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Урок 131. Задачи на работу, мощность, КПД (ч.2)

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Термодинамика

Молекулярная физика

Работа

Решение задач

Ришельевский лицей

Internal Energy

Thermodynamics (Field Of Study)

00:00:14

today we spend the whole day solving problems in the

00:00:17

first lesson the topic will be problems on

00:00:20

calculating internal energy in the

00:00:22

next lesson we will be doing

00:00:24

calculations with work and so the topic is problems on

00:00:35

calculating internal energy problems on

00:00:53

calculating internal energy

00:00:59

homework on both sides at once for tomorrow

00:01:03

kirik for 10th grade problems and with numbers 2

00:01:12

3 4 2 3 4 of a sufficient level page

00:01:21

39

00:01:25

then according to Savchenko problem number 5 6 6 and

00:01:39

problem number 56 15:05 615

00:01:47

two additional words about this problem there you will need to

00:01:49

calculate the energy in an isothermal

00:01:51

process during isothermal expansion,

00:01:53

in fact there is a formula which

00:01:56

allows you to calculate this energy, but outside, a

00:01:58

logarithm is used,

00:01:59

which we have not studied yet, so we

00:02:02

make a graph of the process and using the graph using the

00:02:06

palette method,

00:02:07

we find the work. If anyone is familiar with

00:02:10

logarithms, check your result.

00:02:13

logarithm is how to find out the work in an

00:02:16

isothermal process, but I’m not

00:02:17

telling you yet, but if you will type in

00:02:21

work in an isothermal process in Google, it will

00:02:23

give you a form

00:02:24

with a logarithm and so this is according to the graph,

00:02:35

but it will immediately be clear that you have

00:02:38

graphs as you calculated from it, and

00:02:41

now let’s start with the problem Gelfgat and number

00:02:45

423

00:02:48

number 423 miles per year after turning on

00:03:01

the heating, the air in the room was heated from a

00:03:04

temperature of 7 degrees Celsius to a

00:03:07

temperature of 27 degrees Celsius, how many

00:03:10

times did the internal

00:03:12

energy of the air contained in the room change

00:03:16

after turning on the heating, the air in the

00:03:18

room heated from a temperature of t 17

00:03:23

degrees Celsius to a temperature t2 of 27

00:03:27

degrees Celsius, how many times

00:03:30

did the internal energy change the energy of the air

00:03:33

contained in the room, well, let's

00:03:37

find the ratio y 2 divided by u1

00:03:43

let's remember how

00:03:46

internal energy is calculated, let's assume that

00:03:48

air is an ideal gas, it consists of

00:03:51

molecules of some kind, how many atoms are

00:03:54

in one molecule of air, in fact

00:03:58

there are different molecules, but most of them

00:04:00

of them these are n2 molecules,

00:04:02

then in second place o2 molecules are

00:04:05

diatomic molecules, so we can

00:04:07

assume that air consists of two

00:04:09

atomic molecules and therefore the number of degrees of

00:04:11

freedom is 5 5 and we can calculate the

00:04:17

internal energy using this formula in the

00:04:19

general case, I will now write this form

00:04:22

here as a certificate of equals and secondly,

00:04:26

the mass is divided by the molar mass rt

00:04:32

or you can write it briefly equalized

00:04:35

and secondly not so what do you think

00:04:44

the room is a sealed vessel or not

00:04:50

what your life experience tells you of

00:04:53

course they are not sealed under the threshold of the door

00:04:56

through the cracks in the windows air can escape

00:05:00

from the room and enter it,

00:05:02

if the room were hermetically sealed,

00:05:06

then when we heat the air, the pressure

00:05:08

would increase and since the area of,

00:05:12

for example, glass is very large, it might

00:05:13

even squeeze out the glass, but the

00:05:16

air calmly leaves the room,

00:05:18

this at first glance makes it difficult for us to

00:05:22

solve problems because when the air

00:05:25

heats up, expanding, it leaves the

00:05:28

room and the mass of the air

00:05:30

in the room changes or the

00:05:33

number of molecules of the air

00:05:36

in the room changes, but in fact, this

00:05:40

circumstance, on the contrary, makes it easier for us to

00:05:42

solve the problem

00:05:43

if we recall, in addition to the formula

00:05:46

for internal energy, the equation of the

00:05:48

Perron Mendeleev stake and so on the one hand, y

00:05:51

equals 5 second yurts

00:05:58

for air, the number of degrees of freedom is equal to

00:06:01

5, and on the other hand, the product of

00:06:06

pressure and volume according to

00:06:09

Mendeleev’s equation is equal to not if you

00:06:15

now substitute this product

00:06:18

newerth and here you get internal

00:06:24

energy equals 5 second points. V.

