{"id":34935,"date":"2019-12-04T14:57:11","date_gmt":"2019-12-04T17:57:11","guid":{"rendered":"https:\/\/blog.estrategiavestibulares.com.br\/?p=34935"},"modified":"2021-03-11T11:07:00","modified_gmt":"2021-03-11T14:07:00","slug":"prova-ita-2020-fisica-resolucao-comentada","status":"publish","type":"post","link":"https:\/\/vestibulares.estrategia.com\/portal\/materias\/fisica\/prova-ita-2020-fisica-resolucao-comentada\/","title":{"rendered":"Prova ITA 2020 \u2013 F\u00edsica \u2013 Resolu\u00e7\u00e3o Comentada"},"content":{"rendered":"<p>Fala, pessoal&hellip; Tudo bem? Sou o prof. Toni Burgatto, do Estrat&eacute;gia Vestibulares e Carreiras Militares. Neste artigo, voc&ecirc; vai conferir a resolu&ccedil;&atilde;o das quest&otilde;es da prova de F&iacute;sica do Vestibular ITA 2020. A seguir, voc&ecirc; tamb&eacute;m vai poder baixar a corre&ccedil;&atilde;o da prova que disponibilizei gratuitamente em PDF. Vamos &agrave; resolu&ccedil;&atilde;o!!<div class=\"wp-block-file\">\"&gt;F&iacute;sica &ndash; Prova ITA 2020 &ndash; Resolvida\" class=\"wp-block-file__button\" download&gt;Baixar<\/div><h2 class=\"wp-block-heading\">Prova ITA 2020 &ndash; F&iacute;sica<\/h2><h3 class=\"wp-block-heading\">Quest&atilde;o 01<\/h3><p>Considere uma teoria na qual a for&ccedil;a de intera&ccedil;&atilde;o entre duas &ldquo;cargas generalizadas&rdquo; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;q_1\" alt=\"\\large q_1\" align=\"absmiddle\"> e <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;q_2\" alt=\"\\large q_2\" align=\"absmiddle\"> em universos N-dimensionais &eacute; expressa por <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;F_e=q_1q_2\/(kr%5E%7BN-1%7D)\" alt=\"\\large F_e=q_1q_2\/(kr^{N-1})\" align=\"absmiddle\"> em que k uma constante caracter&iacute;stica do meio. A teoria tamb&eacute;m prev&ecirc; for&ccedil;a entre dois &ldquo;polos generalizados&rdquo; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;p_1\" alt=\"\\large p_1\" align=\"absmiddle\"> e <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;p_2\" alt=\"\\large p_2\" align=\"absmiddle\"> expressa por <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;F_m=p_1p_2\/(%5Cmu&amp;space;r%5E%7BN-1%7D)\" alt=\"\\large F_m=p_1p_2\/(\\mu r^{N-1})\" align=\"absmiddle\">, na qual <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cmu\" alt=\"\\large \\mu\" align=\"absmiddle\"> &eacute; outra constante caracter&iacute;stica do meio. Sabe-se ainda que um polo p pode interagir com uma corrente de carga, t, gerando uma for&ccedil;a <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;F=ip\/(r%5E%7BN-2%7D)\" alt=\"\\large F=ip\/(r^{N-2})\" align=\"absmiddle\">. Em todos os casos, r representa a dist&acirc;ncia entre os entes interagentes. Considerando as grandezas fundamentais massa, comprimento, tempo e corrente de carga, assinale a alternativa que corresponde &agrave; f&oacute;rmula dimensional de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;k%5Cmu\" alt=\"\\large k\\mu\" align=\"absmiddle\">.<\/p><p>a) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;L%5E2T%5E%7B-2%7D\" alt=\"\\large L^2T^{-2}\" align=\"absmiddle\"><\/p><p>b) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;L%5E%7B-2%7DT%5E2\" alt=\"\\large L^{-2}T^2\" align=\"absmiddle\"><\/p><p>c) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;L%5E%7B-2%7DT%5E%7B-2%7D\" alt=\"\\large L^{-2}T^{-2}\" align=\"absmiddle\"><\/p><p>d) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;L%5E%7B1-N%7DT%5E2\" alt=\"\\large L^{1-N}T^2\" align=\"absmiddle\"><\/p><p>e) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;L%5E%7B-2%7DT%5E%7BN-1%7D\" alt=\"\\large L^{-2}T^{N-1}\" align=\"absmiddle\"><\/p><h3 class=\"wp-block-heading\">Resolu&ccedil;&atilde;o Comentada<\/h3><p>Pelo enunciado:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;F=%5Cfrac%7Bq_1%5Ccdot&amp;space;q_2%7D%7Bk%5Ccdot&amp;space;r%5E%7BN-1%7D%7D\" alt=\"\\large F=\\frac{q_1\\cdot q_2}{k\\cdot r^{N-1}}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;F=%5Cfrac%7Bp_1%5Ccdot&amp;space;p_2%7D%7B%5Cmu%5Ccdot&amp;space;r%5E%7BN-1%7D%7D\" alt=\"\\large F=\\frac{p_1\\cdot p_2}{\\mu\\cdot r^{N-1}}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;F=%5Cfrac%7Bi%5Ccdot&amp;space;p%7D%7Br%5E%7BN-2%7D%7D\" alt=\"\\large F=\\frac{i\\cdot p}{r^{N-2}}\" align=\"absmiddle\"><\/p><p>Fazendo-se a an&aacute;lise dimensional das equa&ccedil;&otilde;es, sabendo que:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cleft%5BF%5Cright%5D=M%5Ccdot&amp;space;L%5Ccdot&amp;space;T%5E%7B-2%7D\" alt=\"\\large \\left[F\\right]=M\\cdot L\\cdot T^{-2}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cleft%5Bi%5Cright%5D=%5Cleft%5Bq%5Cright%5D%5Ccdot&amp;space;T%5E%7B-1%7D\" alt=\"\\large \\left[i\\right]=\\left[q\\right]\\cdot T^{-1}\" align=\"absmiddle\"><\/p><p>Tem-se:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M%5Ccdot&amp;space;L%5Ccdot&amp;space;T%5E%7B-2%7D=%5Cleft%5Bq%5Cright%5D%5E2%5Ccdot%5Cleft%5Bk%5Cright%5D%5E%7B-1%7D%5Ccdot&amp;space;L%5E%7B1-N%7D\" alt=\"\\large M\\cdot L\\cdot T^{-2}=\\left[q\\right]^2\\cdot\\left[k\\right]^{-1}\\cdot L^{1-N}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M%5Ccdot&amp;space;L%5Ccdot&amp;space;T%5E%7B-2%7D=%5Cleft%5Bp%5Cright%5D%5E2%5Ccdot%5Cleft%5B%5Cmu%5Cright%5D%5E%7B-1%7D%5Ccdot&amp;space;L%5E%7B1-N%7D\" alt=\"\\large M\\cdot L\\cdot T^{-2}=\\left[p\\right]^2\\cdot\\left[\\mu\\right]^{-1}\\cdot L^{1-N}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M%5Ccdot&amp;space;L%5Ccdot&amp;space;T%5E%7B-2%7D=%5Cleft%5Bq%5Cright%5D%5Ccdot&amp;space;T%5E%7B-1%7D%5Ccdot%5Cleft%5Bp%5Cright%5D%5Ccdot&amp;space;L%5E%7B2-N%7D%5Crightarrow\" alt=\"\\large M\\cdot L\\cdot T^{-2}=\\left[q\\right]\\cdot T^{-1}\\cdot\\left[p\\right]\\cdot L^{2-N}\\rightarrow\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cleft%5Bq%5Cright%5D%5Ccdot%5Cleft%5Bp%5Cright%5D=M%5Ccdot&amp;space;L%5E%7BN-1%7D%5Ccdot&amp;space;T%5E%7B-1%7D\" alt=\"\\large \\left[q\\right]\\cdot\\left[p\\right]=M\\cdot L^{N-1}\\cdot T^{-1}\" align=\"absmiddle\"><\/p><p>Multiplicando as duas primeiras equa&ccedil;&otilde;es:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M%5E2%5Ccdot&amp;space;L%5E2%5Ccdot&amp;space;T%5E%7B-4%7D=%5Cleft(%5Cleft%5Bq%5Cright%5D%5Ccdot%5Cleft%5Bp%5Cright%5D%5Cright)%5E2%5Ccdot%5Cleft(%5Cleft%5Bk%5Cright%5D%5Ccdot%5Cleft%5B%5Cmu%5Cright%5D%5Cright)%5E%7B-1%7D%5Ccdot&amp;space;L%5E%7B2-2N%7D\" alt=\"\\large M^2\\cdot L^2\\cdot T^{-4}=\\left(\\left[q\\right]\\cdot\\left[p\\right]\\right)^2\\cdot\\left(\\left[k\\right]\\cdot\\left[\\mu\\right]\\right)^{-1}\\cdot L^{2-2N}\" align=\"absmiddle\"><\/p><p>Substituindo a rela&ccedil;&atilde;o encontrada anteriormente:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M%5E2%5Ccdot&amp;space;L%5E2%5Ccdot&amp;space;T%5E%7B-4%7D%5Ccdot%5Cleft%5Bk%5Ccdot%5Cmu%5Cright%5D=M%5E2%5Ccdot&amp;space;L%5E%7B2N-2%7D%5Ccdot&amp;space;T%5E%7B-2%7D%5Ccdot&amp;space;L%5E%7B2-2N%7D\" alt=\"\\large M^2\\cdot L^2\\cdot T^{-4}\\cdot\\left[k\\cdot\\mu\\right]=M^2\\cdot L^{2N-2}\\cdot T^{-2}\\cdot L^{2-2N}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfbox%7B%24%5Cleft%5B%5Cmu%5Ccdot&amp;space;k%5Cright%5D=L%5E%7B-2%7D%5Ccdot&amp;space;T%5E2%24%7D\" alt=\"\\large \\fbox{$\\left[\\mu\\cdot k\\right]=L^{-2}\\cdot T^2$}\" align=\"absmiddle\"><\/p><p class=\"has-light-green-cyan-background-color has-background\"><strong>Gabarito: B<\/strong><\/p><h3 class=\"wp-block-heading\">Quest&atilde;o 02<\/h3><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?\" alt=\"\" align=\"absmiddle\">Um sistema de defesa a&eacute;rea testa separadamente dois m&iacute;sseis contra alvos m&oacute;veis que se deslocam com velocidade <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%7B%5Cvec%7Bv%7D%7D_a\" alt=\"\\large {\\vec{v}}_a\" align=\"absmiddle\"> constante ao longo de uma reta distante de d do ponto de lan&ccedil;amento dos m&iacute;sseis. Para atingir o alvo, o m&iacute;ssil 1 executa uma trajet&oacute;ria retil&iacute;nea, enquanto o m&iacute;ssil 2, uma trajet&oacute;ria com velocidade sempre orientada para o alvo. A figura ilustra o instante de disparo de cada m&iacute;ssil, com o alvo passando pela origem do sistema de coordenadas xy. Sendo os m&oacute;dulos das velocidades os m&iacute;sseis iguais entre si, maiores que <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v_a\" alt=\"\\large v_a\" align=\"absmiddle\"> e mantidos constantes, considere as seguintes afirma&ccedil;&otilde;es:<\/p><p>I. Os intervalos de tempo entre o disparo e a colis&atilde;o podem ser iguais para ambos os m&iacute;sseis.