{"id":37695,"date":"2020-01-14T11:37:04","date_gmt":"2020-01-14T14:37:04","guid":{"rendered":"https:\/\/blog.estrategiavestibulares.com.br\/?p=37695"},"modified":"2022-05-06T18:59:25","modified_gmt":"2022-05-06T21:59:25","slug":"prova-fisica-2-fase-unicamp-2020","status":"publish","type":"post","link":"https:\/\/vestibulares.estrategia.com\/portal\/materias\/fisica\/prova-fisica-2-fase-unicamp-2020\/","title":{"rendered":"Resolu\u00e7\u00e3o da prova de F\u00edsica Unicamp 2\u00aa Fase 2020"},"content":{"rendered":"<p>Fala, pessoal. Tudo certo? Sou o prof. Lucas Costa, professor de F&iacute;sica do Estrat&eacute;gia Vestibulares. Escrevo este artigo para, juntos, resolvermos as quest&otilde;es da prova de F&iacute;sica da UNICAMP 2020, 2&ordf; Fase do Vestibular. Vamos nessa?<figure class=\"wp-block-embed aligncenter is-type-video is-provider-youtube wp-block-embed-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<div class=\"cs-embed cs-embed-responsive\"><iframe title=\"Provas Resolvidas f&iacute;sica - UNICAMP 2020\" width=\"1200\" height=\"675\" src=\"https:\/\/www.youtube.com\/embed\/g5zz-YgD_I4?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/div>\n<\/div><\/figure><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\/prova-fisica-2-fase-unicamp-2020\/#Questao-11\" >Quest&atilde;o 11<\/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\/prova-fisica-2-fase-unicamp-2020\/#Questao-12-da-prova-de-Fisica-Unicamp\" >Quest&atilde;o 12 da prova de F&iacute;sica Unicamp<\/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\/prova-fisica-2-fase-unicamp-2020\/#Questao-13\" >Quest&atilde;o 13<\/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\/prova-fisica-2-fase-unicamp-2020\/#Questao-14-da-prova-de-Fisica-Unicamp\" >Quest&atilde;o 14 da prova de F&iacute;sica Unicamp<\/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\/prova-fisica-2-fase-unicamp-2020\/#Questao-15\" >Quest&atilde;o 15<\/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\/prova-fisica-2-fase-unicamp-2020\/#Questao-16-da-prova-de-Fisica-Unicamp\" >Quest&atilde;o 16 da prova de F&iacute;sica Unicamp<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/vestibulares.estrategia.com\/portal\/materias\/fisica\/prova-fisica-2-fase-unicamp-2020\/#Veja-tambem\" >Veja tamb&eacute;m:<\/a><\/li><\/ul><\/nav><\/div>\n<h2 class=\"wp-block-heading\" id=\"h-questao-11\"><span class=\"ez-toc-section\" id=\"Questao-11\"><\/span>Quest&atilde;o 11<span class=\"ez-toc-section-end\"><\/span><\/h2><p>Estudos indicam que uma massa <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;m=1000%5C&amp;space;kg\" alt=\"\\dpi{100} \\large m=1000\\ kg\" align=\"absmiddle\"> de poeira c&oacute;smica, composta por min&uacute;sculas part&iacute;culas, colide com a superf&iacute;cie da Terra a cada intervalo <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5CDelta&amp;space;t=20&amp;space;min\" alt=\"\\dpi{100} \\large \\Delta t=20 min\" align=\"absmiddle\">. Considere, para simplificar, que as part&iacute;culas de poeira t&ecirc;m velocidade m&eacute;dia nula antes de serem arrastadas pela Terra no seu movimento em torno do Sol. Logo ap&oacute;s colidirem com a superf&iacute;cie do nosso planeta, elas passam a se deslocar juntamente com a Terra, com velocidade m&eacute;dia de m&oacute;dulo igual a <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;v_%7BTerra%7D=30%5C&amp;space;km\/s\" alt=\"\\dpi{100} \\large v_{Terra}=30\\ km\/s\" align=\"absmiddle\">. Considere tamb&eacute;m que o movimento da Terra num intervalo <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5CDelta&amp;space;t=20&amp;space;min\" alt=\"\\dpi{100} \\large \\Delta t=20 min\" align=\"absmiddle\"> &eacute; retil&iacute;neo e uniforme.<\/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\/fisica-unicamp-2020-2-fase.jpg\" alt=\"\" class=\"wp-image-37697\"><\/figure><\/div><p>a) Qual &eacute; a densidade da poeira na regi&atilde;o do espa&ccedil;o\natravessada pela Terra? Ver ilustra&ccedil;&atilde;o ao lado.<\/p><p>b) Qual &eacute; o m&oacute;dulo da for&ccedil;a m&eacute;dia aplicada pela Terra sobre a massa de poeira c&oacute;smica que ela intercepta durante um intervalo &#8710;t=20 min?<\/p><p>a) A densidade da poeira &eacute; dada pela raz&atilde;o entre a sua massa e o volume por ela ocupado. A massa &eacute; fornecida e o volume pode ser calculado pelo produto entre a &aacute;rea da se&ccedil;&atilde;o reta e a sua altura, que &eacute; a dist&acirc;ncia percorrida pelo planeta no intervalo de tempo fornecido:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;v=%5Cfrac%7B%5CDelta&amp;space;S%7D%7B%5CDelta&amp;space;t%7D%5CRightarrow&amp;space;%5CDelta&amp;space;S=h=v_%7BTerra%7D%5Ccdot&amp;space;%5CDelta&amp;space;t\" alt=\"\\dpi{100} \\large v=\\frac{\\Delta S}{\\Delta t}\\Rightarrow \\Delta S=h=v_{Terra}\\cdot \\Delta t\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;h=30%5Ccdot%7B10%7D%5E3%5Ccdot20%5Ccdot60=36%5Ccdot%7B10%7D%5E6%5C&amp;space;m\" alt=\"\\dpi{100} \\large