HELP ME PLEASE. ___ 1. An undeformed spring, the stiffness coefficient of which is 40 N / m, was compressed by 5 cm. What was the potential

what is the energy of the spring?

___

HELP ME PLEASE. ___ 1. An undeformed spring, the stiffness coefficient of which is 40 N / m, was compressed by 5 cm.

potential energy of the spring?

2. A body with a mass of 5 kg is located at a height of 12 m above the ground. Calculate its potential energy:

a) relative to the surface of the earth;

b) relative to the roof of the building, the height of which is 4 m.

___
3. An undeformed dynamometer spring was stretched by 10 cm, and its potential energy became 0.4 J. What is the coefficient of spring stiffness?

Two elastic springs, under the action of forces applied to them, lengthened by the same amount. A force was applied to the first spring, with stiffness k1

100 N, and to the second, with stiffness k2, - 50 N. How do the spring stiffnesses compare?

1) convert to si 2.5 t 350mg 10.5g 0.25t 2) it is necessary to determine the spring stiffness of the dynamometer if the distance between

divisions 0 and 1 of its scale is 2 cm.

k=......................

what is the value of the force of gravity acting on the load

G=............................

3) for this task, you need a complete solution to determine the weight of an astronaut with a mass of 100 kg, first on the moon and then on Mars

4) it is necessary to determine the absolute elongation of the spring with a stiffness of 50 N/m if

it is acted upon with a force of 1 n and b) a body of mass 20 g is suspended from it

5) an astronaut, being on the moon, hung a wooden bar with a mass of 1 kg from a spring. the spring lengthened by two cm. then the astronaut, using the same spring, evenly pulled the bar along the horizontal surface. in this case, the spring lengthened by 1 cm

to be determined

spring stiffness ........................

the magnitude of the friction force ..........

how many times the friction force could be greater if the experiment was carried out on Mars

plz need in 4 hours i beg you

6. What is the stiffness of the spring if a force of 2 N stretched it by 4 cm?

7. If the length of the spiral spring is reduced by 3.5 cm, an elastic force equal to 1.4 kN occurs. What will be the elastic force of the spring if its length is reduced by 2.1 cm?
8. When opening the door, the length of the door spring increased by 0.12m; the elastic force of the spring is at the same time 4 N. For which elongation of the spring is the elastic force equal to 10 N?
9. A force of 30 N stretches the spring by 5 cm. What force will stretch the spring by 8 cm?
10. As a result of stretching an undeformed spring 88 mm long, up to 120 mm, an elastic force equal to 120 N arose. Determine the length of this spring when the force acting on it is 90 N.
he is in balance.

Instruction

note

The ruler measures the elongation in centimeters, if you apply the found value without converting to meters, you will get an erroneous calculation of the spring stiffness.

Useful advice

1cm = 0.01m.
4 cm = 4 * 0.01 = 0.04 m.

It is worth every housewife to learn how to determine the hardness of water in order to prevent breakdown of household appliances, damage to linen, and also try to protect your body from undesirable consequences in the form of dry skin and kidney stones. There are several easy ways to find out what the hardness of your tap water is.

Instruction

In general terms, you can get an impression of how hard your water is by tracing the presence on household appliances and. If the water hardness is increased, then the kettle spiral, taps, metal surfaces will be covered with a considerable layer of scale. It has a yellowish gray color and a somewhat crumbly texture upon impact.

Water that has an excess of salts tastes strikingly different from water that has fewer salts. Some people are able to determine the hardness of water by tasting it, as it doesn't taste very good.

In very hard water, soap-based substances completely refuse to lather, so if the soap in your hands does not lather and the shampoo only runs down your hair, it means that the water in your home has exceeded the hardness. If the soap suds are hard to wash off your hands, it means that your tap water is very soft.

But all these definitions of hardness speak only about the approximate state of water, and for accurate indicators it is necessary to use special devices and technologies. For example, aquarists have the so-called water hardness test, which you can find in specialized stores.

There are TDS-meters that measure the level of minerals and salts, as well as the electrical conductivity and hardness of water. The device costs a lot of money, but it is thanks to it that you will know the level of salts contained in your water and, if their amount exceeds the norm, you will be able to start with water hardness in a timely manner with the help of protective equipment.

Related videos

Sources:

• A site about water, its properties and beneficial effects on humans.

