Translated 1.3.3-1.3.15; Fixed GUI image generator

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+ <meta name="description" content="From a hose lying on the ground, water shoots at an angle of 45^\circ to the horizon with an initial velocity of 10 \frac{m}{s}. The cross-sectional area of the hose opening is 5 cm^2 . Determine the mass of the jet in the air.">
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+ <title>From a hose lying on the ground, water shoots at an angle of 45^\circ to the horizon with an initial velocity of 10 \frac{m}{s}. The cross-sectional area of the hose opening is 5 cm^2 . Determine the mass of the jet in the air.</title>
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+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $1.3.12^*.$ From a hose lying on the ground, water shoots at an angle of $45^\circ$ to the horizon with an initial velocity of $10$ $\frac{m}{s}$. The cross-sectional area of the hose opening is $5$ $cm^2$ . Determine the mass of the jet in the air.
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ <p>First, let's determine the flight time of each water drop. This time is:</p>
+$$ t = \frac{2v_0 \cdot \sin \alpha}{g} $$
+<p>There is water in the air, flowing out of the hose in exactly time $t$. The mass of this water is easy to determine: it is the mass of water in a cylinder of height $v_0 t$ and base area $S$.</p>
+$$ m = \rho S v_0 t = \rho S \frac{2v_0^2 \cdot \sin \alpha}{g} $$
+
+$$ m = \frac{2 \rho v_0^2 S \cdot \sin \alpha}{g} = 7.2 \text{ kg}. $$
+
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$x = \frac{2\nu ^{2}}{g(tg\beta +tg\alpha )}$$
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