{"id":821551,"date":"2021-07-25T18:40:24","date_gmt":"2021-07-25T21:40:24","guid":{"rendered":"https:\/\/www.estrategiaconcursos.com.br\/blog\/?p=821551"},"modified":"2021-07-25T18:42:29","modified_gmt":"2021-07-25T21:42:29","slug":"gabarito-extraoficial-itep-rn-fisica","status":"publish","type":"post","link":"https:\/\/www.estrategiaconcursos.com.br\/blog\/gabarito-extraoficial-itep-rn-fisica\/","title":{"rendered":"Gabarito Extraoficial ITEP RN &#8211; F\u00edsica"},"content":{"rendered":"\n<p id=\"block-48b5fcf9-0ddf-4e35-9c5b-1092f2c2616d\">Ol\u00e1, pessoal. Tudo bem?<\/p>\n\n\n\n<p id=\"block-fe2fae61-12a9-4928-831e-799ce15a65e3\">Neste artigo faremos uma breve corre\u00e7\u00e3o da prova de<strong>&nbsp;F\u00edsica <\/strong>aplicada para o cargo de&nbsp;<strong>Perito Criminal <\/strong>do ITEP-RN. Vale ressaltar que essa corre\u00e7\u00e3o \u00e9 EXTRAOFICIAL, sendo necess\u00e1rio aguardar o gabarito preliminar da banca para confirmar o seu desempenho.<\/p>\n\n\n\n<p id=\"block-5139e1b6-4e62-468f-8137-d82dd9d2d565\"><em>Vamos l\u00e1?<\/em><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"block-b7f520bd-8b49-44fe-83d5-f68b89378910\">CORRE\u00c7\u00c3O EXTRAOFICIAL &#8211; CARGO: Perito Criminal<\/h2>\n\n\n\n<p><strong><span class=\"has-inline-color has-vivid-cyan-blue-color\">PROVA 4 &#8211; F\u00cdSICA<\/span><\/strong><\/p>\n\n\n\n<p><strong><em><span class=\"has-inline-color has-vivid-purple-color\">51- Em pequeno diamante est\u00e1 imerso no fundo&#8230;<\/span><\/em><\/strong><\/p>\n\n\n\n<p><strong>A) CERTO \u2013 Para chegar na alternativa correta voc\u00ea precisa calcular o seno do \u00e2ngulo limite que \u00e9 a raz\u00e3o entre os \u00edndices, que \u00e9 a fra\u00e7\u00e3o, e depois calcular a tangente desse \u00e2ngulo, a tangente vai ser o raio da superf\u00edcie circular o cateto oposto conduzido pelo cateto adjacente ao \u00e2ngulo limite que \u00e9 justamente a altura H, vamos considerar o H como altura. J\u00e1 a tangente limite quando se sabe o seno, se sabe a tangente, \u00e9 s\u00f3 usar a rela\u00e7\u00e3o fundamental da trigonometria, vai calcular o cosseno que \u00e9 a raiz quadrada de 1-Sen\u00b2, o seno \u00e9 o \u00edndice de refra\u00e7\u00e3o do ar dividido pelo da \u00e1gua, ap\u00f3s a resolu\u00e7\u00e3o chegamos na alternativa A.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/07\/25183628\/image-541.png\" alt=\"\" class=\"wp-image-821552\" width=\"341\" height=\"184\"\/><\/figure>\n\n\n\n<p>B) ERRADO<\/p>\n\n\n\n<p>C) ERRADO<\/p>\n\n\n\n<p>D) ERRADO<\/p>\n\n\n\n<p>E) ERRADO<\/p>\n\n\n\n<p><strong><em><span class=\"has-inline-color has-vivid-purple-color\">52- O gr\u00e1fico a seguir representa a acelera\u00e7\u00e3o de um&#8230;<\/span><\/em><\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/07\/25183652\/image-542.png\" alt=\"\" class=\"wp-image-821553\" width=\"579\" height=\"270\"\/><\/figure>\n\n\n\n<p><strong><em><span class=\"has-inline-color has-vivid-purple-color\">Sabendo que o m\u00f3vel se encontra em&#8230;<\/span><\/em><\/strong><\/p>\n\n\n\n<p>A) ERRADO \u2013 De 0 a 5 \u00e9 acelera\u00e7\u00e3o constante, e n\u00e3o velocidade.