<br>II. Para que o m&iacute;ssil 1 acerte o alvo &eacute; necess&aacute;rio que o m&oacute;dulo da componente y de sua velocidade seja igual a <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v_a\" alt=\"\\large v_a\" align=\"absmiddle\">.<br>III. Desde o disparo at&eacute; a colis&atilde;o, o m&iacute;ssil 2 executa uma trajet&oacute;ria curva de concavidade positiva com rela&ccedil;&atilde;o ao sistema xy.<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2019\/12\/image13.png\" alt=\"\" class=\"wp-image-34998\" width=\"365\" height=\"223\"><\/figure><\/div><p>Considerando V como verdadeira e F como falsa, as\nafirma&ccedil;&otilde;es I, II e III s&atilde;o, respectivamente,<\/p><p>a) V, V e V.<\/p><p>b) F, F e F.<\/p><p>c) V, F e V.<\/p><p>\nd) F, V e F.\n<\/p><p> e) F, V e V.<\/p><h3 class=\"wp-block-heading\">Resolu&ccedil;&atilde;o Comentada<\/h3><p>I: Falsa. Mas, se considerarmos que a velocidade do alvo pode ser constante e igual a 0 m\/s, a afirmativa &eacute; verdadeira. J&aacute; que nesse caso ambos os m&iacute;sseis percorreriam trajet&oacute;ria retil&iacute;nea, e, ent&atilde;o, demorariam o mesmo tempo para chegar ao alvo j&aacute; que tem velocidades iguais. Assim, implicaria no III como falso.<\/p><p>Entretanto, considerando a velocidade <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v_a%5Cneq0\" alt=\"\\large v_a\\neq0\" align=\"absmiddle\"> , como as trajet&oacute;rias t&ecirc;m comprimentos diferentes e os m&oacute;dulos das velocidades s&atilde;o iguais para ambos misseis, ent&atilde;o o tempo deve ser diferente.<\/p><p>II: Verdadeira. Se pensarmos que <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v_a%5Cneq&amp;space;v_%7BM1y%7D\" alt=\"\\large v_a\\neq v_{M1y}\" align=\"absmiddle\">, ter&iacute;amos que a ordenada do m&iacute;ssel 1 &eacute; sempre diferente da ordenada do alvo. Portanto, eles nunca se encontrariam. Dessa forma, &eacute; condi&ccedil;&atilde;o necess&aacute;ria que <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v_a=v_%7BM1y%7D\" alt=\"\\large v_a=v_{M1y}\" align=\"absmiddle\">.<\/p><p>III: Verdadeira. Do desenho, como <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v_a%5Cgeq0\" alt=\"\\large v_a\\geq0\" align=\"absmiddle\"> e a trajet&oacute;ria do m&iacute;ssel 2 tem sempre <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%7Cv_%7B2x%7D%7C%5Cgeq0\" alt=\"\\large |v_{2x}|\\geq0\" align=\"absmiddle\">, ent&atilde;o, a trajet&oacute;ria do m&iacute;ssel tem concavidade para cima.<\/p><p class=\"has-light-green-cyan-background-color has-background\"><strong>Gabarito: E<\/strong><\/p><h3 class=\"wp-block-heading\"> Quest&atilde;o 03<\/h3><p>Um bloco de massa m sustentado por um par de molas id&ecirc;nticas, paralelas e de constante el&aacute;stica k, desce verticalmente com velocidade constante e de m&oacute;dulo v controlada por um motor, conforme ilustra a figura. Se o motor travar repentinamente, ocorrer&aacute; uma for&ccedil;a de tra&ccedil;&atilde;o m&aacute;xima no cabo com m&oacute;dulo igual a<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2019\/12\/image14.png\" alt=\"\" class=\"wp-image-35001\" width=\"247\" height=\"248\"><\/figure><\/div><p>a) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;mg+%5Csqrt%7B%5Cleft(mg%5Cright)%5E2+2kmv%5E2%7D\" alt=\"\\large mg+\\sqrt{\\left(mg\\right)^2+2kmv^2}\" align=\"absmiddle\"><\/p><p>b) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;mg+%5Csqrt%7B%5Cleft(mg%5Cright)%5E2+kmv%5E2%7D\" alt=\"\\large mg+\\sqrt{\\left(mg\\right)^2+kmv^2}\" align=\"absmiddle\"><\/p><p>c) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;mg+%5Csqrt%7B2kmv%5E2%7D\" alt=\"\\large mg+\\sqrt{2kmv^2}\" align=\"absmiddle\"><\/p><p>d) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;mg+%5Csqrt%7B4kmv%5E2%7D\" alt=\"\\large mg+\\sqrt{4kmv^2}\" align=\"absmiddle\"><\/p><p>e) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;mg+%5Csqrt%7Bkmv%5E2%7D\" alt=\"\\large mg+\\sqrt{kmv^2}\" align=\"absmiddle\"><\/p><h3 class=\"wp-block-heading\">Resolu&ccedil;&atilde;o Comentada<\/h3><p>Pelo enunciado que afirma que a velocidade &eacute; constante, tem-se pelo equil&iacute;brio, na situa&ccedil;&atilde;o inicial:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;2%5Ccdot&amp;space;k%5Ccdot&amp;space;x=m%5Ccdot&amp;space;g\" alt=\"\\large 2\\cdot k\\cdot x=m\\cdot g\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;x=%5Cfrac%7Bm%5Ccdot&amp;space;g%7D%7B2%5Ccdot&amp;space;k%7D\" alt=\"\\large x=\\frac{m\\cdot g}{2\\cdot k}\" align=\"absmiddle\"><\/p><p>Ao travar-se o motor, inicia-se um problema de conserva&ccedil;&atilde;o de energia mec&acirc;nica:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;E_c+E_%7Bp,el%7D=E_%7Bp,el%7D+E_%7Bp,g%7D\" alt=\"\\large E_c+E_{p,el}=E_{p,el}+E_{p,g}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7Bm%5Ccdot&amp;space;v%5E2%7D%7B2%7D+%5Cfrac%7B2%5Ccdot&amp;space;K%5Ccdot&amp;space;x%5E2%7D%7B2%7D=%5Cfrac%7B2%5Ccdot&amp;space;K%5Ccdot%5Cleft(x+y%5Cright)%5E2%7D%7B2%7D-m%5Ccdot&amp;space;g%5Ccdot&amp;space;y\" alt=\"\\large \\frac{m\\cdot v^2}{2}+\\frac{2\\cdot K\\cdot x^2}{2}=\\frac{2\\cdot K\\cdot\\left(x+y\\right)^2}{2}-m\\cdot g\\cdot y\" align=\"absmiddle\"><\/p><p>Assim:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7Bm%5Ccdot&amp;space;v%5E2%7D%7B2%7D+K%5Ccdot&amp;space;x%5E2=K%5Ccdot%5Cleft(x%5E2+2%5Ccdot&amp;space;x%5Ccdot&amp;space;y+y%5E2%5Cright)-m%5Ccdot&amp;space;g%5Ccdot&amp;space;y\" alt=\"\\large \\frac{m\\cdot v^2}{2}+K\\cdot x^2=K\\cdot\\left(x^2+2\\cdot x\\cdot y+y^2\\right)-m\\cdot g\\cdot y\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;m%5Ccdot&amp;space;g%5Ccdot&amp;space;y+%5Cfrac%7Bm%5Ccdot&amp;space;v%5E2%7D%7B2%7D=K%5Ccdot%5Cleft(2%5Ccdot&amp;space;x%5Ccdot&amp;space;y+y%5E2%5Cright)\" alt=\"\\large m\\cdot g\\cdot y+\\frac{m\\cdot v^2}{2}=K\\cdot\\left(2\\cdot x\\cdot y+y^2\\right)\" align=\"absmiddle\"><\/p><p>Substituindo x:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;m%5Ccdot&amp;space;g%5Ccdot&amp;space;y+%5Cfrac%7B%5Cleft(m%5Ccdot&amp;space;v%5E2%5Cright)%7D%7B2%7D=K%5Ccdot%5Cleft(%5Cfrac%7B2%5Ccdot&amp;space;m%5Ccdot&amp;space;g%7D%7B2%5Ccdot&amp;space;K%7D%5Ccdot&amp;space;y+y%5E2%5Cright)\" alt=\"\\large m\\cdot g\\cdot y+\\frac{\\left(m\\cdot v^2\\right)}{2}=K\\cdot\\left(\\frac{2\\cdot m\\cdot g}{2\\cdot K}\\cdot y+y^2\\right)\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7Bm%5Ccdot&amp;space;v%5E2%7D%7B2%7D=K%5Ccdot&amp;space;y%5E2\" alt=\"\\large \\frac{m\\cdot v^2}{2}=K\\cdot y^2\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;y=%5Csqrt%7B%5Cfrac%7Bm%5Ccdot&amp;space;v%5E2%7D%7B2%5Ccdot&amp;space;K%7D%7D\" alt=\"\\large y=\\sqrt{\\frac{m\\cdot v^2}{2\\cdot K}}\" align=\"absmiddle\"><\/p><p>Na situa&ccedil;&atilde;o final:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;T=%5Cleft(2%5Ccdot&amp;space;K%5Cright)%5Ccdot%5Cleft(x+y%5Cright)\" alt=\"\\large T=\\left(2\\cdot K\\right)\\cdot\\left(x+y\\right)\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;T=2%5Ccdot&amp;space;K%5Ccdot%5Cleft(%5Cfrac%7Bm%5Ccdot&amp;space;g%7D%7B2%5Ccdot&amp;space;K%7D+%5Csqrt%7B%5Cfrac%7Bm%5Ccdot&amp;space;v%5E2%7D%7B2%5Ccdot&amp;space;K%7D%7D%5Cright)\" alt=\"\\large T=2\\cdot K\\cdot\\left(\\frac{m\\cdot g}{2\\cdot K}+\\sqrt{\\frac{m\\cdot v^2}{2\\cdot K}}\\right)\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfbox%7B%24T=m%5Ccdot&amp;space;g+%5Csqrt%7B2%5Ccdot&amp;space;m%5Ccdot&amp;space;k%5Ccdot&amp;space;v%5E2%7D%24%7D\" alt=\"\\large \\fbox{$T=m\\cdot g+\\sqrt{2\\cdot m\\cdot k\\cdot v^2}$}\" align=\"absmiddle\"><\/p><p class=\"has-light-green-cyan-background-color has-background\"><strong>Gabarito: C<\/strong><\/p><h3 class=\"wp-block-heading\">Quest&atilde;o 04<\/h3><p>Por uma mangueira de di&acirc;metro <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;D_1\" alt=\"\\large D_1\" align=\"absmiddle\"> flui &aacute;gua a uma velocidade de 360 m\/min, conectando-se na sua extremidade a 30 outras mangueiras iguais entre si, de di&acirc;metro <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;D_2&lt;d_1\" alt=\"\\large D_2&lt;d_1\" align=\"absmiddle\">. Assinale a rela&ccedil;&atilde;o <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;D_2\/D_1\" alt=\"\\large D_2\/D_1\" align=\"absmiddle\"> para que os jatos de &aacute;gua na sa&iacute;da das mangueiras tenham alcance horizontal m&aacute;ximo de 40 m.