{"id":763,"date":"2015-05-13T15:06:27","date_gmt":"2015-05-13T15:06:27","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=763"},"modified":"2024-09-02T17:27:23","modified_gmt":"2024-09-02T17:27:23","slug":"resolucao-comentada-da-ifsul-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-ifsul-2015\/resolucao-comentada-da-ifsul-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada da IFSul 2015"},"content":{"rendered":"

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

01- <\/b><\/span><\/span><\/span>A frequ\u00eancia de propaga\u00e7\u00e3o que independe do meio \u00e9 a mesma para os dois fios (f<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub> = f<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub> = f) e, como est\u00e3o igualmente tracionados T<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub> = T<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub> = T.<\/b><\/span><\/span><\/p>\n

Fio A<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = \u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> \"\" V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub> = \u03bb<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.f<\/b><\/span><\/span> \"\" <\/b><\/span><\/span><\/sub>f = V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/\u03bb<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub> \"\" <\/b><\/span><\/span>f = \u221a(T\/\u03bc<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/\u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>) (I) <\/b><\/span><\/span><\/span><\/p>\n

Fio B<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = \u03bb<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> \"\" V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub> = \u03bb<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.f<\/b><\/span><\/span> \"\" <\/b><\/span><\/span><\/sub>f = V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/\u03bb<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub> \"\" <\/b><\/span><\/span>f = \u221a(T\/3\u03bc<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/2)\u03bb<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>) (II) <\/b><\/span><\/span><\/span><\/p>\n

(I) = (II) <\/b><\/span><\/span>\u221a(T\/\u03bc<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/\u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>) = \u221a(T\/3\u03bc<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/2)\u03bb<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>) \"\" <\/b><\/span><\/span><\/span>(\u03bb<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/(\u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = [\u221a(T\/3\u03bc<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/2)\/\u221a(T\/\u03bc<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/\u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>)]<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \"\" (\u03bb<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/(\u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = (T\/3\u03bc<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/2 X (T\/\u03bc<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>) \"\" (\u03bb<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/(\u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 1\/(3\/2) \"\" <\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/\u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = \u221a(2\/3).<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/b><\/span><\/span><\/span>Um <\/b><\/span><\/span>dinam\u00f4metro ideal<\/b><\/span><\/span><\/span> inserido numa corda ideal (inextens\u00edvel e de massa desprez\u00edvel), indica a intensidade da for\u00e7a de <\/b><\/span><\/span>tra\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>numa das extremidades<\/b><\/span><\/span><\/span>\u00a0da mesma<\/b><\/span><\/span><\/span> (figura abaixo).<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Se tivermos dois ou mais dinam\u00f4metros ligados conforme a figura abaixo e na extremidade livre do dinam\u00f4metro da direita aplicarmos uma for\u00e7a, por exemplo de 20N, cada dinam\u00f4metro indicar\u00e1 20N.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

03- <\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.sen\u03b1<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = n<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.sen \u03b1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> \"\" n<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/n<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub> = sen\u03b1<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/sen\u03b1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub> = 0,707\/0,574 \"\" n<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/n<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub> = 1,232.<\/b><\/span><\/span><\/p>\n

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

04- <\/b><\/span><\/span><\/span> Energia recebida pelo sistema chaleira-\u00e1gua<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>Q = m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>.c<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) + m<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>.c<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>.(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span> = 500.0,23.75 + 1200.1.75 \"\" Q = 8625 + 90000 \"\" <\/b><\/span><\/span>Q = 98635 cal.<\/b><\/span><\/span><\/span><\/p>\n

Energia fornecida pelo fogareiro em 10min = 600s \"\"<\/b><\/span><\/span><\/span> <\/b>Q\u2019 <\/b><\/span><\/span><\/span>= 250×600 <\/b><\/span><\/span>= 150000 cal.<\/b><\/span><\/span><\/span><\/p>\n

Energia perdida para o meio ambie<\/b><\/span><\/span><\/span>nte \"\" <\/b><\/span><\/span>Q<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> <\/b>= 150000 \u2013 98375 <\/b><\/span><\/span>= 51375 cal.<\/b><\/span><\/span><\/span><\/p>\n