00:06:31

the room is not sealed therefore p

00:06:35

equals p atmospheric

00:06:39

dimensions of the room

00:06:42

we will neglect the thermal expansion of the walls as

00:06:45

the temperature increases v equals a constant, it

00:06:50

follows that y one equals 5

00:06:56

second n

00:06:57

atmospheric in the room and y 2 equals

00:07:05

5 second n atmospheric new room that

00:07:12

is, y one equals u2 the internal

00:07:17

energy of the air in the room has

00:07:19

not changed, or if you asked the

00:07:21

ratio of 2 divided by y one equals

00:07:26

one, here is the answer, how does it happen that

00:07:32

we heat the air, its internal energy does

00:07:35

not change, why does this happen,

00:07:37

where does the energy that was

00:07:39

released, for example, due to the operation of the

00:07:42

heating system, go?

00:07:50

air decreased in the room

00:07:53

like this, but for every mole the

00:07:57

internal energy increased

00:08:00

because the temperature became higher,

00:08:02

just part of the air came out and it turns out

00:08:05

that the internal energy of the

00:08:07

air that left the room is

00:08:10

exactly equal to the energy that was

00:08:13

released during the operation of the heating

00:08:15

system, and the internal energy of the

00:08:17

air itself is the room that was like this and

00:08:20

remains like this is a paradox

00:08:22

by the way, this is the technique of joint

00:08:25

use of the Perron Mendeleev stake equation

00:08:27

and the formula for internal

00:08:29

energy very often we will

00:08:31

use when solving problems see

00:08:33

now we will do this the

00:08:41

next problem from the Savchenko problem book

00:08:50

number 562

00:08:54

and number 56

00:08:57

2a Savchenko

00:09:02

we solved problem 562 b there it

00:09:06

was necessary to determine the internal energy of

00:09:08

one kilogram of air

00:09:10

and now the question is what is the internal

00:09:13

energy in joules under normal

00:09:16

conditions of one cubic centimeter of air and

00:09:20

so on up to the feet volume 1 cubic centimeter

00:09:27

we are talking about air

00:09:29

conditions are normal under normal

00:09:31

conditions what the pressure is equal to 10 5

00:09:36

pascals, and the temperature, I don’t know, it will

00:09:40

be needed, it won’t be needed, just in

00:09:42

case, let’s write down for normal conditions

00:09:46

273 kelvin, you can find the internal

00:09:52

energy of air, air, diatomic

00:09:54

molecules and equal to 5, just wondering

00:10:01

how much in one cubic centimeter of

00:10:03

air is this piece like this such

00:10:06

a volume of a check as this piece of chalk, we

00:10:10

remember the standard formula y

00:10:13

equals and the second yurt we are given the volume and

00:10:21

pressure, it is immediately clear that here we also

00:10:24

need to use the equation about the

00:10:26

feather in the middle

00:10:27

of it, namely pe ve equals the yurt again

00:10:36

we replace this product with the

00:10:39

product of volume and pressure and then

00:10:42

we get y

00:10:43

equals and second n in here is the working

00:10:52

formula we calculate y equals 5 second

00:10:57

by 10 in 5 pascals

00:11:02

multiplied by 1 cubic centimeter and

00:11:05

how many cubic meters is 10 minus 6

00:11:09

migal rights 10 minus 6 cubic meters

00:11:13

equals pascal multiplied by

00:11:16

cubic meter this is joule means here

00:11:19

we have one 10 times 5 second, that

00:11:23

is, 2 and a half 0.25

00:11:26

joules, the energy is very modest, but if

00:11:31

you take a kilogram of air, we

00:11:33

calculated it, you get a very impressive

00:11:35

number, just a cubic centimeter is a

00:11:37

very small volume,

00:11:40

move on to the

00:11:50

next problem 567, also according to Savchenko

00:11:54

go to Savchenko 567 listen to

00:12:09

the conditions of the problem in a long

00:12:12

thermally insulated pipe between two

00:12:15

identical pistons of mass m each

00:12:18

there is one mole of monatomic gas

00:12:21

at a temperature t zero at the initial

00:12:25

moment the velocities of the pistons are directed in