<\/p><p>a) 1\/10<\/p><p>b) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Csqrt%7B3\/10%7D\" alt=\"\\large \\sqrt{3\/10}\" align=\"absmiddle\"><\/p><p>c) 4\/5<\/p><p>d) 1\/2<\/p><p>e)&nbsp; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Csqrt%7B2\/3%7D\" alt=\"\\large \\sqrt{2\/3}\" align=\"absmiddle\"><\/p><h3 class=\"wp-block-heading\">Resolu&ccedil;&atilde;o Comentada<\/h3><p>A velocidade inicial ao utilizar-se a mangueira de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;D_1\" alt=\"\\large D_1\" align=\"absmiddle\"> &eacute; de 360 m\/min ou 6 m\/s. Assim, o fluxo &eacute; dado por:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cphi=6%5Ccdot%5Cpi%5Ccdot%5Cleft(%5Cfrac%7BD_1%7D%7B2%7D%5Cright)%5E2\" alt=\"\\large \\phi=6\\cdot\\pi\\cdot\\left(\\frac{D_1}{2}\\right)^2\" align=\"absmiddle\"><\/p><p>O fluxo total que percorre as mangueiras unidas &eacute; o mesmo que na situa&ccedil;&atilde;o inicial. Dessa forma:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cphi=30%5Ccdot&amp;space;v_2%5Ccdot%5Cpi%5Ccdot%5Cleft(%5Cfrac%7BD_2%7D%7B2%7D%5Cright)%5E2\" alt=\"\\large \\phi=30\\cdot v_2\\cdot\\pi\\cdot\\left(\\frac{D_2}{2}\\right)^2\" align=\"absmiddle\"><\/p><p>Para o alcance m&aacute;ximo, sabe-se que o &acirc;ngulo da velocidade com a horizontal deve ser de 45&deg;, dessa forma:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;t_%7Blancamento%7D=2%5Ccdot%5Cfrac%7Bv_2%5Ccdot%5Cfrac%7B%5Csqrt2%7D%7B2%7D%7D%7Bg%7D=%5Cfrac%7Bv_2%5Ccdot%5Csqrt2%7D%7Bg%7D\" alt=\"\\large t_{lancamento}=2\\cdot\\frac{v_2\\cdot\\frac{\\sqrt2}{2}}{g}=\\frac{v_2\\cdot\\sqrt2}{g}\" align=\"absmiddle\"><\/p><p>E, a dist&acirc;ncia percorrida &eacute;:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;d=v_2%5Ccdot%5Cfrac%7B%5Csqrt2%7D%7B2%7D%5Ccdot&amp;space;v_2%5Ccdot%5Cfrac%7B%5Csqrt2%7D%7Bg%7D=%5Cfrac%7Bv_2%5E2%7D%7Bg%7D\" alt=\"\\large d=v_2\\cdot\\frac{\\sqrt2}{2}\\cdot v_2\\cdot\\frac{\\sqrt2}{g}=\\frac{v_2^2}{g}\" align=\"absmiddle\"><\/p><p>Como <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;d=400%5C&amp;space;m\" alt=\"\\large d=400\\ m\" align=\"absmiddle\">:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v_2%5E2=400\" alt=\"\\large v_2^2=400\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v_2=20%5Cfrac%7Bm%7D%7Bs%7D\" alt=\"\\large v_2=20\\frac{m}{s}\" align=\"absmiddle\"><\/p><p>Substituindo na rela&ccedil;&atilde;o do fluxo:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;6%5Ccdot%5Cpi%5Ccdot%5Cleft(%5Cfrac%7BD_1%7D%7B2%7D%5Cright)%5E2=30%5Ccdot20%5Ccdot%5Cpi%5Ccdot%5Cleft(%5Cfrac%7BD_2%7D%7B2%7D%5Cright)%5E2\" alt=\"\\large 6\\cdot\\pi\\cdot\\left(\\frac{D_1}{2}\\right)^2=30\\cdot20\\cdot\\pi\\cdot\\left(\\frac{D_2}{2}\\right)^2\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7B1%7D%7B100%7D%5Ccdot&amp;space;D_1%5E2=D_2%5E2\" alt=\"\\large \\frac{1}{100}\\cdot D_1^2=D_2^2\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7BD_2%7D%7BD_1%7D=%5Csqrt%7B%5Cfrac%7B1%7D%7B100%7D%7D=%5Cfrac%7B1%7D%7B10%7D\" alt=\"\\large \\frac{D_2}{D_1}=\\sqrt{\\frac{1}{100}}=\\frac{1}{10}\" align=\"absmiddle\"><\/p><p class=\"has-light-green-cyan-background-color has-background\"><strong>Gabarito: A<\/strong><\/p><h3 class=\"wp-block-heading\">Quest&atilde;o 05<\/h3><p>Um sat&eacute;lite artificial viaja em dire&ccedil;&atilde;o a um planeta ao longo de uma trajet&oacute;ria parab&oacute;lica. A uma dist&acirc;ncia d desse corpo celeste, propulsores s&atilde;o acionados de modo a, a partir daquele instante, mudar o m&oacute;dulo da velocidade do sat&eacute;lite de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v_p\" alt=\"\\large v_p\" align=\"absmiddle\">&nbsp; para <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v_e\" alt=\"\\large v_e\" align=\"absmiddle\"> e tamb&eacute;m a sua trajet&oacute;ria, que passa a ser el&iacute;ptica em torno do planeta, com semieixo maior a. Sendo a massa do sat&eacute;lite desproporcionalmente menor que a do planeta, a raz&atilde;o <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v_e\/v_p\" alt=\"\\large v_e\/v_p\" align=\"absmiddle\"> &eacute; dada por:<\/p><p>a) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Csqrt%7B%5Cfrac%7Bd%7D%7Ba%7D-%5Cfrac%7B1%7D%7B2%7D%7D\" alt=\"\\large \\sqrt{\\frac{d}{a}-\\frac{1}{2}}\" align=\"absmiddle\"><\/p><p>b) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Csqrt%7B%5Cfrac%7Bd%7D%7B2a%7D%7D\" alt=\"\\large \\sqrt{\\frac{d}{2a}}\" align=\"absmiddle\"><\/p><p>c) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Csqrt%7B1-%5Cfrac%7Bd%7D%7B2a%7D%7D\" alt=\"\\large \\sqrt{1-\\frac{d}{2a}}\" align=\"absmiddle\"><\/p><p>d) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Csqrt%7B1+%5Cfrac%7Bd%7D%7B2a%7D%7D\" alt=\"\\large \\sqrt{1+\\frac{d}{2a}}\" align=\"absmiddle\"><\/p><p>e) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Csqrt%7B1-%5Cfrac%7Bd%7D%7Ba%7D%7D\" alt=\"\\large \\sqrt{1-\\frac{d}{a}}\" align=\"absmiddle\"><\/p><h3 class=\"wp-block-heading\">Resolu&ccedil;&atilde;o Comentada<\/h3><p>Na &oacute;rbita parab&oacute;lica, o sat&eacute;lite tem energia mec&acirc;nica total igual a zero. Portanto, pela equa&ccedil;&atilde;o de conserva&ccedil;&atilde;o de energia mec&acirc;nica em &oacute;rbitas parab&oacute;licas, temos que:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7B-GMm%7D%7Bd%7D+%5Cfrac%7Bmv%5E2%7D%7B2%7D=0\" alt=\"\\large \\frac{-GMm}{d}+\\frac{mv^2}{2}=0\" align=\"absmiddle\"><\/p><p>Logo:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v_p=%5Csqrt%7B%5Cfrac%7B2GM%7D%7Bd%7D%7D\" alt=\"\\large v_p=\\sqrt{\\frac{2GM}{d}}\" align=\"absmiddle\"><\/p><p>Para a &oacute;rbita el&iacute;ptica, temos:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7B-GMm%7D%7Bd%7D+%5Cfrac%7Bmv%5E2%7D%7B2%7D=%5Cfrac%7B-GMm%7D%7B2a%7D\" alt=\"\\large \\frac{-GMm}{d}+\\frac{mv^2}{2}=\\frac{-GMm}{2a}\" align=\"absmiddle\"><\/p><p>Assim:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v_e=%5Csqrt%7B2GM%7D%5Csqrt%7B%5Cfrac%7B1%7D%7Bd%7D-%5Cfrac%7B1%7D%7B2a%7D%7D\" alt=\"\\large v_e=\\sqrt{2GM}\\sqrt{\\frac{1}{d}-\\frac{1}{2a}}\" align=\"absmiddle\"><\/p><p>Portanto, a raz&atilde;o entre as velocidades &eacute; dada por:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7Bv_e%7D%7Bv_p%7D=%5Csqrt%7B1-%5Cfrac%7Bd%7D%7B2a%7D%7D\" alt=\"\\large \\frac{v_e}{v_p}=\\sqrt{1-\\frac{d}{2a}}\" align=\"absmiddle\"><\/p><p class=\"has-light-green-cyan-background-color has-background\"><strong>Gabarito: C<\/strong><\/p><h3 class=\"wp-block-heading\">Quest&atilde;o 06<\/h3><p>Uma pequena esfera com peso de m&oacute;dulo P &eacute; arremessada verticalmente para cima com velocidade de m&oacute;dulo <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V_0\" alt=\"\\large V_0\" align=\"absmiddle\"> a partir do solo. Durante todo o percurso, atua sobre a esfera uma for&ccedil;a de resist&ecirc;ncia do ar de m&oacute;dulo F constante. A dist&acirc;ncia total percorrida pela esfera ap&oacute;s muitas reflex&otilde;es el&aacute;sticas com o solo &eacute; dada aproximadamente por:<\/p><p>a) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7BV_0%5E2%5Cleft(P-F%5Cright)%7D%7B2gF%7D\" alt=\"\\large \\frac{V_0^2\\left(P-F\\right)}{2gF}\" align=\"absmiddle\"><\/p><p>b) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7BV_0%5E2%5Cleft(P+F%5Cright)%7D%7B2gF%7D\" alt=\"\\large \\frac{V_0^2\\left(P+F\\right)}{2gF}\" align=\"absmiddle\"><\/p><p>c) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7B2V_0%5E2P%7D%7BgF%7D\" alt=\"\\large \\frac{2V_0^2P}{gF}\" align=\"absmiddle\"><\/p><p>d) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7BV_0%5E2P%7D%7B2gF%7D\" alt=\"\\large \\frac{V_0^2P}{2gF}\" align=\"absmiddle\"><\/p><p>e) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7BV_0%5E2P%7D%7BgF%7D\" alt=\"\\large \\frac{V_0^2P}{gF}\" align=\"absmiddle\"><\/p><h3 class=\"wp-block-heading\">Resolu&ccedil;&atilde;o Comentada<\/h3><p>Pelo teorema da energia cin&eacute;tica:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5CDelta&amp;space;E_c=%5Ctau_%7BFNC%7D\" alt=\"\\large \\Delta E_c=\\tau_{FNC}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;-%5Cfrac%7Bm%5Ccdot&amp;space;v_0%5E2%7D%7B2%7D=-F%5Ccdot&amp;space;d\" alt=\"\\large -\\frac{m\\cdot v_0^2}{2}=-F\\cdot d\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;d=%5Cfrac%7B%5Cleft(%5Cfrac%7BP%7D%7Bg%7D%5Cright)%5Ccdot&amp;space;v_0%5E2%7D%7B2%5Ccdot&amp;space;F%7D\" alt=\"\\large d=\\frac{\\left(\\frac{P}{g}\\right)\\cdot v_0^2}{2\\cdot F}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;d=%5Cfrac%7BP%5Ccdot&amp;space;v_0%5E2%7D%7B2%5Ccdot&amp;space;g%5Ccdot&amp;space;F%7D\" alt=\"\\large d=\\frac{P\\cdot v_0^2}{2\\cdot g\\cdot F}\" align=\"absmiddle\"><\/p><p class=\"has-light-green-cyan-background-color has-background\"><strong>Gabarito: D<\/strong><\/p><h3 class=\"wp-block-heading\">Quest&atilde;o 07<\/h3><p>A figura ilustra um experimento numa plataforma que, no referencial de um observador externo, se move com velocidade<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cvec%7Bv%7D\" alt=\"\\large \\vec{v}\" align=\"absmiddle\"> constante de m&oacute;dulo compar&aacute;vel ao da velocidade da luz. No instante to, a fonte F emite um pulso de luz de comprimento de onda <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Clambda\" alt=\"\\large \\lambda\" align=\"absmiddle\"> que incide sobre a placa met&aacute;lica A, sendo por ela absorvido e, em consequ&ecirc;ncia, emitindo el&eacute;trons, que s&atilde;o desacelerados pela diferen&ccedil;a de potencial <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V_%7BAB%7D\" alt=\"\\large V_{AB}\" align=\"absmiddle\">. Considerando que os el&eacute;trons atingem a placa ? a partir do instante ?, assinale a alternativa que referencia apenas varia&ccedil;&otilde;es independentes que diminuem o intervalo de tempo <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5CDelta&amp;space;t=t-t_0\" alt=\"\\large \\Delta t=t-t_0\" align=\"absmiddle\"> medido pelo observador.