This lesson is a useful addition to the previous topic. "
The ability to make such things - the thing is not just useful, it is - needless. In all sections of mathematics, from school to the highest. Yes, and in physics too. It is for this reason that the task of this kind is necessarily present in the exam and in OGE. In all levels, both basic and profile.
Actually, the whole theoretical part of such tasks is one single phrase. Universal and simple to disgrace.
Surprising, but remember:
Any equality with letters, any formula is also an equation!
And where the equation is automatically there. So we apply them in a convenient order and - ready-made.) Did you read the previous lesson? Not? However ... then this reference is for you.
Oh, you know? Excellent! Then we use theoretical knowledge in practice.
Let's start with simple.
How to express one variable through the other?
This task is constantly arising when solving systems of equations. For example, there is equality:
3 x. - 2 y. = 5
Here two variables - X and Chert.
Suppose we are asked expressx. throughy..
What does this task mean? It means that we must get some equality where the left of the clean X is on the left. In proud loneliness, without any neighbors and coefficients. And on the right - what happens.
And how do we get such equality? Very simple! With the help of all the same old good identity transformations! So apply them in a convenient us order, step by step getting to pure IKSA.
We analyze the left part of the equation:
3 x. – 2 y. = 5
Here we interfere with the top three before X and - 2 y.. Let's start with - 2owit will be easier.
Throw off - 2ow From the left side to the right. Changing minus on plus, of course. Those. Apply first identical conversion:
3 x. = 5 + 2 y.
Filmed done. Troika remained before X. How to get rid of it? Share both parts for this very troika! Those. cycling second identical conversion.
So Delim:
That's all. we expressed Xreker. On the left - pure X, and on the right - what happened as a result of the "cleansing" of the IKSA.
One could have been first Share both parts on the top three, and then to transfer. But this would lead to the appearance of fractions in the process of transformations, which is not very convenient. And so, the fraction appeared only at the very end.
I remind you that the order of transformations does not play any role. how us Convenient, and do it. The most important thing is not the procedure for the use of identical transformations, but their right!
And you can from the same equality
3 x. – 2 y. = 5
express y throughx.?
Why not? Can! All the same, only this time we are interested in the left pure cheery. So we clean the progress of all too much.
The first thing is getting rid of the expression 3x. Throw it on the right side:
–2 y. = 5 – 3 x.
Double left with a minus. We divide both parts on (-2):
And all things.) We expressedy. through x.Go to more serious tasks.
How to express a variable from the formula?
Not a problem! Similar!If you understand that any formula is also equation.
For example, such a task:
From formula
express the variable with.
Formula - also equation! The task means that through transformations from the proposed formula we need to get some new formula. In which the left will stand clean from, and on the right - what happens, it will work out ...
However ... as we have this very from pull out something?
How-like ... by steps! Clear things that highlight clean from immediately It is impossible: she is sitting in a fraction. And the fraction is multiplied by r.... So, I first clean expression with letter from. The whole fraction entirely. Here you can share both parts of the formula on r..
We get:
Next step must be pulled out from From the numerator of the fraction. How? Easily! Get rid of the fraraty. There is no fraction - there is no numerator.) Multiply both parts of the formula 2:
Elementary remained. We will provide right letter from Proud loneliness. For this variables a. and b. We carry left:
That's all, you can say. It remains to rewrite equality in the usual form, from left to right and - the answer is ready:
It was an easy task. And now the task on the basis of the real version of the EGE:
Batiskaph Locator, evenly immersed vertically down, eats ultrasonic pulses with a frequency of 749 MHz. Batiskop immersion rate is calculated by the formula
where C \u003d 1500 m / s is the speed of sound in water,
f. 0 - frequency of empty pulses (in MHz),
f. - The frequency of the signal reflected from the bottom, recorded by the receiver (in MHz).
Determine the frequency of the reflected signal in MHz, if the battery immersion rate is 2 m / s.