Regra de tr\u00eas:<\/b><\/span><\/span><\/span> 150000 cal \u2013 100% — 51375 cal \u2013 P% \"\" <\/b><\/span><\/span>P<\/b><\/span><\/span><\/span> = 5137500\/150000 <\/b><\/span><\/span>= 34,25%.<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/b><\/span><\/span><\/span>Como o bal\u00e3o <\/b><\/span><\/span>tende a subir<\/b><\/span><\/span><\/span>, o garoto o est\u00e1 segurando, <\/b><\/span><\/span>ele \u00e9 mais leve que o ar que o envolve, <\/b><\/span><\/span><\/span>ent\u00e3o, <\/b><\/span><\/span>a <\/b><\/span><\/span>densidade m\u00e9dia do bal\u00e3o \u00e9 menor que a do ar que a envolve.<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/b><\/span><\/span><\/span>Como <\/b><\/span><\/span>n\u00e3o existe atrito<\/b><\/span><\/span><\/span> o sistema se move no <\/b><\/span><\/span>sentido hor\u00e1rio<\/b><\/span><\/span><\/span> com acelera\u00e7\u00e3o <\/b><\/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>)<\/b><\/span><\/span><\/span> = (24 \u2013 12)\/(6 \u2013 3) \"\" <\/b><\/span><\/span>a = 4 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

07- <\/b><\/span><\/span><\/span>Corrente el\u00e9trica em cada aparelho:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Chuveiro \"\"<\/b><\/span><\/span><\/span> <\/b>P<\/b><\/span><\/span><\/span>CE<\/b><\/span><\/span><\/sub><\/span> = i<\/b><\/span><\/span><\/span>CE<\/b><\/span><\/span><\/sub><\/span>.U<\/b><\/span><\/span><\/span> \"\" 5500 = i<\/b><\/span><\/span>CE<\/b><\/span><\/span><\/sub>.220 \"\" <\/b><\/span><\/span>i<\/b><\/span><\/span><\/span>CE<\/b><\/span><\/span><\/sub><\/span> = 5500\/220 <\/b><\/span><\/span>= 25 A.<\/b><\/span><\/span><\/span><\/p>\n

M\u00e1quina de lavar<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>P<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span> = i<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>.U<\/b><\/span><\/span><\/span> \"\" 3300 = i<\/b><\/span><\/span>CE<\/b><\/span><\/span><\/sub>.220 \"\" <\/b><\/span><\/span>i<\/b><\/span><\/span><\/span>CE<\/b><\/span><\/span><\/sub><\/span> = 3300\/220 <\/b><\/span><\/span>= 15 A<\/b><\/span><\/span><\/span><\/p>\n

Corrente i no circuito, que \u00e9 a mesma que percorre o disjuntor \"\" <\/b><\/span><\/span>i <\/b><\/span><\/span><\/span>= 25 + 15 <\/b><\/span><\/span>= 40 A.<\/b><\/span><\/span><\/span><\/p>\n

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

08- <\/b><\/span><\/span><\/span>Define-se <\/b><\/span><\/span>fluxo magn\u00e9tico pela letra \u03a6 (fi),<\/b><\/span><\/span><\/span> como sendo o <\/b><\/span><\/span>produto<\/b><\/span><\/span><\/span> entre o vetor <\/b><\/span><\/span>indu\u00e7\u00e3o magn\u00e9tica<\/b><\/span><\/span><\/span> \"\", a <\/b><\/span><\/span>\u00e1rea S<\/b><\/span><\/span><\/span> da espira e o <\/b><\/span><\/span>cosseno<\/b><\/span><\/span><\/span> do \u00e2ngulo\u00a0\u03b1\u00a0formado entre\u00a0\"\"\u00a0e\u00a0\"\", ou seja:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe na figura o \u00e2ngulo entre\"\" e \"\" \u00e9 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>, <\/b><\/span><\/span>cos 0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> = 1<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>\u2206S<\/b><\/span><\/span><\/span> = 10.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>x4.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup> <\/b>= 40.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span> m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\u2206\u03a6 = B.\u2206S.cos0<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> = <\/b><\/span><\/span><\/span>5.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.40.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>.1<\/b><\/span><\/span><\/span> \"\" \u2206\u03a6 = <\/b><\/span><\/span><\/span>200.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span> = 2.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span> Wb.<\/b><\/span><\/span><\/span><\/p>\n

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

<\/h2>\n

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

RESOLU\u00c7\u00d5ES 01- A frequ\u00eancia de propaga\u00e7\u00e3o que independe do meio \u00e9 a mesma para os dois fios (fA = fB = f) e, como est\u00e3o igualmente tracionados TA = TB = T. Fio A VA = \u03bbA.fA VA = \u03bbA.f f = VA\/\u03bbA f = \u221a(T\/\u03bcA\/\u03bbA) (I) Fio B VB = \u03bbB.fB VA = \u03bbA.f f = VB\/\u03bbB f =<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":761,"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-763","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/763","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=763"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/763\/revisions"}],"predecessor-version":[{"id":764,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/763\/revisions\/764"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/761"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=763"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}