h=30\\cdot{10}^3\\cdot20\\cdot60=36\\cdot{10}^6\\ m\" align=\"absmiddle\"><\/p><p>Agora podemos calcular o volume do cilindro correspondente ao volume varrido pela Terra:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;V_%7Bcilindro%7D=A_%7Bbase%7D%5Ccdot&amp;space;h\" alt=\"\\dpi{100} \\large V_{cilindro}=A_{base}\\cdot h\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;V_%7Bcilindro%7D=1,25%5Ccdot%7B10%7D%5E%7B14%7D%5Ccdot36%5Ccdot%7B10%7D%5E6=45%5Ccdot%7B10%7D%5E%7B20%7D%5C&amp;space;m%5E3\" alt=\"\\dpi{100} \\large V_{cilindro}=1,25\\cdot{10}^{14}\\cdot36\\cdot{10}^6=45\\cdot{10}^{20}\\ m^3\" align=\"absmiddle\"><\/p><p>Finalmente, podemos calcular a densidade da poeira espacial:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;d=%5Cfrac%7Bm%7D%7BV%7D\" alt=\"\\dpi{100} \\large d=\\frac{m}{V}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;d=%5Cfrac%7B1000%7D%7B45%5Ccdot%7B10%7D%5E%7B20%7D%7D=22,2%5Ccdot%7B10%7D%5E%7B-20%7D\" alt=\"\\dpi{100} \\large d=\\frac{1000}{45\\cdot{10}^{20}}=22,2\\cdot{10}^{-20}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;d=2,%5Cbar%7B2%7D%5Ccdot%7B10%7D%5E%7B-19%7D%5C&amp;space;kg\/m%5E3\" alt=\"\\dpi{100} \\large d=2,\\bar{2}\\cdot{10}^{-19}\\ kg\/m^3\" align=\"absmiddle\"><\/p><p>b) Pelo teorema do impulso, temos:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Cvec%7BI%7D=%5CDelta&amp;space;%5Cvec%7BQ%7D\" alt=\"\\dpi{100} \\large \\vec{I}=\\Delta \\vec{Q}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Cvec%7BF%7D%5Ccdot%5CDelta&amp;space;t=m%5Ccdot&amp;space;%5CDelta&amp;space;%5Cvec%7Bv%7D\" alt=\"\\dpi{100} \\large \\vec{F}\\cdot\\Delta t=m\\cdot \\Delta \\vec{v}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Cvec%7BF%7D=%5Cfrac%7Bm%5Ccdot&amp;space;%5CDelta&amp;space;%5Cvec%7Bv%7D%7D%7B%5CDelta&amp;space;t%7D\" alt=\"\\dpi{100} \\large \\vec{F}=\\frac{m\\cdot \\Delta \\vec{v}}{\\Delta t}\" align=\"absmiddle\"><\/p><p>Devemos assumir que a massa da poeira &eacute; desprez&iacute;vel frente &agrave; massa da Terra. Al&eacute;m disso, devemos assumir que a velocidade inicial da poeira tamb&eacute;m &eacute; desprez&iacute;vel. Substituindo a massa da Terra:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;F=%5Cfrac%7B1000%5Ccdot30%5Ccdot%7B10%7D%5E3%7D%7B20%5Ccdot60%7D=%5Cfrac%7B30%5Ccdot%7B10%7D%5E6%7D%7B12%5Ccdot%7B10%7D%5E2%7D=2,5%5Ccdot%7B10%7D%5E4%5C&amp;space;N=2,5%5C&amp;space;kN\" alt=\"\\dpi{100} \\large F=\\frac{1000\\cdot30\\cdot{10}^3}{20\\cdot60}=\\frac{30\\cdot{10}^6}{12\\cdot{10}^2}=2,5\\cdot{10}^4\\ N=2,5\\ kN\" align=\"absmiddle\"><\/p><h2 class=\"wp-block-heading\" id=\"h-questao-12-da-prova-de-fisica-unicamp\"><span class=\"ez-toc-section\" id=\"Questao-12-da-prova-de-Fisica-Unicamp\"><\/span>Quest&atilde;o 12 da prova de F&iacute;sica Unicamp<span class=\"ez-toc-section-end\"><\/span><\/h2><p>Um dens&iacute;metro de posto de combust&iacute;vel, usado para analisar o etanol, consiste de um tubo de vidro que fica parcialmente submerso no etanol. O peso do tubo &eacute; fixo, de forma que o volume do tubo que fica submerso depende da densidade do etanol. Uma escala na parte superior do tubo indica o valor da densidade medida.<\/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\/fisica-unicamp-2020-2-fase-questao-12.jpg\" alt=\"\" class=\"wp-image-37700\"><\/figure><\/div><p>a) O etanol combust&iacute;vel &eacute; hidratado, ou seja, cont&eacute;m uma porcentagem de &aacute;gua. A figura ao lado ilustra duas medidas de densidade de etanol. A primeira &eacute; de uma amostra de etanol hidratado dentro da especifica&ccedil;&atilde;o, cujo valor &eacute; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Crho_1=0,810%5C&amp;space;g\/cm%5E3\" alt=\"\\dpi{100} \\large \\rho_1=0,810\\ g\/cm^3\" align=\"absmiddle\">. Nessa medida, o volume submerso do dens&iacute;metro &eacute; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;V_1\" alt=\"\\dpi{100} \\large V_1\" align=\"absmiddle\">. A segunda medida, realizada com o mesmo dens&iacute;metro, &eacute; de uma amostra fora da especifica&ccedil;&atilde;o e, nesse caso, o volume submerso do dens&iacute;metro &eacute; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;V_2\" alt=\"\\dpi{100} \\large V_2\" align=\"absmiddle\">. A diferen&ccedil;a dos volumes submersos &eacute; de 10% de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;V_1\" alt=\"\\dpi{100} \\large V_1\" align=\"absmiddle\">, ou seja, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5CDelta&amp;space;V=V_1-V_2=0,1%5C&amp;space;V_1\" alt=\"\\dpi{100} \\large \\Delta V=V_1-V_2=0,1\\ V_1\" align=\"absmiddle\">. Qual &eacute; a densidade <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Crho_2\" alt=\"\\dpi{100} \\large \\rho_2\" align=\"absmiddle\"> da segunda amostra?