<\/p><p>a) aumento de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Clambda\" alt=\"\\large \\lambda\" align=\"absmiddle\">, aumento de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V_%7BAB%7D\" alt=\"\\large V_{AB}\" align=\"absmiddle\">, diminui&ccedil;&atilde;o de v.<\/p><p>b) diminui&ccedil;&atilde;o de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Clambda\" alt=\"\\large \\lambda\" align=\"absmiddle\">, diminui&ccedil;&atilde;o de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V_%7BAB%7D\" alt=\"\\large V_{AB}\" align=\"absmiddle\">, diminui&ccedil;&atilde;o de v.<\/p><p>c) diminui&ccedil;&atilde;o de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Clambda\" alt=\"\\large \\lambda\" align=\"absmiddle\">, aumento de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V_%7BAB%7D\" alt=\"\\large V_{AB}\" align=\"absmiddle\">, diminui&ccedil;&atilde;o de v.<\/p><p>d) diminui&ccedil;&atilde;o de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Clambda\" alt=\"\\large \\lambda\" align=\"absmiddle\">, diminui&ccedil;&atilde;o de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V_%7BAB%7D\" alt=\"\\large V_{AB}\" align=\"absmiddle\">, aumento de v.<\/p><p>e) aumento de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Clambda\" alt=\"\\large \\lambda\" align=\"absmiddle\">, aumento de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V_%7BAB%7D\" alt=\"\\large V_{AB}\" align=\"absmiddle\">, aumento de v.<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2019\/12\/image15.png\" alt=\"\" class=\"wp-image-35014\" width=\"331\" height=\"213\"><\/figure><\/div><h3 class=\"wp-block-heading\">Resolu&ccedil;&atilde;o Comentada<\/h3><p>O tempo medido pelo observador &eacute; dado por:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5CDelta&amp;space;t_%7Bobservador%7D=%5Cfrac%7B%5CDelta&amp;space;t_%7Bplataforma%7D%7D%7B%5Csqrt%7B1-%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%7D%7D\" alt=\"\\large \\Delta t_{observador}=\\frac{\\Delta t_{plataforma}}{\\sqrt{1-\\frac{v^2}{c^2}}}\" align=\"absmiddle\"><\/p><p>Como deseja-se diminuir o tempo medido\npelo observador tem-se duas possibilidades. Pode-se reduzir o tempo medido no\nreferencial da plataforma ou pode diminuir-se o fator de Lorentz. Dessa forma:<\/p><p>&ndash; Para reduzir o tempo medido na\nplataforma:<\/p><p>1) Pode-se aumentar a velocidade de eje&ccedil;&atilde;o do el&eacute;tron. Para isso &eacute; necess&aacute;rio que o el&eacute;tron seja ejetado com uma energia cin&eacute;tica maior. Sabendo que a energia cin&eacute;tica do el&eacute;tron ejetado &eacute; dada por:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;E_c=h%5Ccdot&amp;space;f-%5Cphi\" alt=\"\\large E_c=h\\cdot f-\\phi\" align=\"absmiddle\"><\/p><p>&Eacute; necess&aacute;rio aumentar-se a frequ&ecirc;ncia do f&oacute;ton incidente, e, sabendo que:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;f=%5Cfrac%7Bc%7D%7B%5Clambda%7D\" alt=\"\\large f=\\frac{c}{\\lambda}\" align=\"absmiddle\"><\/p><p>Dessa forma, a diminui&ccedil;&atilde;o do <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Clambda\" alt=\"\\large \\lambda\" align=\"absmiddle\"> ir&aacute; aumentar a velocidade de eje&ccedil;&atilde;o do el&eacute;tron que por sua vez diminui o tempo medido na plataforma e, assim, diminui o tempo medido pelo observador.<\/p><p>2) Pode-se diminuir as for&ccedil;as contr&aacute;rias ao movimento do el&eacute;tron. Neste caso, somente a for&ccedil;a el&eacute;trica por conta do campo el&eacute;trico entre as placas &eacute; contr&aacute;ria ao movimento.<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;q%5Ccdot&amp;space;E=m%5Ccdot&amp;space;a\" alt=\"\\large q\\cdot E=m\\cdot a\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;q%5Ccdot%5Cfrac%7BV_%7BAB%7D%7D%7Bd%7D=m%5Ccdot&amp;space;a\" alt=\"\\large q\\cdot\\frac{V_{AB}}{d}=m\\cdot a\" align=\"absmiddle\"><\/p><p>Assim, a diminui&ccedil;&atilde;o da diferen&ccedil;a de potencial entre as placas acarreta uma diminui&ccedil;&atilde;o da frenagem do el&eacute;tron, permitindo assim, que se atinja a placa em menos tempo.<\/p><p>E, para reduzir o fator de Lorentz:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cgamma=%5Cfrac%7B1%7D%7B%5Csqrt%7B1-%5Cleft(%5Cfrac%7Bv%7D%7Bc%7D%5Cright)%5E2%7D%7D\" alt=\"\\large \\gamma=\\frac{1}{\\sqrt{1-\\left(\\frac{v}{c}\\right)^2}}\" align=\"absmiddle\"><\/p><p>Implica um aumento do denominador. Para isso:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Csqrt%7B1-%5Cleft(%5Cfrac%7Bv_1%7D%7Bc%7D%5Cright)%5E2%7D&lt;%5Csqrt%7B1-%5Cleft(%5Cfrac%7Bv_2%7D%7Bc%7D%5Cright)%5E2%7D\" alt=\"\\large \\sqrt{1-\\left(\\frac{v_1}{c}\\right)^2}&lt;\\sqrt{1-\\left(\\frac{v_2}{c}\\right)^2}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;1-%5Cleft(%5Cfrac%7Bv_1%7D%7Bc%7D%5Cright)%5E2&lt;1-%5Cleft(%5Cfrac%7Bv_2%7D%7Bc%7D%5Cright)%5E2\" alt=\"\\large 1-\\left(\\frac{v_1}{c}\\right)^2&lt;1-\\left(\\frac{v_2}{c}\\right)^2\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cleft(%5Cfrac%7Bv_2%7D%7Bc%7D%5Cright)%5E2&lt;%5Cleft(%5Cfrac%7Bv_1%7D%7Bc%7D%5Cright)%5E2\" alt=\"\\large \\left(\\frac{v_2}{c}\\right)^2&lt;\\left(\\frac{v_1}{c}\\right)^2\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v_2&lt;v_1\" alt=\"\\large v_2&lt;v_1\" align=\"absmiddle\"><\/p><p>Assim, diminuindo-se a velocidade, aumenta-se o denominador, e, por conseguinte, diminui-se o tempo medido pelo operador.<\/p><p class=\"has-light-green-cyan-background-color has-background\"><strong>Gabarito: B<\/strong><\/p><h3 class=\"wp-block-heading\">Quest&atilde;o 08<\/h3><p>Num ambiente controlado, o per&iacute;odo de um p&ecirc;ndulo simples &eacute; medido a uma temperatura T. Sendo <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Calpha=2%5Ccdot%7B10%7D%5E%7B-4%7D%5C&amp;space;%C2%B0C-1\" alt=\"\\large \\alpha=2\\cdot{10}^{-4}\\ &deg;C-1\" align=\"absmiddle\"> o coeficiente de dilata&ccedil;&atilde;o linear do fio do p&ecirc;ndulo, e considerando a aproxima&ccedil;&atilde;o binomial <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cleft(1%5C&amp;space;+%5C&amp;space;x%5Cright)%5En%5Capprox1+nx\" alt=\"\\large \\left(1\\ +\\ x\\right)^n\\approx1+nx\" align=\"absmiddle\">, para <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cleft%7Cx%5Cright%7C%5Cll1\" alt=\"\\large \\left|x\\right|\\ll1\" align=\"absmiddle\"> , pode-se dizer que, com aumento de 10 &deg;?, o per&iacute;odo do p&ecirc;ndulo:<\/p><p>a) aumenta de 0,1%.<\/p><p>b) aumenta de 0,05%.&nbsp;&nbsp;&nbsp; <\/p><p>c) diminui de 0,1%. <\/p><p>d) diminui de 0,05%.