<\/p>\n\n\n\n<p>B) ERRADO \u2013 De 0 a 5 o correto \u00e9 25 metros.<\/p>\n\n\n\n<p>C) ERRADO \u2013 De 5 a 10 a acelera\u00e7\u00e3o \u00e9 0 e n\u00e3o a velocidade.<\/p>\n\n\n\n<p><strong>D) CERTO.<\/strong><\/p>\n\n\n\n<p>E) ERRADO.<\/p>\n\n\n\n<p><strong><em><span class=\"has-inline-color has-vivid-purple-color\">53- No sistema Internacional de Unidade (SI), a unidade&#8230;<\/span><\/em><\/strong><\/p>\n\n\n\n<p>A) ERRADO.<\/p>\n\n\n\n<p>B) ERRADO.<\/p>\n\n\n\n<p>C) ERRADO.<\/p>\n\n\n\n<p>D) ERRADO.<\/p>\n\n\n\n<p><strong>E) CERTO. Nessa quest\u00e3o voc\u00ea deve analisar item por item para encontrar a resposta correta, tem que fazer massa x velocidade\u00b2 que resulta em energia, e verificar que em baixo ficaria carga, que \u00e9 os amperes x segundos, em cima energia, energia dividido por carga \u00e9 diferen\u00e7a de potencial por isso que \u00e9 a alternativa correta.<\/strong><\/p>\n\n\n\n<p><strong><em><span class=\"has-inline-color has-vivid-purple-color\">54- Uma crian\u00e7a de 20 kg cai da janela do 6\u00b0 andar de um edif\u00edcio, que se encontra a uma altura de 20 m&#8230;<\/span><\/em><\/strong><\/p>\n\n\n\n<p><strong>A) CERTO. O correto \u00e9 72 Km\/h, ou seja, menor que 100 Km\/h.<\/strong><\/p>\n\n\n\n<p>B) ERRADO.<\/p>\n\n\n\n<p>C) ERRADO.<\/p>\n\n\n\n<p>D) ERRADO.<\/p>\n\n\n\n<p>E) ERRADO.<\/p>\n\n\n\n<p><strong><em><span class=\"has-inline-color has-vivid-purple-color\">55- Uma pessoa precisa receber soro&#8230;<\/span><\/em><\/strong><\/p>\n\n\n\n<p>A) ERRADO.<\/p>\n\n\n\n<p>B) ERRADO.<\/p>\n\n\n\n<p>C) ERRADO.<\/p>\n\n\n\n<p>D) ERRADO.<\/p>\n\n\n\n<p><strong>E) CERTO \u2013 Para chegar na alternativa correta \u00e9 s\u00f3 fazer a equival\u00eancia de 12 cm de merc\u00fario com os mil\u00edmetros de \u00e1gua, vamos considerar que a densidade do soro \u00e9 igual ao da \u00e1gua, voc\u00ea vai encontrar 1632 mm.<\/strong><\/p>\n\n\n\n<p><strong><em><span class=\"has-inline-color has-vivid-purple-color\">56- Considere o circuito el\u00e9trico da figura a seguir&#8230;<\/span><\/em><\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/07\/25183745\/image-543.png\" alt=\"\" class=\"wp-image-821554\" width=\"359\" height=\"338\"\/><\/figure>\n\n\n\n<p><strong><em><span class=\"has-inline-color has-vivid-purple-color\">Em rela\u00e7\u00e3o a esse circuito el\u00e9trico&#8230;<\/span><\/em><\/strong><\/p>\n\n\n\n<p>A) ERRADO. O correto \u00e9 20 \u03a9.<\/p>\n\n\n\n<p>B) ERRADO. A corrente n\u00e3o \u00e9 1,5, o correto \u00e9 0,75 A.&nbsp;&nbsp;&nbsp;&nbsp;<\/p>\n\n\n\n<p><strong>C) CERTO.<\/strong><\/p>\n\n\n\n<p>D) ERRADO. O correto \u00e9 15V.<\/p>\n\n\n\n<p>E) ERRADO. O correto \u00e9 45 W.<\/p>\n\n\n\n<p><strong><em><span class=\"has-inline-color has-vivid-purple-color\">57- A temperatura do ar no interior de um laborat\u00f3rio&#8230;<\/span><\/em><\/strong><\/p>\n\n\n\n<p>A) ERRADO.<\/p>\n\n\n\n<p>B) ERRADO.<\/p>\n\n\n\n<p>C) ERRADO.<\/p>\n\n\n\n<p>D) ERRADO.