<\/p><p>b) Num posto de combust&iacute;vel, a gasolina &eacute; bombeada do reservat&oacute;rio subterr&acirc;neo at&eacute; o tanque do ve&iacute;culo, numa altura <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;h=3,0%5C&amp;space;m\" alt=\"\\dpi{100} \\large h=3,0\\ m\" align=\"absmiddle\"> acima do n&iacute;vel superior do reservat&oacute;rio. A gasolina, que &eacute; sempre retirada da parte superior do reservat&oacute;rio, encontra-se inicialmente parada e &eacute; despejada no tanque do ve&iacute;culo a uma velocidade <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;v=0,8%5C&amp;space;m\/s\" alt=\"\\dpi{100} \\large v=0,8\\ m\/s\" align=\"absmiddle\">.<\/p><p>Dado: <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Crho_%7Bgasolina%7D=0,75%5C&amp;space;g\/cm%5E3\" alt=\"\\dpi{100} \\large \\rho_{gasolina}=0,75\\ g\/cm^3\" align=\"absmiddle\">.<\/p><h3>Coment&aacute;rios<\/h3><p>a) Nas duas situa&ccedil;&otilde;es, temos o equil&iacute;brio entre a for&ccedil;a peso do dens&iacute;metro e o empuxo exercido pelo etanol.<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%5Cvec%7BP%7D=%7B%5Cvec%7BE%7D%7D_1%5C%5C%5Cvec%7BP%7D=%7B%5Cvec%7BE%7D%7D_2%5C%5C%5Cend%7Bmatrix%7D%5Cright.\" alt=\"\\dpi{100} \\large \\left\\{\\begin{matrix}\\vec{P}={\\vec{E}}_1\\\\\\vec{P}={\\vec{E}}_2\\\\\\end{matrix}\\right.\" align=\"absmiddle\"><\/p><p>Como o peso do dens&iacute;metro n&atilde;o se altera, podemos escrever:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;E_1=E_2\" alt=\"\\dpi{100} \\large E_1=E_2\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Crho_1%5Ccdot&amp;space;V_1%5Ccdot&amp;space;g=%5Crho_2%5Ccdot&amp;space;V_2%5Ccdot&amp;space;g\" alt=\"\\dpi{100} \\large \\rho_1\\cdot V_1\\cdot g=\\rho_2\\cdot V_2\\cdot g\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Crho_1%5Ccdot&amp;space;V_1%5Ccdot%5Ccancel&amp;space;g=%5Crho_2%5Ccdot&amp;space;V_2%5Ccdot%5Ccancel&amp;space;g\" alt=\"\\dpi{100} \\large \\rho_1\\cdot V_1\\cdot\\cancel g=\\rho_2\\cdot V_2\\cdot\\cancel g\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Crho_1%5Ccdot&amp;space;V_1=%5Crho_2%5Ccdot&amp;space;V_2\" alt=\"\\dpi{100} \\large \\rho_1\\cdot V_1=\\rho_2\\cdot V_2\" align=\"absmiddle\"><\/p><p>Pela rela&ccedil;&atilde;o entre os volumes fornecidas no enunciado, temos:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;V_1-V_2=0,1V_1\" alt=\"\\dpi{100} \\large V_1-V_2=0,1V_1\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;-V_2=0,1V_1-V_1\" alt=\"\\dpi{100} \\large -V_2=0,1V_1-V_1\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;V_2=V_1-0,1V_1=0,9V_1\" alt=\"\\dpi{100} \\large V_2=V_1-0,1V_1=0,9V_1\" align=\"absmiddle\"><\/p><p>Voltando para a express&atilde;o anterior:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Crho_1%5Ccdot&amp;space;V_1=%5Crho_2%5Ccdot&amp;space;V_2\" alt=\"\\dpi{100} \\large \\rho_1\\cdot V_1=\\rho_2\\cdot V_2\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Crho_1%5Ccdot&amp;space;V_1=%5Crho_2%5Ccdot0,9V_1\" alt=\"\\dpi{100} \\large \\rho_1\\cdot V_1=\\rho_2\\cdot0,9V_1\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Crho_1%5Ccdot%5Ccancel%7BV_1%7D=%5Crho_2%5Ccdot0,9%5Ccdot%5Ccancel%7BV_1%7D\" alt=\"\\dpi{100} \\large \\rho_1\\cdot\\cancel{V_1}=\\rho_2\\cdot0,9\\cdot\\cancel{V_1}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Crho_1=%5Crho_2%5Ccdot0,9%5CRightarrow%5Crho_2=%5Cfrac%7B%5Crho_1%7D%7B0,9%7D\" alt=\"\\dpi{100} \\large \\rho_1=\\rho_2\\cdot0,9\\Rightarrow\\rho_2=\\frac{\\rho_1}{0,9}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Crho_2=%5Cfrac%7B0,810%7D%7B0,9%7D=0,900%5C&amp;space;g\/cm%5E3\" alt=\"\\dpi{100} \\large \\rho_2=\\frac{0,810}{0,9}=0,900\\ g\/cm^3\" align=\"absmiddle\"><\/p><p>b) A varia&ccedil;&atilde;o da energia mec&acirc;nica ser&aacute; dada pela soma da varia&ccedil;&atilde;o da energia cin&eacute;tica e da energia potencial gravitacional adquirida pela gasolina. Perceba que devemos desprezar a varia&ccedil;&atilde;o da altura do fluido no tanque, visto que o volume retirado &eacute; desprez&iacute;vel frente ao total nele armazenado.<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5CDelta&amp;space;E_%7Bmec%7D=%5CDelta&amp;space;E_%7Bcinetica%7D+%5CDelta&amp;space;E_%7Bpot%5C&amp;space;gravitacional%7D\" alt=\"\\dpi{100} \\large \\Delta E_{mec}=\\Delta E_{cinetica}+\\Delta E_{pot\\ gravitacional}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5CDelta&amp;space;E_%7Bmec%7D=%5Cfrac%7Bm_%7Btotal%7D%5Ccdot&amp;space;v%5E%7B2%7D%7D%7B2%7D+m_%7Btotal%7D%5Ccdot&amp;space;g%5Ccdot&amp;space;%5CDelta&amp;space;h\" alt=\"\\dpi{100} \\large \\Delta E_{mec}=\\frac{m_{total}\\cdot v^{2}}{2}+m_{total}\\cdot g\\cdot \\Delta h\" align=\"absmiddle\"><\/p><p>A massa da gasolina pode ser calculada pelo produto entre o volume bombeado e a sua massa espec&iacute;fica:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;d_%7Bgas%7D=%5Cfrac%7Bm_%7Bgas%7D%7D%7BV_%7Bgas%7D%7D%5CRightarrow&amp;space;m_%7Bgasolina%7D=d_%7Bgas%7D%5Ccdot&amp;space;V_%7Bgas%7D\" alt=\"\\dpi{100} \\large d_{gas}=\\frac{m_{gas}}{V_{gas}}\\Rightarrow m_{gasolina}=d_{gas}\\cdot V_{gas}\" align=\"absmiddle\"><\/p><p>Note que <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;V=40%5C&amp;space;l=40%5C&amp;space;dm%5E3=40%5Ccdot%7B10%7D%5E3%5C&amp;space;cm%5E3\" alt=\"\\dpi{100} \\large