The table shows the values ​​of physical quantities. Physical quantities and units of measurement

Basic SI units

Magnitude

Unit

Designation

Name

Russian

international

Electric current strength

Thermodynamic temperature

The power of light

Quantity of substance

Derived SI units with their own names

Unit

Derived unit expression

Magnitude

Name

Designation

Through other SI units

Through SI major and supplementary units

Pressure

m -1 ChkgChs -2

Energy, work, amount of heat

m 2 ChkgChs -2

Power, energy flow

m 2 ChkgChs -3

Amount of electricity, electric charge

Electrical voltage, electrical potential

m 2 ChkgChs -3 ChA -1

Electrical capacity

m -2 Chkg -1 Ch 4 Ch 2

Electrical resistance

m 2 ChkgChs -3 ChA -2

Electrical conductivity

m -2 Chkg -1 Ch 3 Ch 2

Magnetic induction flux

m 2 ChkgChs -2 ChA -1

Magnetic induction

kgHs -2 HA -1

Inductance

m 2 ChkgChs -2 ChA -2

Light flow

Illumination

m 2 ChkdChsr

Radioactive source activity

becquerel

Absorbed radiation dose

Thermodynamic

temperature

GRAPHICS

PHYSICAL QUANTITIES

Kinetic energy of cargo

Load speed

Spring stiffness

Kinetic energy of cargo

Load speed

Spring stiffness

COORDINATE

SPEED

Where X 0 – initial coordinate of the body; v x– projection of the velocity vector onto the selected axis; a x– projection of the acceleration vector onto the selected axis; t– movement time.

For body A we write: initial coordinate X 0 = 10 m; v x= –5 m/s; a x= 4 m/s 2. Then the equation for the projection of velocity versus time will be:

v x= v 0x + a x t (2)

For our case vx = 4t – 5.

For body B we write, taking into account formula (1): X 0 = 5 m; v x= 0 m/s; a x= –8 m/s 2 . Then we write the equation for the projection of velocity versus time for body B v x = –8t.

Where k – Boltzmann constant, T – gas temperature in Kelvin. From the formula it is clear that the dependence of the average kinetic energy on temperature is direct, that is, the number of times the temperature changes, the number of times the average kinetic energy of the thermal motion of molecules changes.

Answer: 4 times.

Task 9

In a certain process, the gas gave up an amount of heat of 35 J, and the internal energy of the gas in this process increased by 10 J. How much work was done on the gas by external forces?

Solution

The problem statement deals with the work of external forces on the gas. Therefore, it is better to write the first law of thermodynamics in the form:

∆U = Q + A v.s (1),

Where ∆ U= 10 J – change in the internal energy of the gas; Q= –35 J – the amount of heat given off by the gas, A v.s – work of external forces.

Let's substitute the numerical values ​​into formula (1) 10 = –35 + A v.s; Therefore, the work done by external forces will be equal to 45 J.

Answer: 45 J.

The partial pressure of water vapor at 19° C was equal to 1.1 kPa. Find the relative humidity of the air if the saturated vapor pressure at this temperature is 2.2 kPa?

Solution

By definition of relative air humidity

φ – relative air humidity, in percent; P v.p – partial pressure of water vapor, P n.p. – saturated vapor pressure at a given temperature.

Let's substitute the numerical values ​​into formula (1).

Answer: 50%.

The change in state of a fixed amount of monatomic ideal gas occurs according to the cycle shown in the figure.

Establish a correspondence between processes and physical quantities (∆ U– change in internal energy; A– gas work), which characterize them.

For each position from the first column, select the corresponding position from the second column and write the selected numbers in the table using the corresponding letters.

Solution

This graph can be rearranged in axes PV or deal with what is given. In section 1–2, isochoric process V= const; Pressure and temperature rise. Gas does not do work. That's why A= 0, The change in internal energy is greater than zero. Consequently, the physical quantities and their changes are correctly written under number 4) Δ U > 0; A= 0. Section 2–3: isobaric process, P= const; temperature increases and volume increases. The gas expands, gas work A>0. Therefore, transition 2–3 corresponds to entry number 1) Δ U > 0; A > 0.

An ideal monatomic gas located in a cylinder under a heavy piston (friction between the surface of the piston and the cylinder can be neglected) is slowly heated from 300 K to 400 K. The external pressure does not change. Then the same gas is heated again from 400 K to 500 K, but with the piston fixed (the piston does not move).

