surface charge density on the plane is 2 μC/m 2 and a linear charge density in the plane of 2 μC/m. .

1.12. A thin straight rod 15 cm long is uniformly charged with a linear density of 0.10 mC/m. On the continuation of the axis of the rod at a distance of 10 cm from the nearest end, there is a point charge of 10 nC. Determine the force of interaction between the rod and the charge. .

1.13. A thin rod 20 cm long carries a uniformly distributed electric charge. On the continuation of the axis of the rod, at a distance of 10 cm from the near end, there is a point charge of 40 nC, which interacts with the rod with a force of 6 μN. Determine the linear charge density on the rod. .

1.14. Two point charges q 1 \u003d 4 nC and q 2 \u003d -2 nC are located at a distance of 60 cm from each other. Determine the field strength

in a point midway between the charges. .

1.15. What is the field strength at a point located in the middle between point charges q 1 = 4nC and q 2 = 2 nC? The distance between charges is 60 cm.

1.16. q 1 \u003d 10 nC and q 2 \u003d -8 nC, at a distance of 8 cm to the right of the negative charge. The distance between the charges is 20 cm.

1.17. Determine the field strength at a point located on a straight line connecting the charges q 1 \u003d 10 nC and q 2 \u003d -8 nC, at a distance of 8 cm to the left of the negative charge. The distance between the charges is 20 cm.

Preparation for the OGE and the Unified State Examination

Secondary general education

Line UMK A. V. Grachev. Physics (10-11) (basic, advanced)

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

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

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

We analyze the tasks of the exam in physics (Option C) with the teacher.

Lebedeva Alevtina Sergeevna, teacher of physics, work experience 27 years. Diploma of the Ministry of Education of the Moscow Region (2013), Gratitude of the Head of the 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 of different levels of complexity: basic, advanced and high. Basic level tasks are simple tasks that test the assimilation of the most important physical concepts, models, phenomena and laws. Advanced level tasks 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 for the application of one or two laws (formulas) on any of the topics of a school physics course. In work 4, tasks of part 2 are tasks of a high level of complexity and test the ability to use the laws and theories of physics in a changed or new situation. The fulfillment of such tasks requires the application of knowledge from two three sections of physics at once, i.e. high level of training. This option is fully consistent with the demo version of the USE in 2017, the 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 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. The geometric sum of the forces acting on a body is equal to the product of the body's mass 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.

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

Decision. Let's imagine the 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 two correct 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, it is necessary to recall simple mechanisms, namely blocks: a movable and a 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 make a schematic drawing and indicate the forces acting on the load: the force of the thread tension 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 slides off a fixed rough inclined plane with an 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; 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 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 us write this 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 in the opposite direction relative to the selected axis); the projection of gravity is positive and equal to mg x = mg sinα (4) from a right triangle. 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 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

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 saturation vapor pressure 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. As matter cooled, its internal energy decreased. The results of temperature measurements allow to determine the temperature at which the substance begins to crystallize. As long as a substance changes from a liquid state to a solid state, 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, flown into the gap between the poles of an electromagnet, has a speed perpendicular to the magnetic field induction vector, as shown in the figure. Where is the Lorentz force acting on the proton directed relative to the 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 should enter the palm perpendicularly, the 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.

The modulus of the electric field strength in a flat air capacitor with a capacity of 50 μF 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 - the absolute refractive index of glass, the medium where the light goes; n 1 is the absolute refractive index of the first medium where the light comes from. 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. The module of the EMF of induction that occurs in the circuit is 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 the current strength on time in an electric circuit whose inductance is 1 mH, determine the self-induction EMF module 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.

We 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. In 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 the electric charge of the 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 in the 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 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 a certain wavelength. 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. Indicate first the number of the experiment in which a diffraction grating with a shorter period was used, and then the number of the experiment in which a diffraction grating with a 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 of the 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 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 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 cross-sectional area. 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.

The period of oscillation of a mathematical pendulum on the surface of the Earth is 1.2 times greater than the period of its oscillation on some planet. What is the gravitational acceleration modulus 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.

A straight conductor with a length of 1 m, 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 a direct current is passed through it is 120 J. How many times should the strength of the current flowing through the coil winding be increased 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 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. 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. In accordance with Lenz's rule, the magnetic field created by the inductive current of the loop must 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 the modulus of the Archimedean force 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 of 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). The 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 an ideal oscillatory circuit, the amplitude of current oscillations in the inductor I m= 5 mA, and the amplitude of the voltage across the capacitor U m= 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		U m 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 the 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.