V=40\\ l=40\\ dm^3=40\\cdot{10}^3\\ cm^3\" align=\"absmiddle\">:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;m_%7Bgasolina%7D=0,75%5Ccdot40%5Ccdot%7B10%7D%5E3=30%5Ccdot%7B10%7D%5E3g=30%5C&amp;space;kg\" alt=\"\\dpi{100} \\large m_{gasolina}=0,75\\cdot40\\cdot{10}^3=30\\cdot{10}^3g=30\\ kg\" align=\"absmiddle\"><\/p><p>Voltando &agrave; express&atilde;o anterior:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5CDelta&amp;space;E_%7Bmec%7D=%5Cfrac%7B30%5Ccdot%5Cleft(0,8%5Cright)%5E2%7D%7B2%7D+30%5Ccdot10%5Ccdot3,0\" alt=\"\\dpi{100} \\large \\Delta E_{mec}=\\frac{30\\cdot\\left(0,8\\right)^2}{2}+30\\cdot10\\cdot3,0\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5CDelta&amp;space;E_%7Bmec%7D=9,6+900=909,6%5C&amp;space;J\" alt=\"\\dpi{100} \\large \\Delta E_{mec}=9,6+900=909,6\\ J\" align=\"absmiddle\"><\/p><h2 class=\"wp-block-heading\" id=\"h-questao-13\"><span class=\"ez-toc-section\" id=\"Questao-13\"><\/span>Quest&atilde;o 13<span class=\"ez-toc-section-end\"><\/span><\/h2><p>Rel&ecirc;s s&atilde;o dispositivos eletromec&acirc;nicos usados para abrir e fechar contatos el&eacute;tricos atrav&eacute;s da deflex&atilde;o de uma l&acirc;mina met&aacute;lica (armadura) que &eacute; atra&iacute;da pelo campo magn&eacute;tico gerado por uma bobina, conforme ilustra a Figura 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\/fisica-unicamp-2020-2-fase-questao-13.jpg\" alt=\"\" class=\"wp-image-37706\"><\/figure><\/div><p>a) No rel&ecirc; da Figura A, a constante el&aacute;stica da mola presa &agrave; armadura &eacute; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;k=1500%5C&amp;space;N\/m\" alt=\"\\dpi{100} \\large k=1500\\ N\/m\" align=\"absmiddle\">. Quando a bobina &eacute; ligada, qual &eacute; a energia potencial da mola, se ela for distendida de &#8710;x=0,8 mm em rela&ccedil;&atilde;o &agrave; sua posi&ccedil;&atilde;o de equil&iacute;brio?<\/p><p>b) Resistores LDR (Resistor Dependente de Luz) apresentam alta resist&ecirc;ncia el&eacute;trica na aus&ecirc;ncia de luz, e baixa resist&ecirc;ncia quando iluminados. Um uso frequente desses resistores se verifica no acionamento de rel&ecirc;s. A Figura B (no espa&ccedil;o de resposta) fornece a resist&ecirc;ncia do LDR do circuito da Figura C em fun&ccedil;&atilde;o da intensidade luminosa. Qual &eacute; a tens&atilde;o no LDR quando a intensidade de luz solar nele incidente &eacute; igual a <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;I=0,5%5C&amp;space;W\/m%5E2%5C\" alt=\"\\dpi{100} \\large I=0,5\\ W\/m^2\\\" align=\"absmiddle\">?<\/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\/fisica-unicamp-2020-2-fase-questao-13-1-1024x457.jpg\" alt=\"\" class=\"wp-image-37708\"><\/figure><\/div><h3 class=\"wp-block-heading\" id=\"h-comentarios\">Coment&aacute;rios<\/h3><p>a) A energia potencial da mola pode ser calculada por:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;E_%7Bel%7D=%5Cfrac%7Bk%5Ccdot%5Cleft(%5CDelta&amp;space;x%5Cright)%5E2%7D%7B2%7D\" alt=\"\\dpi{100} \\large E_{el}=\\frac{k\\cdot\\left(\\Delta x\\right)^2}{2}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;E_%7Bel%7D=%5Cfrac%7B1500%5Ccdot%5Cleft(0,8%5Ccdot%7B10%7D%5E%7B-3%7D%5Cright)%5E2%7D%7B2%7D=%5Cfrac%7B1,5%5Ccdot%7B10%7D%5E3%5Ccdot0,64%5Ccdot%7B10%7D%5E%7B-6%7D%7D%7B2%7D\" alt=\"\\dpi{100} \\large E_{el}=\\frac{1500\\cdot\\left(0,8\\cdot{10}^{-3}\\right)^2}{2}=\\frac{1,5\\cdot{10}^3\\cdot0,64\\cdot{10}^{-6}}{2}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;E_%7Bel%7D=0,48%5Ccdot%7B10%7D%5E%7B-3%7D=4,8%5Ccdot%7B10%7D%5E%7B-4%7D%5C&amp;space;J\" alt=\"\\dpi{100} \\large E_{el}=0,48\\cdot{10}^{-3}=4,8\\cdot{10}^{-4}\\ J\" align=\"absmiddle\"><\/p><p>b) Pelo gr&aacute;fico, devemos extrair a resist&ecirc;ncia do LDR dada a intensidade de luz solar <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;I=0,5%5C&amp;space;W\/m%5E2\" alt=\"\\dpi{100} \\large I=0,5\\ W\/m^2\" align=\"absmiddle\">:<\/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\/fisica-unicamp-2020-2-fase-questao-13-GABARITO.jpg\" alt=\"\" class=\"wp-image-37710\"><\/figure><\/div><p>Sendo <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;R_%7BLDR%7D=7,0%5Ccdot%7B10%7D%5E3%5C&amp;space;%5COmega\" alt=\"\\dpi{100} \\large R_{LDR}=7,0\\cdot{10}^3\\ \\Omega\" align=\"absmiddle\">, podemos usar a primeira lei de Ohm para determinarmos a corrente do circuito. Note que os dois resistores est&atilde;o ligados em s&eacute;rie, assim a resist&ecirc;ncia equivalente &eacute; dada pela soma dos dois:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;R_%7Beq%7D=R_1+R_%7BLDR%7D=3,0%5Ccdot%7B10%7D%5E3+7,0%5Ccdot%7B10%7D%5E3=10%5Ccdot%7B10%7D%5E3%5COmega\" alt=\"\\dpi{100} \\large R_{eq}=R_1+R_{LDR}=3,0\\cdot{10}^3+7,0\\cdot{10}^3=10\\cdot{10}^3\\Omega\" align=\"absmiddle\"><\/p><p>Aplicando a primeira lei:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Cvarepsilon=R_%7Beq%7D%5Ccdot&amp;space;i%5CRightarrow&amp;space;i=%5Cfrac%7B%5Cvarepsilon%7D%7BR_%7Beq%7D%7D=%5Cfrac%7B5%7D%7B10%5Ccdot%7B10%7D%5E3%5C&amp;space;%7D=0,5%5Ccdot%7B10%7D%5E%7B-3%7DA=0,5%5C&amp;space;mA\" alt=\"\\dpi{100} \\large \\varepsilon=R_{eq}\\cdot i\\Rightarrow i=\\frac{\\varepsilon}{R_{eq}}=\\frac{5}{10\\cdot{10}^3\\ }=0,5\\cdot{10}^{-3}A=0,5\\ mA\" align=\"absmiddle\"><\/p><p>A tens&atilde;o no LDR pode ser calculada de maneira similar:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;V_%7BLDR%7D=R_%7BLDR%7D%5Ccdot&amp;space;i\" alt=\"\\dpi{100} \\large V_{LDR}=R_{LDR}\\cdot i\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;V_%7BLDR%7D=7,0%5Ccdot%7B10%7D%5E3%5Ccdot0,5%5Ccdot%7B10%7D%5E%7B-3%7D=3,5%5C&amp;space;V\" alt=\"\\dpi{100} \\large V_{LDR}=7,0\\cdot{10}^3\\cdot0,5\\cdot{10}^{-3}=3,5\\ V\" align=\"absmiddle\"><\/p><h2 class=\"wp-block-heading\" id=\"h-questao-14-da-prova-de-fisica-unicamp\"><span class=\"ez-toc-section\" id=\"Questao-14-da-prova-de-Fisica-Unicamp\"><\/span>Quest&atilde;o 14 da prova de F&iacute;sica Unicamp<span class=\"ez-toc-section-end\"><\/span><\/h2><p>Filtros &oacute;pticos t&ecirc;m muitas aplica&ccedil;&otilde;es: &oacute;culos de sol, equipamentos fotogr&aacute;ficos, equipamentos de prote&ccedil;&atilde;o individual (EPI) em atividades profissionais, etc. A densidade &oacute;ptica de um filtro (OD) &eacute; definida por <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;OD=-%5Clog_%7B10%7D%7BT%7D\" alt=\"\\dpi{100} \\large OD=-\\log_{10}{T}\" align=\"absmiddle\">, sendo T a transmit&acirc;ncia &oacute;ptica, que &eacute; dada pela raz&atilde;o entre a intensidade luminosa transmitida e a intensidade incidente. Nas m&aacute;scaras de soldador, bem como naquelas usadas para a observa&ccedil;&atilde;o direta do Sol durante um eclipse, s&atilde;o necess&aacute;rios filtros de densidades &oacute;pticas muito elevadas, ou seja, filtros que transmitem muito pouca luz, tanto na regi&atilde;o vis&iacute;vel (de 400 nm a 700 nm) quanto no ultravioleta e no infravermelho.<\/p><p>a) No espa&ccedil;o de resposta, apresenta-se um gr&aacute;fico da densidade &oacute;ptica em fun&ccedil;&atilde;o do comprimento de onda <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Clambda\" alt=\"\\dpi{100} \\large \\lambda\" align=\"absmiddle\"> para v&aacute;rios filtros, sendo que para cada um deles a densidade &oacute;ptica na regi&atilde;o vis&iacute;vel &eacute; aproximadamente constante. Quanto vale a transmit&acirc;ncia para <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Clambda=900%5C&amp;space;nm\" alt=\"\\dpi{100} \\large \\lambda=900\\ nm\" align=\"absmiddle\"> do filtro de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;OD%5C&amp;space;%5Csim&amp;space;%5C&amp;space;0,4\" alt=\"\\dpi{100} \\large OD\\ \\sim \\ 0,4\" align=\"absmiddle\">&nbsp;na regi&atilde;o vis&iacute;vel?<\/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\/fisica-unicamp-2020-2-fase-questao-14.jpg\" alt=\"\" class=\"wp-image-37714\"><\/figure><\/div><p>b) A &aacute;gua &eacute; um bom filtro &oacute;ptico no infravermelho pr&oacute;ximo, e tem um pico de absor&ccedil;&atilde;o em comprimentos de onda ligeiramente inferiores a <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;3,0%5C&amp;space;%5Cmu&amp;space;m\" alt=\"\\dpi{100} \\large 3,0\\ \\mu m\" align=\"absmiddle\">. A energia do f&oacute;ton &eacute; dada por <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;E=hf\" alt=\"\\dpi{100} \\large E=hf\" align=\"absmiddle\">, em que <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;h=6,6%5Ccdot%7B10%7D%5E%7B-34%7D%5C&amp;space;J%5Ccdot&amp;space;s\" alt=\"\\dpi{100} \\large h=6,6\\cdot{10}^{-34}\\ J\\cdot s\" align=\"absmiddle\"> &eacute; a constante de Planck, e f &eacute; a frequ&ecirc;ncia da onda eletromagn&eacute;tica. Quanto vale a energia do f&oacute;ton absorvido no comprimento de onda <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Clambda=3,0%5C&amp;space;%5Cmu&amp;space;m\" alt=\"\\dpi{100} \\large \\lambda=3,0\\ \\mu m\" align=\"absmiddle\">?<\/p><p>*A velocidade da luz no v&aacute;cuo vale <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;c=3,0%5Ccdot%7B10%7D%5E8%5C&amp;space;m\/s\" alt=\"\\dpi{100} \\large c=3,0\\cdot{10}^8\\ m\/s\" align=\"absmiddle\">.<\/p><h3>Coment&aacute;rios<\/h3><p>a) Pelo gr&aacute;fico, se o filtro possui <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;OD%5Ccong4\" alt=\"\\dpi{100} \\large OD\\cong4\" align=\"absmiddle\">, na regi&atilde;o do vis&iacute;vel, sabemos se tratar da terceira curva de cima para baixo. Podemos usar o mesmo gr&aacute;fico para determinarmos a densidade &oacute;ptica para o comprimento de onda <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Clambda=900%5C&amp;space;nm\" alt=\"\\dpi{100} \\large \\lambda=900\\ nm\" align=\"absmiddle\">:<\/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\/fisica-unicamp-2020-2-fase-questao-14-gabarito.jpg\" alt=\"\" class=\"wp-image-37716\"><\/figure><\/div><p>Sabendo que OD=1, podemos calcular a transmit&acirc;ncia &oacute;ptica:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;OD=-%5Clog_%7B10%7D%7BT%7D\" alt=\"\\dpi{100} \\large