Boxer | 180 days

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    <img src="boxer.png" alt="Boxer | 180 дней" class="product-image">
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        <h1 class="product-title">Boxer | 180 дней</h1>
        <p class="product-description">Boxer | 180 дней – это цифровой товар, который позволит вам безопасно и анонимно использовать интернет в течение полугода. С помощью этого продукта вы сможете обходить блокировки и ограничения, а также защитить свои данные от нежелательных глаз.</p>
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            <li>180 дней использования</li>
            <li>Безлимитный трафик</li>
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            <li>Обход блокировок и ограничений</li>
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Product description: "Boxer | 180 days" is a digital product that provides secure and anonymous Internet access for six months. This product will allow you to bypass blocks and restrictions and protect your data. With "Boxer | 180 days" you will receive unlimited traffic and complete anonymity on the Internet. This product is suitable for people who value their privacy and security on the Internet. The product price is $29.99. You can buy it right now by clicking on the "Buy" button.

"Boxer | 180 days" is a digital product that provides secure and anonymous Internet access for 180 days. With "Boxer | 180 days" you can bypass blocks and restrictions, and protect your data and privacy. The product provides unlimited traffic and complete anonymity on the Internet. "Boxer | 180 days" is suitable for people who value their security and privacy on the Internet. The product price is $29.99. You can purchase it now by clicking on the "Buy" button.

This product is an electronic textbook that will help you solve a physics problem about a compressed spring.

From the description of the problem, we know that the spring is already compressed by 10 cm and that for additional compression to 15 cm, an elastic force at the end of compression equal to 150 N is required.

To solve the problem, we can use the formula to calculate the work done by the elastic force when the spring is deformed:

$$W = \frac{1}{2} k x^2,$$

where $W$ is the work done by the elastic force, $k$ is the elasticity coefficient of the spring and $x$ is the amount of deformation of the spring.

To find the work that needs to be done to compress the spring to 15 cm, we need to calculate the difference between the work done when compressing the spring to 15 cm and the work done when compressing the spring to 10 cm:

$$W_{15} - W_{10} = \frac{1}{2} k (15^2 - 10^2),$$

where $W_{15}$ is the work done by the elastic force when the spring is compressed to 15 cm, $W_{10}$ is the work done by the elastic force when the spring is compressed to 10 cm.

To calculate the work value, we need to find the elasticity coefficient $k$. To do this, we can use Hooke's law:

$$F = kx,$$

where $F$ is the elastic force acting on the spring, $k$ is the elasticity coefficient of the spring and $x$ is the amount of deformation of the spring.

From the conditions of the problem we know that the elastic force acting on the spring when it is compressed to 10 cm is equal to 150 N. We also know that when the spring is compressed to 10 cm, its deformation is 10 cm. Substituting these values ​​​​into Hooke's law, we we can find the elasticity coefficient:

$$k = \frac{F}{x} = \frac{150}{10} = 15\ N/cm.$$

Now we can substitute the value of the elasticity coefficient into the formula for calculating the work and calculate it:

$$W_{15} - W_{10} = \frac{1}{2} \cdot 15 \cdot (15^2 - 10^2) = 1125\ Дж.$$

Thus, to compress the spring to 15 cm, we need to do 1125 J of work.

In the tutorial you will find a detailed description of the formula for calculating the work done by the elastic force when the spring is deformed, a step-by-step solution to a specific problem about a compressed spring, including the calculation of the elastic coefficient, the work done by the elastic force, as well as illustrations and diagrams to help you better understand the physical process described in the problem. In addition, the textbook may provide other examples of solving problems on elasticity and deformation, as well as general information about Hooke's law and its application in physics. The manual may also contain tasks for you to solve on your own so that you can consolidate your knowledge and skills in this topic.

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