00:12:27

one direction and are equal to 3 v and at

00:12:30

what is the maximum temperature

00:12:32

the gas pistons heat up do not carry out

00:12:36

the mass of gas in comparison with the mass of the pistons,

00:12:38

neglect, write down the brief conditions, let's

00:12:44

think about what is happening here, so between

00:12:48

two identical pistons of mass emma

00:12:53

there is one mole of monatomic gas and is

00:13:02

equal to three at a temperature t zero at the

00:13:09

initial moment of speed the pistons are

00:13:11

directed in one direction and are equal to 3 c

00:13:15

and c yes what is the maximum temperature

00:13:21

the gas will heat up, let's just

00:13:29

designate this maximum

00:13:31

temperature t make a drawing of a pipe

00:13:42

in it there are two pistons 1 has a mass m 2 has

00:13:53

a mass and

00:13:54

the pipe is smooth the pistons move in one

00:14:03

direction and one of the pistons has a speed

00:14:07

3 you and the other one c yes what the maximum

00:14:10

temperature the gas will heat up if both

00:14:14

pistons move, for example, to the right and the gas is

00:14:16

heated as a result of this process,

00:14:19

then which of these two pistons has the greater

00:14:22

speed, which of them is the 3rd,

00:14:24

the left one is 3, because heating

00:14:28

occurs as a result of compression, the

00:14:30

work is done by the pistons

00:14:33

above the gas this leads to an increase in

00:14:35

internal energy and therefore temperature,

00:14:37

therefore this pistons have a speed of 3 in

00:14:41

this in

00:14:50

between the pistons there is gas at the initial

00:14:53

moment having a temperature t zero

00:14:55

here of this gas 1 mole what else is known well, in

00:15:02

fact,

00:15:07

we will show everything here gas molecules and so why does the

00:15:12

gas heat up?

00:15:14

If the gas heats up touch me, its

00:15:17

internal energy internal energy

00:15:19

can be changed in two ways either by

00:15:21

doing work on the gas or through

00:15:25

heat transfer,

00:15:26

here, according to the conditions of the problem, everyone is

00:15:28

thermally insulated, which means the change in

00:15:31

internal energy occurs due to the

00:15:33

work done,

00:15:34

this work is done by pistons, they

00:15:37

move with at different speeds and the

00:15:40

left rear piston catches up with the front ones as

00:15:46

a result the gas is compressed, what can be

00:15:48

said about the speeds of the pistons when the gas

00:15:53

has a maximum temperature

00:15:56

Andrey it will be the same, will the

00:16:00

pistons touch at the same time, no, they

00:16:03

rush, so we understand with you that

00:16:07

when the distance between the pistons becomes

00:16:10

the smallest, the gas temperature will be

00:16:13

the largest, but if the distance has become

00:16:16

the smallest, this means that both

00:16:18

pistons at this moment in time are moving at the

00:16:20

same speed,

00:16:21

let’s denote this speed with a stroke and show the

00:16:24

state of the system at this moment in time

00:16:31

pipe 2 pistons,

00:16:37

the mass of each m, they have already approached,

00:16:41

each of them has a speed of stroke

00:16:48

it is not difficult to guess that this one will slow down

00:16:51

into a stroke and this one will accelerate due to the

00:16:55

gas pressure, there is a compressed gas at a

00:17:02

temperature t which you need to find something to

00:17:09

catch on the work that the pistons do

00:17:14

on the gas due to what it is

00:17:18

done because any work is

00:17:20

done due to energy due to the

00:17:24

kinetic energy of the pistons this means

00:17:29

how much the kinetic energy of the pistons has decreased, the

00:17:32

internal

00:17:36

energy of the gas will increase, and why is it not necessary to take into

00:17:40

account the fact that the gas itself is ordered and

00:17:43

moves along with the pistons; it does not

00:17:45

just move chaotically, but in general the entire

00:17:48

mass of gas also rushes along with the

00:17:50

pistons; why is this contribution not needed?

