{"id":37373,"date":"2020-01-07T14:31:22","date_gmt":"2020-01-07T17:31:22","guid":{"rendered":"https:\/\/blog.estrategiavestibulares.com.br\/?p=37373"},"modified":"2021-03-11T17:07:49","modified_gmt":"2021-03-11T20:07:49","slug":"fisica-2a-fase-da-fuvest-2020","status":"publish","type":"post","link":"https:\/\/vestibulares.estrategia.com\/portal\/materias\/fisica\/fisica-2a-fase-da-fuvest-2020\/","title":{"rendered":"Como resolver as quest\u00f5es de F\u00edsica da 2\u00aa Fase da Fuvest 2020"},"content":{"rendered":"<p>Fala, pessoal. Tudo certo? Sou o prof. Lucas Costa, professor de F&iacute;sica do Estrat&eacute;gia Vestibulares. Hoje, vamos resolver as quest&otilde;es de F&iacute;sica da prova da 2&ordf; Fase do Vestibular da Fuvest 2020.<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 counter-hierarchy ez-toc-counter ez-toc-transparent ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\"><p class=\"ez-toc-title\" style=\"cursor:inherit\">Navegue pelo conte\u00fado<\/p>\n<\/div><nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/vestibulares.estrategia.com\/portal\/materias\/fisica\/fisica-2a-fase-da-fuvest-2020\/#Questao-01\" >Quest&atilde;o 01<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/vestibulares.estrategia.com\/portal\/materias\/fisica\/fisica-2a-fase-da-fuvest-2020\/#Questao-02\" >Quest&atilde;o 02<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/vestibulares.estrategia.com\/portal\/materias\/fisica\/fisica-2a-fase-da-fuvest-2020\/#Questao-03\" >Quest&atilde;o 03<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/vestibulares.estrategia.com\/portal\/materias\/fisica\/fisica-2a-fase-da-fuvest-2020\/#Questao-04\" >Quest&atilde;o 04<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/vestibulares.estrategia.com\/portal\/materias\/fisica\/fisica-2a-fase-da-fuvest-2020\/#Questao-05\" >Quest&atilde;o 05<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/vestibulares.estrategia.com\/portal\/materias\/fisica\/fisica-2a-fase-da-fuvest-2020\/#Questao-06\" >Quest&atilde;o 06<\/a><\/li><\/ul><\/nav><\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Questao-01\"><\/span>Quest&atilde;o 01<span class=\"ez-toc-section-end\"><\/span><\/h2><p>Uma pessoa produz oscila&ccedil;&otilde;es peri&oacute;dicas em uma longa corda formada por duas por&ccedil;&otilde;es de materiais diferentes 1 e 2, nos quais a velocidade de propaga&ccedil;&atilde;o das ondas &eacute;, respectivamente, de 5 m\/s e 4 m\/s. Segurando a extremidade feita do material 1, a pessoa abaixa e levanta sua m&atilde;o regularmente, completando um ciclo a cada 0,5 s, de modo que as ondas propagam&#8208;se do material 1 para o material 2, conforme mostrado na figura. Despreze eventuais efeitos de reflex&atilde;o das ondas.<\/p><figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2020\/01\/image21.png\" alt=\"\" class=\"wp-image-37376\"><\/figure><p>a) Circule, dentre os vetores na folha de respostas,\naquele que melhor representa a velocidade do ponto P da corda no instante\nmostrado na figura.&nbsp; <\/p><p>b) Calcule a frequ&ecirc;ncia e o comprimento de onda no\nmaterial 1.<\/p><p>c) Calcule a frequ&ecirc;ncia e o comprimento de onda no material 2.<\/p><h3 class=\"wp-block-heading\">Gabarito<\/h3><p>a) Vertical e para baixo.<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2020\/01\/image22.png\" alt=\"\" class=\"wp-image-37378\"><\/figure><\/div><div class=\"wp-block-image\"><figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2020\/01\/image23.png\" alt=\"\" class=\"wp-image-37379\" width=\"525\" height=\"163\"><\/figure><\/div><p>b) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cmathbit%7Bf%7D=%5Cmathbf%7B2%7D%5C&amp;space;%5Cmathbit%7BHz%7D%5C&amp;space;e%5C&amp;space;%5Cmathbit%7B%5Clambda%7D_%5Cmathbf%7B1%7D=%5Cmathbf%7B2%7D,%5Cmathbf%7B5%7D%5C&amp;space;%5Cmathbit%7Bm%7D\" alt=\"\\large \\mathbit{f}=\\mathbf{2}\\ \\mathbit{Hz}\\ e\\ \\mathbit{\\lambda}_\\mathbf{1}=\\mathbf{2},\\mathbf{5}\\ \\mathbit{m}\" align=\"absmiddle\"><\/p><p>A frequ&ecirc;ncia &eacute; dada pelo inverso do per&iacute;odo, que &eacute; o tempo que a pessoa leva para completar um ciclo.<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;f=%5Cfrac%7B1%7D%7BT%7D=%5Cfrac%7B1%7D%7B0,5%7D=2%5C&amp;space;Hz\" alt=\"\\large f=\\frac{1}{T}=\\frac{1}{0,5}=2\\ Hz\" align=\"absmiddle\"><\/p><p>Podemos usar a equa&ccedil;&atilde;o fundamental da ondulat&oacute;ria para determinarmos o comprimento de onda na corda feita com o material 1:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v=%5Clambda%5Ccdot&amp;space;f%5CRightarrow%5Clambda=%5Cfrac%7Bv%7D%7Bf%7D\" alt=\"\\large v=\\lambda\\cdot f\\Rightarrow\\lambda=\\frac{v}{f}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Clambda_1=%5Cfrac%7B5%7D%7B2%7D=2,5%5C&amp;space;m\" alt=\"\\large \\lambda_1=\\frac{5}{2}=2,5\\ m\" align=\"absmiddle\"><\/p><p>c) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cmathbit%7Bf%7D=%5Cmathbf%7B2%7D%5C&amp;space;%5Cmathbit%7BHz%7D%5C&amp;space;e%5C&amp;space;%5Cmathbit%7B%5Clambda%7D_%5Cmathbf%7B2%7D=%5Cmathbf%7B2%7D%5C&amp;space;%5Cmathbit%7Bm%7D\" alt=\"\\large \\mathbit{f}=\\mathbf{2}\\ \\mathbit{Hz}\\ e\\ \\mathbit{\\lambda}_\\mathbf{2}=\\mathbf{2}\\ \\mathbit{m}\" align=\"absmiddle\"><\/p><p>A frequ&ecirc;ncia depende somente da fonte emissora. Como a pessoa &eacute; a mesma, e o tempo que ela leva para completar cada ciclo tamb&eacute;m, a frequ&ecirc;ncia da onda no material 2 tamb&eacute;m vale 2 Hz.<\/p><p>Podemos calcular o comprimento de onda de maneira an&aacute;loga:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Clambda_2=%5Cfrac%7B4%7D%7B2%7D=2%5C&amp;space;m\" alt=\"\\large \\lambda_2=\\frac{4}{2}=2\\ m\" align=\"absmiddle\"><\/p><h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Questao-02\"><\/span>Quest&atilde;o 02<span class=\"ez-toc-section-end\"><\/span><\/h2><p>Um mol de um g&aacute;s ideal monoat&ocirc;mico &eacute; resfriado adiabaticamente de uma temperatura inicial T<sub>1<\/sub> at&eacute; uma temperatura final T<sub>1<\/sub>\/3.