Nowadays, finding the necessary normative act is not difficult. Any law, ranging from federal to regional, can be found using the Internet. However, not every edition found on the network law and may be relevant. Sites of proven, constantly updated legal systems are preferred for such purposes.

You will need

• - a computer;
• - Internet access;
• - at least the general meaning of the name of the law, but the more accurate information, the better.

Instruction

The simplest option seems to be to drive an approximate name in the corresponding line of a search engine.
At the same time, there will be a lot of options, but there is no one hundred percent that the first lines with the results will open the way for you to the current edition law a. Therefore, it is preferable to immediately turn to the services of reference systems sites, the most authoritative of which are "Consultant Plus" and "Garant". By the way, and among the first in the issue when searching for this or that law and they are likely to be present.

On the main page of the site of each of these systems there is a search function. Enter in it at least the approximate name of the regulatory act of interest and feel free to click on the search button.
In response, you will usually be given several options, from which it will not be particularly difficult to choose the right one. If you get into an outdated version law a, the system will inform about this and offer to switch to the current edition.
Another convenience with such systems is that if the text law but refers to others, in the text there is a hyperlink to them.
Along with the federal ones, both systems can also help in finding many law ov regional level.

Along with "Consultant" and "Guarantor" law You can search the State Duma of the Russian Federation (it promptly reflects the situation with a change in the status of each, starting from submission for consideration in the first reading and ending with signing by the President of the Russian Federation), and departments (concerning the competence of each of them), regional law odative and executive. Local authorities can be helpful when looking for regional law ov and other acts.
All federal acts that come into force must be published in Rossiyskaya Gazeta and posted on its official website.

Springs- this is a component of the car's suspension, which protect the car not only from road roughness, but also provide the desired body height above the road, which greatly affects the vehicle's handling, comfort and load capacity. As a result of tests for each car, the optimal rigidity suspension springs for certain driving conditions.

Instruction

When "breakdowns" occur, the suspension is considered too soft. In such situations, motorists become unstable in. Ideally, the spring force should be equal to a value that prevents excessive body roll.
Stiffer springs are required by cars that are prepared for racing. In different types of races of the same car, it involves installing springs with different rigidity Yu. When passing any turns, pay attention to body roll, which, with properly selected springs, should be no more than two or three degrees.

For the front and rear suspension, select springs according to stiffness in pairs. However, it is not immediately possible to achieve the desired suspension height, because the spring shrinks and can “lose” at the moment, which is very bad. This is due to a lack of bearing capacity even at full compression, but with rigidity yu, providing the desired height of the suspension. It is always easy to determine: between the coils of the spring there should be a gap of less than 4 mm.

Choose springs so that when filled, the gap between the coils of the springs is slightly more than 6.5 mm. It is advisable to install the softest springs, although they will roll the car within acceptable limits. It is usually incorrect to use stiff springs, relying on the fact that they reduce the roll of the car, improving handling.

Check rigidity springs according to the product code or according to the marks applied (by stamping or paint). Also define rigidity springs can be using hand, floor scales and a measuring ruler in kilograms per centimeter.
A wooden block (thickness not less than 12 mm) of a larger end area of ​​the spring is placed on household floor scales, and a spring is installed on top. Then a second piece of wood and the length of the spring are placed on top of the spring. Using a press, the spring is compressed to a certain value (for example, 30 mm) and the readings of the scales are taken, thereby calculating rigidity.

note

The pressing force on the spring is measured according to the readings of the scales, but this method of determining the stiffness of the springs is dangerous, since the spring can fly off a fairly large distance.

Cutting the compression coil springs used in today's passenger cars is the most common way to lower ride height or stiffen the springs for a more comfortable ride.

Instruction

Work on the springs of motor characteristics is best done in a car service by a qualified specialist. This operation greatly affects the quality of the move, and if you are not confident in your abilities, it is better not to try to do such work yourself.

If you decide to trim, remove, because in order to cut off the springs, they must first be released. Do it like this: lift the desired part of the machine, remove the wheel and clean the bolts holding the bottom of the rack. Like, they are very dirty, so you will have to fill them with oil and leave them to soak for a while. Loosen the bolts and knock them down carefully so as not to accidentally damage the brake disc. Unscrew the last top fastener and take out the stand assembly. Now you can disconnect the springs.