<\/p><p>e) permanece inalterado. <\/p><h3 class=\"wp-block-heading\">Resolu&ccedil;&atilde;o Comentada<\/h3><p>O per&iacute;odo de oscila&ccedil;&atilde;o de um p&ecirc;ndulo simples &eacute; dado por:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;T=2%5Ccdot%5Cpi%5Ccdot%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%7D\" alt=\"\\large T=2\\cdot\\pi\\cdot\\sqrt{\\frac{l}{g}}\" align=\"absmiddle\"><\/p><p>Portanto, na situa&ccedil;&atilde;o inicial:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;T=2%5Ccdot%5Cpi%5Ccdot%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%7D\" alt=\"\\large T=2\\cdot\\pi\\cdot\\sqrt{\\frac{l}{g}}\" align=\"absmiddle\"><\/p><p>E, na situa&ccedil;&atilde;o final:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;T%5Cprime=2%5Ccdot%5Cpi%5Ccdot%5Csqrt%7B%5Cfrac%7Bl%5E%5Cprime%7D%7Bg%7D%7D\" alt=\"\\large T\\prime=2\\cdot\\pi\\cdot\\sqrt{\\frac{l^\\prime}{g}}\" align=\"absmiddle\"><\/p><p>Onde:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;l%5E%5Cprime=l%5Ccdot%5Cleft(1+%5Calpha%5Ccdot%5CDelta%5Ctheta%5Cright)\" alt=\"\\large l^\\prime=l\\cdot\\left(1+\\alpha\\cdot\\Delta\\theta\\right)\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;l%5E%5Cprime=l%5Ccdot%5Cleft(1+2%5Ccdot%7B10%7D%5E%7B-4%7D%5Ccdot10%5Cright)=l%5Ccdot%5Cleft(1+2%5Ccdot%7B10%7D%5E%7B-3%7D%5Cright)\" alt=\"\\large l^\\prime=l\\cdot\\left(1+2\\cdot{10}^{-4}\\cdot10\\right)=l\\cdot\\left(1+2\\cdot{10}^{-3}\\right)\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;T=2%5Ccdot%5Cpi%5Ccdot%5Csqrt%7B%5Cfrac%7Bl%5Ccdot%5Cleft(1+2%5Ccdot%7B10%7D%5E%7B-3%7D%5Cright)%7D%7Bg%7D%7D\" alt=\"\\large T=2\\cdot\\pi\\cdot\\sqrt{\\frac{l\\cdot\\left(1+2\\cdot{10}^{-3}\\right)}{g}}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;T%5E%5Cprime=2%5Ccdot%5Cpi%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%7D%5Ccdot%5Cleft(1+2%5Ccdot%7B10%7D%5E%7B-3%7D%5Cright)%5E%5Cfrac%7B1%7D%7B2%7D%5Ccong\" alt=\"\\large T^\\prime=2\\cdot\\pi\\sqrt{\\frac{l}{g}}\\cdot\\left(1+2\\cdot{10}^{-3}\\right)^\\frac{1}{2}\\cong\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;2%5Ccdot%5Cpi%5Ccdot%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%7D%5Ccdot%5Cleft(1+1%5Ccdot%7B10%7D%5E%7B-3%7D%5Cright)=\" alt=\"\\large 2\\cdot\\pi\\cdot\\sqrt{\\frac{l}{g}}\\cdot\\left(1+1\\cdot{10}^{-3}\\right)=\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;T%5Ccdot%5Cleft(1+1%5Ccdot%7B10%7D%5E%7B-3%7D%5Cright)\" alt=\"\\large T\\cdot\\left(1+1\\cdot{10}^{-3}\\right)\" align=\"absmiddle\"><\/p><p>Dessa forma, o aumento foi de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;1%5Ccdot%7B10%7D%5E%7B-3%7D\" alt=\"\\large 1\\cdot{10}^{-3}\" align=\"absmiddle\"> do per&iacute;odo original, ou, de outra forma, 0,1%.<\/p><p class=\"has-light-green-cyan-background-color has-background\"><strong>Gabarito: A<\/strong><\/p><h3 class=\"wp-block-heading\">Quest&atilde;o 09<\/h3><p>Uma certa quantidade de g&aacute;s com temperatura inicial <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;T_0\" alt=\"\\large T_0\" align=\"absmiddle\">, press&atilde;o <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;P_0\" alt=\"\\large P_0\" align=\"absmiddle\"> e volume <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V_0\" alt=\"\\large V_0\" align=\"absmiddle\"> , &eacute; aquecida por uma corrente el&eacute;trica que flui por um fio de platina num intervalo de tempo <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5CDelta&amp;space;t\" alt=\"\\large \\Delta t\" align=\"absmiddle\">. Esse procedimento &eacute; feito duas vezes: primeiro, co volume constante <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V_0\" alt=\"\\large V_0\" align=\"absmiddle\"> e press&atilde;o variando de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;P_0\" alt=\"\\large P_0\" align=\"absmiddle\"> para <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?%5Clarge&amp;space;P_1\" alt=\"\\large P_1\" align=\"absmiddle\"> e, a seguir, com press&atilde;o constante <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;P_0\" alt=\"\\large P_0\" align=\"absmiddle\"> e com volume variando de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V_0\" alt=\"\\large V_0\" align=\"absmiddle\"> para <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V_1\" alt=\"\\large V_1\" align=\"absmiddle\">. Assinale a alternativa que explicita a rela&ccedil;&atilde;o <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;C_P\/C_V\" alt=\"\\large C_P\/C_V\" align=\"absmiddle\"> do g&aacute;s.<\/p><p>a) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7B%5Cfrac%7BP_0%7D%7BP_1%7D-1%7D%7B%5Cfrac%7BV_0%7D%7BV_1%7D-1%7D\" alt=\"\\large \\frac{\\frac{P_0}{P_1}-1}{\\frac{V_0}{V_1}-1}\" align=\"absmiddle\"><\/p><p>b) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7B%5Cfrac%7BP_1%7D%7BP_0%7D-1%7D%7B%5Cfrac%7BV_1%7D%7BV_0%7D-1%7D\" alt=\"\\large \\frac{\\frac{P_1}{P_0}-1}{\\frac{V_1}{V_0}-1}\" align=\"absmiddle\"><\/p><p>c) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7B2%5Cfrac%7BP_0%7D%7BP_1%7D-1%7D%7B%5Cfrac%7BV_0%7D%7BV_1%7D-1%7D\" alt=\"\\large \\frac{2\\frac{P_0}{P_1}-1}{\\frac{V_0}{V_1}-1}\" align=\"absmiddle\"><\/p><p>d) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7B2%5Cfrac%7BP_1%7D%7BP_0%7D-1%7D%7B%5Cfrac%7BV_1%7D%7BV_0%7D-1%7D\" alt=\"\\large \\frac{2\\frac{P_1}{P_0}-1}{\\frac{V_1}{V_0}-1}\" align=\"absmiddle\"><\/p><p>e) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7B%5Cfrac%7BP_1%7D%7BP_0%7D-1%7D%7B%5Cfrac%7B2V_1%7D%7BV_0%7D-1%7D\" alt=\"\\large \\frac{\\frac{P_1}{P_0}-1}{\\frac{2V_1}{V_0}-1}\" align=\"absmiddle\"><\/p><h3 class=\"wp-block-heading\">Resolu&ccedil;&atilde;o Comentada<\/h3><p>calor cedido pelo fio de platina em ambos os casos &eacute; dado por:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;Q=R_%7Bfio%7D%5Ccdot&amp;space;i%5E2%5Ccdot&amp;space;t\" alt=\"\\large Q=R_{fio}\\cdot i^2\\cdot t\" align=\"absmiddle\"><\/p><p>Assim, o calor recebido pelo g&aacute;s em ambos os casos foi id&ecirc;ntico, logo:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;n%5Ccdot&amp;space;c_v%5Ccdot%5CDelta&amp;space;T_1=n%5Ccdot&amp;space;c_p%5Ccdot%5CDelta&amp;space;T_2\" alt=\"\\large n\\cdot c_v\\cdot\\Delta T_1=n\\cdot c_p\\cdot\\Delta T_2\" align=\"absmiddle\"><\/p><p>Pela transforma&ccedil;&atilde;o gasosa da isoc&oacute;rica:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7BP_0%7D%7BT_0%7D=%5Cfrac%7BP_1%7D%7BT_1%7D%5Crightarrow&amp;space;T_1=T_0%5Ccdot%5Cfrac%7BP_1%7D%7BP_0%7D\" alt=\"\\large \\frac{P_0}{T_0}=\\frac{P_1}{T_1}\\rightarrow T_1=T_0\\cdot\\frac{P_1}{P_0}\" align=\"absmiddle\"><\/p><p>Pela transforma&ccedil;&atilde;o gasosa da isob&aacute;rica:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7BV_0%7D%7BT_0%7D=%5Cfrac%7BV_1%7D%7BT_2%7D%5Crightarrow&amp;space;T_2=T_0%5Ccdot%5Cfrac%7BV_1%7D%7BV_0%7D\" alt=\"\\large \\frac{V_0}{T_0}=\\frac{V_1}{T_2}\\rightarrow T_2=T_0\\cdot\\frac{V_1}{V_0}\" align=\"absmiddle\"><\/p><p>Assim:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;n%5Ccdot&amp;space;c_v%5Ccdot%5Cleft(T_1-T_0%5Cright)=n%5Ccdot&amp;space;c_p%5Ccdot%5Cleft(T_2-T_0%5Cright)\" alt=\"\\large n\\cdot c_v\\cdot\\left(T_1-T_0\\right)=n\\cdot c_p\\cdot\\left(T_2-T_0\\right)\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;c_v%5Ccdot&amp;space;T_0%5Ccdot%5Cleft(%5Cfrac%7BP_1%7D%7BP_0%7D-1%5Cright)=c_p%5Ccdot&amp;space;T_0%5Ccdot%5Cleft(%5Cfrac%7BV_1%7D%7BV_0%7D-1%5Cright)\" alt=\"\\large c_v\\cdot T_0\\cdot\\left(\\frac{P_1}{P_0}-1\\right)=c_p\\cdot T_0\\cdot\\left(\\frac{V_1}{V_0}-1\\right)\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7Bc_p%7D%7Bc_v%7D=%5Cfrac%7B%5Cfrac%7BP_1%7D%7BP_0%7D-1%7D%7B%5Cfrac%7BV_1%7D%7BV_0%7D-1%7D\" alt=\"\\large \\frac{c_p}{c_v}=\\frac{\\frac{P_1}{P_0}-1}{\\frac{V_1}{V_0}-1}\" align=\"absmiddle\"><\/p><p class=\"has-light-green-cyan-background-color has-background\"><strong>Gabarito: B<\/strong><\/p><h3 class=\"wp-block-heading\">Quest&atilde;o 10<\/h3><div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2019\/12\/image16.png\" alt=\"\" class=\"wp-image-35027\" width=\"370\" height=\"250\"><\/figure><\/div><p>Ao redor de um cilindro de massa m, raio a e comprimento b, s&atilde;o enroladas sim&eacute;trica e longitudinalmente N espiras. Estas s&atilde;o dispostas paralelamente a um plano inclinado onde se encontra um cilindro, que n&atilde;o desliza devido ao atrito com a superf&iacute;cie do plano. Considerando a exist&ecirc;ncia de um campo magn&eacute;tico uniforme e vertical <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cvec%7BB%7D\" alt=\"\\large \\vec{B}\" align=\"absmiddle\"> na regi&atilde;o, assinale a intensidade da corrente i que deve circular nas espiras para que o conjunto permane&ccedil;a em repouso na posi&ccedil;&atilde;o indicada pela figura.