<\/p><p>Com base nessas informa&ccedil;&otilde;es, responda:<\/p><p>a) O g&aacute;s sofreu expans&atilde;o ou compress&atilde;o ao final do processo? Justifique sua resposta.<\/p><p>b) Encontre o valor do trabalho realizado pelo g&aacute;s nesse processo em termos da constante universal dos gases ideais R e de T<sub>1<\/sub>.<\/p><p>c) Encontre a raz&atilde;o entre as press&otilde;es final e inicial do g&aacute;s ap&oacute;s o processo.<br>Note e adote:<\/p><p>Em um processo adiab&aacute;tico, n&atilde;o h&aacute; troca de calor com o ambiente. <br>Energia interna por mol de um g&aacute;s ideal monoat&ocirc;mico: <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;U=3RT\/2\" alt=\"\\large U=3RT\/2\" align=\"absmiddle\">.<\/p><p>Para o processo adiab&aacute;tico em quest&atilde;o, vale a rela&ccedil;&atilde;o <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;PV%5E%7B5\/3%7D=constante\" alt=\"\\large PV^{5\/3}=constante\" align=\"absmiddle\">.<\/p><h3 class=\"wp-block-heading\">Gabarito<\/h3><p>a) Em uma transforma&ccedil;&atilde;o adiab&aacute;tica, vale a Lei de Poisson-Laplace:<\/p><figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2020\/01\/image24-1024x121.png\" alt=\"\" class=\"wp-image-37387\"><\/figure><p>Esse tipo de transforma&ccedil;&atilde;o normalmente &eacute; observada uma curva que vai de uma isoterma a outra.<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2020\/01\/image25.png\" alt=\"\" class=\"wp-image-37388\"><figcaption>Figura 07.8 &ndash; Uma expans&atilde;o adiab&aacute;tica de um g&aacute;s. Note que VB &gt; VA.<\/figcaption><\/figure><\/div><p>Isso significa que a press&atilde;o, o volume e a temperatura do g&aacute;s variam. Em uma contra&ccedil;&atilde;o adiab&aacute;tica ocorre diminui&ccedil;&atilde;o do volume, aumento de press&atilde;o e aumento de temperatura, por outro lado, em uma expans&atilde;o adiab&aacute;tica o volume aumenta, e a press&atilde;o e a temperatura diminuem.<\/p><p>A quest&atilde;o pede que demonstremos:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;p_1%5Ccdot&amp;space;V_1%5E%5Cgamma=p_2%5Ccdot&amp;space;V_2%5E%5Cgamma\" alt=\"\\large p_1\\cdot V_1^\\gamma=p_2\\cdot V_2^\\gamma\" align=\"absmiddle\"><\/p><p>Devemos substituir a press&atilde;o por sua rela&ccedil;&atilde;o com a equa&ccedil;&atilde;o de Clapeyron<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;p%5Ccdot&amp;space;V=n%5Ccdot&amp;space;R%5Ccdot&amp;space;T%5CRightarrow&amp;space;p=%5C&amp;space;%5Cfrac%7Bn%5Ccdot&amp;space;R%5Ccdot&amp;space;T%7D%7BV%7D\" alt=\"\\large p\\cdot V=n\\cdot R\\cdot T\\Rightarrow p=\\ \\frac{n\\cdot R\\cdot T}{V}\" align=\"absmiddle\"><\/p><p>Fazendo a substitui&ccedil;&atilde;o na express&atilde;o anterior, temos:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;p_1%5Ccdot&amp;space;V_1%5E%5Cgamma=p_2%5Ccdot&amp;space;V_2%5E%5Cgamma\" alt=\"\\large p_1\\cdot V_1^\\gamma=p_2\\cdot V_2^\\gamma\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7Bn%5Ccdot&amp;space;R%5Ccdot&amp;space;T_1%7D%7BV_1%7D%5Ccdot&amp;space;V_1%5E%5Cgamma=%5Cfrac%7Bn%5Ccdot&amp;space;R%5Ccdot&amp;space;T_2%7D%7BV_2%7D%5Ccdot&amp;space;V_2%5E%5Cgamma\" alt=\"\\large \\frac{n\\cdot R\\cdot T_1}{V_1}\\cdot V_1^\\gamma=\\frac{n\\cdot R\\cdot T_2}{V_2}\\cdot V_2^\\gamma\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7B%5Ccancel&amp;space;n&amp;space;%5Ccdot&amp;space;%5Ccancel&amp;space;R&amp;space;%5Ccdot&amp;space;T_%7B1%7D%7D%7BV_%7B1%7D%7D%5Ccdot&amp;space;V_%7B1%7D%5E%7B%5Cgamma&amp;space;%7D=%5Cfrac%7B%5Ccancel&amp;space;n&amp;space;%5Ccdot&amp;space;%5Ccancel&amp;space;R&amp;space;%5Ccdot&amp;space;T_%7B2%7D%7D%7BV_%7B2%7D%7D%5Ccdot&amp;space;V_%7B2%7D%5E%7B%5Cgamma&amp;space;%7D\" alt=\"\\large \\frac{\\cancel n \\cdot \\cancel R \\cdot T_{1}}{V_{1}}\\cdot V_{1}^{\\gamma }=\\frac{\\cancel n \\cdot \\cancel R \\cdot T_{2}}{V_{2}}\\cdot V_{2}^{\\gamma }\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7BT_1%7D%7BV_1%7D%5Ccdot&amp;space;V_1%5E%5Cgamma=%5Cfrac%7BT_2%7D%7BV_2%7D%5Ccdot&amp;space;V_2%5E%5Cgamma\" alt=\"\\large \\frac{T_1}{V_1}\\cdot V_1^\\gamma=\\frac{T_2}{V_2}\\cdot V_2^\\gamma\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;T_1%5Ccdot&amp;space;V_1%5E%7B%5Cgamma-1%7D=T_2%5Ccdot&amp;space;V_2%5E%7B%5Cgamma-1%7D\" alt=\"\\large T_1\\cdot V_1^{\\gamma-1}=T_2\\cdot V_2^{\\gamma-1}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;T_1%5Ccdot&amp;space;V_1%5E%7B%5Cgamma-1%7D=%5Cfrac%7BT_1%7D%7B3%7D%5Ccdot&amp;space;V_2%5E%7B%5Cgamma-1%7D\" alt=\"\\large T_1\\cdot V_1^{\\gamma-1}=\\frac{T_1}{3}\\cdot V_2^{\\gamma-1}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;3%5Ccdot&amp;space;V_1%5E%7B%5Cgamma-1%7D=V_2%5E%7B%5Cgamma-1%7D\" alt=\"\\large 3\\cdot V_1^{\\gamma-1}=V_2^{\\gamma-1}\" align=\"absmiddle\"><\/p><p>Pelo &ldquo;Note e adote&rdquo; devemos perceber que <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cgamma=5\/3\" alt=\"\\large \\gamma=5\/3\" align=\"absmiddle\">, logo <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cgamma-1=2\/3\" alt=\"\\large \\gamma-1=2\/3\" align=\"absmiddle\">:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V_2=V_1%5Ccdot3%5E%7B%5Cleft(%5Cfrac%7B1%7D%7B%5Cgamma-1%7D%5Cright)%7D=V_1%5Ccdot3%5E%7B%5Cleft(%5Cfrac%7B3%7D%7B2%7D%5Cright)%7D\" alt=\"\\large V_2=V_1\\cdot3^{\\left(\\frac{1}{\\gamma-1}\\right)}=V_1\\cdot3^{\\left(\\frac{3}{2}\\right)}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V_2=%5Csqrt%7B27%7D%5Ccdot&amp;space;V_1\" alt=\"\\large V_2=\\sqrt{27}\\cdot V_1\" align=\"absmiddle\"><\/p><p>Como <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;V_2&gt;V_1%E2%80%B3%20alt=%E2%80%9D%5Clarge%20V_2&gt;V_1%E2%80%B3%20align=%E2%80%9Dabsmiddle%E2%80%9D&gt;,%20temos%20uma%20expans%C3%A3o%20gasosa.