Determine how much you want to shorten the spring. For a slight decrease in ride height, it will be enough to cut off only 1.5 turns. Each car has its own characteristics, but usually 2 turns are cut off from the front springs, and the rear ones are cut off by a maximum of 3. It is better to consult a specialist in advance so as not to cut off the excess.

It is best to shorten the springs with a grinder, but you can get by with a regular hacksaw, although it will take more time. Make a mark in the right place with a marker and cut or saw off the excess length. You need to cut at the top, and assemble the rack with the cut end up. In the place where you shortened the spring, it has too much angle. This must be eliminated with a gas wrench or a large piece using great physical effort.

Install the spring in place and assemble all parts of the car in reverse order.

Related videos

What motorists go to to improve the quality of the ride. Among the many ingenious tricks, there is also a change in the ground clearance inherent in the design of the car. This can be done by changing the size of the screw springs shock absorber, that is, simply speaking, cutting it. You can carry out such a "surgical intervention" on your own. The main thing is to think carefully about the consequences of such an operation.

You will need

• - angle grinder ("Bulgarian");
• - hacksaw for metal;
• - a set of car wrenches.

Instruction

Deciding to perform pruning by force, first free springs by removing the stand. Support each side of the car in turn with a jack. Disconnect the wheels. Remove the bolts that secure the bottom of the rack. After that disconnect springs. Fold all fasteners carefully in one place, after cleaning them of dirt.

Decide how much you need to cut springs. For this, consult a car service specialist. To significantly change the clearance, you will need to cut one and a half to two turns. When in doubt, shorten first springs one turn and try them out. If necessary, the procedure can be repeated. cutting off springs for more turns at once, you, of course, will not be able to restore them to the required level later, so think carefully before taking up the tool.

Direct cutting of metal springs produce with the help of an angle grinder ("grinder"). If it is not available, use a hacksaw. Pre-markup in the right place. Cutting should be marked at the top of the product. This will reduce the negative consequences of the deformation of the updated springs.

Repeat the same operations for all springs, trying to end up with all of them being the same size. It is especially important that the size of the undercut springs coincides with the axes of the car in order to prevent loss of controllability due to even minimal distortion of the structure.

To avoid gross errors, use the capabilities of an automotive service department to trim the springs. A qualified one will allow you to assess how desirable it is for your vehicle, and will perform it at the highest professional level. Inept trimming of the springs may require their complete replacement in the future, and, consequently, unforeseen financial costs.

• - jack;
• - a set of keys;
• - ball joint puller;
• - tie rod end puller;
• - device for compressing the spring;
• - anti-recoil stops;
• - safety supports.

Instruction

Park the vehicle on a level surface and place wheel chocks under the rear wheels. The VAZ-2106 car is rear-wheel drive, so you can turn on the speed for reliability. Remove, but do not remove, the front wheel-to-hub bolts. Now jack up the side of the car and remove the wheel bolts completely. Place a safety stand under the vehicle and lower the side of the machine onto it. In the role of a support, there can also be a strong stump of suitable sizes, and several wooden blocks stacked together.

Remove the pin from the steering rod and use a 22 wrench to unscrew the nut that secures the rod pin. After that, carefully put on the tie rod puller so as not to damage the boot. If during disassembly the anther was damaged, then it must be replaced. Screw in the puller bolt, lightly tap it with a hammer from time to time. This is the only way the tie rod end will come off the cone. When you remove the rod, pull it to the side so that it does not interfere. Now the hub rotates freely on ball bearings. To remove the spring, it is necessary to disassemble the lower arm, as it rests against it.

Remove the nut that secures the shock absorber rod to the body. Then unscrew the two nuts that secure the shock absorber bracket to the lower arm. The shock absorber is pulled out from below; for convenience, its rod should be pushed into the body. Now you can put a puller on the spring and compress it. Try to put on the puller so that the two sides of the spring are compressed evenly. Both parts of the puller must be opposite each other. But after compressing the spring, further dismantling can be carried out. First you need to unscrew the mount of the anti-roll bar. Then you need to remove the lower ball joint. This can be done in two different ways.