OD=-\\log_{10}{T}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Clog_%7B10%7D%7BT%7D=-OD\" alt=\"\\dpi{100} \\large \\log_{10}{T}=-OD\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;T=%7B10%7D%5E%7B-OD%7D=%7B10%7D%5E%7B-1%7D=0,1\" alt=\"\\dpi{100} \\large T={10}^{-OD}={10}^{-1}=0,1\" align=\"absmiddle\"><\/p><p>b) A energia do f&oacute;ton, como dito no enunciado, &eacute; calculada por:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;E=h%5Ccdot&amp;space;f\" alt=\"\\dpi{100} \\large E=h\\cdot f\" align=\"absmiddle\"><\/p><p>Podemos escrever a frequ&ecirc;ncia em fun&ccedil;&atilde;o da velocidade e do comprimento de onda, fazendo uso da equa&ccedil;&atilde;o fundamental da ondulat&oacute;ria:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;v=%5Clambda%5Ccdot&amp;space;f%5CRightarrow&amp;space;f=%5Cfrac%7Bv%7D%7B%5Clambda%7D\" alt=\"\\dpi{100} \\large v=\\lambda\\cdot f\\Rightarrow f=\\frac{v}{\\lambda}\" align=\"absmiddle\"><\/p><p>Voltando &agrave; express&atilde;o da energia do f&oacute;ton:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;E=h%5Ccdot&amp;space;f=h%5Ccdot%5Cfrac%7Bv%7D%7B%5Clambda%7D\" alt=\"\\dpi{100} \\large E=h\\cdot f=h\\cdot\\frac{v}{\\lambda}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;E=6,6%5Ccdot%7B10%7D%5E%7B-34%7D%5Ccdot%5Cfrac%7B3,0%5Ccdot%7B10%7D%5E8%7D%7B3,0%5Ccdot%7B10%7D%5E%7B-6%7D%7D=6,6%5Ccdot%7B10%7D%5E%7B-34%7D%5Ccdot%5Cfrac%7B%5Ccancel%7B3,0%7D%5Ccdot%7B10%7D%5E8%7D%7B%5Ccancel%7B3,0%7D%5Ccdot%7B10%7D%5E%7B-6%7D%7D\" alt=\"\\dpi{100} \\large E=6,6\\cdot{10}^{-34}\\cdot\\frac{3,0\\cdot{10}^8}{3,0\\cdot{10}^{-6}}=6,6\\cdot{10}^{-34}\\cdot\\frac{\\cancel{3,0}\\cdot{10}^8}{\\cancel{3,0}\\cdot{10}^{-6}}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;E=6,6%5Ccdot%7B10%7D%5E%7B-34%7D%5Ccdot1%5Ccdot%7B10%7D%5E%7B14%7D=6,6%5Ccdot%7B10%7D%5E%7B-20%7D%5C&amp;space;J\" alt=\"\\dpi{100} \\large E=6,6\\cdot{10}^{-34}\\cdot1\\cdot{10}^{14}=6,6\\cdot{10}^{-20}\\ J\" align=\"absmiddle\"><\/p><h2 class=\"wp-block-heading\" id=\"h-questao-15\"><span class=\"ez-toc-section\" id=\"Questao-15\"><\/span>Quest&atilde;o 15<span class=\"ez-toc-section-end\"><\/span><\/h2><p>As vidra&ccedil;as de um arranha-c&eacute;u em Londres, conhecido como &ldquo;Walkie Talkie&rdquo;, reproduzem a forma de um espelho c&ocirc;ncavo. Os raios solares refletidos pelo edif&iacute;cio provocaram danos em ve&iacute;culos e com&eacute;rcios pr&oacute;ximos.<\/p><p>a) Considere um objeto em frente e ao longo do eixo do espelho c&ocirc;ncavo de raio de curvatura R=1,0 m, conforme mostra a figura no espa&ccedil;o de resposta. Complete os raios luminosos na figura. Em seguida, calcule a dist&acirc;ncia d do objeto ao v&eacute;rtice do espelho (ponto O), de forma que a intensidade de raios solares, incidentes paralelamente ao eixo do espelho, seja m&aacute;xima na posi&ccedil;&atilde;o do objeto.<\/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\/fisica-unicamp-2020-2-fase-questao-15.jpg\" alt=\"\" class=\"wp-image-37721\"><\/figure><\/div><p>b) Um objeto met&aacute;lico de massa m=200 g e calor espec&iacute;fico <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;c=480%5C&amp;space;J%5Cleft(kg%5Ccdot%5C&amp;space;%5Cdegree&amp;space;C%5Cright)\" alt=\"\\dpi{100} \\large c=480\\ J\\left(kg\\cdot\\ \\degree C\\right)\" align=\"absmiddle\"> absorve uma pot&ecirc;ncia P=60 W de radia&ccedil;&atilde;o solar focalizada por um espelho c&ocirc;ncavo. Desprezando as perdas de calor por radia&ccedil;&atilde;o, condu&ccedil;&atilde;o e convec&ccedil;&atilde;o, calcule a varia&ccedil;&atilde;o de temperatura do objeto ap&oacute;s &#8710;t=32 s de exposi&ccedil;&atilde;o a essa radia&ccedil;&atilde;o.<\/p><h3>Coment&aacute;rios<\/h3><p>a) Devemos nos lembrar das propriedades dos raios que chegam at&eacute; um espelho esf&eacute;rico. Raios paralelos ao eixo principal devem ser refletidos em dire&ccedil;&atilde;o ao foco do espelho e raios que incidem no v&eacute;rtice s&atilde;o refletidos com mesmo &acirc;ngulo de incid&ecirc;ncia:<\/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\/fisica-unicamp-2020-2-fase-questao-15-1-1024x358.jpg\" alt=\"\" class=\"wp-image-37722\"><\/figure><\/div><div class=\"wp-block-image\"><figure class=\"aligncenter\"><img decoding=\"async\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2020\/01\/fisica-unicamp-2020-2-fase-questao-15-2-1024x426.jpg\" alt=\"\" class=\"wp-image-37723\"><\/figure><\/div><p>Para a quest&atilde;o:<\/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\/fisica-unicamp-2020-2-fase-questao-15-3-1024x669.jpg\" alt=\"\" class=\"wp-image-37725\"><\/figure><\/div><p>Os raios paralelos convergem para o foco. Dessa forma, para que a intensidade dos raios seja m&aacute;xima, o objeto deve estar posicionado a uma dist&acirc;ncia de 0,5 m do v&eacute;rtice do espelho, que corresponde &agrave; metade do seu raio.