Compare the work done by the gas, the change in internal energy and the amount of heat received by the gas in the first and second processes.

For each quantity, determine the corresponding nature of the change:

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

Write it down to the table selected numbers for each physical quantity. The numbers in the answer may be repeated.

Solution

If a gas is slowly heated in a cylinder with a loose heavy piston, then at a constant external pressure the process can be considered isobaric (gas pressure does not change)

Therefore, the gas work can be calculated using the formula:

A = P · ( V 2 – V 1), (1)

Where A– gas work in an isobaric process; P– gas pressure; V 1 – volume of gas in the initial state; V 2 – volume of gas in the final state.

The change in internal energy of an ideal monatomic gas is calculated by the formula:

Where v– amount of substance; R– universal gas constant; ∆ T– change in gas temperature.

∆T= T 2 – T 1 = 400 K – 300 K = 100 K.

According to the first law of thermodynamics, the amount of heat received by the gas is equal to

Q = ∆U + A (3)

Q = 150v R + P(V 2 – V 1) (4);

If a gas is heated in a cylinder with a fixed piston, then the process can be considered isochoric (the volume of the gas does not change). In an isochoric process, an ideal gas does not do any work (the piston does not move).

A z = 0 (5)

The change in internal energy is equal to:

Answer: 232.

An uncharged piece of dielectric was introduced into the electric field (see figure). It was then divided into two equal parts (dashed line) and then removed from the electric field. What charge will each part of the dielectric have?

1. The charge on both parts is zero;
2. The left side is positively charged, the right side is negatively charged;
3. The left side is negatively charged, the right side is positively charged;
4. Both parts are negatively charged;
5. Both parts are positively charged.

Solution

If you introduce a dielectric (a substance in which there are no free electric charges) into an electric field under normal conditions, then the phenomenon of polarization is observed. In dielectrics, charged particles are not able to move throughout the entire volume, but can only move short distances relative to their constant positions, the electric charges in dielectrics are bound. If the dielectric is removed from the field, then the charge on both parts is zero.

The oscillatory circuit consists of a capacitor with a capacity C and inductor coils L. How will the frequency and wavelength of the oscillating circuit change if the area of ​​the capacitor plates is halved? For each quantity, determine the corresponding nature of the change:

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

Write it down to the table selected numbers for each physical quantity. The numbers in the answer may be repeated.

Solution

The problem talks about an oscillatory circuit. By determining the period of oscillations occurring in the circuit , wavelength is related to frequency

Where v– oscillation frequency. By determining the capacitance of a capacitor

C = ε 0 ε S/d (3),

where ε 0 is the electrical constant, ε is the dielectric constant of the medium. According to the conditions of the problem, the area of ​​the plates is reduced. Consequently, the capacitance of the capacitor decreases. From formula (1) we see that the period of electromagnetic oscillations arising in the circuit will decrease. Knowing the relationship between the period and frequency of oscillations

The graph shows how the magnetic field induction changes over time in a conducting circuit. In what period of time will an induced current appear in the circuit?

Solution

By definition, an induced current in a conducting closed circuit occurs under the condition of a change in the magnetic flux passing through this circuit.

Law of electromagnetic induction, where Ɛ – induced emf, ∆Φ – change in magnetic flux, ∆ t the period of time during which changes occur.

According to the conditions of the problem, the magnetic flux will change if the magnetic field induction changes. This occurs in a time interval from 1 s to 3 s. The contour area does not change. Therefore, the induced current occurs in the case

1. By the time t= 1 s change in magnetic flux through the circuit is greater than zero.
2. The induced current in the circuit occurs in the range from ( t= 1 s to t= 3 s)
3. The module of the inductive emf arising in the circuit is 10 mV.
4. change in magnetic flux through the circuit from t = 3 s to t = 4 s less than zero.
5. The induction current is zero at intervals from ( t= 0 s to t= 1 s) and from ( t= 3 s to t= 4 s)

Answer: 2.5.