Use a 22 wrench to unscrew the nut that secures the ball pin to the hub. But you will have to use a puller for ball joints. It must be used in the same way as the steering puller. Try not to damage the ball boot, and if it is already damaged, be sure to replace it. But it will be easier to unscrew the three bolts that secure the ball housing to the lever. This can be done with a 13 socket and socket wrench. After the hub is separated from the lever, the latter must be lowered down to remove the spring. If the spring does not come out, you will have to remove the lever completely. To do this, unscrew the two nuts that secure the lever to the body. Just keep in mind that under the rectangular bolt there are metal washers that adjust the camber. The spring on the second side is removed in the same way.

The concept of torsion stiffness, it turns out that this is the ability of a body to resist twisting. This characteristic is often used in relation to bicycle forks. There this moment is categorically important.

After all, it turns out that in the case of low torsional stiffness (or torsion), a bicycle fork, when exposed to a load on one side, will cause the fork to break and turn out.

To understand the situation, imagine a bicycle fork. The fork secures the so-called bushing. As long as the bushing is fixed evenly, all forces are distributed evenly. Now let's imagine that the wheel got into the mud or a hole, and the cyclist twists the steering wheel in the opposite direction. The moment of force that appears on the sleeve is distributed over the fork legs. These pants are starting to curl into a figure eight.

If there is enough torsional rigidity, then the fork will perfectly cope with such loading. If the balance between the strength of the material and the twisting moment is disturbed, i.e. the fork turns out at such an angle that the shoulder on which the force acts increases, then a break will occur. Accordingly, if this happens at high speed, the cyclist is likely to fall.

It is the ability of the body not to twist is determined by the concept of torsion rigidity. This characteristic applies to both the bicycle frame and other solids.

Sooner or later, when studying a physics course, pupils and students are faced with problems on the elastic force and Hooke's law, in which the coefficient of spring stiffness appears. What is this quantity, and how is it related to the deformation of bodies and Hooke's law?

First, let's define the basic terms that will be used in this article. It is known that if you act on a body from the outside, it will either gain acceleration or deform. Deformation is a change in the size or shape of a body under the influence of external forces. If the object is fully restored after the termination of the load, then such a deformation is considered elastic; if the body remains in an altered state (for example, bent, stretched, compressed, etc.), then the deformation is plastic.

Examples of plastic deformations are:

• clay crafting;
• bent aluminum spoon.

In its turn, elastic deformations will be considered:

• elastic band (you can stretch it, after which it will return to its original state);
• spring (after compression, it straightens again).

As a result of elastic deformation of a body (in particular, a spring), an elastic force arises in it, equal in absolute value to the applied force, but directed in the opposite direction. The elastic force for a spring will be proportional to its elongation. Mathematically, this can be written like this:

where F is the elastic force, x is the distance by which the length of the body has changed as a result of stretching, k is the stiffness coefficient that we need. The above formula is also a special case of Hooke's law for a thin tensile rod. In general form, this law is formulated as follows: "The deformation that has arisen in an elastic body will be proportional to the force that is applied to this body." It is valid only in those cases when we are talking about small deformations (tension or compression is much less than the length of the original body).

Determination of the stiffness factor

Stiffness factor(it also has the names of the coefficient of elasticity or proportionality) is most often written with the letter k, but sometimes you can see the designation D or c. Numerically, the stiffness will be equal to the magnitude of the force that stretches the spring per unit length (in the case of SI, by 1 meter). The formula for finding the elasticity coefficient is derived from a special case of Hooke's law:

The greater the value of rigidity, the greater will be the resistance of the body to its deformation. The Hooke coefficient also shows how stable the body is to the action of an external load. This parameter depends on the geometric parameters (wire diameter, number of turns and winding diameter from the wire axis) and on the material from which it is made.

The unit of stiffness in SI is N/m.

System Stiffness Calculation

There are more complex tasks in which total stiffness calculation required. In such tasks, the springs are connected in series or in parallel.

Serial connection of the spring system

When connected in series, the overall rigidity of the system is reduced. The formula for calculating the coefficient of elasticity will be as follows:

1/k = 1/k1 + 1/k2 + … + 1/ki,

where k is the total stiffness of the system, k1, k2, …, ki are the individual stiffnesses of each element, i is the total number of all springs involved in the system.

Parallel connection of the spring system

When the springs are connected in parallel, the value of the total coefficient of elasticity of the system will increase. The calculation formula will look like this:

k = k1 + k2 + … + ki.

Measuring the stiffness of a spring empirically - in this video.

