{"id":2834,"date":"2017-04-12T04:39:04","date_gmt":"2017-04-12T04:39:04","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2834"},"modified":"2024-09-03T17:03:02","modified_gmt":"2024-09-03T17:03:02","slug":"resolucao-comentada-das-questoes-de-fisica-da-ufla-mg-2017","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2017\/ufla-2017\/resolucao-comentada-das-questoes-de-fisica-da-ufla-mg-2017\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da UFLA-MG – 2017"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da <\/span><\/span><\/span><\/p>\n

UFLA-MG – 2017<\/span><\/span><\/span><\/p>\n

01-<\/span><\/span><\/span> Pelo enunciado<\/b><\/span><\/span><\/span>: O <\/b><\/span><\/span><\/span>Sol irradia energia <\/b><\/span><\/span><\/span>para o espa\u00e7o sideral. <\/b><\/span><\/span><\/span>Essa energia tem origem <\/b><\/span><\/span><\/span>em seu colapso gravitacional, em <\/b><\/span><\/span><\/span>raz\u00e3o de seu <\/b><\/span><\/span><\/span>potencial gravitacional.<\/b><\/span><\/span><\/span><\/p>\n

Nesse processo<\/b><\/span><\/span><\/span>, os \u00edons de hidrog\u00eanio <\/b><\/span><\/span><\/span>(pr\u00f3tons) <\/b><\/span><\/span><\/span>contidos no seu interior <\/b><\/span><\/span><\/span>adquirem <\/b><\/span><\/span><\/span>velocidades muito altas (energia cin\u00e9tica)<\/b><\/span><\/span><\/span>, <\/b><\/span><\/span><\/span>que os levam a <\/b><\/span><\/span><\/span>atingir <\/b><\/span><\/span><\/span>temperaturas de milh\u00f5es de graus (energia t\u00e9rmica)<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span><\/p>\n

Com isso, t\u00eam <\/b><\/span><\/span><\/span>in\u00edcio rea\u00e7\u00f5es exot\u00e9rmicas <\/b><\/span><\/span><\/span>de fus\u00e3o nuclear, nas quais <\/b><\/span><\/span><\/span>n\u00facleos de hidrog\u00eanio s\u00e3o fundidos<\/b><\/span><\/span><\/span>, gerando <\/b><\/span><\/span><\/span>n\u00facleos de He (H\u00e9lio) e <\/b><\/span><\/span><\/span>propiciando a produ\u00e7\u00e3o da radia\u00e7\u00e3o solar,<\/b><\/span><\/span><\/span> que \u00e9 emitida para o espa\u00e7o.<\/b><\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

02- <\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

03- <\/span><\/span><\/span><\/p>\n

Fun\u00e7\u00e3o (equa\u00e7\u00e3o) hor\u00e1ria da velocidade do MUV<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>O<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>m\u00f3vel<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>da figura abaixo tem\u00a0<\/b><\/span><\/span><\/span>velocidade V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0no instante t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>velocidade V num instante posterior t.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Admitindo<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/sub><\/span>= 0 como a origem dos tempos<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span><\/span>a = (V \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\/(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0\"\"\u00a0a = (V \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\/t\u00a0\"\"\u00a0V \u2013 V<\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/sub><\/span>= a.t\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a.t, sendo <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o<\/b><\/span><\/span><\/span> do m\u00f3vel.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 0 \"\" V = a.t.<\/b><\/span><\/span><\/span><\/p>\n

R- A<\/span><\/span><\/span><\/p>\n

04- <\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>velocidade das nanosondas corresponde a 20% da velocidade da luz <\/b><\/span><\/span><\/span>e ter\u00e1 <\/b><\/span><\/span><\/span>valor<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>V = 20% de 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> m\/s \"\" <\/b><\/span><\/span><\/span>V = 0,2×3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> = 0,6.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> m\/s \"\" <\/b><\/span><\/span><\/span>V = 6.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span> m\/s.<\/b><\/span><\/span><\/span><\/p>\n

Um <\/b><\/span><\/span><\/span>ano-luz \u00e9 a dist\u00e2ncia que a luz<\/b><\/span><\/span><\/span> pode viajar em <\/b><\/span><\/span><\/span>um ano com velocidade de 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> m\/s.<\/b><\/span><\/span><\/span> Por exemplo, <\/b><\/span><\/span><\/span>em<\/span><\/span><\/span>\u00a0<\/span><\/span><\/span>um ano-luz, a dist\u00e2ncia percorrida <\/span><\/span><\/span>seria de <\/span><\/span><\/span>aproximadamente 9.460.800.000.000 (nove\u00a0trilh\u00f5es, quatrocentos e sessenta\u00a0bilh\u00f5es\u00a0e oitocentos milh\u00f5es) de\u00a0quil\u00f4metros, ou, aproximadamente 9.10<\/span><\/span><\/span>15<\/span><\/span><\/sup><\/span> m.<\/span><\/span><\/span><\/p>\n

V =\"\"<\/span><\/span><\/span> \"\" <\/span><\/span><\/span>\u2206S = 4,4×9.10<\/span><\/span><\/span>15<\/span><\/span><\/sup><\/span> = 39,6.10<\/span><\/span><\/span>15<\/span><\/span><\/sup><\/span> m \"\" 6.10<\/span><\/span><\/span>7<\/span><\/span><\/sup><\/span> = \"\" \"\" \u2206t =\"\" \"\" <\/span><\/span><\/span>\u2206t = 6,6.10<\/span><\/span><\/span>8<\/span><\/span><\/sup><\/span> s<\/span><\/span><\/span>. <\/span><\/span><\/span><\/p>\n

