{"id":948,"date":"2015-06-29T00:56:15","date_gmt":"2015-06-29T00:56:15","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=948"},"modified":"2024-09-02T17:24:46","modified_gmt":"2024-09-02T17:24:46","slug":"resolucao-comentada-da-ifg-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-ifg-2015\/resolucao-comentada-da-ifg-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada da IFG 2015"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada da prova do vestibular da IFG-GO-2015 pelo fisicaevestibular.com.br<\/b><\/span><\/span><\/span><\/p>\n

01- <\/b><\/span><\/span><\/span>\u2206t =<\/b><\/span><\/span><\/span> 2h50min = 2 + 50\/60 = (120 + 50)\/60 <\/b><\/span><\/span>= 17\/6 h.<\/b><\/span><\/span><\/span><\/p>\n

\u2206S = 203 km \"\" V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = \u2206S\/\u2206t<\/b><\/span><\/span><\/span> = 203\/(17\/6) = 203×6\/17 <\/b><\/span><\/span>= 70,1 km\/h.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = ma<\/b><\/span><\/span><\/span> =20.3 <\/b><\/span><\/span>= 60 N (vertical e para cima)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

03- <\/b><\/span><\/span><\/span>Velocidade<\/b><\/span><\/span><\/span> com que a jaca <\/b><\/span><\/span>chega at\u00e9 a cabe\u00e7a<\/b><\/span><\/span><\/span> da pessoa <\/b><\/span><\/span>percorrendo h <\/b><\/span><\/span><\/span>= 6,8 \u2013 1,8 <\/b><\/span><\/span>= 5 m, com V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span> <\/b>0<\/b><\/span><\/span><\/span> e num local onde <\/b><\/span><\/span>a = g = 10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>\"\" <\/b><\/span><\/span>equa\u00e7\u00e3o de Torricelli<\/b><\/span><\/span><\/span>numa queda livre \"\" <\/b><\/span><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 2.g.h<\/b><\/span><\/span><\/span> = 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> + 2.10.5 = 100 \"\" <\/b><\/span><\/span>V =<\/b><\/span><\/span><\/span> \u221a(100) = <\/b><\/span><\/span>10 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

Sendo o choque <\/b><\/span><\/span>inel\u00e1stico a jaca de massa m = 8 kg fica presa \u00e0 cabe\u00e7a da pessoa<\/b><\/span><\/span><\/span> e sua <\/b><\/span><\/span>velocidade varia<\/b><\/span><\/span><\/span> em <\/b><\/span><\/span>m\u00f3dulo<\/b><\/span><\/span><\/span> de <\/b><\/span><\/span>\u0399\u2206V\u0399<\/b><\/span><\/span><\/span> = \u03990 – 10\u0399 \"\" \u0399\u2206V\u0399 = <\/b><\/span><\/span>\u0399- 10\u0399<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>\u2206V = 10 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da <\/b><\/span><\/span>acelera\u00e7\u00e3o<\/b><\/span><\/span><\/span> <\/b>a <\/b><\/span><\/span><\/span>da jaca durante o choque quando sua velocidade varia de \u2206V = 10 m\/s em \u2206t = 0,1 s \"\" a = \u2206V\/\u2206t = 10\/0,1 \"\" a = 100 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

For\u00e7a que a jaca de <\/b><\/span><\/span>massa 8 kg<\/b><\/span><\/span><\/span> troca com a cabe\u00e7a \"\" <\/b><\/span><\/span>F = m.a<\/b><\/span><\/span><\/span> = 8.100 <\/b><\/span><\/span>= 800 N.<\/b><\/span><\/span><\/span><\/p>\n

Essa for\u00e7a <\/b><\/span><\/span>de 800 N corresponde a uma massa de P =mg<\/b><\/span><\/span><\/span> \"\" 800 = m.10 \"\" <\/b><\/span><\/span>m = 80 kg<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

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

04- <\/b><\/span><\/span><\/span>Para o <\/b><\/span><\/span>Ciclo de Carnot<\/b><\/span><\/span><\/span>, o rendimento \u00e9 uma <\/b><\/span><\/span>fun\u00e7\u00e3o exclusiva das temperaturas absolutas de ambas fontes: quentes e frias.<\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span>rendimento<\/b><\/span><\/span><\/span> m\u00e1ximo para uma m\u00e1quina t\u00e9rmica \u00e9 igual \u00e0:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sendo que <\/b><\/span><\/span>t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 a temperatura da fonte fria medida em Kelvin, et<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> \u00a0\u00e9 a temperatura da fonte quente tamb\u00e9m medida em Kelvin.<\/b><\/span><\/span><\/span><\/p>\n

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

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

06- <\/b><\/span><\/span><\/span>Corrente el\u00e9trica em cada aparelho, sendo que todos est\u00e3o submetidos \u00e0 ddp de U = 220 V, pois devem estar em paralelo com a ddp da entrada que \u00e9 U = 220 V.<\/b><\/span><\/span><\/p>\n

Batedeira \"\" P = i.U \"\" i<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub> = P\/U = 150\/220 = 0,68 A.<\/b><\/span><\/span><\/p>\n