<\/p><p>a) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7Bmg%7D%7B2bB%7D\" alt=\"\\large \\frac{mg}{2bB}\" align=\"absmiddle\"><\/p><p>b) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7BNmg%7D%7B2aB%7D\" alt=\"\\large \\frac{Nmg}{2aB}\" align=\"absmiddle\"><\/p><p>c) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7BNmg%7D%7BbB%7D\" alt=\"\\large \\frac{Nmg}{bB}\" align=\"absmiddle\"><\/p><p>d) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7Bmg%7D%7B2aBN%7D\" alt=\"\\large \\frac{mg}{2aBN}\" align=\"absmiddle\"><\/p><p>e) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7Bmg%7D%7B2bBN%7D\" alt=\"\\large \\frac{mg}{2bBN}\" align=\"absmiddle\"><\/p><h3 class=\"wp-block-heading\">Resolu&ccedil;&atilde;o Comentada<\/h3><p>Para o equil&iacute;brio de momentos do cilindro, temos que:<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2019\/12\/image17.png\" alt=\"\" class=\"wp-image-35029\"><\/figure><\/div><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Ctau_%7BF_%7Bmag%7D%7D=%5Ctau_%7BPeso%7D\" alt=\"\\large \\tau_{F_{mag}}=\\tau_{Peso}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Ctau_%7Bpeso%7D=magsen(%5Ctheta)\" alt=\"\\large \\tau_{peso}=magsen(\\theta)\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Ctau_%7BF_%7Bmag%7D%7D=2NBiabsen(%5Ctheta)\" alt=\"\\large \\tau_{F_{mag}}=2NBiabsen(\\theta)\" align=\"absmiddle\"><\/p><p>Igualando as duas equa&ccedil;&otilde;es, temos que:&nbsp;<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Ctau_%7Bpeso%7D=%5Ctau_%7BF_%7Bmag%7D%7D\" alt=\"\\large \\tau_{peso}=\\tau_{F_{mag}}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;m%5Ccdot&amp;space;g%5Ccdot&amp;space;a%5Ccdot&amp;space;sen%5Cleft(%5Ctheta%5Cright)=2%5Ccdot&amp;space;N%5Ccdot&amp;space;B%5Ccdot&amp;space;i%5Ccdot&amp;space;a%5Ccdot&amp;space;b%5Ccdot&amp;space;sen(%5Ctheta)\" alt=\"\\large m\\cdot g\\cdot a\\cdot sen\\left(\\theta\\right)=2\\cdot N\\cdot B\\cdot i\\cdot a\\cdot b\\cdot sen(\\theta)\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;i=%5Cfrac%7Bm%5Ccdot&amp;space;g%7D%7B2%5Ccdot&amp;space;N%5Ccdot&amp;space;B%5Ccdot&amp;space;b%7D\" alt=\"\\large i=\\frac{m\\cdot g}{2\\cdot N\\cdot B\\cdot b}\" align=\"absmiddle\"><\/p><p class=\"has-light-green-cyan-background-color has-background\"><strong>Gabarito: E<\/strong><\/p><h3 class=\"wp-block-heading\">Quest&atilde;o 11<\/h3><p>O som produzido pelo alto-falante F (fonte) ilustrado na figura tem frequ&ecirc;ncia de 10 kHz e chega a um microfone M atrav&eacute;s de dois caminhos diferentes. As ondas sonoras viajam simultaneamente pelo tubo esquerdo FXM, de comprimento fixo, e pelo tubo direito FYM, cujo comprimento pode ser alterado movendo-se em M. Quando a se&ccedil;&atilde;o deslizante do caminho FYM &eacute; puxada para fora por 0,025 m, a intensidade sonora detectada pelo microfone passa de um m&aacute;ximo para um m&iacute;nimo. Assinale o m&oacute;dulo da velocidade do som no interior do tubo.<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2019\/12\/image18.png\" alt=\"\" class=\"wp-image-35032\" width=\"365\" height=\"198\"><\/figure><\/div><p>a) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;5,0%5Ccdot%7B10%7D%5E2%5C&amp;space;m\/s\" alt=\"\\large 5,0\\cdot{10}^2\\ m\/s\" align=\"absmiddle\"><\/p><p>b) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;2,5%5Ccdot%7B10%7D%5E2%5C&amp;space;m\/s\" alt=\"\\large 2,5\\cdot{10}^2\\ m\/s\" align=\"absmiddle\"><\/p><p>c) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;1,0%5Ccdot%7B10%7D%5E3%5C&amp;space;m\/s\" alt=\"\\large 1,0\\cdot{10}^3\\ m\/s\" align=\"absmiddle\"><\/p><p>d) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;2,0%5Ccdot%7B10%7D%5E3%5C&amp;space;m\/s\" alt=\"\\large 2,0\\cdot{10}^3\\ m\/s\" align=\"absmiddle\"><\/p><p>e) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;3,4%5Ccdot%7B10%7D%5E2%5C&amp;space;m\/s\" alt=\"\\large 3,4\\cdot{10}^2\\ m\/s\" align=\"absmiddle\"><\/p><h3 class=\"wp-block-heading\">Resolu&ccedil;&atilde;o Comentada<\/h3><p>O m&iacute;nimo de intensidade &eacute; sentido gra&ccedil;as ao deslocamento da\nsec&ccedil;&atilde;o deslizante FYM. Como, para que ocorra diferen&ccedil;a de fase devemos ter:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;n_%7Bimpar%7D%5Cfrac%7B%5Clambda%7D%7B2%7D=2d\" alt=\"\\large n_{impar}\\frac{\\lambda}{2}=2d\" align=\"absmiddle\"><\/p><p>Em que 2d &eacute; dist&acirc;ncia que o som percorre a mais pelo lado FYM com rela&ccedil;&atilde;o ao lado FXM. Assim, substituindo <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v=%5Clambda&amp;space;f\" alt=\"\\large v=\\lambda f\" align=\"absmiddle\"> e <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;f=%7B10%7D%5E4Hz\" alt=\"\\large f={10}^4Hz\" align=\"absmiddle\">, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;d=0,025m\" alt=\"\\large d=0,025m\" align=\"absmiddle\">, obtemos que:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;n_%7Bimpar%7D%5Cfrac%7Bv%7D%7B2f%7D=2d\" alt=\"\\large n_{impar}\\frac{v}{2f}=2d\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;n_%7Bimpar%7D%5Cfrac%7Bv%7D%7B2%5Ccdot%7B10%7D%5E4%7D=0,05\" alt=\"\\large n_{impar}\\frac{v}{2\\cdot{10}^4}=0,05\" align=\"absmiddle\"><\/p><p>&nbsp;Assim:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v=%5Cfrac%7B%7B10%7D%5E3%7D%7Bn_%7Bimpar%7D%7D\" alt=\"\\large v=\\frac{{10}^3}{n_{impar}}\" align=\"absmiddle\"><\/p><p>De acordo com as alternativas, a &uacute;nica que pode ser satisfeita &eacute; a C.<\/p><p class=\"has-light-green-cyan-background-color has-background\"><strong> Gabarito: C <\/strong><\/p><h3 class=\"wp-block-heading\">Quest&atilde;o 12<\/h3><p>Considere o circuito da figura no qual h&aacute; uma chave el&eacute;trica, um reostato linear de comprimento total de 20 cm, uma fonte de tens&atilde;o <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V=1,5%5C&amp;space;V\" alt=\"\\large V=1,5\\ V\" align=\"absmiddle\"> e um capacitor de capacit&acirc;ncia <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;C=10%5C&amp;space;%5Cmu&amp;space;F\" alt=\"\\large C=10\\ \\mu F\" align=\"absmiddle\"> conectado a um ponto intermedi&aacute;rio do reostato, de modo a manter contato el&eacute;trico e permitir seu carregamento. A resist&ecirc;ncia R entre uma das extremidades do reostato e o ponto de contato el&eacute;trico, a uma dist&acirc;ncia as, varia segundo o gr&aacute;fico ao lado.<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2019\/12\/image19.png\" alt=\"\" class=\"wp-image-35037\"><\/figure><\/div><p>Com a chave fechada e no regime estacion&aacute;rio, a carga no capacitor &eacute; igual a<\/p><p>a) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;1,5%5C&amp;space;mC\" alt=\"\\large 1,5\\ mC\" align=\"absmiddle\"><\/p><p>b) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;75%5C&amp;space;%5Cmu&amp;space;C\" alt=\"\\large 75\\ \\mu C\" align=\"absmiddle\"><\/p><p>c) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;75x%5Cmu&amp;space;C\/cm\" alt=\"\\large 75x\\mu C\/cm\" align=\"absmiddle\"><\/p><p>d) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;15x%5Cmu&amp;space;C\/cm\" alt=\"\\large 15x\\mu C\/cm\" align=\"absmiddle\"><\/p><p>e) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;7,5%5C&amp;space;%5Cmu&amp;space;C\" alt=\"\\large 7,5\\ \\mu C\" align=\"absmiddle\"><\/p><h3 class=\"wp-block-heading\">Resolu&ccedil;&atilde;o Comentada<\/h3><p>O circuito da figura para uma situa&ccedil;&atilde;o gen&eacute;rica fica:<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2019\/12\/image20.png\" alt=\"\" class=\"wp-image-35039\" width=\"364\" height=\"164\"><\/figure><\/div><p>Onde:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V=1,5%5C&amp;space;V\" alt=\"\\large V=1,5\\ V\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;R=60%5C&amp;space;%5COmega\" alt=\"\\large R=60\\ \\Omega\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;R_x=3%5Ccdot&amp;space;x%5C&amp;space;(cm)%5Ccdot%5COmega\" alt=\"\\large R_x=3\\cdot x\\ (cm)\\cdot\\Omega\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;C=10%5C&amp;space;%5Cmu&amp;space;F\" alt=\"\\large C=10\\ \\mu F\" align=\"absmiddle\"><\/p><p>A corrente que percorre o circuito no regime estacion&aacute;rio &eacute; de:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;i=%5Cfrac%7BV%7D%7BR_x+R-R_x%7D=%5Cfrac%7BV%7D%7BR%7D=%5Cfrac%7B1,5%7D%7B60%7D=%5Cfrac%7B1%7D%7B40%7D\" alt=\"\\large i=\\frac{V}{R_x+R-R_x}=\\frac{V}{R}=\\frac{1,5}{60}=\\frac{1}{40}\" align=\"absmiddle\"><\/p><p>A diferen&ccedil;a de potencial entre as extremidades da resist&ecirc;ncia <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;R_x\" alt=\"\\large R_x\" align=\"absmiddle\"> fica:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V_x=R_x%5Ccdot&amp;space;i=3%5Ccdot&amp;space;x%5C&amp;space;(cm)%5Ccdot%5Cfrac%7B1%7D%7B40%7D=%5Cfrac%7B3%5Ccdot&amp;space;x%5C&amp;space;(cm)%7D%7B40%7D\" alt=\"\\large V_x=R_x\\cdot i=3\\cdot x\\ (cm)\\cdot\\frac{1}{40}=\\frac{3\\cdot x\\ (cm)}{40}\" align=\"absmiddle\"><\/p><p>E, a carga do capacitor &eacute; dada por:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;Q=C%5Ccdot&amp;space;U\" alt=\"\\large Q=C\\cdot U\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;Q=10%5C&amp;space;%5Cmu&amp;space;F%5Ccdot%5Cfrac%7B3%5Ccdot&amp;space;x%7D%7B40%7D=%5Cfrac%7B3%7D%7B4%7D%5Ccdot&amp;space;x%5C&amp;space;(cm)%5C&amp;space;%5Cmu&amp;space;C\" alt=\"\\large Q=10\\ \\mu F\\cdot\\frac{3\\cdot x}{40}=\\frac{3}{4}\\cdot x\\ (cm)\\ \\mu C\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfbox%7B%24Q=0,75%5Ccdot&amp;space;x%5C&amp;space;(cm)%5C&amp;space;%5Cmu&amp;space;C%24%7D\" alt=\"\\large \\fbox{$Q=0,75\\cdot x\\ (cm)\\ \\mu C$}\" align=\"absmiddle\"><\/p><p class=\"has-luminous-vivid-amber-background-color has-background\"><strong>Gabarito: Sem Alternativa<\/strong><\/p><h3 class=\"wp-block-heading\">Quest&atilde;o 13<\/h3><p>Tr&ecirc;s esferas id&ecirc;nticas de