00:17:52

take into account the kinetic energy, let's

00:17:58

also read the conditions of the problem with the mass of the gas

00:18:00

compared to the mass of the piston, neglect

00:18:03

that is, we assume that the gas is so small

00:18:06

that its contribution to the total kinetic

00:18:09

energy is negligible, then we can write the

00:18:14

kinetic initial

00:18:18

minus and the kinetic prime and the

00:18:22

kinetic 0 minus and the kinetic

00:18:24

simply without a primer like this this is what is this

00:18:29

how much has the kinetic energy decreased

00:18:32

so much the

00:18:34

internal

00:18:36

energy in the case has increased so now let's

00:18:42

look at all this in detail and the

00:18:45

kinetic zero

00:18:46

initial reserve the kinetic energy of the pistons

00:18:49

is the kinetic energy of this

00:18:51

piston plus the kinetic energy of this

00:18:53

guy this will be emma at 3 v squared in

00:18:59

half plus m squared in half

00:19:08

equals 9 plus 110

00:19:14

equals 5 m squared is the reserve of

00:19:19

kinetic energy of both pistons at the

00:19:21

beginning at the end and the kinetic equals

00:19:28

twice

00:19:29

since there are 2 pistons per m.v.

00:19:33

dash square in half 2 pistons each have

00:19:39

kinetic energy such as

00:19:41

we designated the speed as a dash, it

00:19:43

turns out equal to m in a dash square

00:19:49

internal energy

00:19:51

delta u equals the final value of the

00:19:56

internal energy minus the initial

00:19:58

of course this is when the temperature

00:20:01

is maximum that is, it will be a

00:20:04

monatomic gas three second not although there one

00:20:10

mole, but the dimension must be preserved,

00:20:12

so I write

00:20:15

yurt

00:20:17

minus the initial internal energy three

00:20:21

second yurts 0 and so

00:20:29

now we substitute all this here, we

00:20:35

get 5 m squared minus m a stroke

00:20:46

squared equals 3 second new era

00:20:51

three second is not a reproach

00:20:54

on p minus t zero what we

00:21:03

know here what we don’t know we are asked

00:21:05

to find t t 0 is not is mass is in is

00:21:11

missing wasd them

00:21:20

how can you find your stroke rods

00:21:27

smooth

00:21:28

pipe smooth pistons throats friction no

00:21:33

gravity can also be considered not clearly

00:21:45

arithmetic mean from wherever

00:21:47

it worked out but good yes they

00:21:55

interact with each other cass like a

00:21:58

spring before there are external forces acting between them

00:22:03

there is no what is the name of a

00:22:07

system that is not acted upon by external

00:22:09

forces closed for a closed system

00:22:14

some conservation laws are satisfied

00:22:17

you can call them the law of conservation of

00:22:21

momentum let's use the law of

00:22:24

conservation of momentum system the pistons are

00:22:26

closed, therefore the total momentum of the bodies

00:22:31

forming a closed system remains

00:22:33

unchanged for any interactions

00:22:35

between these bodies, here the interaction

00:22:37

occurs through the gas, the momentum of the gas is

00:22:39

neglected according to the conditions of the problem, the mass of the

00:22:41

gas is small, therefore we can write,

00:22:43

according to the law of conservation of momentum, 3 m in

00:22:51

this is the momentum of the first piston plus m.v .

