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://www.ejercicios-fyq.com/local/cache-vignettes/L144xH25/siteon0-da713.png?1758361862 https://ejercicios-fyq.com/ 25 144 https://www.ejercicios-fyq.com/Principle-of-dimensional-homogeneity-8304 https://www.ejercicios-fyq.com/Principle-of-dimensional-homogeneity-8304 2024-09-11T10:46:47Z text/html es F_y_Q Dimensions SOLVED <p>What is the principle of dimensional homogeneity?</p> - <a href="https://www.ejercicios-fyq.com/Units-and-magnitudes" rel="directory">Units and magnitudes</a> / <a href="https://www.ejercicios-fyq.com/Dimensions" rel="tag">Dimensions</a>, <a href="https://www.ejercicios-fyq.com/SOLVED" rel="tag">SOLVED</a> <div class='rss_texte'><p>What is the <i>principle of dimensional homogeneity</i>?</p></div> <hr /> <div <div class='rss_ps'><p>This principle states that equations relating physical quantities must be consistent, meaning that the same dimensions must appear on both sides of the equation. The best way to understand this is through an example. <br/> <br/> The speed at which an object moves under the influence of acceleration is: <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/e0b460db1a312f718d70061752740edb.png' style="vertical-align:middle;" width="117" height="19" alt="\color[RGB]{2,112,20}{\bm{v = v_0 + at}}" title="\color[RGB]{2,112,20}{\bm{v = v_0 + at}}" /> (Ec. 1) <br/> <br/> Speed is the ratio of distance to the time taken to cover it. In terms of dimensions, it is expressed as: <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/ea54684dba160033962f928c8fcfaee1.png' style="vertical-align:middle;" width="88" height="55" alt="\color[RGB]{0,112,192}{\bm{[v] = \frac{[L]}{[t]}}}" title="\color[RGB]{0,112,192}{\bm{[v] = \frac{[L]}{[t]}}}" /> <br/> <br/> Acceleration is the ratio of the change in speed to the time taken for that change, so it can be expressed as: <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/c77f5d751b5213f8ec8fc9dc4121f887.png' style="vertical-align:middle;" width="91" height="55" alt="\color[RGB]{0,112,192}{\bm{[a] = \frac{[L]}{[t]^2}}}" title="\color[RGB]{0,112,192}{\bm{[a] = \frac{[L]}{[t]^2}}}" /> <br/> <br/> If you apply this to equation (Eq. 1), you will have: <br/> <br/> <p class="spip" style="text-align: center;"><img src='https://www.ejercicios-fyq.com/local/cache-TeX/6b3f5fdd7ad7d86f719f461d86c8bc2e.png' style="vertical-align:middle;" width="388" height="55" alt="\frac{[L]}{[t]} = \frac{[L]}{[t]} + \frac{[L]}{[t]^2}\cdot [t]\ \to\ \color[RGB]{192,0,0}{\bm{\frac{[L]}{[t]} = \frac{[L]}{[t]} + \frac{[L]}{[t]}}}" title="\frac{[L]}{[t]} = \frac{[L]}{[t]} + \frac{[L]}{[t]^2}\cdot [t]\ \to\ \color[RGB]{192,0,0}{\bm{\frac{[L]}{[t]} = \frac{[L]}{[t]} + \frac{[L]}{[t]}}}" /></p> <br/> As you can see, the dimensions are the same on both sides of the equation.</math></p></div>