&lt;\/p&gt;%0A&lt;p&gt;b)%20Pela%20primeira%20lei%20da%20Termodin%C3%A2mica,%20temos:&lt;\/p&gt;%0A&lt;p&gt;&lt;img%20src=\" https: alt=\"\\large &#8710;U=Q-W\" align=\"absmiddle\"><\/p><p>Como se trata de uma transforma&ccedil;&atilde;o adiab&aacute;tica, Q=0, da&iacute;:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5CDelta&amp;space;U=-W%5CRightarrow&amp;space;W=-%5CDelta&amp;space;U\" alt=\"\\large \\Delta U=-W\\Rightarrow W=-\\Delta U\" align=\"absmiddle\"><\/p><p>Pela express&atilde;o fornecida no &ldquo;Note e adote&rdquo;:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;W=%5Cfrac%7B-3%5Ccdot&amp;space;R%5Ccdot&amp;space;%5CDelta&amp;space;T%7D%7B2%7D=%5Cfrac%7B3%5Ccdot&amp;space;R%5Ccdot&amp;space;-%5CDelta&amp;space;T%7D%7B2%7D=%5Cfrac%7B3%5Ccdot&amp;space;R%5Ccdot%5Cleft(T_1-T_1\/3%5Cright)%7D%7B2%7D\" alt=\"\\large W=\\frac{-3\\cdot R\\cdot \\Delta T}{2}=\\frac{3\\cdot R\\cdot -\\Delta T}{2}=\\frac{3\\cdot R\\cdot\\left(T_1-T_1\/3\\right)}{2}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;W=%5Cfrac%7B3%5Ccdot&amp;space;R%5Ccdot%5Cfrac%7B2%7D%7B3%7D%5Ccdot&amp;space;T_1%7D%7B2%7D=R%5Ccdot&amp;space;T_1\" alt=\"\\large W=\\frac{3\\cdot R\\cdot\\frac{2}{3}\\cdot T_1}{2}=R\\cdot T_1\" align=\"absmiddle\"><\/p><p>c) Devemos usar novamente a rela&ccedil;&atilde;o para uma transforma&ccedil;&atilde;o adiab&aacute;tica:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;p_1%5Ccdot&amp;space;V_1%5E%5Cgamma=p_2%5Ccdot&amp;space;V_2%5E%5Cgamma\" alt=\"\\large p_1\\cdot V_1^\\gamma=p_2\\cdot V_2^\\gamma\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;p_i%5Ccdot&amp;space;V_1%5E%7B5\/3%7D=p_f%5Ccdot&amp;space;V_2%5E%7B5\/3%7D\" alt=\"\\large p_i\\cdot V_1^{5\/3}=p_f\\cdot V_2^{5\/3}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;p_i%5Ccdot&amp;space;V_1%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D=p_f%5Ccdot%5Cleft(3%5E%5Cfrac%7B3%7D%7B2%7D%5Ccdot&amp;space;V_1%5Cright)%5E%5Cfrac%7B5%7D%7B3%7D\" alt=\"\\large p_i\\cdot V_1^{\\frac{5}{3}}=p_f\\cdot\\left(3^\\frac{3}{2}\\cdot V_1\\right)^\\frac{5}{3}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;p_i%5Ccdot&amp;space;V_1%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D=p_f%5Ccdot&amp;space;3%5E%5Cfrac%7B5%7D%7B2%7D%5Ccdot&amp;space;V_1%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D\" alt=\"\\large p_i\\cdot V_1^{\\frac{5}{3}}=p_f\\cdot 3^\\frac{5}{2}\\cdot V_1^{\\frac{5}{3}}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;p_i%5Ccdot&amp;space;V_1%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D=p_f%5Ccdot&amp;space;3%5E%5Cfrac%7B5%7D%7B2%7D%5Ccdot&amp;space;V_1%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D\" alt=\"\\large p_i\\cdot V_1^{\\frac{5}{3}}=p_f\\cdot 3^\\frac{5}{2}\\cdot V_1^{\\frac{5}{3}}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;p_f=%5Cfrac%7Bp_i%7D%7B3%5E%5Cfrac%7B5%7D%7B2%7D%7D\" alt=\"\\large p_f=\\frac{p_i}{3^\\frac{5}{2}}\" align=\"absmiddle\"><\/p><h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Questao-03\"><\/span>Quest&atilde;o 03<span class=\"ez-toc-section-end\"><\/span><\/h2><p>A tomografia por emiss&atilde;o de p&oacute;sitrons (PET) &eacute; uma\nt&eacute;cnica de imagem por contraste na qual se utilizam marcadores com\nradionucl&iacute;deos emissores de p&oacute;sitrons. O radionucl&iacute;deo mais utilizado em PET &eacute;\no is&oacute;topo 18 do fl&uacute;or, que decai para um n&uacute;cleo de oxig&ecirc;nio&#8208;18, emitindo um\np&oacute;sitron. O n&uacute;mero de is&oacute;topos de fl&uacute;or&#8208;18 decai de forma exponencial, com um\ntempo de meia&#8208;vida de aproximadamente 110 minutos.<\/p><p>A imagem obtida pela t&eacute;cnica de PET &eacute; decorrente da\ndetec&ccedil;&atilde;o de dois f&oacute;tons emitidos em sentidos opostos devido &agrave; aniquila&ccedil;&atilde;o, por\num el&eacute;tron, do p&oacute;sitron resultante do decaimento. A detec&ccedil;&atilde;o &eacute; feita por um\nconjunto de detectores montados num arranjo radial. Ao colidir com um dos\ndetectores, o f&oacute;ton gera cargas no material do detector, as quais, por sua vez,\nresultam em um sinal el&eacute;trico registrado no computador do equipamento de\ntomografia. A intensidade do sinal &eacute; proporcional ao n&uacute;mero de n&uacute;cleos de\nfl&uacute;or&#8208;18 existentes no in&iacute;cio do processo.<\/p><p>a) Ap&oacute;s a realiza&ccedil;&atilde;o de uma imagem PET, o m&eacute;dico\npercebeu um problema no funcionamento do equipamento e o reparo durou 3h40min.\nCalcule a raz&atilde;o entre a intensidade do sinal da imagem obtida ap&oacute;s o reparo do\nequipamento e a da primeira imagem.<\/p><p>b) Calcule a energia de cada f&oacute;ton gerado pelo processo de aniquila&ccedil;&atilde;o el&eacute;tron&#8208;p&oacute;sitron considerando que o p&oacute;sitron e o el&eacute;tron estejam praticamente em repouso. Esta &eacute; a energia m&iacute;nima poss&iacute;vel para esse f&oacute;ton.<\/p><p>c) A carga el&eacute;trica gerada dentro do material do detector pela absor&ccedil;&atilde;o do f&oacute;ton &eacute; proporcional &agrave; energia desse f&oacute;ton. Sabendo&#8208;se que &eacute; necess&aacute;ria a energia de 3<em>eV<\/em> para gerar o equivalente &agrave; carga de um el&eacute;tron no material, estime a carga total gerada quando um f&oacute;ton de energia 600<em>keV<\/em> incide no detector.<\/p><p>Note e adote:<\/p><p>O el&eacute;tron e o p&oacute;sitron, sua antipart&iacute;cula, possuem massas iguais e cargas de sinais opostos. Rela&ccedil;&atilde;o de Einstein para a energia de repouso de uma part&iacute;cula: E = mc<sup>2<\/sup>.<\/p><p>Carga do el&eacute;tron <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;=1,6%5Ccdot%7B10%7D%5E%7B-19%7D%5C&amp;space;C\" alt=\"\\large =1,6\\cdot{10}^{-19}\\ C\" align=\"absmiddle\"><\/p><p>Massa do el&eacute;tron: <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;m=9%5Ccdot%7B10%7D%5E%7B-31%7D%5C&amp;space;kg\" alt=\"\\large m=9\\cdot{10}^{-31}\\ kg\" align=\"absmiddle\"><\/p><p>Velocidade da luz: <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;c=3%5Ccdot%7B10%7D%5E8%5C&amp;space;m\/s\" alt=\"\\large c=3\\cdot{10}^8\\ m\/s\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;1%5C&amp;space;eV=1,6%5Ccdot%7B10%7D%5E%7B-19%7D%5C&amp;space;J\" alt=\"\\large 1\\ eV=1,6\\cdot{10}^{-19}\\ J\" align=\"absmiddle\"><\/p><p>&ldquo;Tempo de meia&#8208;vida&rdquo;: tempo necess&aacute;rio para que o n&uacute;mero de n&uacute;cleos radioativos caia para metade do valor inicial.