https://ejercicios-fyq.com/ Ejercicios Resueltos, Situaciones de aprendizaje y VÍDEOS de Física y Química para Secundaria y Bachillerato es SPIP - www.spip.net https://ejercicios-fyq.com/local/cache-vignettes/L144xH25/siteon0-da713.png?1758361862 https://ejercicios-fyq.com/ 25 144 https://ejercicios-fyq.com/Application-of-Torricelli-s-theorem-8284 https://ejercicios-fyq.com/Application-of-Torricelli-s-theorem-8284 2024-08-12T02:53:41Z text/html es F_y_Q Torricelli's theorem Horizontal launch SOLVED <p>A cylindrical container is filled with a liquid up to a height of one meter from the base of the container. Then, a hole is made at a point 80 cm below the liquid level: <br class='autobr' /> a) What is the exit velocity of the liquid through the hole? <br class='autobr' /> b) At what distance from the container will the first drop of liquid that touches the ground fall?</p> - <a href="https://ejercicios-fyq.com/Higher-level-physics-university" rel="directory">Higher level physics (university)</a> / <a href="https://ejercicios-fyq.com/Torricelli-s-theorem" rel="tag">Torricelli's theorem</a>, <a href="https://ejercicios-fyq.com/Horizontal-launch" rel="tag">Horizontal launch</a>, <a href="https://ejercicios-fyq.com/SOLVED" rel="tag">SOLVED</a> <div class='rss_texte'><p>A cylindrical container is filled with a liquid up to a height of one meter from the base of the container. Then, a hole is made at a point 80 cm below the liquid level:</p> <p>a) What is the exit velocity of the liquid through the hole?</p> <p>b) At what distance from the container will the first drop of liquid that touches the ground fall?</p></div> <hr /> <div <div class='rss_ps'><p>a) The exit velocity of the liquid through the hole is given by the expression: <br/> <br/> <img src='https://ejercicios-fyq.com/local/cache-TeX/6b8ef53b9924732e6103f56f9f68c5d6.png' style="vertical-align:middle;" width="141" height="26" alt="\color[RGB]{2,112,20}{\bm{v= \sqrt{2\cdot g\cdot h}}}" title="\color[RGB]{2,112,20}{\bm{v= \sqrt{2\cdot g\cdot h}}}" /> <br/> <br/> Since the hole is made at a distance of 0.8 m from the liquid level: <br/> <br/> <p class="spip" style="text-align: center;"><img src='https://ejercicios-fyq.com/local/cache-TeX/4255c299f7ff628166fc0ce03791540a.png' style="vertical-align:middle;" width="324" height="52" alt="v = \sqrt{2\cdot 9.8\ \frac{m}{s^2}\cdot 0.8\ m} = \fbox{\color[RGB]{192,0,0}{\bm{3.96\ \frac{m}{s}}}}" title="v = \sqrt{2\cdot 9.8\ \frac{m}{s^2}\cdot 0.8\ m} = \fbox{\color[RGB]{192,0,0}{\bm{3.96\ \frac{m}{s}}}}" /></p> <br/> b) To calculate the distance at which the first drop falls, you must consider that it follows a motion similar to a horizontal launch. In this case, the position with respect to the X and Y axes follows the equations: <br/> <br/> <img src='https://ejercicios-fyq.com/local/cache-TeX/031243fec7947d73e589855882668bd5.png' style="vertical-align:middle;" width="105" height="54" alt="\left {\color[RGB]{2,112,20}{\bf x = v\cdot t}} \atop {\color[RGB]{2,112,20}{\bm{y = \frac{1}{2}gt^2}}}} \right \}" title="\left {\color[RGB]{2,112,20}{\bf x = v\cdot t}} \atop {\color[RGB]{2,112,20}{\bm{y = \frac{1}{2}gt^2}}}} \right \}" /> <br/> <br/> Since you know that the drop starts at a height of 0.2 m: <br/> <br/> <img src='https://ejercicios-fyq.com/local/cache-TeX/8cc8244ea60608b15a7b13fca5849665.png' style="vertical-align:middle;" width="291" height="65" alt="t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2\cdot 0.2\ \cancel{m}}{9.8\ \frac{\cancel{m}}{s^2}}}= \color[RGB]{0,112,192}{\bf 0.2\ s}" title="t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2\cdot 0.2\ \cancel{m}}{9.8\ \frac{\cancel{m}}{s^2}}}= \color[RGB]{0,112,192}{\bf 0.2\ s}" /> <br/> <br/> To find the horizontal position, substitute this time: <br/> <br/> <p class="spip" style="text-align: center;"><img src='https://ejercicios-fyq.com/local/cache-TeX/43a81991bb4b60becdd55b81eb5a1c8b.png' style="vertical-align:middle;" width="343" height="45" alt="x = v\cdot t = 3.96\ \frac{m}{\cancel{s}}\cdot 0.2\ \cancel{s}= \fbox{\color[RGB]{192,0,0}{\bf 0.79\ m}}" title="x = v\cdot t = 3.96\ \frac{m}{\cancel{s}}\cdot 0.2\ \cancel{s}= \fbox{\color[RGB]{192,0,0}{\bf 0.79\ m}}" /></p> </math></p></div>