"Mall Buka", yes ... But the letters are a lyrics, and the general essence is still the same. First of all, it is necessary to express this very frequency of the reflected signal (i.e., the letter f.) From the formula proposed to us. That's what we do. We look at the formula:
Directly, naturally, letter f. I will not pull out in any way, she has been crushed again into the fraction. And in the numerator and the denominator. Therefore, the most logical step will get rid of the fraraty. And there - it will be visible. For this apply second Conversion - multiply both parts for a denominator.
We get:
And here is the next rake. Please pay attention to brackets of both parts! It is often in these most brackets and bugs in such tasks. More precisely, not in the brackets themselves, but in their absence.)
Brackets on the left mean that the letter v. multiply on the whole denominator. And not on his individual pieces ...
To the right, after multiplication, fraction disappeared And the lonely numerator remained. Which, again, all full multiplied on Bukovka from. What is expressed by brackets in the right part.)
But now brackets can be revealed:
Excellent. The process goes.) Now Bukovka f. Left began common factor. We carry it out for braces:
There is nothing left. We divide both parts on the bracket (v.- c.) And - the thing is in the hat!
In principle, everything is ready. Variable f. already expressed. But you can additionally "combat" the resulting expression - to make f. 0 Behind the bracket in the numerator and cut the entire fraction on (-1), thereby getting rid of extra minuses:
Here is such an expression. But now you can substitute numeric data. We get:
Answer: 751 MHz
That's all. I hope the overall idea is understandable.
We make elementary identity transformations in order to retire the variable you are interested in. The main thing here is not a sequence of actions (it can be any), but their correctness.
In these two lessons, only two basic identity transformations of equations are considered. They work always. That they are basic. In addition to this couple, there are still many other transformations, which will also be identical, but not always, but only under certain conditions.
For example, the construction of both parts of the equation (or formula) into a square (or vice versa, the extraction of the root from both parts) will be identical conversion if both parts of the equation obviously non-negative.
Or, let's say, the logarithmation of both parts of the equation will be identical conversion, if both parts obviously positive.Etc…
Such transformations will be considered in relevant topics.
And here and now - examples for training for elementary basic transformations.
Simple task:
From formula
express the variable a and find its value whenS.=300, V. 0 =20, t.=10.
The task is more comprehensive:
The average speed of the skier (in km / h) at a distance of two circles is calculated by the formula:
WhereV. 1 andV. 2 - average speeds (in km / h) on the first and second circles, respectively. What was the average speed of the skier in the second round, if it is known that the first circle of the skier ran at a speed of 15 km / h, and the average speed over the entire distance was equal to 12 km / h?
Task based on a real OGE version:
Centripetal acceleration when driving around the circle (in m / s 2) can be calculated by the formulaa.\u003d ω 2.R.where ω is the angular velocity (in s -1), andR. - Radius of the circle. Using this formula, find the radiusR. (in meters) if the angular velocity is 8.5 s -1, and the centripetal acceleration is 289 m / s 2.
Task based on the real version of the profile exam:
To the source with EDC ε \u003d 155 V and internal resistancer.\u003d 0.5 ohms want to connect the load with resistanceR. Ohm. The voltage on this load expressed in volts is given by the formula:
With what load resistance voltage on it will be 150 V? Answer express in Omah.
Answers (in disorder): 4; fifteen; 2; 10.
And so where numbers, kilometers per hour, meters, Omms are somehow themselves ...)
Taking advantage of the first start of thermodynamics in differential form (9.2), we obtain an expression for the heat capacity of an arbitrary process:
Imagine a complete differential internal energy through private derivatives by parameters and:
After that, formula (9.6) will rewrite in the form of
The ratio (9.7) has an independent value, since it determines the heat capacity in any thermodynamic process and for any macroscopic system, if the caloric and thermal equation of the state is known.
Consider the process at constant pressure and we get a common relationship between and.
Based on the resulting formula, it is easy to find a link between heat-loads and in perfect gas. We will deal with this. However, the answer is already known, we were actively used at 7.5.