<\/p><h3>Gabarito<\/h3><p>a) Como o texto diz que &ldquo;A intensidade do sinal &eacute; proporcional ao n&uacute;mero de n&uacute;cleos de fl&uacute;or&#8208;18 existentes no in&iacute;cio do processo&rdquo;, podemos relacionar a intensidade com o que sobrou, ap&oacute;s a degrada&ccedil;&atilde;o dos n&uacute;cleos radioativos. Podemos relacionar essa quantidade com o tempo de meia-vida:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;N=%5Cfrac%7BN_0%7D%7B2%5En%7D\" alt=\"\\large N=\\frac{N_0}{2^n}\" align=\"absmiddle\"><\/p><p>O n&uacute;mero de meias-vidas n pode ser calculado pela raz&atilde;o entre o tempo total decorrido e o tempo de meia-vida, desde que esses estejam em uma mesma unidade de tempo:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;n=%5Cfrac%7Bt%7D%7B%5Cfrac%7Bt_1%7D%7B2%7D%7D=%5Cfrac%7B3h%5C&amp;space;40%5C&amp;space;min%7D%7B110%5C&amp;space;min%7D=%5Cfrac%7B220%5C&amp;space;min%7D%7B110%5C&amp;space;min%7D=2\" alt=\"\\large n=\\frac{t}{\\frac{t_1}{2}}=\\frac{3h\\ 40\\ min}{110\\ min}=\\frac{220\\ min}{110\\ min}=2\" align=\"absmiddle\"><\/p><p>Voltando &agrave; express&atilde;o anterior:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;N=%5Cfrac%7BN_0%7D%7B2%5En%7D=%5Cfrac%7BN_0%7D%7B2%5E2%7D=%5Cfrac%7BN_0%7D%7B4%7D\" alt=\"\\large N=\\frac{N_0}{2^n}=\\frac{N_0}{2^2}=\\frac{N_0}{4}\" align=\"absmiddle\"><\/p><p>Isso nos permite concluir que a raz&atilde;o pedida vale 1\/4.<\/p><p>b) Devemos usar a equa&ccedil;&atilde;o de Einstein fornecida no &ldquo;Note e adote&rdquo;, com o cuidado de nos lembrar que se trata de duas part&iacute;culas envolvidas na aniquila&ccedil;&atilde;o: o el&eacute;tron e o p&oacute;sitron e que s&atilde;o formados dois f&oacute;tons:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;E=m%5Ccdot&amp;space;c%5E2\" alt=\"\\large E=m\\cdot c^2\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;2%5Ccdot&amp;space;E=%5Cleft(2%5Ccdot9%5Ccdot%7B10%7D%5E%7B-31%7D%5Cright)%5Ccdot%5Cleft(3%5Ccdot%7B10%7D%5E8%5Cright)%5E2\" alt=\"\\large 2\\cdot E=\\left(2\\cdot9\\cdot{10}^{-31}\\right)\\cdot\\left(3\\cdot{10}^8\\right)^2\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;E=9%5Ccdot%7B10%7D%5E%7B-31%7D%5Ccdot9%5Ccdot%7B10%7D%5E%7B16%7D\" alt=\"\\large E=9\\cdot{10}^{-31}\\cdot9\\cdot{10}^{16}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;E=81%5Ccdot%7B10%7D%5E%7B-15%7D=8,1%5Ccdot%7B10%7D%5E%7B-14%7D%5C&amp;space;J\" alt=\"\\large E=81\\cdot{10}^{-15}=8,1\\cdot{10}^{-14}\\ J\" align=\"absmiddle\"><\/p><p>c) O enunciado diz que &ldquo;A carga el&eacute;trica gerada dentro do material do detector pela absor&ccedil;&atilde;o do f&oacute;ton &eacute; proporcional &agrave; energia desse f&oacute;ton&rdquo;. Devemos relacionar os dados fornecidos no enunciado:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;3%5C&amp;space;eV&amp;space;%5C&amp;space;%5C&amp;space;%5C&amp;space;%5C&amp;space;----------&amp;space;%5C&amp;space;%5C&amp;space;%5C&amp;space;%5C&amp;space;1%5C&amp;space;qe&amp;space;%5C%5C&amp;space;%5C%5C&amp;space;600%5Ccdot&amp;space;10%5E%7B3%7D%5C&amp;space;eV&amp;space;%5C&amp;space;%5C&amp;space;%5C&amp;space;%5C&amp;space;----------&amp;space;%5C&amp;space;%5C&amp;space;%5C&amp;space;%5C&amp;space;q\" alt=\"\\large 3\\ eV \\ \\ \\ \\ ---------- \\ \\ \\ \\ 1\\ qe \\\\ \\\\ 600\\cdot 10^{3}\\ eV \\ \\ \\ \\ ---------- \\ \\ \\ \\ q\" align=\"absmiddle\"><\/p><p>Como as grandezas s&atilde;o diretamente proporcionais, podemos efetuar a multiplica&ccedil;&atilde;o cruzada dos valores:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;3%5Ccdot&amp;space;q=1%5Ccdot&amp;space;600%5Ccdot%7B10%7D%5E3\" alt=\"\\large 3\\cdot q=1\\cdot 600\\cdot{10}^3\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;q=200%5Ccdot%7B10%7D%5E3=2%5Ccdot%7B10%7D%5E5%5C&amp;space;qe\" alt=\"\\large q=200\\cdot{10}^3=2\\cdot{10}^5\\ qe\" align=\"absmiddle\"><\/p><p>Devemos substituir a carga do el&eacute;tron para determinarmos a carga em coulomb:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;q=2%5Ccdot%7B10%7D%5E5%5Ccdot1,6%5Ccdot%7B10%7D%5E%7B-19%7D=3,2%5Ccdot%7B10%7D%5E%7B-14%7D%5C&amp;space;C\" alt=\"\\large q=2\\cdot{10}^5\\cdot1,6\\cdot{10}^{-19}=3,2\\cdot{10}^{-14}\\ C\" align=\"absmiddle\"><\/p><h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Questao-04\"><\/span>Quest&atilde;o 04<span class=\"ez-toc-section-end\"><\/span><\/h2><p>Em um ambiente do qual se retirou praticamente todo o ar, as placas de um capacitor est&atilde;o arranjadas paralelamente e carregadas com cargas de mesma magnitude Q e sinais contr&aacute;rios, produzindo, na regi&atilde;o entre as placas, um campo el&eacute;trico que pode ser considerado uniforme, com m&oacute;dulo igual a <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%7B10%7D%5E6%5C&amp;space;V\/m\" alt=\"\\large {10}^6\\ V\/m\" align=\"absmiddle\">. Uma part&iacute;cula carregada negativamente, com carga de m&oacute;dulo igual a <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%7B10%7D%5E%7B-9%7D%5C&amp;space;C\" alt=\"\\large {10}^{-9}\\ C\" align=\"absmiddle\">, &eacute; lan&ccedil;ada 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\"> igual a <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;100%5C&amp;space;m\/s\" alt=\"\\large 100\\ m\/s\" align=\"absmiddle\"> ao longo da linha que passa exatamente pelo centro da regi&atilde;o entre as placas, como mostrado na figura. A dist&acirc;ncia d entre as placas &eacute; igual a <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;1%5C&amp;space;mm\" alt=\"\\large 1\\ mm\" align=\"absmiddle\">. Despreze os efeitos gravitacionais.<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2020\/01\/image26.png\" alt=\"\" class=\"wp-image-37396\"><\/figure><\/div><p>a) Aponte, entre as trajet&oacute;rias 1 e 2 mostradas na figura, aquela que mais se aproxima do movimento da part&iacute;cula na regi&atilde;o entre as placas.