The square frame is located in a uniform magnetic field in the plane of the magnetic induction lines (see figure). The direction of the current in the frame is shown by arrows. How is the force acting on the side directed? ab frames from the external magnetic field? (right, left, up, down, towards the observer, away from the observer)

Solution

The ampere force acts on the current-carrying frame from the magnetic field. The direction of the Ampere force vector is determined by the mnemonic rule of the left hand. We direct the four fingers of the left hand along the side current ab, induction vector IN, should enter the palm, then the thumb will show the direction of the Ampere force vector.

Answer: to the observer.

A charged particle flies at a certain speed into a uniform magnetic field perpendicular to the field lines. From a certain point in time, the magnetic field induction module increased. The charge of the particle has not changed.

How did the force acting on a moving particle in a magnetic field, the radius of the circle along which the particle moves, and the kinetic energy of the particle change after increasing the magnetic field induction modulus?

For each quantity, determine the corresponding nature of the change:

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

Write it down to the table selected numbers for each physical quantity. The numbers in the answer may be repeated.

Solution

A particle moving in a magnetic field is acted upon by the magnetic field by the Lorentz force. The Lorentz force modulus can be calculated using the formula:

F l = B · q· v sinα (1),

Where B– magnetic field induction, q– particle charge, v– particle speed, α – angle between the speed vector and the magnetic induction vector.

In our case, the particle flies in perpendicular to the lines of force, α = 90°, sin90 = 1.

From formula (1) it is clear that with increasing magnetic field induction, the force acting on a particle moving in a magnetic field increases.

The formula for the radius of the circle along which a charged particle moves is:

Where m – particle mass. Consequently, with increasing field induction, the radius of the circle decreases.

The Lorentz force does not do any work on a moving particle, since the angle between the force vector and the displacement vector (the displacement vector is directed along the velocity vector) is 90°.

Therefore, kinetic energy, regardless of the value of the magnetic field induction does not change.

Answer: 123.

Along a section of a DC circuit with resistance R current flows I. Establish a correspondence between physical quantities and formulas by which they can be calculated. For each position from the first column, select the corresponding position from the second column and write down the selected numbers in the table under the corresponding letters.

Where P– electric current power, A– work of electric current, t– the time during which an electric current flows through a conductor. The work, in turn, is calculated

A = I Ut (2),

Where I – electric current strength, U – tension in the area,

As a result of the reaction of the nucleus and α particle, a proton and a nucleus appeared:

Solution

Let's write the nuclear reaction for our case:

As a result of this reaction, the law of conservation of charge and mass number is satisfied. Z = 13 + 2 – 1 = 14; M = 27 + 4 – 1 = 30.

Therefore, the core is number 3)

The half-life of the substance is 18 minutes, the initial mass is 120 mg. What will be the mass of the substance after 54 minutes, the answer expressed in mg?

Solution

The task is to use the law of radioactive decay. It can be written in the form

Answer: 15 mg.

The photocathode of the photocell is illuminated with ultraviolet light of a certain frequency. How do the work function of the photocathode material (substance), the maximum kinetic energy of photoelectrons, and the red limit of the photoelectric effect change if the frequency of light is increased?

For each quantity, determine the corresponding nature of the change:

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

Write it down to the table selected numbers for each physical quantity. The numbers in the answer may be repeated.

Solution

It is useful to recall the definition of the photoelectric effect. This is the phenomenon of interaction of light with matter, as a result of which the energy of photons is transferred to the electrons of the substance. There are external and internal photoeffects. In our case we are talking about the external photoelectric effect. When, under the influence of light, electrons are ejected from a substance. The work function depends on the material from which the photocathode of the photocell is made, and does not depend on the frequency of the light. Therefore, as the frequency of ultraviolet light incident on the photocathode increases, the work function does not change.

Let's write Einstein's equation for the photoelectric effect:

hv = A out + E to (1),

hv– energy of a photon incident on the photocathode, A out – work function, E k is the maximum kinetic energy of photoelectrons emitted from the photocathode under the influence of light.

From formula (1) we express

E k = hv – A out (2),

therefore, as the frequency of ultraviolet light increases the maximum kinetic energy of photoelectrons increases.

red border

Answer: 313.

Water is poured into the beaker. Select the correct value for the volume of water, taking into account that the measurement error is equal to half the scale division.

Solution

The task tests the ability to record the readings of a measuring device, taking into account a given measurement error. Let's determine the price of the scale division

The measurement error according to the condition is equal to half the division value, i.e.