Robert Majer equation
Express the private derivatives in the right-hand side of equation (9.8), using thermal and caloric equations recorded for one mole of perfect gas. The internal energy of the perfect gas depends only on temperature and does not depend on the volume of the gas, therefore
From thermal equation easy to get
Substitute (9.9) and (9.10) in (9.8), then
Finally write
I hope, learned (9.11). Yes, of course, this is the Majer equation. Recall once again that the Mayer equation is valid only for the perfect gas.
9.3. Polytropic processes in perfect gas
As noted above, the first beginning of thermodynamics can be used to derive the equations of the processes occurring in Gaza. A large practical application finds a class of processes called polytropic. Politropic called the process passing at constant heat capacity .
The equation of the process is defined by the functional link of two macroscopic parameters describing the system. On the corresponding coordinate plane, the equation of the process is clearly represented as a graph - the process curve. The curve depicting the polytropic process is called polytropy. The polytropic process equation for any substance can be obtained based on the first start of thermodynamics using its thermal and caloric equations of the state. We will demonstrate how this is done on the example of the output of the equation of the process for the perfect gas.
The output of the equation of the polytropic process in perfect gas
The requirement of constancy of heat capacity in the process allows you to record the first beginning of the thermodynamics in the form of
Using the Mayer equation (9.11) and the equation of the state of the ideal gas, we obtain the following expression for
Dividing equation (9.12) on T and substituting (9.13) to express
Sharing () on, we find
Integration (9.15), we get
This is polytropic equation in variables
Excluding from equation (), with the help of equality We obtain the polytropic equation in variables
The parameter is called a polytropic indicator that can be taken according to () a variety of values, positive and negative, integers and fractional. For formula (), many processes are hidden. Famous isobaric, isochhore and isothermal processes are special cases of polytropic.
This class of processes also refers adiabatic or adiabatic process . Adiabatar is called a process passing without heat exchange (). You can implement such a process in two ways. The first method assumes the presence of a thermal insulating shell system capable of changing its volume. The second is to implement such a rapid process, in which the system does not have time to exchange the amount of heat with the environment. The process of propagation of sound in Gaza can be considered adiabat due to its high speed.
It follows from the determination of heat capacity that in the adiabatic process. According to
where is the adiabat rate.
In this case, the polytropy equation takes the form
The adiabatic process equation (9.20) is also called the Poisson equation, so the parameter is often referred to as Poisson's constant. Permanent is an important gasein characteristic. From experience it follows that its values \u200b\u200bfor different gases lie in the range of 1.30 ÷ 1.67, therefore, on the form of the processes of the adiabat "drops" more cool than isotherm.
Graphs of polytropic processes for different values \u200b\u200bare presented in Fig. 9.1.
In fig. 9.1 Process charts are numbered in accordance with Table. 9.1.
To remove the formula of complex, it is necessary first of all by analyzing the analysis, from which elements the substance consists and in which elements included in each other are connected to each other. Usually the composition is complex expressed in percent, but it can be expressed in any other numbers indicating between the weights of the elements forming this substance. For example, the composition of aluminum oxide containing 52.94% aluminum and 47.06% oxygen will be fully defined if we say that they are connected in a weight ratio of 9: 8, i.e. that for 9 weight. h. Aluminum accounts for 8 weight. oxygen. It is clear that the ratio of 9: 8 should be equal to 52.94: 47.06.
Knowing the weighting composition of the complex and atomic weights of forming it elements, it is not difficult to find the relative number of atoms of each element in the molecule of the resulting substance and thus install its simple formula.
For example, it is necessary to derive calcium chloride formula containing 36% calcium and 64% chlorine. Atomic weight calcium 40, chlorine 35.5.
Denote by calcium atoms in calcium chloride molecule through x,and the number of chlorine atoms through y Since the calcium atom weighs 40, and the chlorine atom 35.5 oxygen units, the total weight of calcium atoms included in the calcium chloride molecule will be equal to 40 x,and the weight of chlorine atoms 35.5 y The ratio of these numbers obviously should be equal to the ratio of calcium and chlorine weights in any amount of calcium chloride. But the last ratio is 36: 64.