<\/p><p>b) Sabendo que a massa da part&iacute;cula &eacute; igual a <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;10%5C&amp;space;%5Cmu&amp;space;g\" alt=\"\\large 10\\ \\mu g\" align=\"absmiddle\">, determine a que dist&acirc;ncia horizontal x a part&iacute;cula atingir&aacute; uma das placas, supondo que elas sejam suficientemente longas.<\/p><p>c) Quais seriam o sentido e o m&oacute;dulo de um eventual campo magn&eacute;tico a ser aplicado na regi&atilde;o entre as placas, perpendicularmente ao plano da p&aacute;gina, para que a part&iacute;cula, em vez de seguir uma trajet&oacute;ria curva, permane&ccedil;a movendo&#8208;se na mesma dire&ccedil;&atilde;o e no mesmo sentido com que foi lan&ccedil;ada?<\/p><h3>Gabarito<\/h3><p>a) Como a carga &eacute; negativa, temos que a for&ccedil;a el&eacute;trica ser&aacute; vertical e para cima, fazendo com que a part&iacute;cula descreva a trajet&oacute;ria 1. Devemos nos lembrar que o enunciado fala para desconsiderarmos efeitos gravitacionais, ou seja, devemos desconsiderar a a&ccedil;&atilde;o da for&ccedil;a peso.<\/p><p>b) Devemos calcular a acelera&ccedil;&atilde;o vertical, sabendo que a for&ccedil;a resultante &eacute; a for&ccedil;a el&eacute;trica:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;F_r=m%5Ccdot&amp;space;a\" alt=\"\\large F_r=m\\cdot a\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;F_%7Bel%7D=m%5Ccdot&amp;space;a_y%5CRightarrow&amp;space;a_y=%5Cfrac%7Bq%5Ccdot&amp;space;E%7D%7Bm%7D\" alt=\"\\large F_{el}=m\\cdot a_y\\Rightarrow a_y=\\frac{q\\cdot E}{m}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;a_y=%5Cfrac%7B%7B10%7D%5E%7B-9%7D%5Ccdot%7B10%7D%5E6%7D%7B10%5Ccdot%7B10%7D%5E%7B-6%7D%5Ccdot%7B10%7D%5E%7B-3%7D%7D=%5Cfrac%7B%7B10%7D%5E%7B-3%7D%7D%7B%7B10%7D%5E%7B-8%7D%7D=%7B10%7D%5E5%5C&amp;space;m\/s%5E2\" alt=\"\\large a_y=\\frac{{10}^{-9}\\cdot{10}^6}{10\\cdot{10}^{-6}\\cdot{10}^{-3}}=\\frac{{10}^{-3}}{{10}^{-8}}={10}^5\\ m\/s^2\" align=\"absmiddle\"><\/p><p>Podemos calcular o tempo para que a part&iacute;cula chegue at&eacute; a placa positiva do capacitor usando a equa&ccedil;&atilde;o da posi&ccedil;&atilde;o para o MRUV. Note que a velocidade vertical inicial &eacute; nula e que a dist&acirc;ncia a ser percorrida &eacute; a metade da dist&acirc;ncia entre as placas:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5CDelta&amp;space;S_y=v_0%5Ccdot&amp;space;t+%5Cfrac%7Ba%5Ccdot&amp;space;t%5E2%7D%7B2%7D\" alt=\"\\large \\Delta S_y=v_0\\cdot t+\\frac{a\\cdot t^2}{2}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7B1%5Ccdot%7B10%7D%5E%7B-3%7D%7D%7B2%7D=%5Cfrac%7B%7B10%7D%5E5%5Ccdot&amp;space;t%5E2%7D%7B2%7D\" alt=\"\\large \\frac{1\\cdot{10}^{-3}}{2}=\\frac{{10}^5\\cdot t^2}{2}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;t%5E2=%7B10%7D%5E%7B-8%7D%5CRightarrow&amp;space;t=%7B10%7D%5E%7B-4%7D%5C&amp;space;s\" alt=\"\\large t^2={10}^{-8}\\Rightarrow t={10}^{-4}\\ s\" align=\"absmiddle\"><\/p><p>Sabemos que o deslocamento horizontal &eacute; uniforme. De posse do tempo, podemos calcular a dist&acirc;ncia horizontal percorrida:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5CDelta&amp;space;_%7Bx%7D=v_x%5Ccdot&amp;space;t=100%5Ccdot%7B10%7D%5E%7B-4%7D=%7B10%7D%5E%7B-2%7D%5C&amp;space;m=1%5C&amp;space;cm\" alt=\"\\large \\Delta _{x}=v_x\\cdot t=100\\cdot{10}^{-4}={10}^{-2}\\ m=1\\ cm\" align=\"absmiddle\"><\/p><p>c) Para que a part&iacute;cula permane&ccedil;a movendo-se na mesma dire&ccedil;&atilde;o e sentido com que foi lan&ccedil;ada, a resultante vertical das for&ccedil;as deve ser nula. Para que isso aconte&ccedil;a, a for&ccedil;a magn&eacute;tica dever&aacute; ser vertical e para baixo.<\/p><p>Nessa hip&oacute;tese, para a part&iacute;cula carregada negativamente e com velocidade horizontal e para a direita, precisaremos do campo magn&eacute;tico entrando no plano do papel. Isso pode ser verificado pela regra da m&atilde;o direita:<\/p><figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2020\/01\/image27.png\" alt=\"\" class=\"wp-image-37399\"><\/figure><p>Podemos calcular o m&oacute;dulo do campo magn&eacute;tico igualando o m&oacute;dulo da for&ccedil;a magn&eacute;tica ao m&oacute;dulo da for&ccedil;a el&eacute;trica:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;F_%7Bmag%7D=F_%7Bel%7D\" alt=\"\\large F_{mag}=F_{el}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;q%5Ccdot&amp;space;v%5Ccdot&amp;space;B=q%5Ccdot&amp;space;E\" alt=\"\\large q\\cdot v\\cdot B=q\\cdot E\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;B=%5Cfrac%7BE%7D%7Bv%7D=%5Cfrac%7B%7B10%7D%5E6%7D%7B100%7D=%7B10%7D%5E4%5C&amp;space;T\" alt=\"\\large B=\\frac{E}{v}=\\frac{{10}^6}{100}={10}^4\\ T\" align=\"absmiddle\"><\/p><h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Questao-05\"><\/span>Quest&atilde;o 05<span class=\"ez-toc-section-end\"><\/span><\/h2><p>Em janeiro de 2019, a sonda chinesa <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;Chang'\" alt=\"\\large Chang'\" align=\"absmiddle\"> e 4 fez o primeiro pouso suave de um objeto terrestre no lado oculto da Lua, reavivando a discuss&atilde;o internacional sobre programas de explora&ccedil;&atilde;o lunar.<\/p><p>Considere que a trajet&oacute;ria de uma sonda com destino &agrave; Lua passa por um ponto P, localizado a <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;2\/3%5C&amp;space;d_%7BTL%7D\" alt=\"\\large 2\/3\\ d_{TL}\" align=\"absmiddle\"> do centro da Terra e a <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;1\/3%5C&amp;space;d_%7BTL%7D\" alt=\"\\large 1\/3\\ d_{TL}\" align=\"absmiddle\"> do centro da Lua, sendo <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;d_%7BTL%7D\" alt=\"\\large d_{TL}\" align=\"absmiddle\"> a dist&acirc;ncia entre os centros da Terra e da Lua.<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2020\/01\/image28.png\" alt=\"\" class=\"wp-image-37402\"><\/figure><\/div><p>a) Considerando que a massa da Terra &eacute; cerca de 82 vezes maior que a massa da Lua, determine a raz&atilde;o <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;F_T\/F_L\" alt=\"\\large F_T\/F_L\" align=\"absmiddle\"> entre os m&oacute;dulos da for&ccedil;a gravitacional que a Terra e a Lua, respectivamente, exercem sobre a sonda no ponto P.