00:22:56

this is the impulse 2 of the piston equals to prime

00:23:02

plus to r prime I don’t write vectors

00:23:09

because all the pistons move in one direction and

00:23:11

what is the ratio that connects the modules the

00:23:14

same ratio connects the projections onto the

00:23:16

axis from here the mass will be reduced and it turns out on the

00:23:22

left 4 in the right 2

00:23:26

prime that is, it turns out that a prime

00:23:31

equals simply 2 in the

00:23:34

arithmetic mean indeed

00:23:36

Andrey was right but this is his intuition, his

00:23:39

intuition told him now we

00:23:41

used the laws of conservation

00:23:42

of momentum,

00:23:43

we substitute here we get 5 m

00:23:49

squared minus m by 2 v squared

00:23:56

equals 3 second nes p-n t minus t 0

00:24:03

and here we have 4 mv square here 5 mv

00:24:09

square means we can write m squared

00:24:14

equals 3 second yurts minus t 0 all that

00:24:22

remains is algebra

00:24:31

from here we first find the parenthesis you

00:24:35

minus t zero equals 2m squared

00:24:42

divided by 3 by 3 new era 3 newer and the

00:24:49

last step is to find the

00:24:51

maximum temperature t equals t 0

00:24:57

+ 2 m squared divisor by 3 not a

00:25:07

problem solved numerical value

00:25:09

not required here this is how mechanics

00:25:13

interact with molecular physics

00:25:15

the laws of mechanics can be easily

00:25:17

used when solving problems at home

00:25:20

this problem

00:25:21

566 of the same type to a little simpler we

00:25:24

took it more difficult I will give a

00:25:27

simpler task is proposed the next task is

00:25:41

also Savchenko 5-6 4 number 5 6 4 Savchenko

00:25:54

something similar we have already encountered but

00:25:58

nevertheless this is a completely different

00:25:59

task in a vessel with a capacity

00:26:03

of one is located one there for gas at

00:26:06

pressure p1 and temperature t1 and in a vessel with

00:26:10

a capacity of 2

00:26:11

monatomic gas at pressure p2 and

00:26:14

temperature t2

00:26:15

what pressure and what temperature

00:26:18

will be in these vessels after they are

00:26:21

connected the vessels are thermally insulated so

00:26:25

we make two drawings

00:26:28

two vessels are connected to each other by a closed valve

00:26:45

here pressure p1

00:26:50

here volume of one temperature t one

00:26:55

valve is closed gas 1 there and and equals 3

00:27:02

then in a vessel of capacity 2 there is

00:27:08

also a monatomic gas at pressure p2

00:27:13

volume of 2

00:27:14

temperature t2 what pressure and what

00:27:20

temperature will be in these vessels

00:27:22

after they are connected the

00:27:35

valve is open

00:27:40

what can we say about the pressures on the left and

00:27:43

right?