<\/p><p>b) Devemos usar a defini&ccedil;&atilde;o da pot&ecirc;ncia, em conjunto com a equa&ccedil;&atilde;o fundamental da calorimetria:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;Pot=%5Cfrac%7BE%7D%7B%5CDelta&amp;space;t%7D\" alt=\"\\dpi{100} \\large Pot=\\frac{E}{\\Delta t}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;Pot=%5Cfrac%7Bm%5Ccdot&amp;space;c%5Ccdot%5CDelta&amp;space;%5Ctheta%7D%7B%5CDelta&amp;space;t%7D\" alt=\"\\dpi{100} \\large Pot=\\frac{m\\cdot c\\cdot\\Delta \\theta}{\\Delta t}\" align=\"absmiddle\"><\/p><p>Podemos isolar o tempo e substituir os valores fornecidos:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5CDelta&amp;space;%5Ctheta=%5Cfrac%7BPot%5Ccdot&amp;space;%5CDelta&amp;space;t%7D%7Bm%5Ccdot&amp;space;c%7D=%5Cfrac%7B60%5Ccdot32%7D%7B200%5Ccdot%7B10%7D%5E%7B-3%7D%5Ccdot480%5C&amp;space;%5C&amp;space;%7D=%5Cfrac%7B60%5Ccdot32%7D%7B2%5Ccdot%7B10%7D%5E%7B-1%7D%5Ccdot4,8%5Ccdot%7B10%7D%5E2%5C&amp;space;%7D\" alt=\"\\dpi{100} \\large \\Delta \\theta=\\frac{Pot\\cdot \\Delta t}{m\\cdot c}=\\frac{60\\cdot32}{200\\cdot{10}^{-3}\\cdot480\\ \\ }=\\frac{60\\cdot32}{2\\cdot{10}^{-1}\\cdot4,8\\cdot{10}^2\\ }\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5CDelta&amp;space;%5Ctheta=%5Cfrac%7B1920%7D%7B2%5Ccdot%7B10%7D%5E%7B-1%7D%5Ccdot4,8%5Ccdot%7B10%7D%5E2%5C&amp;space;%7D=%5Cfrac%7B1920%7D%7B96%7D=20%5C&amp;space;%C2%B0C\" alt=\"\\dpi{100} \\large \\Delta \\theta=\\frac{1920}{2\\cdot{10}^{-1}\\cdot4,8\\cdot{10}^2\\ }=\\frac{1920}{96}=20\\ &deg;C\" align=\"absmiddle\"><\/p><h2 class=\"wp-block-heading\" id=\"h-questao-16-da-prova-de-fisica-unicamp\"><span class=\"ez-toc-section\" id=\"Questao-16-da-prova-de-Fisica-Unicamp\"><\/span>Quest&atilde;o 16 da prova de F&iacute;sica Unicamp<span class=\"ez-toc-section-end\"><\/span><\/h2><p>Julho de 2019 marcou o cinquenten&aacute;rio da chegada do homem &agrave; Lua com a miss&atilde;o Apollo 11. As caminhadas dos astronautas em solo lunar, com seus demorados saltos, s&atilde;o imagens emblem&aacute;ticas dessa aventura humana.<\/p><p>a) A acelera&ccedil;&atilde;o da gravidade na superf&iacute;cie da Lua &eacute; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;g_L=1,6%5C&amp;space;m\/s%5E2\" alt=\"\\dpi{100} \\large g_L=1,6\\ m\/s^2\" align=\"absmiddle\">. Calcule o tempo de queda de um corpo solto a partir do repouso de uma altura de 1,8 m com rela&ccedil;&atilde;o &agrave; superf&iacute;cie lunar.<\/p><p>b) A espectrometria de massas &eacute; uma t&eacute;cnica que pode ser usada na identifica&ccedil;&atilde;o de mol&eacute;culas da atmosfera e do solo lunar. A figura ao lado mostra a trajet&oacute;ria (no plano do papel) de uma determinada mol&eacute;cula ionizada (carga <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;q=1,6%5Ccdot%7B10%7D%5E%7B-19%7DC\" alt=\"\\dpi{100} \\large q=1,6\\cdot{10}^{-19}C\" align=\"absmiddle\">) que entra na regi&atilde;o de campo magn&eacute;tico do espectr&ocirc;metro, sombreada na figura, com velocidade de m&oacute;dulo <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;V=3,2%5Ccdot%7B10%7D%5E5m\/s\" alt=\"\\dpi{100} \\large V=3,2\\cdot{10}^5m\/s\" align=\"absmiddle\">. O campo magn&eacute;tico &eacute; uniforme e perpendicular ao plano do papel, dirigido de baixo para cima, e tem m&oacute;dulo B=0,4T. Como ilustra a figura, na regi&atilde;o de campo magn&eacute;tico a trajet&oacute;ria &eacute; circular de raio R=36 cm, e a for&ccedil;a centr&iacute;peta &eacute; dada pela for&ccedil;a magn&eacute;tica de Lorentz, cujo m&oacute;dulo vale F=qVB. Qual &eacute; a massa m da mol&eacute;cula?<\/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\/fisica-unicamp-2020-2-fase-questao-16.jpg\" alt=\"\" class=\"wp-image-37720\"><\/figure><\/div><h3 class=\"wp-block-heading\" id=\"h-comentarios-1\">Coment&aacute;rios<\/h3><p>a) Podemos usar a equa&ccedil;&atilde;o da posi&ccedil;&atilde;o em fun&ccedil;&atilde;o do tempo para determinarmos o tempo de queda. Sabemos que a velocidade inicial &eacute; nula, e que a acelera&ccedil;&atilde;o da gravidade age a favor durante a queda:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5CDelta&amp;space;S=v_0%5Ccdot&amp;space;t+%5Cfrac%7Ba%5Ccdot&amp;space;t%5E2%7D%7B2%7D\" alt=\"\\dpi{100} \\large \\Delta S=v_0\\cdot t+\\frac{a\\cdot t^2}{2}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5CDelta&amp;space;S=%5Ccancel%7Bv_0%7D%5Ccdot&amp;space;t+%5Cfrac%7Ba%5Ccdot&amp;space;t%5E2%7D%7B2%7D\" alt=\"\\dpi{100} \\large \\Delta S=\\cancel{v_0}\\cdot t+\\frac{a\\cdot t^2}{2}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5CDelta&amp;space;S=%5Cfrac%7Ba%5Ccdot&amp;space;t%5E2%7D%7B2%7D%5CRightarrow&amp;space;t%5E2=%5Cfrac%7B2%5Ccdot&amp;space;%5CDelta&amp;space;S%7D%7Ba%7D\" alt=\"\\dpi{100} \\large \\Delta S=\\frac{a\\cdot t^2}{2}\\Rightarrow t^2=\\frac{2\\cdot \\Delta S}{a}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;t=%5Csqrt%7B%5Cfrac%7B2%5Ccdot&amp;space;%5CDelta&amp;space;S%7D%7Ba%7D%7D=%5Csqrt%7B%5Cfrac%7B2%5Ccdot1,8%7D%7B1,6%7D%7D=%5Csqrt%7B%5Cfrac%7B3,6%7D%7B1,6%7D%7D=%5Csqrt%7B%5Cfrac%7B36%7D%7B16%7D%7D=%5Cfrac%7B6%7D%7B4%7D=1,5%5C&amp;space;s\" alt=\"\\dpi{100} \\large t=\\sqrt{\\frac{2\\cdot \\Delta S}{a}}=\\sqrt{\\frac{2\\cdot1,8}{1,6}}=\\sqrt{\\frac{3,6}{1,6}}=\\sqrt{\\frac{36}{16}}=\\frac{6}{4}=1,5\\ s\" align=\"absmiddle\"><\/p><p>b) Sabemos que a for&ccedil;a magn&eacute;tica atua como resultante centr&iacute;peta. Desse modo, podemos igualar as duas rela&ccedil;&otilde;es e isolar a massa:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;F_%7Bcp%7D=F_%7Bmag%7D\" alt=\"\\dpi{100} \\large F_{cp}=F_{mag}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Cfrac%7Bm%5Ccdot&amp;space;v%5E2%7D%7BR%7D=q%5Ccdot&amp;space;v%5Ccdot&amp;space;B\" alt=\"\\dpi{100} \\large \\frac{m\\cdot v^2}{R}=q\\cdot v\\cdot B\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Cfrac%7Bm%5Ccdot&amp;space;v%5E%7B%5Ccancel%7B2%7D%7D%7D%7BR%7D=q%5Ccdot%5Ccancel%7Bv%7D%5Ccdot&amp;space;B\" alt=\"\\dpi{100} \\large \\frac{m\\cdot v^{\\cancel{2}}}{R}=q\\cdot\\cancel{v}\\cdot B\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;%5Cfrac%7Bm%5Ccdot&amp;space;v%7D%7BR%7D=q%5Ccdot&amp;space;B\" alt=\"\\dpi{100} \\large \\frac{m\\cdot v}{R}=q\\cdot B\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;m=%5Cfrac%7Bq%5Ccdot&amp;space;B%5Ccdot&amp;space;R%7D%7Bv%7D\" alt=\"\\dpi{100} \\large m=\\frac{q\\cdot B\\cdot R}{v}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;m=%5Cfrac%7B1,6%5Ccdot%7B10%7D%5E%7B-19%7D%5Ccdot0,4%5Ccdot36%5Ccdot%7B10%7D%5E%7B-2%7D%7D%7B3,2%5Ccdot%7B10%7D%5E5%7D\" alt=\"\\dpi{100} \\large m=\\frac{1,6\\cdot{10}^{-19}\\cdot0,4\\cdot36\\cdot{10}^{-2}}{3,2\\cdot{10}^5}\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B100%7D&amp;space;%5Clarge&amp;space;m=%5Cfrac%7B%7B10%7D%5E%7B-21%7D%5Ccdot0,4%5Ccdot36%7D%7B2%5Ccdot%7B10%7D%5E5%7D=7,2%5Ccdot%7B10%7D%5E%7B-26%7D%5C&amp;space;kg\" alt=\"\\dpi{100} \\large m=\\frac{{10}^{-21}\\cdot0,4\\cdot36}{2\\cdot{10}^5}=7,2\\cdot{10}^{-26}\\ kg\" align=\"absmiddle\"><\/p><p>Com isso, encerramos a nossa corre&ccedil;&atilde;o da prova de F&iacute;sica Unicamp 2&ordf; Fase  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 F&iacute;sica UNICAMP 2020 &ndash; 2&ordf; Fase\" 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>\n\n\n\n\n\n\n\n<a id=\"cta\" class=\"cta-imagem\" href=\"https:\/\/vestibulares.estrategia.com\/curso\/pacote-para-unicamp\/\" target=\"blank\">\n                <img decoding=\"async\" width=\"100%\" src=\"https:\/\/vestibulares.estrategia.com\/portal\/wp-content\/uploads\/2021\/12\/Unicamp.jpg\" alt=\"CTA Curso Vestibular Unicamp\" title=\"CTA Curso Vestibular Unicamp\">\n        <\/a>\n\n\n\n<\/p><h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Veja-tambem\"><\/span>Veja tamb&eacute;m:<span class=\"ez-toc-section-end\"><\/span><\/h2><ul class=\"wp-block-list\"><li><a href=\"https:\/\/vestibulares.estrategia.com\/portal\/fisica\/fisica-enem\/\">O que mais cai em F&iacute;sica no Enem?<\/a>&nbsp;<\/li><li><a href=\"https:\/\/vestibulares.estrategia.com\/portal\/fisica\/ordem-de-grandeza\/\">Ordem de grandeza: conhe&ccedil;a os conceitos e a nota&ccedil;&atilde;o cient&iacute;fica<\/a>&nbsp;<\/li><li><a href=\"https:\/\/vestibulares.estrategia.com\/portal\/fisica\/diagrama-de-fases\/\">Diagrama de Fases: o que &eacute; e como interpretar<\/a>&nbsp;<\/li><li><a href=\"https:\/\/vestibulares.estrategia.com\/portal\/fisica\/formulas-fisica-enem\/\">F&oacute;rmulas de F&iacute;sica mais comuns do Enem e outros vestibulares<\/a>&nbsp;<\/li><li><a href=\"https:\/\/vestibulares.estrategia.com\/portal\/fisica\/calorimetria-no-enem\/\">Calorimetria no Enem: como cai?<\/a>&nbsp;<\/li><li><a href=\"https:\/\/vestibulares.estrategia.com\/portal\/fisica\/hidrostatica\/\">Hidrost&aacute;tica: press&atilde;o, densidade e f&oacute;rmulas<\/a>&nbsp;&nbsp;<\/li><li><a href=\"https:\/\/vestibulares.estrategia.com\/portal\/fisica\/optica\/\">Como cai a &oacute;ptica no Enem<\/a>&nbsp;<\/li><li><a href=\"https:\/\/vestibulares.estrategia.com\/portal\/fisica\/eletricidade\/\">Eletricidade: conhe&ccedil;a as principais f&oacute;rmulas e conceitos<\/a><\/li><li><a href=\"https:\/\/vestibulares.estrategia.com\/portal\/fisica\/prova-fuvest-2020-fisica\/\"><\/a><a href=\"https:\/\/vestibulares.estrategia.com\/portal\/fisica\/prova-fuvest-2020-fisica\/\">Resolu&ccedil;&atilde;o Comentada da p<\/a>rova de <a href=\"https:\/\/vestibulares.estrategia.com\/portal\/fisica\/prova-fuvest-2020-fisica\/\">F&iacute;sica<\/a> <a href=\"https:\/\/vestibulares.estrategia.com\/portal\/fisica\/prova-fuvest-2020-fisica\/\">FUVEST 2020<\/a><\/li><\/ul><\/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. Escrevo este artigo&hellip;\n","protected":false},"author":18,"featured_media":37732,"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-37695","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>Resolu\u00e7\u00e3o da prova de F\u00edsica Unicamp 2\u00aa Fase 2020<\/title>\n<meta name=\"description\" content=\"O Estrat\u00e9gia Vestibulares traz para voc\u00ea a resolu\u00e7\u00e3o da prova de F\u00edsica da 2\u00aa Fase da Unicamp 2020! 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