<\/p><p>Ao chegar pr&oacute;ximo &agrave; Lua, a sonda foi colocada em uma &oacute;rbita lunar circular a uma altura igual ao raio da <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;Lua&amp;space;(R_L)\" alt=\"\\large Lua (R_L)\" align=\"absmiddle\">, acima de sua superf&iacute;cie, como mostra a figura. Desprezando os efeitos da for&ccedil;a gravitacional da Terra e de outros corpos celestes ao longo da &oacute;rbita da sonda,<\/p><p>b) determine a velocidade orbital da sonda em torno da Lua em termos da constante gravitacional G, da massa da <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?%5Clarge&amp;space;Lua%5C&amp;space;M_L\" alt=\"\\large Lua\\ M_L\" align=\"absmiddle\"> e do raio da <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;Lua%5C&amp;space;R_L\" alt=\"\\large Lua\\ R_L\" align=\"absmiddle\">;<\/p><p>c) determine a varia&ccedil;&atilde;o da energia mec&acirc;nica da nave quando a altura da &oacute;rbita, em rela&ccedil;&atilde;o &agrave; superf&iacute;cie da Lua, &eacute; reduzida para <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;0,5%5C&amp;space;R_L\" alt=\"\\large 0,5\\ R_L\" align=\"absmiddle\">. Expresse seu resultado em termos de G, R_L, M_L e da massa da sonda <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;m_S\" alt=\"\\large m_S\" align=\"absmiddle\">.<\/p><p>Note e adote:<\/p><p>O m&oacute;dulo da for&ccedil;a gravitacional entre dois objetos de massas M e m separados por uma dist&acirc;ncia d &eacute; dado por <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;F=GMm\/d%5E2\" alt=\"\\large F=GMm\/d^2\" align=\"absmiddle\"><\/p><p>A energia potencial gravitacional correspondente &eacute; dada por <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;U=-GMd\/d\" alt=\"\\large U=-GMd\/d\" align=\"absmiddle\">.<br>Assuma a dist&acirc;ncia da Terra &agrave; Lua como sendo constante.<\/p><h3>Gabarito<\/h3><p>a) Podemos calcular a raz&atilde;o pedida usando a rela&ccedil;&atilde;o fornecida no &ldquo;Note e adote&rdquo;:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7BF_T%7D%7BF_L%7D=%5Cfrac%7B%5Cfrac%7BG%5Ccdot&amp;space;M_T%5Ccdot&amp;space;m_S%7D%7B%5Cleft(%5Cfrac%7B2%7D%7B3%7Dd_%7BTL%7D%5Cright)%5E2%7D%7D%7B%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B%5Cleft(%5Cfrac%7B1%7D%7B3%7Dd_%7BTL%7D%5Cright)%5E2%7D%7D=%5Cfrac%7B%5Cfrac%7BG%5Ccdot&amp;space;M_T%5Ccdot&amp;space;m_S%7D%7B%5Cfrac%7B4%7D%7B9%7D%5Ccdot%5Cleft(d_%7BTL%7D%5Cright)%5E2%7D%7D%7B%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B%5Cfrac%7B1%7D%7B9%7D%5Ccdot%5Cleft(d_%7BTL%7D%5Cright)%5E2%7D%7D=%5Cfrac%7B%5Cfrac%7BG%5Ccdot&amp;space;M_T%5Ccdot&amp;space;m_S%7D%7B%5Cfrac%7B4%7D%7B9%7D%5Ccdot%5Cleft(d_%7BTL%7D%5Cright)%5E2%7D%7D%7B%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B%5Cfrac%7B1%7D%7B9%7D%5Ccdot%5Cleft(d_%7BTL%7D%5Cright)%5E2%7D%7D\" alt=\"\\large \\frac{F_T}{F_L}=\\frac{\\frac{G\\cdot M_T\\cdot m_S}{\\left(\\frac{2}{3}d_{TL}\\right)^2}}{\\frac{G\\cdot M_L\\cdot m_S}{\\left(\\frac{1}{3}d_{TL}\\right)^2}}=\\frac{\\frac{G\\cdot M_T\\cdot m_S}{\\frac{4}{9}\\cdot\\left(d_{TL}\\right)^2}}{\\frac{G\\cdot M_L\\cdot m_S}{\\frac{1}{9}\\cdot\\left(d_{TL}\\right)^2}}=\\frac{\\frac{G\\cdot M_T\\cdot m_S}{\\frac{4}{9}\\cdot\\left(d_{TL}\\right)^2}}{\\frac{G\\cdot M_L\\cdot m_S}{\\frac{1}{9}\\cdot\\left(d_{TL}\\right)^2}}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7BF_T%7D%7BF_L%7D=%5Cfrac%7B%5Cfrac%7BM_T%7D%7B4%7D%7D%7B%5Cfrac%7BM_L%7D%7B1%7D%7D=%5Cfrac%7B%5Cfrac%7BM_T%7D%7B4%7D%7D%7BM_L%7D=%5Cfrac%7BM_T%7D%7B4%5Ccdot&amp;space;M_L%7D\" alt=\"\\large \\frac{F_T}{F_L}=\\frac{\\frac{M_T}{4}}{\\frac{M_L}{1}}=\\frac{\\frac{M_T}{4}}{M_L}=\\frac{M_T}{4\\cdot M_L}\" align=\"absmiddle\"><\/p><p>Como a massa da Terra &eacute; 82 vezes maior que a massa da Lua:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7BF_T%7D%7BF_L%7D=%5Cfrac%7B82%5Ccdot&amp;space;M_L%7D%7B4%5Ccdot&amp;space;M_L%7D=%5Cfrac%7B82%5Ccdot&amp;space;M_L%7D%7B4%5Ccdot&amp;space;M_L%7D=%5Cfrac%7B41%7D%7B2%7D\" alt=\"\\large \\frac{F_T}{F_L}=\\frac{82\\cdot M_L}{4\\cdot M_L}=\\frac{82\\cdot M_L}{4\\cdot M_L}=\\frac{41}{2}\" align=\"absmiddle\"><\/p><p>b) Quando a sonda descreve uma trajet&oacute;ria circular em torno da Lua, temos que a for&ccedil;a gravitacional atua como resultante centr&iacute;peta:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;F_%7Bgrav%7D=F_%7Bcp%7D\" alt=\"\\large F_{grav}=F_{cp}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B%5Cleft(2%5Ccdot&amp;space;R_L%5Cright)%5E2%7D=%5Cfrac%7Bm_S%5Ccdot&amp;space;v%5E2%7D%7B2%5Ccdot&amp;space;R_L%7D\" alt=\"\\large \\frac{G\\cdot M_L\\cdot m_S}{\\left(2\\cdot R_L\\right)^2}=\\frac{m_S\\cdot v^2}{2\\cdot R_L}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B4%5Ccdot%5Cleft(R_L%5Cright)%5E2%7D=%5Cfrac%7Bm_S%5Ccdot&amp;space;v%5E2%7D%7B%7B2%5Ccdot&amp;space;R%7D_L%7D\" alt=\"\\large \\frac{G\\cdot M_L\\cdot m_S}{4\\cdot\\left(R_L\\right)^2}=\\frac{m_S\\cdot v^2}{{2\\cdot R}_L}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;v=%5Csqrt%7B%5Cfrac%7BG%5Ccdot&amp;space;M_L%7D%7B2%5Ccdot&amp;space;R_L%7D%7D\" alt=\"\\large v=\\sqrt{\\frac{G\\cdot M_L}{2\\cdot R_L}}\" align=\"absmiddle\"><\/p><p>c) E energia mec&acirc;nica da nave &eacute; dada pela soma da energia potencial gravitacional e da energia cin&eacute;tica. Para a situa&ccedil;&atilde;o do sat&eacute;lite a uma dist&acirc;ncia <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;R_L\" alt=\"\\large R_L\" align=\"absmiddle\"> da superf&iacute;cie da Lua:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;EM=Epot+Ec\" alt=\"\\large EM=Epot+Ec\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;EM_%7BR_L%7D=%5Cfrac%7B-G%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B2%5Ccdot&amp;space;R_L%7D+%5Cfrac%7Bm_S%5Ccdot&amp;space;v%5E2%7D%7B2%7D\" alt=\"\\large EM_{R_L}=\\frac{-G\\cdot M_L\\cdot m_S}{2\\cdot R_L}+\\frac{m_S\\cdot v^2}{2}\" align=\"absmiddle\"><\/p><p>Podemos escrever a energia cin&eacute;tica em fun&ccedil;&atilde;o dos elementos da alternativa