Added Luis's English solutions of 4.2.6, 4.2.7 and 4.2.14

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+ <header style="text-align:center;">
+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+
+ <p>
+ $4.2.7$
+ A body in water weighs three times less than in air. What is the density of the body?
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ Let's consider the following figure
+ </p>
+ <center>
+ <figure>
+ <img src="draw.png"
+ loading="lazy" alt="4.2.7" width="400" />
+ <figcaption>
+ Force analysis
+ </figcaption>
+ </figure>
+ </center>
+ <p>
+ These normal forces $N_i$ is what the equipment measures as weight in each case.
+ From Newton Second Law for air
+ $$N_a = mg$$
+ Since $m=\rho~V$,
+ $$N_a = \rho~V~g \;(1)$$
+ Applying Newton Second Law for water
+ $$mg = N_w + F_A$$
+ $$\rho~V~g = N_w + \rho_w~V~g$$
+ $$N_w = (\rho-\rho_w)~V~g \;(2)$$
+ Dividing $(1)$ by $(2)$, taking account that $\frac{N_a}{N_w} = 3$
+ $$\rho = \frac{3}{2}\rho_w$$
+ as $\rho_w = 1$ g/cm$^3$
+ $$\boxed{\rho = 1.5~\rm{g/cm^3}}$$
+ </p>
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ BSc. Luis Daniel Fernández Quintana<br>
+ Physics Department (FCNE)<br>
+ Universidad de Oriente, Cuba<br>
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