Calculation of the stiffness coefficient by experimental method

With the help of a simple experiment, you can independently calculate, what will be the Hooke coefficient. For the experiment you will need:

• ruler;
• spring;
• cargo with a known mass.

The sequence of actions for experience is as follows:

1. It is necessary to fix the spring vertically, hanging it from any convenient support. The bottom edge must remain free.
2. Using a ruler, its length is measured and written as x1.
3. At the free end, you need to hang a load with a known mass m.
4. The length of the spring is measured in the loaded state. Denoted by x2.
5. Absolute elongation is calculated: x = x2-x1. In order to get the result in the international system of units, it is better to immediately convert it from centimeters or millimeters to meters.
6. The force that caused the deformation is the force of gravity of the body. The formula for calculating it is F = mg, where m is the mass of the load used in the experiment (translated into kg), and g is the free acceleration value, which is approximately 9.8.
7. After the calculations, it remains to find only the stiffness coefficient itself, the formula of which was indicated above: k = F / x.

Examples of tasks for finding stiffness

Task 1

A force F = 100 N acts on a spring 10 cm long. The length of the stretched spring is 14 cm. Find the stiffness coefficient.

1. We calculate the length of the absolute elongation: x = 14-10 = 4 cm = 0.04 m.
2. According to the formula, we find the stiffness coefficient: k = F / x = 100 / 0.04 = 2500 N / m.

Answer: the spring stiffness will be 2500 N/m.

Task 2

A load of mass 10 kg, when suspended on a spring, stretched it by 4 cm. Calculate how long another load of mass 25 kg will stretch it.

1. Let's find the force of gravity that deforms the spring: F = mg = 10 9.8 = 98 N.
2. Let's determine the coefficient of elasticity: k = F/x = 98 / 0.04 = 2450 N/m.
3. Calculate the force with which the second load acts: F = mg = 25 9.8 = 245 N.
4. According to Hooke's law, we write the formula for absolute elongation: x = F/k.
5. For the second case, we calculate the stretching length: x = 245 / 2450 = 0.1 m.

Answer: in the second case, the spring will stretch by 10 cm.

Were you good at physics at school? Do you know the basic physical laws and could you simply take and calculate, for example, the stiffness of a spring? Let's start with theoretical knowledge. The stiffness of a spring is a coefficient that relates the elongation of an elastic body and the elastic force that arises as a result of this elongation. The stiffness of the spring is also called the coefficient of elasticity or Hooke's coefficient, since the stiffness of the spring refers specifically to Hooke's law. What is the force of elasticity, which is mentioned in this law? The elastic force is the force that occurs when the body deforms and opposes this deformation.

mathematical method

How to determine the stiffness of a spring, or, in the terminology of such a science as physics, the coefficient of spring stiffness? To do this, you need to know a simple formula by which the spring stiffness is calculated. This formula, or rather Hooke's law, looks like this: F=|kx|, where k is the coefficient of elasticity of the spring, x is the elongation of the spring, or, as it is also called, the amount of deformation of the spring. And the value indicated by the letter F, respectively, is the elastic force, which we calculate. To find out what the stiffness of the spring is, it is necessary to measure the other two quantities indicated in the formula, using standard mathematical laws. The next step is to simply solve the equation in one unknown.

Experimental Method

To understand how to find the stiffness of the spring, or rather, to determine the coefficient of spring stiffness empirically, the following manipulations should be performed. You need to deform the body by applying force to it. The simplest type of deformation is compression or tension. The stiffness coefficient shows exactly what force must be applied to the body in order to elastically deform it per unit length. We are now talking about elastic deformation, when the body takes its original shape after the impact on it. In order to conduct this visual experiment, you will need the following things:

• calculator,
• a pen,
• notebook,
• spring,
• ruler,
• cargo.

So, fasten one end of the spring vertically, and leave the other free. Measure the length of the spring and write down the result in a notebook (this will be the value x1). Hang a hundred-gram weight from the free end of the spring and again measure the length of the spring, write down the value (x2). Calculate the absolute elongation of the spring (difference between x1 and x2). For small compressions and tensions, the elastic force is proportional to the deformation. Here we already apply Hooke's Law, according to which Fupr = |kx|, where k is the stiffness coefficient. In order to find the stiffness coefficient we need, we need to divide the tensile force by the elongation of the spring. We find the tensile force as follows: Fupr \u003d - N \u003d -mg. This implies that mg = kx. So, k = mg/x. Then everything is simple: substitute the values ​​\u200b\u200bthat you know into the formula and find what the stiffness of the spring is equal to.