anterior:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7Bm_S%5Ccdot&amp;space;v%5E2%7D%7B2%5Ccdot&amp;space;R_L%7D=%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B4%5Ccdot%5Cleft(R_L%5Cright)%5E2%7D\" alt=\"\\large \\frac{m_S\\cdot v^2}{2\\cdot R_L}=\\frac{G\\cdot M_L\\cdot m_S}{4\\cdot\\left(R_L\\right)^2}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7Bm_S%5Ccdot&amp;space;v%5E2%7D%7B2%7D=%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B4%5Ccdot&amp;space;R_L%7D\" alt=\"\\large \\frac{m_S\\cdot v^2}{2}=\\frac{G\\cdot M_L\\cdot m_S}{4\\cdot R_L}\" align=\"absmiddle\"><\/p><p>Voltando&nbsp; &agrave; express&atilde;o anterior:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;EM_%7BR_L%7D=%5Cfrac%7B-G%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B2%5Ccdot&amp;space;R_L%7D+%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B4%5Ccdot&amp;space;R_L%7D\" alt=\"\\large EM_{R_L}=\\frac{-G\\cdot M_L\\cdot m_S}{2\\cdot R_L}+\\frac{G\\cdot M_L\\cdot m_S}{4\\cdot R_L}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;EM_%7BR_L%7D=%5Cfrac%7B-2%5Ccdot&amp;space;G%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B4%5Ccdot&amp;space;R_L%7D+%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B4%5Ccdot&amp;space;R_L%7D\" alt=\"\\large EM_{R_L}=\\frac{-2\\cdot G\\cdot M_L\\cdot m_S}{4\\cdot R_L}+\\frac{G\\cdot M_L\\cdot m_S}{4\\cdot R_L}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;EM_%7BR_L%7D=%5Cfrac%7B-G%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B4%5Ccdot&amp;space;R_L%7D\" alt=\"\\large EM_{R_L}=\\frac{-G\\cdot M_L\\cdot m_S}{4\\cdot R_L}\" align=\"absmiddle\"><\/p><p>De maneira an&aacute;loga, para a dist&acirc;ncia reduzida para <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;0,5%5C&amp;space;R_L\" alt=\"\\large 0,5\\ R_L\" align=\"absmiddle\">:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;EM_%7B%7B0,5R%7D_L%7D=%5Cfrac%7B-G%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B1,5%5Ccdot&amp;space;R_L%7D+%5Cfrac%7Bm_S%5Ccdot&amp;space;v_%7B%7B0,5R%7D_L%7D%5E2%7D%7B2%7D\" alt=\"\\large EM_{{0,5R}_L}=\\frac{-G\\cdot M_L\\cdot m_S}{1,5\\cdot R_L}+\\frac{m_S\\cdot v_{{0,5R}_L}^2}{2}\" align=\"absmiddle\"><\/p><p>E para o termo da cin&eacute;tica:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7Bm_S%5Ccdot&amp;space;v%5E2%7D%7B%7B%5Cfrac%7B3%7D%7B2%7DR%7D_L%7D=%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B%5Cleft(%5Cfrac%7B3%7D%7B2%7D%5Ccdot&amp;space;R_L%5Cright)%5E2%7D\" alt=\"\\large \\frac{m_S\\cdot v^2}{{\\frac{3}{2}R}_L}=\\frac{G\\cdot M_L\\cdot m_S}{\\left(\\frac{3}{2}\\cdot R_L\\right)^2}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7Bm_S%5Ccdot&amp;space;v%5E2%7D%7B1%7D=%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B%5Cfrac%7B3%7D%7B2%7D%5Ccdot&amp;space;R_L%7D\" alt=\"\\large \\frac{m_S\\cdot v^2}{1}=\\frac{G\\cdot M_L\\cdot m_S}{\\frac{3}{2}\\cdot R_L}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;%5Cfrac%7Bm_S%5Ccdot&amp;space;v%5E2%7D%7B2%7D=%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B3%5Ccdot&amp;space;R_L%7D\" alt=\"\\large \\frac{m_S\\cdot v^2}{2}=\\frac{G\\cdot M_L\\cdot m_S}{3\\cdot R_L}\" align=\"absmiddle\"><\/p><p>Voltando:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;EM_%7B%7B0,5R%7D_L%7D=%5Cfrac%7B-G%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B%5Cfrac%7B3%7D%7B2%7D%5Ccdot&amp;space;R_L%7D+%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B3%5Ccdot&amp;space;R_L%7D\" alt=\"\\large EM_{{0,5R}_L}=\\frac{-G\\cdot M_L\\cdot m_S}{\\frac{3}{2}\\cdot R_L}+\\frac{G\\cdot M_L\\cdot m_S}{3\\cdot R_L}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;EM_%7B%7B0,5R%7D_L%7D=-%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B3%5Ccdot&amp;space;R_L%7D\" alt=\"\\large EM_{{0,5R}_L}=-\\frac{G\\cdot M_L\\cdot m_S}{3\\cdot R_L}\" align=\"absmiddle\"><\/p><p>E a diferen&ccedil;a ser&aacute; de:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;EM_%7B%7B0,5R%7D_L%7D-EM_%7BR_L%7D=-%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B3%5Ccdot&amp;space;R_L%7D-%5Cleft(%5Cfrac%7B-G%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B4%5Ccdot&amp;space;R_L%7D%5Cright)\" alt=\"\\large EM_{{0,5R}_L}-EM_{R_L}=-\\frac{G\\cdot M_L\\cdot m_S}{3\\cdot R_L}-\\left(\\frac{-G\\cdot M_L\\cdot m_S}{4\\cdot R_L}\\right)\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;EM_%7BR_L%7D-EM_%7B%7B0,5R%7D_L%7D=%5Cfrac%7B-G%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B3%5Ccdot&amp;space;R_L%7D+%5Cfrac%7BG%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B4%5Ccdot&amp;space;R_L%7D\" alt=\"\\large EM_{R_L}-EM_{{0,5R}_L}=\\frac{-G\\cdot M_L\\cdot m_S}{3\\cdot R_L}+\\frac{G\\cdot M_L\\cdot m_S}{4\\cdot R_L}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;EM_%7BR_L%7D-EM_%7B%7B0,5R%7D_L%7D=%5Cfrac%7B-4%5Ccdot&amp;space;G%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B12%5Ccdot&amp;space;R_L%7D+%5Cfrac%7B3%5Ccdot&amp;space;G%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B12%5Ccdot&amp;space;R_L%7D\" alt=\"\\large EM_{R_L}-EM_{{0,5R}_L}=\\frac{-4\\cdot G\\cdot M_L\\cdot m_S}{12\\cdot R_L}+\\frac{3\\cdot G\\cdot M_L\\cdot m_S}{12\\cdot R_L}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;EM_%7BR_L%7D-EM_%7B%7B0,5R%7D_L%7D=%5Cfrac%7B-G%5Ccdot&amp;space;M_L%5Ccdot&amp;space;m_S%7D%7B12%5Ccdot&amp;space;R_L%7D\" alt=\"\\large EM_{R_L}-EM_{{0,5R}_L}=\\frac{-G\\cdot M_L\\cdot m_S}{12\\cdot R_L}\" align=\"absmiddle\"><\/p><h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Questao-06\"><\/span>Quest&atilde;o 06<span class=\"ez-toc-section-end\"><\/span><\/h2><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M=60%5C&amp;space;kg\" alt=\"\\large M=60\\ kg\" align=\"absmiddle\">Uma equilibrista de massa M desloca&#8208;se sobre uma t&aacute;bua uniforme de comprimento L e massa m apoiada (sem fixa&ccedil;&atilde;o) sobre duas colunas separadas por uma dist&acirc;ncia <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;D%5C&amp;space;(D&lt;l)\" alt=\"\\large D\\ (D&lt;l)\" align=\"absmiddle\"> de modo que o centro da t&aacute;bua esteja equidistante das colunas. O ponto de apoio da equilibrista est&aacute; a uma dist&acirc;ncia <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;d%5C&amp;space;(tal%5C&amp;space;que%5C&amp;space;D\/2&lt;d&lt;l\/2)\" alt=\"\\large d\\ (tal\\ que\\ D\/2&lt;d&lt;l\/2)\" align=\"absmiddle\"> do centro da t&aacute;bua, como mostra a figura.