We write the final result in the form:

V= (100 ± 5) ml

The conductors are made of the same material. Which pair of conductors should be chosen in order to experimentally discover the dependence of the wire resistance on its diameter?

Solution

The task states that the conductors are made of the same material, i.e. their resistivities are the same. Let’s remember what values ​​the conductor resistance depends on and write the formula for calculating the resistance:

Where R– conductor resistance, p– resistivity material, l– length of the conductor, S– cross-sectional area of ​​the conductor. In order to identify the dependence of the conductor on the diameter, you need to take conductors of the same length, but different diameters. Loan that the cross-sectional area of ​​a conductor is defined as the area of ​​a circle:

Where d– conductor diameter. Therefore, answer option: 3.

A projectile with a mass of 40 kg, flying in a horizontal direction at a speed of 600 m/s, breaks into two parts with masses of 30 kg and 10 kg. Most of it moves in the same direction at a speed of 900 m/s. Determine the numerical value and direction of the velocity of the smaller part of the projectile. In response, write down the magnitude of this speed.

At the moment of shell explosion (∆ t→ 0) the effect of gravity can be neglected and the projectile can be considered as a closed system. According to the law of conservation of momentum: the vector sum of the momentum of the bodies included in a closed system remains constant for any interactions of the bodies of this system with each other. For our case we write:

m= m 1 1 + m 2 2 (1)

– projectile speed; m- mass of the projectile before bursting; 1 – speed of the first fragment; m 1 – mass of the first fragment; m 2 – mass of the second fragment; 2 – speed of the second fragment.

Let us choose the positive direction of the X axis, which coincides with the direction of the projectile velocity, then in the projection onto this axis we write equation (1):

mv x = m 1 v 1 x + m 2 v 2x (2)

Let us express from formula (2) the projection of the velocity vector of the second fragment.

The smaller part of the projectile at the moment of explosion has a speed of 300 m/s, directed in the direction opposite to the initial movement of the projectile.

Answer: 300 m/s.

In a calorimeter, 50 g of water and 5 g of ice are in thermal equilibrium. What must be the minimum mass of a bolt having a specific heat capacity of 500 J/kg K and a temperature of 339 K so that all the ice melts after it is lowered into the calorimeter? Neglect heat losses. Give the answer in grams.

Solution

To solve the problem, it is important to remember the heat balance equation. If there are no losses, then heat transfer of energy occurs in the system of bodies. As a result, the ice melts. Initially, water and ice were in thermal equilibrium. This means that the initial temperature was 0 ° C or 273 K. Remember the conversion from degrees Celsius to degrees Kelvin. T = t+ 273. Since the condition of the problem asks about the minimum mass of the bolt, the energy should only be enough to melt the ice.

With b m b ( t b – 0) = λ m l (1),

where λ is the specific heat of fusion, m l – mass of ice, m b – bolt mass.

Let us express from formula (1)

Answer: 50 g.

In the circuit shown in the figure, the ideal ammeter shows 6 A. Find the emf of the source if its internal resistance is 2 ohms.

Solution

We carefully read the problem statement and understand the diagram. There is one element in it that may be overlooked. This is a blank wire between the 1 ohm and 3 ohm resistors. If the circuit is closed, then the electric current will pass through this wire with the least resistance and through the 5 ohm resistor.

Then we write Ohm’s law for the complete circuit in the form:

where is the current strength in the circuit, ε is the source emf, R– load resistance, r– internal resistance. From formula (1) we express the emf

ε = I (R + r) (2)

ε = 6 A (5 Ohm + 2 Ohm) = 42 V.

Answer: 42 V.

In the chamber from which the air was pumped out, an electric field was created with a intensity and magnetic field with induction . The fields are homogeneous and the vectors are mutually perpendicular. A proton flies into the chamber p, the velocity vector of which is perpendicular to the intensity vector and the magnetic induction vector. The magnitudes of the electric field strength and magnetic field induction are such that the proton moves in a straight line. Explain how the initial part of the proton trajectory will change if the magnetic field induction is increased. In your answer, indicate what phenomena and patterns you used to explain. Neglect the influence of gravity.