Definition

The force that occurs as a result of the deformation of the body and trying to return it to its original state is called elastic force.

Most often it is denoted by $(\overline(F))_(upr)$. The elastic force appears only when the body is deformed and disappears if the deformation disappears. If, after removing the external load, the body completely restores its size and shape, then such a deformation is called elastic.

R. Hooke, a contemporary of I. Newton, established the dependence of the elastic force on the magnitude of the deformation. Hooke doubted the validity of his conclusions for a long time. In one of his books, he gave an encrypted formulation of his law. Which meant: "Ut tensio, sic vis" in Latin: what is the stretch, such is the strength.

Consider a spring subject to a tensile force ($\overline(F)$) that is directed vertically downwards (Fig. 1).

The force $\overline(F\ )$ is called the deforming force. Under the influence of a deforming force, the length of the spring increases. As a result, an elastic force ($(\overline(F))_u$) appears in the spring, balancing the force $\overline(F\ )$. If the deformation is small and elastic, then the elongation of the spring ($\Delta l$) is directly proportional to the deforming force:

\[\overline(F)=k\Delta l\left(1\right),\]

where in the coefficient of proportionality is called the stiffness of the spring (coefficient of elasticity) $k$.

Rigidity (as a property) is a characteristic of the elastic properties of a body that is being deformed. Rigidity is considered the ability of a body to resist an external force, the ability to maintain its geometric parameters. The greater the stiffness of the spring, the less it changes its length under the influence of a given force. The stiffness coefficient is the main characteristic of stiffness (as a property of a body).

The coefficient of spring stiffness depends on the material from which the spring is made and its geometric characteristics. For example, the stiffness coefficient of a coiled coil spring, which is wound from round wire and subjected to elastic deformation along its axis, can be calculated as:

where $G$ is the shear modulus (value depending on the material); $d$ - wire diameter; $d_p$ - spring coil diameter; $n$ is the number of coils of the spring.

The unit of measure for the stiffness coefficient in the International System of Units (SI) is the newton divided by the meter:

\[\left=\left[\frac(F_(upr\ ))(x)\right]=\frac(\left)(\left)=\frac(H)(m).\]

The stiffness coefficient is equal to the amount of force that must be applied to the spring to change its length per unit distance.

Spring stiffness formula

Let $N$ springs be connected in series. Then the stiffness of the entire joint is equal to:

\[\frac(1)(k)=\frac(1)(k_1)+\frac(1)(k_2)+\dots =\sum\limits^N_(\ i=1)(\frac(1) (k_i)\left(3\right),)\]

where $k_i$ is the stiffness of the $i-th$ spring.

When the springs are connected in series, the stiffness of the system is determined as:

Examples of problems with a solution

Example 1

Exercise. The spring in the absence of load has a length $l=0.01$ m and a stiffness equal to 10 $\frac(N)(m).\ $What will be the stiffness of the spring and its length if the force acting on the spring is $F$= 2 N ? Assume that the deformation of the spring is small and elastic.

Solution. The stiffness of the spring under elastic deformations is a constant value, which means that in our problem:

Under elastic deformations, Hooke's law is fulfilled:

From (1.2) we find the elongation of the spring:

\[\Delta l=\frac(F)(k)\left(1.3\right).\]

The length of the stretched spring is:

Calculate the new length of the spring:

Answer. 1) $k"=10\ \frac(Н)(m)$; 2) $l"=0.21$ m

Example 2

Exercise. Two springs with stiffnesses $k_1$ and $k_2$ are connected in series. What will be the elongation of the first spring (Fig. 3) if the length of the second spring is increased by $\Delta l_2$?

Solution. If the springs are connected in series, then the deforming force ($\overline(F)$) acting on each of the springs is the same, that is, it can be written for the first spring:

For the second spring we write:

If the left parts of expressions (2.1) and (2.2) are equal, then the right parts can also be equated:

From equality (2.3) we obtain the elongation of the first spring:

\[\Delta l_1=\frac(k_2\Delta l_2)(k_1).\]

Answer.$\Delta l_1=\frac(k_2\Delta l_2)(k_1)$