<\/p>\n\n\n\n<p><strong>E) CERTO \u2013 Para resolver essa quest\u00e3o basta trabalhar com a formula com a Lei de Fourier, que \u00e9 o fluxo de calor:&nbsp;Q = k x A x (TD-TE)\/l e fazendo as contas encontramos 2400W.<\/strong><\/p>\n\n\n\n<p><strong><em><span class=\"has-inline-color has-vivid-purple-color\">58- O gr\u00e1fico a seguir representa a posi\u00e7\u00e3o em fun\u00e7\u00e3o do tempo&#8230;<\/span><\/em><\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/07\/25183809\/image-544.png\" alt=\"\" class=\"wp-image-821555\" width=\"602\" height=\"296\"\/><\/figure>\n\n\n\n<p><strong>Em rela\u00e7\u00e3o a esse movimento&#8230;<\/strong><\/p>\n\n\n\n<p>A) ERRADO. A amplitude correta \u00e9 de 2 m.<\/p>\n\n\n\n<p><strong>B) CERTO.<\/strong><\/p>\n\n\n\n<p>C) ERRADO. O per\u00edodo correto \u00e9 de 4 segundos.<\/p>\n\n\n\n<p>D) ERRADO. A frequ\u00eancia correta \u00e9 de \u00bc Hz.<\/p>\n\n\n\n<p>E) ERRADO. O correto \u00e9 \u03c0 m\/s.<\/p>\n\n\n\n<p><strong><em><span class=\"has-inline-color has-vivid-purple-color\">59- Um proj\u00e9til de massa m e velocidade v atinge um bloco de madeira&#8230;<\/span><\/em><\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/07\/25183836\/image-545.png\" alt=\"\" class=\"wp-image-821556\" width=\"535\" height=\"345\"\/><\/figure>\n\n\n\n<p><strong><em><span class=\"has-inline-color has-vivid-purple-color\">Desprezando-se a resist\u00eancia do ar e as perdas de energia&#8230;<\/span><\/em><\/strong><\/p>\n\n\n\n<p>A) ERRADO.<\/p>\n\n\n\n<p>B) ERRADO.<\/p>\n\n\n\n<p><strong>C) CERTO \u2013 Para resolver essa quest\u00e3o basta conservar a quantidade de movimento da colis\u00e3o e da energia na subida do p\u00eandulo.<\/strong><\/p>\n\n\n\n<p>D) ERRADO.<\/p>\n\n\n\n<p>E) ERRADO.<\/p>\n\n\n\n<p><strong><em><span class=\"has-inline-color has-vivid-purple-color\">60- Uma pessoa precisa de 1 L de \u00e1gua a 50\u00ba C para fazer uma&#8230;<\/span><\/em><\/strong><\/p>\n\n\n\n<p>A) ERRADO.<\/p>\n\n\n\n<p>B) ERRADO.<\/p>\n\n\n\n<p>C) ERRADO.<\/p>\n\n\n\n<p><strong>D) CERTO \u2013 \u00c9 uma quest\u00e3o de calorimetria, deve-se equilibrar o calor cedido e o calor recebido da \u00e1gua quente e da \u00e1gua fria, voc\u00ea precisa saber a quantidade das duas \u00e1guas, a soma das massas tem que dar 1000 gramas, mas para saber o valor de cada uma \u00e9 preciso saber o equil\u00edbrio t\u00e9rmico, e assim descobre que 5&nbsp; vezes a massa M1 \u00e9 igual a 3 vezes a massa M2, 5 vezes a massa de \u00e1gua quente \u00e9 igual a 3 vezes a massa de \u00e1gua fria, ent\u00e3o com essa rela\u00e7\u00e3o e a soma delas igual a 1000, resolvendo a quest\u00e3o se descobre que o M1 \u00e9 igual a 375, e o M2 \u00e9 igual a 625 gramas, ou seja massa de \u00e1gua. Ent\u00e3o 625 seria a \u00e1gua fria e 375 a \u00e1gua quente.<\/strong><\/p>\n\n\n\n<p>E) ERRADO.<\/p>\n\n\n\n<hr class=\"wp-block-separator\" id=\"block-e77dbb11-95b7-4583-aa6e-b5b33da0622f\"\/>\n\n\n\n<p id=\"block-46827b9f-4be3-46ac-9cc4-7838a5f4714a\">Esperamos que tenham ido bem!<\/p>\n\n\n\n<p id=\"block-cccbad17-4719-4490-9836-e4abb435144f\">Abra\u00e7os.