{"id":850,"date":"2015-05-14T18:51:37","date_gmt":"2015-05-14T18:51:37","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=850"},"modified":"2024-09-03T13:21:13","modified_gmt":"2024-09-03T13:21:13","slug":"resolucao-comentada-ifpe-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-ifpe-2015\/resolucao-comentada-ifpe-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada IFPE 2015"},"content":{"rendered":"

RESOLU\u00c7\u00d5ES<\/span><\/span><\/span><\/p>\n

01- <\/span><\/span><\/span>No tri\u00e2ngulo hachurado \"\" <\/b><\/span><\/span>sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> = h\/d<\/b><\/span><\/span><\/span> \"\" 1\/2 = 6\/d \"\" <\/b><\/span><\/span>d <\/b><\/span><\/span><\/span>= 6×2 <\/b><\/span><\/span>= 12m.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/span><\/span><\/span>Dados \"\" <\/b><\/span><\/span>F = 2 N<\/b><\/span><\/span><\/span>, <\/b><\/span><\/span>S <\/b><\/span><\/span><\/span>= 0,08 mm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> = 8.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup> <\/b>= 8.10<\/b><\/span><\/span><\/span>-8<\/b><\/span><\/span><\/sup><\/span> m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> e <\/b><\/span><\/span>P = ?.<\/b><\/span><\/span><\/span><\/p>\n

P = F\/S<\/b><\/span><\/span><\/span> = 2\/8.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup> = 0,25.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup> N\/m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> <\/b>= 25.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span> Pa (N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>).<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/span><\/span><\/span>Q = m<\/b><\/span><\/span><\/span>ferro<\/b><\/span><\/span><\/sub><\/span>.c<\/b><\/span><\/span><\/span>ferro<\/b><\/span><\/span><\/sub><\/span>.(\u03b8 \u2013 \u03b8<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) + m<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.c<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.(\u03b8 \u2013 \u03b8<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span> = 10.460.(100 \u2013 10) + 20.4200.(100 \u2013 10) = 4600.90 + 84000.90 \"\" <\/b><\/span><\/span>Q<\/b><\/span><\/span><\/span> = 414000 + 7560000 = 7974000J<\/b><\/span><\/span>= 7974 kJ.<\/b><\/span><\/span><\/span><\/p>\n

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

04- <\/span><\/span><\/span>Quando chega o <\/b><\/span><\/span>final da tarde de um dia de inverno<\/b><\/span><\/span><\/span> em uma grande cidade, os raios solares tornam-se mais difusos e fr\u00e1geis, assim o <\/b><\/span><\/span>solo da cidade se resfria rapidamente<\/b><\/span><\/span><\/span>. E conseq\u00fcentemente, o <\/b><\/span><\/span>ar<\/b><\/span><\/span><\/span> <\/b>pr\u00f3ximo do solo se resfria rapidamente<\/b><\/span><\/span><\/span>. Aquele <\/b><\/span><\/span>ar quente<\/b><\/span><\/span><\/span> <\/b>( mais leve)<\/b><\/span><\/span><\/span> que ainda est\u00e1 na atmosfera <\/b><\/span><\/span>continua a subir, mas o ar frio pr\u00f3ximo ao solo, por ser mais denso e pesado, fica parado<\/b><\/span><\/span><\/span>. Assim a <\/b><\/span><\/span>temperatura cai ainda mais<\/b><\/span><\/span><\/span> e os <\/b><\/span><\/span>poluentes<\/b><\/span><\/span><\/span>, que normalmente s\u00e3o “levados” <\/b><\/span><\/span>pelo ar quente, acabam retidos na camada mais baixa da atmosfera.<\/b><\/span><\/span><\/span><\/p>\n

Quando ocorre a invers\u00e3o t\u00e9rmica, <\/b><\/span><\/span>os poluentes<\/b><\/span><\/span><\/span>, que normalmente iriam se dispersar acompanhando o ar quente liberado na terra, <\/b><\/span><\/span>ficam presos.<\/b><\/span><\/span><\/span> A <\/b><\/span><\/span>faixa cinza alaranjada<\/b><\/span><\/span><\/span> da figura da direita acima \u00e9 a <\/b><\/span><\/span>conseq\u00fc\u00eancia vis\u00edvel do fen\u00f4meno.<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>Colocando as for\u00e7as<\/b><\/span><\/span><\/span> no sistema:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

