Translated 1.5.11-1.5.20

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+ <meta name="description" content="The reel of tape is played back during time t at the film drawing speed v. The initial radius of the reel (with film) is R, and the final radius (without film) is r. What is the thickness of the film?">
+ <meta name="author" content="Aliaksandr Melnichenka">
+ <meta name="date" content="2023-10" scheme="YYYY-MM">
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+ <header style="text-align:center;">
+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#1.5">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $1.5.20.$ The reel of tape is played back during time $t$ at the film drawing speed $v$. The initial radius of the reel (with film) is $R$, and the final radius (without film) is $r$. What is the thickness of the film?
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <center>
+<figure>
+<img src="https://savchenkosolutions.com/1/1.5.20/draw.png"
+loading="lazy" width="150" />
+<figcaption>
+Film reel
+</figcaption>
+</figure>
+</center>
+
+<p>Film wound on a reel takes up an area</p>
+<p>
+$$ S = \pi (R^2 - r^2) $$
+</p>
+<p>If the film is completely unwound at a speed of $v$ in time $t$, then the length of the film </p>
+<p>
+$$ l = vt $$
+</p>
+<p>In cross-section, the film is a rectangle whose area is $S$ and length is $l$. Then the width of this rectangle (it is also the thickness of the film)</p>
+<p>
+$$ \fbox{$d = \frac{S}{l} = \frac{\pi (R^2 - r^2)}{vt}$} $$
+</p>
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$d = \frac{\pi (R^2 - r^2)}{vt}$$
+ </p>
+
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