Equating both relationships, we get:
40x: 35,5y \u003d 36:64
Then free from coefficients at unknown h.and w.by dividing the first members of the proportion to 40, and the second to 35.5:
The numbers 0.9 and 1.8 express the relative number of atoms in the calcium chloride molecule, but they are fraction, whereas in the molecule can only contain an integer number of atoms. To express the attitude h.:w.two whole numbers, we divide both members of the second relationship to the smallest of them. Receive
x: w. = 1:2
Consequently, in the calcium chloride molecule, two chlorine atoms fall on one calcium atom. This condition satisfies a number of formulas: SASL 2, CA 2 CL 4, CA 3 SL 6, etc. Since we do not have data to judge which of the written formulas corresponds to the valid atomic composition of calcium chloride molecule, we will focus on the simplest Of these, SASL 2, indicating the smallest possible number of atoms in the calcium chloride molecule.
However, the arbitrariness in the choice of formula disappears if the substance is also known for its molecular compositionweight. In this case, it is not difficult to derive the formula that expresses the true composition of the molecule. Let us give an example.
The analysis establishes that glucose contains 4.5 weight. C carbon 0.75 weight. h. hydrogen and 6 weight. oxygen. Its molecular weight was found equal to 180. It is required to derive the glucose formula.
As in the previous case, we find the ratio between the number of carbon atoms (atomic weight 12), hydrogen and oxygen in the glucose molecule. Denote by carbon atoms through x, hydrogen w.and oxygen through z,we compile proportion:
2x : y: 16Z \u003d 4.5: 0.75: 6
from
Dividing all three members of the second half of equality by 0.375, we get:
h. : y:z \u003d. 1: 2: 1
Therefore, the simplest glucose formula will be CH 2 O. But the calculated on it would be 30, whereas in reality glucose 180, i.e., six times more. Obviously, for glucose you need to adopt the C 6 H 12 O 6 formula.
Formulas based, except for analysis data, also on the determination of molecular weight and indicating the actual number of atoms in the molecule are called true or molecular formulas; The formulas derived only from these analysis are called the simplest or empirical.
Having become acquainted with the withdrawal of chemical formulas, "it is easy to understand how accurate molecular weights are established. As we mentioned, the existing methods for determining molecular weights in most cases do not give quite accurate results. But, knowing at least an approximate and percentage of the substance, it is possible to establish its formula expressing the atomic composition of the molecule. Since the weight of the molecule is equal to the sum of the weights of the formatives of its atoms, to resolve the weight of atoms included in the molecule, we will determine its weight in oxygen units, i.e. the molecular weight of the substance. The accuracy of the found molecular weight will be the same as the accuracy of atomic scales.
Finding the chemical compound formula in many cases can be significantly simplified if you use the concept of ovalence elements.
Recall that the element's valence is called the property of its atoms to attach or replace a certain number of atoms of another element.
What is valence
The element is determined by the number showing how many hydrogen atoms(or Another monovalent element) joins or replaces the atom of this item.
The concept of valence is applied not only to individual atoms, but also by integer groups of atoms, which are included in the chemical compounds and participating as one in chemical reactions. Such groups of atoms received the name of radicals. In inorganic chemistry, the most important radicals are: 1) a water residue, or hydroxyl; 2) acid residues; 3) Basic balances.
The aqueous residue, or hydroxyl, it turns out if the water molecules take away one hydrogen atom. In the water molecule, hydroxyl is associated with one hydrogen atom, therefore, the group is monovalent.
Acid residues are called groups of atoms (aidek and one atom), "remaining" from acid molecules, if one or more of the atoms of hydrogen substituted from them from them from them. These groups are determined by the number of hydrogen atoms taken. For example, it gives two acid residues - one bivalent SO 4 and other monovalent HSO 4, which is part of various acidic salts. Phosphoric acid 3 PO 4 can give three acid residues: trivalent PO 4, bivalent HRO 4 and monovalent
H 2 PO 4 etc.