massa m, carga el&eacute;trica Q e dimens&otilde;es desprez&iacute;veis, s&atilde;o presas pelas extremidades de fios isolantes e inextens&iacute;veis de comprimento l. As demais pontas dos fios s&atilde;o fixadas a um ponto P, que sustenta as massas. Na condi&ccedil;&atilde;o de equil&iacute;brio do sistema, verifica-se que o &acirc;ngulo entre um dos fios e a dire&ccedil;&atilde;o vertical &eacute; &theta;, conforme mostra a figura. Sendo <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cvarepsilon_0\" alt=\"\\large \\varepsilon_0\" align=\"absmiddle\"> permissividade el&eacute;trica do meio, o valor da carga el&eacute;trica Q &eacute; dada por<\/p><p>a) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;l%5C&amp;space;%5Csqrt%7B12%5Cpi%5Cvarepsilon_0mg%5C&amp;space;sen%5Ctheta%5C&amp;space;cos%5Ctheta%7D\" alt=\"\\large l\\ \\sqrt{12\\pi\\varepsilon_0mg\\ sen\\theta\\ cos\\theta}\" align=\"absmiddle\"><\/p><p>b) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;l%5C&amp;space;%5Csqrt%7B4%5Cpi%5Cvarepsilon_0mg%5C&amp;space;tg%5Ctheta%5C&amp;space;%5Csqrt3%7D\" alt=\"\\large l\\ \\sqrt{4\\pi\\varepsilon_0mg\\ tg\\theta\\ \\sqrt3}\" align=\"absmiddle\"><\/p><p>c) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;l%5C&amp;space;sen%5Ctheta%5Csqrt%7B4%5Cpi%5Cvarepsilon_0mg%5C&amp;space;tg%5Ctheta%5C&amp;space;%5Csqrt3%7D\" alt=\"\\large l\\ sen\\theta\\sqrt{4\\pi\\varepsilon_0mg\\ tg\\theta\\ \\sqrt3}\" align=\"absmiddle\"><\/p><p>d) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;l%5C&amp;space;sen%5Ctheta%5Csqrt%7B%5C&amp;space;%5Cfrac%7B4%5Cpi%5Cvarepsilon_0mg%5C&amp;space;tg%5Ctheta%7D%7B%5Csqrt3%7D%7D\" alt=\"\\large l\\ sen\\theta\\sqrt{\\ \\frac{4\\pi\\varepsilon_0mg\\ tg\\theta}{\\sqrt3}}\" align=\"absmiddle\"><\/p><p>e) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;l%5C&amp;space;sen%5Ctheta%5Csqrt%7B4%5Cpi%5Cvarepsilon_0mg%5C&amp;space;tg%5Ctheta%5C&amp;space;%7D\" alt=\"\\large l\\ sen\\theta\\sqrt{4\\pi\\varepsilon_0mg\\ tg\\theta\\ }\" align=\"absmiddle\"><\/p><div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2019\/12\/image21.png\" alt=\"\" class=\"wp-image-35042\" width=\"286\" height=\"245\"><\/figure><\/div><h3 class=\"wp-block-heading\">Resolu&ccedil;&atilde;o Comentada<\/h3><p>De acordo com o enunciado, na condi&ccedil;&atilde;o de equil&iacute;brio, temos a seguinte configura&ccedil;&atilde;o espacial das cargas:<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2019\/12\/image22.png\" alt=\"\" class=\"wp-image-35044\"><\/figure><\/div><p>Pelo equil&iacute;brio das for&ccedil;as, temos:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;tg(%5Ctheta&amp;space;)=%5Cfrac%7B2%5Ccdot&amp;space;F_%7Be%7D%5Ccdot&amp;space;cos%5C&amp;space;60%5Cdegree%7D%7Bm%5Ccdot&amp;space;g%7D\" alt=\"\\large tg(\\theta )=\\frac{2\\cdot F_{e}\\cdot cos\\ 60\\degree}{m\\cdot g}\" align=\"absmiddle\"><\/p><p>Pela geometria no tri&acirc;ngulo da base formado pelas cargas, temos:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;l%5Ccdot&amp;space;sen%5Cleft(%5Ctheta%5Cright)=MN=x%5Ccdot%5Cfrac%7B%5Csqrt3%7D%7B2%7D%5Ccdot%5Cfrac%7B2%7D%7B3%7D\" alt=\"\\large l\\cdot sen\\left(\\theta\\right)=MN=x\\cdot\\frac{\\sqrt3}{2}\\cdot\\frac{2}{3}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;x=l%5Ccdot%5Csqrt3%5Ccdot&amp;space;sen%5Cleft(%5Ctheta%5Cright)\" alt=\"\\large x=l\\cdot\\sqrt3\\cdot sen\\left(\\theta\\right)\" align=\"absmiddle\"><\/p><p>Portanto:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;tg%5Cleft(%5Ctheta%5Cright)=%5Cfrac%7B2%5Ccdot%5Cfrac%7B1%7D%7B4%5Cpi%5Ccdot%5Cvarepsilon_0%7D%5Ccdot%5Cfrac%7BQ%5Ccdot&amp;space;Q%7D%7Bx%5E2%7D%5Ccdot%5Cfrac%7B%5Csqrt3%7D%7B2%7D%7D%7Bm%5Ccdot&amp;space;g%7D\" alt=\"\\large tg\\left(\\theta\\right)=\\frac{2\\cdot\\frac{1}{4\\pi\\cdot\\varepsilon_0}\\cdot\\frac{Q\\cdot Q}{x^2}\\cdot\\frac{\\sqrt3}{2}}{m\\cdot g}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;Q=%5Csqrt%7B%5Cfrac%7Bm%5Ccdot&amp;space;g%5Ccdot4%5Cpi%5Cvarepsilon_0%5Ccdot&amp;space;t&amp;space;g%5Cleft(%5Ctheta%5Cright)%7D%7B%5Csqrt3%7D%7D%5Ccdot&amp;space;l%5Ccdot%5Csqrt3%5Ccdot&amp;space;sen%5Cleft(%5Ctheta%5Cright)\" alt=\"\\large Q=\\sqrt{\\frac{m\\cdot g\\cdot4\\pi\\varepsilon_0\\cdot t g\\left(\\theta\\right)}{\\sqrt3}}\\cdot l\\cdot\\sqrt3\\cdot sen\\left(\\theta\\right)\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfbox%7B%24q=l%5Ccdot&amp;space;s&amp;space;e&amp;space;n%5Cleft(%5Ctheta%5Cright)%5Ccdot%5Csqrt%7B4%5Cpi%5Cvarepsilon_0%5Ccdot&amp;space;m%5Ccdot&amp;space;g%5Ccdot&amp;space;t&amp;space;g%5Cleft(%5Ctheta%5Cright)%5Csqrt3%7D%24%7D\" alt=\"\\large \\fbox{$q=l\\cdot s e n\\left(\\theta\\right)\\cdot\\sqrt{4\\pi\\varepsilon_0\\cdot m\\cdot g\\cdot t g\\left(\\theta\\right)\\sqrt3}$}\" align=\"absmiddle\"><\/p><p class=\"has-light-green-cyan-background-color has-background\"><strong>Gabarito: C<\/strong><\/p><h3 class=\"wp-block-heading\">Quest&atilde;o 14<\/h3><p>Dois raios luminosos paralelos e sim&eacute;tricos em rela&ccedil;&atilde;o ao eixo &oacute;ptico, interdistantes de 2mm, devem ser focados em um ponto P no interior de um bloco transparente, a 1 mm de sua superf&iacute;cie, confirme mostra a figura. Para tal, utiliza-se uma lente delgada convergente com dist&acirc;ncia focal de 1 mm. Considerando que o bloco tem &iacute;ndice de refra&ccedil;&atilde;o <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;n=%5C&amp;space;%5Csqrt2\" alt=\"\\large n=\\ \\sqrt2\" align=\"absmiddle\">, a dist&acirc;ncia L entre o v&eacute;rtice V da lente e a superf&iacute;cie do bloco deve ser ajustada para:<\/p><p>a)<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;1mm\" alt=\"\\large 1mm\" align=\"absmiddle\"><\/p><p>b) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7Dmm\" alt=\"\\large \\frac{\\sqrt{2}}{2}mm\" align=\"absmiddle\"><\/p><p>c) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cleft&amp;space;(&amp;space;1-%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D&amp;space;%5Cright&amp;space;)mm\" alt=\"\\large \\left ( 1-\\frac{\\sqrt{2}}{2} \\right )mm\" align=\"absmiddle\"><\/p><p>d) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?%5Clarge&amp;space;%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7Dmm\" alt=\"\\large \\frac{\\sqrt{3}}{3}mm\" align=\"absmiddle\"><\/p><p>e) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?%5Clarge&amp;space;%5Cleft&amp;space;(&amp;space;1-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B3%7D&amp;space;%5Cright&amp;space;)mm\" alt=\"\\large \\left ( 1-\\frac{\\sqrt{3}}{3} \\right )mm\" align=\"absmiddle\"><\/p><h3 class=\"wp-block-heading\">Resolu&ccedil;&atilde;o Comentada<\/h3><p>Segundo as condi&ccedil;&otilde;es do problema, temos:<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2019\/12\/image23.png\" alt=\"\" class=\"wp-image-35048\" width=\"529\" height=\"403\"><\/figure><\/div><p><span style=\"font-size: 13.0pt; mso-bidi-font-size: 12.0pt; font-family: 'Calibri',sans-serif; mso-fareast-font-family: Cambria; mso-fareast-theme-font: minor-latin; mso-bidi-font-family: 'Times New Roman'; mso-bidi-theme-font: minor-bidi; mso-ansi-language: PT-BR; mso-fareast-language: EN-US; mso-bidi-language: AR-SA;\">Pela geometria, temos <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5CDelta&amp;space;AFD%5Csim&amp;space;%5CDelta&amp;space;ACB\" alt=\"\\large \\Delta AFD\\sim \\Delta ACB\" align=\"absmiddle\">, ent&atilde;o:<\/span><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;AD=DF\" alt=\"\\large AD=DF\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?%5Clarge&amp;space;D%5Cwidehat%7BA%7DF=%5Cfrac%7B%5Cpi&amp;space;%7D%7B4%7D\" alt=\"\\large D\\widehat{A}F=\\frac{\\pi }{4}\" align=\"absmiddle\"><\/p><p>Aplicando a Lei de Snell em C, temos:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;n_1%5Ccdot&amp;space;sen%5Cleft(%5Ctheta_1%5Cright)=n_2%5Ccdot&amp;space;sen%5Cleft(%5Ctheta_2%5Cright)\" alt=\"\\large n_1\\cdot sen\\left(\\theta_1\\right)=n_2\\cdot sen\\left(\\theta_2\\right)\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;1%5Ccdot&amp;space;sen%5Cleft(%5Cfrac%7B%5Cpi%7D%7B4%7D%5Cright)=%5Csqrt2%5Ccdot&amp;space;sen%5Cleft(%5Ctheta_2%5Cright)\" alt=\"\\large 1\\cdot sen\\left(\\frac{\\pi}{4}\\right)=\\sqrt2\\cdot sen\\left(\\theta_2\\right)\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;sen%5Cleft(%5Ctheta_2%5Cright)=%5Cfrac%7B1%7D%7B2%7D\" alt=\"\\large sen\\left(\\theta_2\\right)=\\frac{1}{2}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Ctheta_2=30%5Cdegree\" alt=\"\\large \\theta_2=30\\degree\" align=\"absmiddle\"><\/p><p>No tri&acirc;ngulo <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5CDelta&amp;space;CEP\" alt=\"\\large \\Delta CEP\" align=\"absmiddle\">, temos:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?