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Ol\u00e1, pessoal. Tudo bem? Neste artigo faremos uma breve corre\u00e7\u00e3o da prova de&nbsp;F\u00edsica aplicada para o cargo de&nbsp;Perito Criminal do ITEP-RN. Vale ressaltar que essa corre\u00e7\u00e3o \u00e9 EXTRAOFICIAL, sendo necess\u00e1rio aguardar o gabarito preliminar da banca para confirmar o seu desempenho. Vamos l\u00e1? CORRE\u00c7\u00c3O EXTRAOFICIAL &#8211; CARGO: Perito Criminal PROVA 4 &#8211; F\u00cdSICA 51- Em [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":821447,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"post_tipo":"article","footnotes":""},"categories":[220190],"tags":[],"tax_estado":[219983],"class_list":["post-821551","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-policial-peritos","tax_estado-rn"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.2 (Yoast SEO v27.2) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>Gabarito Extraoficial ITEP RN - F\u00edsica<\/title>\n<meta name=\"description\" content=\"Confira neste artigo o gabarito e a corre\u00e7\u00e3o extraoficial das provas de F\u00edsica do concurso ITEP-RN.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.estrategiaconcursos.com.br\/blog\/gabarito-extraoficial-itep-rn-fisica\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Gabarito Extraoficial ITEP RN - F\u00edsica\" \/>\n<meta property=\"og:description\" content=\"Confira neste artigo o gabarito e a corre\u00e7\u00e3o extraoficial das provas de F\u00edsica do concurso ITEP-RN.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.estrategiaconcursos.com.br\/blog\/gabarito-extraoficial-itep-rn-fisica\/\" \/>\n<meta property=\"og:site_name\" content=\"Estrat\u00e9gia Concursos\" \/>\n<meta property=\"article:author\" content=\"https:\/\/www.facebook.com\/EstrategiaConcursos\" \/>\n<meta property=\"article:published_time\" content=\"2021-07-25T21:40:24+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2021-07-25T21:42:29+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/dhg1h5j42swfq.cloudfront.net\/2021\/07\/25155728\/iteprn.jpg\" \/>\n\t<meta property=\"og:image:width\" content=\"1200\" \/>\n\t<meta property=\"og:image:height\" content=\"628\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"Coordena\u00e7\u00e3o\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@EstratConcursos\" \/>\n<meta name=\"twitter:site\" content=\"@EstratConcursos\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"Coordena\u00e7\u00e3o\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. tempo de leitura\" \/>\n\t<meta name=\"twitter:data2\" content=\"5 minutos\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"NewsArticle\",\"@id\":\"https:\/\/www.estrategiaconcursos.com.br\/blog\/gabarito-extraoficial-itep-rn-fisica\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.estrategiaconcursos.com.br\/blog\/gabarito-extraoficial-itep-rn-fisica\/\"},\"author\":{\"name\":\"Coordena\u00e7\u00e3o\",\"@id\":\"https:\/\/www.estrategiaconcursos.com.br\/blog\/#\/schema\/person\/b4a388c97673c5f483d72185668afa1f\"},\"headline\":\"Gabarito Extraoficial ITEP RN &#8211; 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