Ano em segundos \"\"<\/b><\/span><\/span><\/span> <\/b>1 minuto = <\/b><\/span><\/span><\/span>60 segundos<\/b><\/span><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span> <\/b>1 hora <\/b><\/span><\/span><\/span>= 60 x 60 <\/b><\/span><\/span><\/span>= 3600 segundos<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>1 dia <\/b><\/span><\/span><\/span>= 24h x 3600 <\/b><\/span><\/span><\/span>= 86.400 segundos<\/b><\/span><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span> <\/b>365 dias = 31.536.000 segundos<\/b><\/span><\/span><\/span>\u00a0\"\" 1 ano tem aproximadamente 3,15.10<\/b><\/span><\/span><\/span>7 <\/b><\/span><\/span><\/sup><\/span>s.<\/b><\/span><\/span><\/span><\/p>\n

\u2206t = \"\"<\/span><\/span><\/span> \"\"<\/span><\/span><\/span>21 anos<\/span><\/span><\/span><\/p>\n

R- C<\/span><\/span><\/span><\/p>\n

05- <\/span><\/span><\/span>Como o<\/b><\/span><\/span> <\/b>tempo de propaga\u00e7\u00e3o do som \u00e9 ignorado <\/b><\/span><\/span><\/span>trata-se de <\/b><\/span><\/span><\/span>uma queda livre com V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 0.<\/b><\/span><\/span><\/span><\/p>\n

Corpo abandonado de certa altura h do solo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

No caso do exerc\u00edcio \"\" <\/b><\/span><\/span><\/span>h = \"\"<\/b><\/span><\/span><\/span> \"\" h = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>h = 20 m.<\/b><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

06- <\/span><\/span><\/span><\/p>\n

Plano inclinado com atrito<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

C\u00e1lculo\u00a0<\/b><\/span><\/span><\/span>da intensidade de P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>\u00a0e P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>utilizando o\u00a0<\/b><\/span><\/span><\/span>tri\u00e2ngulo ret\u00e2ngulo\u00a0<\/b><\/span><\/span><\/span>da figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A <\/b><\/span><\/span><\/span>rea\u00e7\u00e3o normal <\/b><\/span><\/span><\/span>pedida tem <\/b><\/span><\/span><\/span>intensidade P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span> = N = P.cos60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> <\/b>= mgcos60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> = 20.10.\"\" \"\" <\/b><\/span><\/span><\/span>N = 100 N.<\/b><\/span><\/span><\/span><\/p>\n

R- A<\/b><\/span><\/span><\/span><\/p>\n

07- <\/b><\/span><\/span><\/span>Como <\/b><\/span><\/span><\/span>n\u00e3o existe atrito<\/b><\/span><\/span><\/span> e, em <\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> ele desce acelerando sob a\u00e7\u00e3o da acelera\u00e7\u00e3o da gravidade g, <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>alternativa correta s\u00f3 pode ser a D. <\/b><\/span><\/span><\/span><\/p>\n

R- D<\/span><\/span><\/span><\/p>\n

08- <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- C <\/span><\/span><\/span><\/p>\n

09- \"\"<\/b><\/span><\/span><\/span> = 40 min<\/b><\/span><\/span><\/span> = \"\"<\/b><\/span><\/span><\/span> = \"\" <\/b><\/span><\/span><\/span>h.<\/b><\/span><\/span><\/span><\/p>\n

Autom\u00f3vel \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" 100 = \"\" \"\" <\/b><\/span><\/span><\/span>\u2206S<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> =\"\"<\/b><\/span><\/span><\/span> <\/b>km. <\/b><\/span><\/span><\/span><\/p>\n

\u00d4nibus \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = \"\" \"\"<\/b><\/span><\/span><\/span> 75 = \"\" \"\" <\/b><\/span><\/span><\/span>\u2206S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> =\"\" km. <\/b><\/span><\/span><\/span><\/p>\n

Diferen\u00e7a da dist\u00e2ncia percorrida <\/b><\/span><\/span><\/span>ap\u00f3s 40 min \"\" <\/b><\/span><\/span><\/span>d <\/b><\/span><\/span><\/span>= \"\" – \"\" \"\" <\/b><\/span><\/span><\/span>d = \"\" km.<\/b><\/span><\/span><\/span><\/p>\n

Tempo que o \u00f4nibus demora <\/b><\/span><\/span><\/span>para percorrer <\/b><\/span><\/span><\/span>essa dist\u00e2ncia d = \"\" km com V = 100 km\/h \"\"<\/b><\/span><\/span><\/span> 100 =\"\"<\/b><\/span><\/span><\/span><\/p>\n

\"\" = \"\" = \"\" h = \"\"<\/b><\/span><\/span><\/span> min \"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> = 10 min.<\/b><\/span><\/span><\/span><\/p>\n

R- A<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da UFLA-MG – 2017 01- Pelo enunciado: O Sol irradia energia para o espa\u00e7o sideral. Essa energia tem origem em seu colapso gravitacional, em raz\u00e3o de seu potencial gravitacional. Nesse processo, os \u00edons de hidrog\u00eanio (pr\u00f3tons) contidos no seu interior adquirem velocidades muito altas (energia cin\u00e9tica), que os levam a atingir temperaturas de milh\u00f5es<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2831,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-2834","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2834","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=2834"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2834\/revisions"}],"predecessor-version":[{"id":2835,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2834\/revisions\/2835"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2831"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2834"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}