Cafeteria \"\" P = i.U \"\" i<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub> = P\/U = 600\/220 = 2,72 A.<\/b><\/span><\/span><\/p>\n

Churrasqueira \"\" P = i.U \"\" i<\/b><\/span><\/span>ch<\/b><\/span><\/span><\/sub> = P\/U = 3000\/220 = 13,63 A.<\/b><\/span><\/span><\/p>\n

Espremedor \"\" P = i.U \"\" i<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub> = P\/U = 200\/220 = 0,90 A.<\/b><\/span><\/span><\/p>\n

Forno \"\" P = i.U \"\" i<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span><\/sub> = P\/U = 2000\/220 = 9,09 A.<\/b><\/span><\/span><\/span><\/p>\n

Geladeira \"\" P = i.U \"\" i<\/b><\/span><\/span>g<\/b><\/span><\/span><\/sub> = P\/U = 500\/220 = 2,27 A.<\/b><\/span><\/span><\/p>\n

Grill \"\" P = i.U \"\" i<\/b><\/span><\/span><\/span>gr<\/b><\/span><\/span><\/span><\/sub> = P\/U = 1200\/220 = 5,45 A.<\/b><\/span><\/span><\/span><\/p>\n

Liquidificador \"\" P = i.U \"\" i<\/b><\/span><\/span>li<\/b><\/span><\/span><\/sub> = P\/U = 800\/220 = 3.63 A.<\/b><\/span><\/span><\/p>\n

M\u00e1quina \"\" P = i.U \"\" i<\/b><\/span><\/span>b=m<\/b><\/span><\/span><\/sub> = P\/U = 1500\/220 = 6,81 A.<\/b><\/span><\/span><\/p>\n

Torradeira \"\" P = i.U \"\" i<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub> = P\/U = 2500\/220 = 11,36 A.<\/b><\/span><\/span><\/p>\n

A <\/b><\/span><\/span>soma das correntes n\u00e3o pode ultrapassar 20 A.<\/b><\/span><\/span><\/span><\/p>\n

Geladeira (2,27 A) + churrasqueira (13,63 A) + cafeteira (2,72 A) + espremedor (0,90 A) = 19,52 A.<\/b><\/span><\/span><\/span><\/p>\n

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

07- <\/b><\/span><\/span><\/span>LEI DE LENZ<\/span><\/span><\/span><\/p>\n

A lei de Lenz<\/b><\/span><\/span><\/span> se refere ao <\/b><\/span><\/span>sentido da corrente el\u00e9trica induzida<\/b><\/span><\/span><\/span> afirmando que a corrente el\u00e9trica induzida, sempre tem sentido oposto as linhas do campo magn\u00e9tico indutor: <\/b><\/span><\/span>\u201c O sentido da corrente el\u00e9trica induzida \u00e9 tal que, por seus efeitos, op\u00f5e-se \u00e0 causa que lhe deu origem\u201d<\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span>p\u00f3lo sul do im\u00e3 se aproxima da espira ou do solen\u00f3ide<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span>corrente induzida<\/b><\/span><\/span><\/span> deve ter um sentido que vai <\/b><\/span><\/span>originar na espira um p\u00f3lo que deve se opor ao p\u00f3lo sul que se aproxima.<\/b><\/span><\/span><\/span><\/p>\n

Assim, a face superior da espira deve ser um p\u00f3lo sul.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sabendo que a face <\/b><\/span><\/span>superior da espira \u00e9 um p\u00f3lo sul<\/b><\/span><\/span><\/span> (onde chegam as linhas de indu\u00e7\u00e3o) e a face <\/b><\/span><\/span>inferior \u00e9 um p\u00f3lo norte<\/b><\/span><\/span><\/span> (de onde saem as linhas de indu\u00e7\u00e3o), aplica-se a <\/b><\/span><\/span>regra da m\u00e3o direita (polegar no sentido da corrente e a \u201cfechada\u201d da m\u00e3o passando por dentro da espira, fornece o sentido das linhas de indu\u00e7\u00e3o).<\/b><\/span><\/span><\/span><\/p>\n

Assim, para que a face superior da espira seja um p\u00f3lo sul e a direita um p\u00f3lo norte, a <\/b><\/span><\/span>corrente el\u00e9trica induzida deve ter o sentido da figura acima.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

\u00a0Voltar para os Exerc\u00edcios<\/span><\/a><\/h2>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada da prova do vestibular da IFG-GO-2015 pelo fisicaevestibular.com.br 01- \u2206t = 2h50min = 2 + 50\/60 = (120 + 50)\/60 = 17\/6 h. \u2206S = 203 km Vm = \u2206S\/\u2206t = 203\/(17\/6) = 203×6\/17 = 70,1 km\/h. R- B 02- FR = ma =20.3 = 60 N (vertical e para cima) R- B 03- Velocidade com que a<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":946,"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-948","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/948","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=948"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/948\/revisions"}],"predecessor-version":[{"id":951,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/948\/revisions\/951"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/946"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=948"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}