We will call the main balances; Atoms or groups of atoms, "remaining" from the base molecules, if one or more hydroxyls are mentally taken away from them. For example, consistently take away from the Fe (OH) 3 hydroxyl molecule, we obtain the following main residues: FE (OH) 2, Feoh and Fe. They are determined by the number of taken away hydroxyl groups: FE (OH) 2 is monovalent; Fe (Oh) -dvuvaleten; Fe - trivalent.
The main residues containing hydroxyl groups are included in the so-called basic salts. The latter can be considered as grounds in which part of the hydroxyls with substituted substitutional residues. So, when replacing two hydroxyls QFE (OH) 3, the acid residue of SO 4 turns out the main salt of FeOHSO 4, when replacing one hydroxyl in Bi (OH) 3
the acid residue NO 3 obtains the main salt Bi (OH) 2 NO 3, etc.
The knowledge of the valence of individual elements and radicals allows in simple cases to quickly draw up the formulas of very many chemical compounds, which frees the chemist from the need to mechanically memorize them.
Chemical formulas
Example 1. Create calcium hydrocarbonate formula - acidic acidic acid salt.
The composition of this salt should include calcium atoms and monovalent acid residues of the NSO 3. Since bivalent, then one calcium atom must take two acid residues. Consequently, the salt formula will be Ca (NSO 3).
In each task, physics is required from the formula to express the unknown, the next step to substitute numerical values \u200b\u200band get an answer, in some cases it is only necessary to express an unknown value. The methods of removing unknown from the formula a lot. If you see the pages of the Internet, then we will see a lot of recommendations on this. This suggests that a unified approach to solving this problem, the scientific community has not yet developed, and the methods that are used as shifting experience in school - they are all ineffective. Up to 90% of student graduation classes do not know how to express the unknown. Those who know how to do it - perform bulky transformations. Very strange, but physicists, mathematics, chemists have different approaches, explaining the methods of transferring parameters through the sign of equality (offer the rules of the triangle, cross or proportions other) can be said that they have a different culture of working with formulas. It can be submitted what happens to most students who occur with different interpretations of solving this problem, sequentially visiting the lessons of these items. This situation describes a typical dialogue on the network:
Teach express from the formulas of the magnitude. Grade 10, I am ashamed to know how from one formula to do another.
Yes, do not worry - this is the problem of many of my classmates, although I and 9 CL. The teachers show this most often by the triangle method, but it seems to me that it is inconvenient, and it's easy to get confused. I will show the easiest way I use ...
Suppose the formula is given:
Well, more simple .... you need from this formula to find time. You take into this formula to substitute numbers only different, based on the algebra. Suppose:
and you probably clearly can be seen that to find time in algebraic expression 5 you need 45/9. Those moving to physics: T \u003d S / V
Most students have a psychological block. Often, students note that when reading the textbook, difficulties first arouse those fragments of the text in which many formulas that "long conclusions do not even understand", but at the same time there is a sense of inferiority, disbelief in their forces.
I suggest the following solution to this problem - most students can still solve examples and, therefore, arrange the procedure. We use this ability.
1. In that part of the formula where the variable is to be expressed, it is necessary to break the order of actions, and we will not do in onemasters that do not contain this desired value.
2. Then, in the reverse sequence of calculations, transfer the elements of the formula to another part of the formula (through the equality sign) with the opposite action ("minus" - "plus", "divide" - "multiply", "erection in the square" - "Extraction of the square root" ).
That is, we will find the last action in the expression and suffer unrocked or a polynomial that performs this action, through the sign of equality is first, but already with the opposite action. Thus, consistently, finding the last action in the expression, transfer from one part of equality to another all known values. In conclusion, we rewrite the formula so that the unknown variable stood on the left.
We get a clear algorithm of work, we know exactly how many transformations must be executed. We can use already known formulas for training, we can invent yours. To begin work on the assimilation of this algorithm, a presentation was created.
Experience with students shows that this method is well perceived by them. The reaction of teachers on my presentation at the Festival "Teacher's Profile School" also talks about the positive grain laid down in this work.