<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2020\/01\/image29.png\" alt=\"\" class=\"wp-image-37403\"><\/figure><\/div><p>a) Considerando que a t&aacute;bua est&aacute; em equil&iacute;brio, fa&ccedil;a um diagrama indicando todas as for&ccedil;as que atuam <strong><span style=\"text-decoration: underline;\">sobre a t&aacute;bua<\/span><\/strong> e seus respectivos pontos de aplica&ccedil;&atilde;o.<\/p><p>b) Calcule o torque resultante exercido pelos pesos da equilibrista e da t&aacute;bua em rela&ccedil;&atilde;o ao ponto A (ponto de apoio da t&aacute;bua na coluna mais pr&oacute;xima da equilibrista). Escreva sua resposta em termos de grandezas mencionadas no enunciado <em>(M, L, m, D, d)<\/em> e da acelera&ccedil;&atilde;o da gravidade g.<\/p><p>c) Calcule a dist&acirc;ncia m&aacute;xima <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;d_%7Bmax%7D\" alt=\"\\large d_{max}\" align=\"absmiddle\"> da equilibrista ao centro da t&aacute;bua para que o conjunto permane&ccedil;a em equil&iacute;brio est&aacute;tico.<br>Considere os seguintes dados: comprimento da t&aacute;bua: <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;L=5%5C&amp;space;m\" alt=\"\\large L=5\\ m\" align=\"absmiddle\">; massa da t&aacute;bua: <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;m=20%5C&amp;space;kg\" alt=\"\\large m=20\\ kg\" align=\"absmiddle\">, massa da equilibrista: , dist&acirc;ncia entre as colunas: <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;D=3%5C&amp;space;m\" alt=\"\\large D=3\\ m\" align=\"absmiddle\">.<\/p><p><strong>Note e adote:<\/strong><\/p><p>Despreze as espessuras da t&aacute;bua e da coluna.<\/p><p>Use <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;g=10%5C&amp;space;m\/s%5E2\" alt=\"\\large g=10\\ m\/s^2\" align=\"absmiddle\"><\/p><h3 class=\"wp-block-heading\">Gabarito<\/h3><p>a)<\/p><div class=\"wp-block-image\"><figure class=\"aligncenter\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2020\/01\/image30.png\" alt=\"\" class=\"wp-image-37404\"><\/figure><\/div><p>b) Devemos nos lembrar que a conven&ccedil;&atilde;o adota torque no sentido anti-hor&aacute;rio como positivo. Com isso, o torque gerado pela equilibrista ser&aacute; negativo e pelo peso da t&aacute;bua positivo. O torque resultante ser&aacute; dado por:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M_%7Bres%7D=m%5Ccdot&amp;space;g%5Ccdot%5Cfrac%7BD%7D%7B2%7D-M%5Ccdot&amp;space;g%5Ccdot%5Cleft(d-%5Cfrac%7BD%7D%7B2%7D%5Cright)\" alt=\"\\large M_{res}=m\\cdot g\\cdot\\frac{D}{2}-M\\cdot g\\cdot\\left(d-\\frac{D}{2}\\right)\" align=\"absmiddle\"><\/p><p>c) Na situa&ccedil;&atilde;o da dist&acirc;ncia m&aacute;xima, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;N_1\" alt=\"\\large N_1\" align=\"absmiddle\"> tender&aacute; a zero:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M_%7Bres%7D=0\" alt=\"\\large M_{res}=0\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;m%5Ccdot&amp;space;g%5Ccdot%5Cfrac%7BD%7D%7B2%7D=M%5Ccdot&amp;space;g%5Ccdot%5Cleft(d-%5Cfrac%7BD%7D%7B2%7D%5Cright)\" alt=\"\\large m\\cdot g\\cdot\\frac{D}{2}=M\\cdot g\\cdot\\left(d-\\frac{D}{2}\\right)\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;m%5Ccdot%5Cfrac%7BD%7D%7B2%7D=M%5Ccdot&amp;space;d-M%5Ccdot%5Cfrac%7BD%7D%7B2%7D\" alt=\"\\large m\\cdot\\frac{D}{2}=M\\cdot d-M\\cdot\\frac{D}{2}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;M%5Ccdot&amp;space;d=m%5Ccdot%5Cfrac%7BD%7D%7B2%7D+M%5Ccdot%5Cfrac%7BD%7D%7B2%7D\" alt=\"\\large M\\cdot d=m\\cdot\\frac{D}{2}+M\\cdot\\frac{D}{2}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;d=%5Cfrac%7B%5Cleft(m+M%5Cright)%5Ccdot&amp;space;D%7D%7B2%5Ccdot&amp;space;M%7D\" alt=\"\\large d=\\frac{\\left(m+M\\right)\\cdot D}{2\\cdot M}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Clarge&amp;space;d=%5Cfrac%7B%5Cleft(20+60%5Cright)%5Ccdot3%7D%7B2%5Ccdot60%7D=%5Cfrac%7B80%5Ccdot3%7D%7B120%7D=%5Cfrac%7B240%7D%7B120%7D=2%5C&amp;space;m\" alt=\"\\large d=\\frac{\\left(20+60\\right)\\cdot3}{2\\cdot60}=\\frac{80\\cdot3}{120}=\\frac{240}{120}=2\\ m\" align=\"absmiddle\"><\/p><p>Com isso, encerramos a nossa corre&ccedil;&atilde;o de F&iacute;sica da 2&ordf; Fase da UNESP 2020. Se ficou alguma d&uacute;vida, pode entrar em contato comigo atrav&eacute;s do nosso F&oacute;rum de D&uacute;vidas ou pelas minhas redes sociais. Vou deitar esta corre&ccedil;&atilde;o dispon&iacute;vel para download. Voc&ecirc; vai poder baixar de forma gratuita no link a seguir.<\/p><div class=\"wp-block-file\">\"&gt;Resolu&ccedil;&atilde;o FUVEST 2020 &ndash; 2&ordf; Fase &ndash; F&iacute;sica\" class=\"wp-block-file__button\" download&gt;Baixar<\/div><p>Abra&ccedil;os<\/p><p>Prof. Lucas Costa<\/p><p><strong>Instagram:<\/strong>&nbsp;<a href=\"https:\/\/instagram.com\/prof.lucascosta\" target=\"_blank\">@prof.lucascosta<\/a><\/p><p class=\"has-text-align-center has-light-green-cyan-background-color has-background has-medium-font-size\"><a aria-label=\"CURSOS PARA FIVEST (abre numa nova aba)\" href=\"https:\/\/estrategiavestibulares.com.br\/vestibulares\/fuvest\/\" target=\"_blank\">CURSOS PARA FUVEST<\/a><\/p><\/p>\n","protected":false},"excerpt":{"rendered":"Fala, pessoal. Tudo certo? Sou o prof. Lucas Costa, professor de F&iacute;sica do Estrat&eacute;gia Vestibulares. Hoje, vamos resolver&hellip;\n","protected":false},"author":18,"featured_media":40454,"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-37373","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>Como resolver as quest\u00f5es de F\u00edsica da 2\u00aa Fase da Fuvest 2020<\/title>\n<meta name=\"description\" content=\"O Estrat\u00e9gia Vestibulares traz para voc\u00ea um guia de como resolver as quest\u00f5es de f\u00edsica da segunda fase do vestibular Unesp! 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