Solution

In solving the problem, it is necessary to focus on the initial motion of the proton and the change in the nature of the motion after a change in the magnetic field induction. The proton is acted upon by a magnetic field by the Lorentz force, the modulus of which is equal to F l = qvB and an electric field with a force whose modulus is equal to F e = qE. Since the proton charge is positive, then e is codirectional with the voltage vector electric field. (See figure) Since the proton initially moved rectilinearly, these forces were equal in magnitude according to Newton’s second law.

With increasing magnetic field induction, the Lorentz force will increase. The resultant force in this case will be different from zero and directed towards the greater force. Namely in the direction of the Lorentz force. The resultant force imparts an acceleration to the proton directed to the left; the proton's trajectory will be curvilinear, deviating from the original direction.

The body slides without friction along an inclined chute, forming a “dead loop” with a radius R. From what height should the body begin to move in order not to break away from the chute at the top point of the trajectory?

Solution

We are given a problem about the unevenly variable motion of a body in a circle. During this movement, the position of the body in height changes. It is easier to solve the problem using the equations of the law of conservation of energy and the equations of Newton’s second law normal to the trajectory of motion. We made a drawing. Let's write down the formula for the law of conservation of energy:

A = W 2 – W 1 (1),

Where W 2 and W 1 – total mechanical energy in the first and second positions. For the zero level, select the position of the table. We are interested in two positions of the body - this is the position of the body at the initial moment of movement, the second is the position of the body at the top point of the trajectory (this is point 3 in the figure). During movement, two forces act on the body: gravity = and ground reaction force. The work of gravity is taken into account in the change in potential energy, the force does not do work, so it is perpendicular to the displacement everywhere. A = 0 (2)

To position 1: W 1 = mgh(3), where m- body mass; g- acceleration of gravity; h– the height from which the body begins to move.

In position 2 (point 3 in the figure):

v 2 + 4gR – 2gh = 0 (5)

At the top point of the loop, two forces act on the body, according to Newton's second law

Solving equations (5) and (7) we obtain h= 2.5 R

Answer: 2.5 R.

Air volume in the room V = 50 m 3 has a temperature t = 27° C and relative air humidity φ 1 = 30%. How long τ must a humidifier operate, spraying water with a productivity of μ = 2 kg/h, so that the relative humidity in the room increases to φ 2 = 70%. Saturated water vapor pressure at t = 27°C equals p n = 3665 Pa. The molar mass of water is 18 g/mol.

Solution

When starting to solve problems on steam and humidity, it is always useful to keep in mind the following: If the temperature and pressure (density) of the saturating steam are given, then its density (pressure) is determined from the Mendeleev-Clapeyron equation. Write down the Mendeleev-Clapeyron equation and the relative humidity formula for each state.

For the first case, at φ 1 = 30%, we express the partial pressure of water vapor from the formula:

Where T = t+ 273 (K), R– universal gas constant. Let us express the initial mass of steam contained in the room using equations (2) and (3):

The time that the humidifier should operate can be calculated using the formula

let's substitute (4) and (5) into (6)

Let's substitute the numerical values ​​and get that the humidifier should work for 15.5 minutes.

Answer: 15.5 min.

Determine the emf of the source if, when connecting a resistor with a resistance R voltage at source terminals U 1 = 10 V, and when connecting a resistor 5 R voltage U 2 = 20 V.

Solution

Let's write down the equations for two cases.

Ɛ = I 1 R + I 1 r (1)

U 1 = I 1 R (2)

Where r– internal resistance of the source, Ɛ – emf of the source.

Ɛ = I 2 5R + I 2 r(3)

U 2 = I 2 5R (4)

Taking into account Ohm's law for a section of the circuit, we rewrite equations (1) and (3) in the form:

The last substitution for calculating the EMF. Let's substitute formula (7) into (5)

Answer: 27 V.

When a plate made of some material is illuminated with light with a frequency v 1 = 8 1014 Hz and then v 2 = 6 · 1014 Hz it was found that the maximum kinetic energy of electrons changed by a factor of 3. Determine the work function of electrons from this metal.

Solution

If the frequency of the light quantum causing the photoelectric effect decreases, then the kinetic energy also decreases. Therefore, the kinetic energy in the second case will also be three times less. Let's write Einstein's equation for the photoelectric effect for two cases.

hv 1 = A + E to (1)

for the first frequency of light

formula for kinetic energy.

From equation (1) we express the work function and substitute expression (3) instead of kinetic energy

The final expression will look like:

Answer: 2 eV.