00:28:05

we open the tap, thermal

00:28:09

equilibrium occurs if there are too many

00:28:11

fast molecules here and they don’t get here and

00:28:15

heat up the gas here in thermal

00:28:19

equilibrium the temperatures here will also be

00:28:21

the same you stroke here floor 3 here

00:28:24

and so we are given we are given p 1 in 1 p1

00:28:33

p2 in two t2 gas 1 tons and we need to find p

00:28:42

prime and t prime we had a similar

00:28:46

problem

00:28:47

but it was solved in a completely different way

00:28:50

because it was said that the temperature was the

00:28:53

same all the time, that is, there

00:28:56

the temperature did not change, but here the system is

00:28:59

thermally insulated then what is

00:29:01

preserved here if the system

00:29:03

thermally insulated internal energy

00:29:07

pistons here gas does not move

00:29:09

any work does not perform

00:29:12

thermal insulation ensures the absence of

00:29:15

heat transfer means the internal energy

00:29:18

remains the same so let's

00:29:21

denote the internal energy of this portion of

00:29:23

gas in one of this portion of gas u2 of

00:29:27

this portion of gas in one stroke of this

00:29:32

portion u2 stroke and Let's write according to the

00:29:36

conservation law for the

00:29:37

gene

00:29:44

y 1 plus 2 equals y one stroke plus

00:29:51

two strokes

00:29:54

then I remember the formulas for calculating the

00:29:56

internal energy we have such a

00:29:59

formula will equalize three second gas

00:30:02

one-ton yurt

00:30:07

there is another version that we have already

00:30:11

used twice this is not rt this is foam in

00:30:14

then we can write y equals the

00:30:19

second three pv which of these formulas is more convenient

00:30:25

to use the right right one because

00:30:28

we are given pressures and here and here and

00:30:33

here one unknown is also included, the

00:30:35

pressure at the end means we use the

00:30:37

right formula then as u1 we

00:30:40

write the three second p 1 in 1 plus u two three

00:30:49

second n 2 in 2 these all values are known

00:30:54

equals 3 second n stroke in one plus

00:31:02

three second n dash 2

00:31:04

everything from here can be found by stroke by three

00:31:10

second divided term by term and you get p 1 v1 + p

00:31:17

2 in 2 equals here the prime can be

00:31:21

taken out of the bracket

00:31:23

here it will be one plus you barely

00:31:28

so the b-prime

00:31:35

equals the fraction in the numerator p 1 v1 + p 2

00:31:43

v2 divided by v1 + v 2 yes we

00:31:59

neglect the volume of this tube the

00:32:04

first part of the problem is solved

00:32:07

the law of conservation of energy we used

00:32:10

but the temperature we never found this

00:32:15

for it’s probably useless to use these formulas already

00:32:18

what can we do to

00:32:21

find the temperature

00:32:27

let’s use the law of conservation

00:32:30

of matter the total number of molecules here has

00:32:35

not changed the number of moles in this

00:32:39

vessel we denote not one here not u2

00:32:44

here not one stroke here not u2 prime

00:32:48

means we can write that no

00:32:52

plus me 2 equals no prime plus

00:32:56

her two primes

00:32:58

this is what we did when we were told that the

00:33:02

process is isothermal we needed to find the

00:33:04

pressure remember when we connected two vessels

00:33:06

now the

00:33:09

law of conservation of matter continues to work and now

00:33:17

let’s use by the Clapeyron

00:33:18

Mendeleev equation, namely pe ve equals not p

00:33:27

volume and pressure we already know everywhere,

00:33:31

which means we can somehow

00:33:34

express the temperature from here and so let’s

00:33:36

express from here the amount of substance does not

00:33:40

equal p in divided by

00:33:43

mouths now we use this formula

00:33:48

to record these four values not u1

00:33:53

this will be p 1 in 1 divided by r t1 plus

00:34:02

not y 2 p 2 in 2 by r

00:34:09

t2 is equal to the right not really their pressure p

00:34:15

prime we know it if that on

00:34:19

in 1 divided by p a t prime + p prime on

00:34:29

v2

00:34:31

on rts stroke you stroke on r can

00:34:39

be shortened, we can write down what is left on the

00:34:48

left left n 1 in 1 divided by d1 plus

00:34:56

d2 in 2 divided by t2 on the right left

00:35:02

b-stroke divided by you stroke by the sum of

00:35:06

volumes I put it out of brackets in one plus

00:35:11

[ __ ] so now so that we don't have double

00:35:14

fractions double fractions let's

00:35:17

bring it to a common denominator here

00:35:19

we get p 1 in 1 t2 plus p 2

00:35:30

in 2d 1 divided by t1 and t2 equals

00:35:38

b-prime over r prime on v1 + v 2

00:35:45

from here we find a test of them

00:35:53

you the stroke is equal to then the stroke will go

00:35:58

here in the numerator and this whole fraction will go

00:36:02

to the right side with the numerator and

00:36:05

denominator changing places and so here we

00:36:07

will have a stroke multiplied by v1 + and v2

00:36:14

now this is all that will be here in the numerator

00:36:16

in the denominator p 1 in 1 t2 plus p

00:36:24

2 in two t1 and this t 1 t2

00:36:29

will go to the right in the numerator t1 and t2

00:36:35

this is not the answer yet because now we need to

00:36:38

substitute this stroke here,

00:36:44

then we will get that stroke equals

00:36:48

/ and this denominator p 1 in 1 t2 plus p 2

00:36:57

in two t1 then look when we

00:37:03

substitute primes in the numerator we

00:37:06

must write p 1 v1 + p 2 in 2 write

00:37:09

p 1 v1 + p 2 in 2 in the denominator we must

00:37:15

write in one plus in 2,

00:37:17

write it down here, but there is exactly the

00:37:20

same one in the numerator, they will cancel out,

00:37:23

so I won’t write anything except

00:37:27

m1 and m2, here is the answer

00:37:37

since the formula is cumbersome and does

00:37:40

not require us to do numerical calculations, let’s at

00:37:43

least check the dimension here we have

00:37:47

joules joules yes here joule multiplied

00:37:53

by kelvin plus joule multiplied by

00:37:56

kelvin joules will be reduced inverse

00:37:57

terms multiplied by kelvins squared

00:38:01

turns out in kelvins

00:38:03

well here you can immediately see that the answer

00:38:05

is obtained colleagues the problem is solved yes you do

00:38:16

n’t have to substitute this,

00:38:19

but by the way this is a lot for you if

00:38:21

you need to find a number you’re this

00:38:23

you can use the number and substitute it here,

00:38:25

in fact, it’s not very good for you and

00:38:28

will simplify the problem, all the same, when

00:38:31

we performed the substitution, this one was

00:38:33

reduced by one plus by 2, so we take a

00:38:36

break, then we solve the problems again

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