%5Clarge&amp;space;tg30%5Cdegree=%5Cfrac%7BCE%7D%7BPE%7D\" alt=\"\\large tg30\\degree=\\frac{CE}{PE}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7B%5Csqrt3%7D%7B3%7D=%5Cfrac%7BCE%7D%7B1%7D%5CRightarrow&amp;space;CE=%5Cfrac%7B%5Csqrt3%7D%7B3%7D%5C&amp;space;mm\" alt=\"\\large \\frac{\\sqrt3}{3}=\\frac{CE}{1}\\Rightarrow CE=\\frac{\\sqrt3}{3}\\ mm\" align=\"absmiddle\"><\/p><p>Portanto:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;AB+BD=1%5C&amp;space;mm\" alt=\"\\large AB+BD=1\\ mm\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;AB+CE=1%5C&amp;space;mm\" alt=\"\\large AB+CE=1\\ mm\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;L+%5Cfrac%7B%5Csqrt3%7D%7B3%7D=1\" alt=\"\\large L+\\frac{\\sqrt3}{3}=1\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;L=1-%5Cfrac%7B%5Csqrt3%7D%7B3%7D%5C&amp;space;mm\" alt=\"\\large L=1-\\frac{\\sqrt3}{3}\\ mm\" align=\"absmiddle\"><\/p><p class=\"has-light-green-cyan-background-color has-background\"><strong>Gabarito: E<\/strong><\/p><h3 class=\"wp-block-heading\">Quest&atilde;o 15<\/h3><p>Considere um sistema de tr&ecirc;s m&aacute;quinas t&eacute;rmicas <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M_1,M_2%5C&amp;space;e%5C&amp;space;M_3\" alt=\"\\large M_1,M_2\\ e\\ M_3\" align=\"absmiddle\"> acopladas, tal que o rejeito energ&eacute;tico de uma &eacute; aproveitado pela seguinte. Sabe-se que a cada ciclo, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M_1\" alt=\"\\large M_1\" align=\"absmiddle\"> recebe 800kJ de calor de uma fonte quente a 300K e rejeita 600kJ, dos quais 150kJ s&atilde;o aproveitados por <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M_2\" alt=\"\\large M_2\" align=\"absmiddle\"> para realiza&ccedil;&atilde;o de trabalho. Por fim, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M_3\" alt=\"\\large M_3\" align=\"absmiddle\"> aproveita o rejeito de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M_2\" alt=\"\\large M_2\" align=\"absmiddle\"> e descarta 360kJ em uma fonte fria a 6K. S&atilde;o feitas as seguintes afirma&ccedil;&otilde;es:<\/p><p>I. &Eacute; inferior a 225K a temperada da fonte fria de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M_3\" alt=\"\\large M_3\" align=\"absmiddle\">.<\/p><p class=\"Questes-Enunciado\">II. O rendimento do sistema &eacute; de 55%.<\/p><p class=\"Questes-Enunciado\">III. O rendimento do sistema corresponde a 80% do rendimento de uma m&aacute;quina de Carnot operando entre as mesmas temperaturas.<\/p><p class=\"Questes-Enunciado\">Conclui-se ent&atilde;o que<\/p><p class=\"Questes-Enunciado\">a) somente a afirma&ccedil;&atilde;o I est&aacute; incorreta.<\/p><p class=\"Questes-Enunciado\">b) somente a afirma&ccedil;&atilde;o II est&aacute; incorreta.<\/p><p class=\"Questes-Enunciado\">c) somente a afirma&ccedil;&atilde;o III est&aacute; incorreta.<\/p><p class=\"Questes-Enunciado\">d) todas as afirma&ccedil;&otilde;es est&atilde;o corretas.<\/p><p class=\"Questes-Enunciado\">e) as afirma&ccedil;&otilde;es I e III est&atilde;o incorretas.<\/p><h3 class=\"wp-block-heading\">Resolu&ccedil;&atilde;o Comentada<\/h3><p>De acordo com as informa&ccedil;&otilde;es do enunciado, temos:<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2019\/12\/image24.png\" alt=\"\" class=\"wp-image-35054\" width=\"428\" height=\"549\"><\/figure><\/div><p>O calor transmitido entre a fonte quente que est&aacute; a 300 K e a m&aacute;quina t&eacute;rmica <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M_1\" alt=\"\\large M_1\" align=\"absmiddle\"> se transforma em trabalho e em calor que &eacute; rejeitado para a m&aacute;quina t&eacute;rmica <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M_2\" alt=\"\\large M_2\" align=\"absmiddle\">. A m&aacute;quina t&eacute;rmica 2 aproveita o calor rejeitado por <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M_1\" alt=\"\\large M_1\" align=\"absmiddle\"> e o converte em trabalho e calor rejeitado para <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M_3\" alt=\"\\large M_3\" align=\"absmiddle\">. Por fim, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M_3\" alt=\"\\large M_3\" align=\"absmiddle\"> utiliza este calor rejeitado por <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M_2\" alt=\"\\large M_2\" align=\"absmiddle\"> e o converte em trabalho e calor rejeitado para a fonte fria que est&aacute; a 6K.<\/p><p>Assim, pelas equa&ccedil;&otilde;es de conserva&ccedil;&atilde;o de energia, temos:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;Q_Q=W_1+Q_%7BM2%7D\" alt=\"\\large Q_Q=W_1+Q_{M2}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;Q_%7BM2%7D=W_2+Q_%7BM3%7D\" alt=\"\\large Q_{M2}=W_2+Q_{M3}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;Q_%7BM3%7D=W_3+Q_f\" alt=\"\\large Q_{M3}=W_3+Q_f\" align=\"absmiddle\"><\/p><p>Como Sabemos que&nbsp;<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;Q_Q=800%5C&amp;space;kJ,Q_%7BM2%7D=600%5C&amp;space;kJ,W_2=150%5C&amp;space;kJeQ_F=360%5C&amp;space;kJ\" alt=\"\\large Q_Q=800\\ kJ,Q_{M2}=600\\ kJ,W_2=150\\ kJeQ_F=360\\ kJ\" align=\"absmiddle\"><\/p><p>chegamos que<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;W_1=200%5C&amp;space;kJ,Q_%7BM3%7D=450%5C&amp;space;kJ%5C&amp;space;e%5C&amp;space;W_3=90%5C&amp;space;kJ\" alt=\"\\large W_1=200\\ kJ,Q_{M3}=450\\ kJ\\ e\\ W_3=90\\ kJ\" align=\"absmiddle\"><\/p><p>Dessa forma, podemos analisar as afirmativas:<\/p><p>I:&nbsp; <strong>Verdadeira<\/strong>. O rendimento de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M_1\" alt=\"\\large M_1\" align=\"absmiddle\"> &eacute;:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Ceta_1=%5Cfrac%7BW_1%7D%7BQ_q%7D=%5Cfrac%7B200%7D%7B800%7D=%5Cfrac%7B1%7D%7B4%7D\" alt=\"\\large \\eta_1=\\frac{W_1}{Q_q}=\\frac{200}{800}=\\frac{1}{4}\" align=\"absmiddle\"><\/p><p>Portanto <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;T_f%5Cle225K\" alt=\"\\large T_f\\le225K\" align=\"absmiddle\"> . Como Carnot &eacute; uma m&aacute;quina te&oacute;rica, na pr&aacute;tica, a temperatura da fonte fria de 1 deve ser inferior a 225 K.<\/p><p>II:&nbsp; <strong>Verdadeira<\/strong>. Para o sistema, temos:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Ceta=%5Cfrac%7B%5Csum&amp;space;W_i%7D%7BQ_Q%7D=%5Cfrac%7B440%7D%7B800%7D=0,55=55%5C&amp;space;%5C%\" alt=\"\\large \\eta=\\frac{\\sum W_i}{Q_Q}=\\frac{440}{800}=0,55=55\\ \\%\" align=\"absmiddle\"><\/p><p>III: <strong>Falsa<\/strong>. Considerando o rendimento de Carnot para as mesmas temperaturas, temos:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Ceta=1-%5Cfrac%7BT_f%7D%7BT_q%7D=1-%5Cfrac%7B6%7D%7B300%7D=%5Cfrac%7B49%7D%7B50%7D\" alt=\"\\large \\eta=1-\\frac{T_f}{T_q}=1-\\frac{6}{300}=\\frac{49}{50}\" align=\"absmiddle\"><\/p><p>Agora, fazendo <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;0,8%5Ccdot%5Ceta=0,784&gt;0,55%E2%80%B3%20alt=%E2%80%9D%5Clarge%200,8%5Ccdot%5Ceta=0,784&gt;0,55%E2%80%B3%20align=%E2%80%9Dabsmiddle%E2%80%9D&gt;&lt;\/p&gt;%0A%0A%0A&lt;p%20class=\" has-light-green-cyan-background-color has-background><strong>Gabarito: C<\/strong><\/p><p>&Eacute; isso, pessoal! Espero que tenham curtido a resolu&ccedil;&atilde;o da prova de F&iacute;sica da prova da 1&ordf; Fase do Vestibular ITA 2020.&nbsp;Sigam-me nas redes sociais. T&ecirc;m muitas dicas l&aacute;. Mande uma mensagem, caso tenha tido alguma d&uacute;vida. Abra&ccedil;os!<\/p><p><strong>Instagram:<\/strong> <a href=\"https:\/\/www.instagram.com\/proftoniburgatto\/\" target=\"_blank\" rel=\"noopener noreferrer\">@proftoniburgatto<\/a><\/p><p class=\"has-text-align-center has-luminous-vivid-amber-background-color has-background has-medium-font-size\"><a href=\"https:\/\/estrategiavestibulares.com.br\/vestibulares\/itaime\/\" target=\"_blank\">CURSOS ITA<\/a><\/p><\/p>\n","protected":false},"excerpt":{"rendered":"Fala, pessoal&hellip; Tudo bem? Sou o prof. Toni Burgatto, do Estrat&eacute;gia Vestibulares e Carreiras Militares. Neste artigo, voc&ecirc;&hellip;\n","protected":false},"author":18,"featured_media":35058,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"wl_entities_gutenberg":"","footnotes":""},"categories":[28],"tags":[],"wl_entity_type":[732],"class_list":{"0":"post-34935","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-fisica","8":"wl_entity_type-article"},"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v25.9 (Yoast SEO v25.9) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Prova ITA 2020 \u2013 F\u00edsica \u2013 Resolu\u00e7\u00e3o Comentada<\/title>\n<meta name=\"description\" content=\"Voc\u00ea participou da prova do ITA 2020? 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