06- <\/span><\/span><\/span>sen\u03b8 <\/b><\/span><\/span><\/span>= 1,5\/2.5 <\/b><\/span><\/span><\/span>= 0,6<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>parcela do peso<\/b><\/span><\/span><\/span> respons\u00e1vel pela <\/b><\/span><\/span><\/span>descida<\/b><\/span><\/span><\/span> do sistema \"\" <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = mgsen\u03b8<\/b><\/span><\/span><\/span> = m.10.0,6 \"\" <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = 6m.<\/b><\/span><\/span><\/span><\/p>\n

Pit\u00e1goras \"\" <\/b><\/span><\/span><\/span>2,5<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = b<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 1,5<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \"\" b<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 6,25 \u2013 2,25 =4 \"\" <\/b><\/span><\/span><\/span>b = 2 \"\" cos\u03b8 = <\/b><\/span><\/span><\/span>2\/2,5<\/b><\/span><\/span><\/span> = 0,8.<\/b><\/span><\/span><\/span><\/p>\n

For\u00e7a de atrito<\/b><\/span><\/span><\/span>, contr\u00e1ria ao movimento \"\" <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span> = \u03bc.m.g.cos\u03b8<\/b><\/span><\/span><\/span> = 0,5.m.10.0,8 \"\" <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span> = 4m.<\/b><\/span><\/span><\/span><\/p>\n

Segunda lei de Newton \"\" <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = ma<\/b><\/span><\/span><\/span> \"\" P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> \u2013 F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span> = ma \"\" 6m \u2013 4m = ma \"\" <\/b><\/span><\/span><\/span>a = 2 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

07- <\/span><\/span><\/span>O <\/b><\/span><\/span><\/span>per\u00edodo T <\/b><\/span><\/span><\/span>do movimento do barco corresponde ao <\/b><\/span><\/span><\/span>tempo que ele demora para efetuar um \u201csobe e desce\u201d completo que, no caso \u00e9 de T = (2 + 2) = 4 s.<\/b><\/span><\/span><\/span><\/p>\n

f = 1\/T<\/b><\/span><\/span><\/span> = 1\/4 <\/b><\/span><\/span><\/span>= 0,25 s.<\/b><\/span><\/span><\/span><\/p>\n

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

08- <\/span><\/span><\/span>A <\/b><\/span><\/span>catarata<\/b><\/span><\/span><\/span> ocorre, principalmente, com a <\/b><\/span><\/span>idade<\/b><\/span><\/span><\/span>, ou na presen\u00e7a de outros fatores de risco como <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

contus\u00f5es, doen\u00e7as do<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>olho<\/b><\/span><\/span><\/span><\/a>, problemas org\u00e2nicos (como<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>diabetes<\/b><\/span><\/span><\/span><\/a>, por exemplo), efeitos colaterais<\/b><\/span><\/span><\/span> <\/b>de certas <\/b><\/span><\/span><\/span>medica\u00e7\u00f5es ou patologias cong\u00eanitas<\/b><\/span><\/span><\/span>, <\/b><\/span><\/span>n\u00e3o sendo um defeito de vis\u00e3o.<\/b><\/span><\/span><\/span>
\n<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

09- <\/span><\/span><\/span>C\u00e1lculo da <\/b><\/span><\/span>resist\u00eancia<\/b><\/span><\/span><\/span> do resistor equivalente e da <\/b><\/span><\/span>corrente el\u00e9trica em cada resistor<\/b><\/span><\/span><\/span> com a chave fechada (figura abaixo):<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = U\/i<\/b><\/span><\/span><\/span> \"\" 10 = 20\/i<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub> \"\" <\/b><\/span><\/span>i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 2 A.<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo de <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> e i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> (figura abaixo):<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

<\/h2>\n

Voltar para exerc\u00edcios.<\/span><\/a><\/h2>\n","protected":false},"excerpt":{"rendered":"

RESOLU\u00c7\u00d5ES 01- No tri\u00e2ngulo hachurado sen30o = h\/d 1\/2 = 6\/d d = 6×2 = 12m. R- B 02- Dados F = 2 N, S = 0,08 mm2 = 8.10-2.10-6 = 8.10-8 m2 e P = ?. P = F\/S = 2\/8.10-8 = 0,25.108 N\/m2 = 25.106 Pa (N\/m2). R- B 03- Q = mferro.cferro.(\u03b8 \u2013 \u03b8o) + m\u00e1gua.c\u00e1gua.(\u03b8 \u2013<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":817,"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-850","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/850","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=850"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/850\/revisions"}],"predecessor-version":[{"id":851,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/850\/revisions\/851"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/817"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=850"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}