{"id":428,"date":"2015-02-11T02:22:53","date_gmt":"2015-02-11T02:22:53","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=428"},"modified":"2024-09-02T17:21:25","modified_gmt":"2024-09-02T17:21:25","slug":"resolucao-comentada-da-fuvest-2a-fase-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-fuvest-2a-fase-2015\/resolucao-comentada-da-fuvest-2a-fase-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada da FUVEST 2a Fase – 2015"},"content":{"rendered":"

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

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

a)<\/b><\/span><\/span><\/span><\/p>\n

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b) Veja figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

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02- <\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span><\/span> Do <\/b><\/span><\/span>gr\u00e1fico fornecido:<\/b><\/span><\/span><\/span><\/p>\n

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c<\/b><\/span><\/span><\/span>) \u00c9 <\/b><\/span><\/span>pedida a temperatura <\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span> <\/b>sendo <\/b><\/span><\/span>fornecido<\/b><\/span><\/span><\/span> que, <\/b><\/span><\/span>nessa temperatura, a corrente el\u00e9trica \u00e9 i = 5 A.<\/b><\/span><\/span><\/span><\/p>\n

Como <\/b><\/span><\/span>V \u00e9 constante<\/b><\/span><\/span><\/span> e calculado em (b), de valor <\/b><\/span><\/span>V = 120 V<\/b><\/span><\/span><\/span> voc\u00ea pode <\/b><\/span><\/span>aplicar a primeira lei de Ohm:<\/b><\/span><\/span><\/span><\/p>\n

R = V \/ i<\/b><\/span><\/span><\/span> = <\/b><\/span><\/span>120 \/ 5<\/b><\/span><\/span><\/span> — <\/b><\/span><\/span>R = 24 \u2126<\/b><\/span><\/span><\/span> <\/b>(resist\u00eancia quando a temperatura \u00e9 T<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>).<\/b><\/span><\/span><\/span><\/p>\n

Olhando no <\/b><\/span><\/span>gr\u00e1fico<\/b><\/span><\/span><\/span> voc\u00ea observa que, <\/b><\/span><\/span>quando R = 24 \u2126, a temperatura \u00e9 de T<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span> = 220 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C.<\/b><\/span><\/span><\/span><\/p>\n

d)<\/b><\/span><\/span><\/span> A <\/b><\/span><\/span>pot\u00eancia dissipada<\/b><\/span><\/span><\/span> pelo resistor na <\/b><\/span><\/span>temperatura T<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span> <\/b>(quando i = 5 A e V = constante = 120V)<\/b><\/span><\/span><\/span> pode ser calculada por:<\/b><\/span><\/span><\/p>\n

P = i.U<\/b><\/span><\/span><\/span><\/span> = <\/b><\/span><\/span><\/span>5.120<\/b><\/span><\/span><\/span><\/span> — <\/b><\/span><\/span><\/span>P = 600 W.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

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04- <\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span><\/span> o <\/b><\/span><\/span>di\u00e2metro D <\/b><\/span><\/span><\/span>da imagem do Sol;<\/b><\/span><\/span><\/span><\/p>\n

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05- <\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span><\/span> a energia <\/b><\/span><\/span>cin\u00e9tica E e o m\u00f3dulo Q da quantidade de movimento<\/b><\/span><\/span><\/span> da crian\u00e7a, <\/b><\/span><\/span>na metade do percurso<\/b><\/span><\/span><\/span>;<\/b><\/span><\/span><\/p>\n

Como <\/b><\/span><\/span>n\u00e3o existem for\u00e7as dissipativas<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span>energia mec\u00e2nica \u00e9 a mesma em qualquer ponto da trajet\u00f3ria<\/b><\/span><\/span><\/span><\/b><\/p>\n

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b)<\/b><\/span><\/span><\/span> o <\/b><\/span><\/span>m\u00f3dulo F da for\u00e7a de contato<\/b><\/span><\/span><\/span> <\/b>entre a crian\u00e7a e o escorregador<\/b><\/span><\/span><\/span><\/p>\n

As <\/b><\/span><\/span>for\u00e7as <\/b><\/span><\/span><\/span>que agem sobre um corpo apoiado sobre um plano inclinado s\u00e3o seu <\/b><\/span><\/span>peso\"\",<\/b><\/span><\/span><\/span> vertical e para baixo e a for\u00e7a normal\u00a0\"\"<\/b><\/span><\/span>, perpendicular \u00e0 superf\u00edcie de contato entre o bloco e o plano.<\/b><\/span><\/span><\/span><\/p>\n

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\u03b1\u00a0<\/b><\/span><\/span><\/span>\u00e9 o <\/b><\/span><\/span>\u00e2ngulo de inclina\u00e7\u00e3o do plano<\/b><\/span><\/span><\/span>. Como\u00a0\"\"\u00a0e\u00a0\"\"\u00a0n\u00e3o tem a mesma dire\u00e7\u00e3o, vamos decompor o peso\u00a0\"\"\u00a0em duas parcelas:<\/b><\/span><\/span><\/p>\n

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\"\" \"\" parcela do peso paralela \u00e0 superf\u00edcie do plano inclinado<\/b><\/span><\/span><\/span> <\/b>(respons\u00e1vel pela tentativa de descida do bloco).<\/b><\/span><\/span><\/span><\/p>\n

\"\" \"\" \u00a0parcela do peso perpendicular \u00e0 superf\u00edcie do plano inclinado <\/b><\/span><\/span><\/span>(trocada entre o bloco e o plano inclinado).<\/b><\/span><\/span><\/span><\/p>\n

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sen\u03b1<\/b><\/span><\/span><\/span> = <\/b><\/span><\/span>cateto oposto\/hipotenusa<\/b><\/span><\/span><\/span>\u00a0 \"\" sen\u03b1=P<\/b><\/span><\/span>P<\/b><\/span><\/span><\/sub>\/P\u00a0 —\u00a0 <\/b><\/span><\/span>P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=P.sen\u03b1<\/b><\/span><\/span><\/span><\/p>\n

cos\u03b1 <\/b><\/span><\/span><\/span>= <\/b><\/span><\/span>cateto adjacente\/hipotenusa<\/b><\/span><\/span><\/span>\u00a0 \"\"\u00a0 cos\u03b1=P<\/b><\/span><\/span>N<\/b><\/span><\/span><\/sub>\/P\u00a0 \"\"\u00a0 <\/b><\/span><\/span>N = P<\/b><\/span><\/span><\/span>N <\/b><\/span><\/span><\/sub><\/span>= P.cos\u03b1<\/b><\/span><\/span><\/span> <\/b>(parcela do peso perpendicular \u00e0 superf\u00edcie do plano inclinado ,for\u00e7a que comprime o bloco contra o plano ou for\u00e7a de contato entre<\/b><\/span><\/span><\/span><\/p>\n

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o bloco, no exerc\u00edcio, crian\u00e7a e o plano)<\/b><\/span><\/span><\/span><\/p>\n

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c) o <\/b><\/span><\/span>m\u00f3dulo da acelera\u00e7\u00e3o<\/b><\/span><\/span><\/span> da crian\u00e7a.<\/b><\/span><\/span><\/p>\n

Veja de <\/b><\/span><\/span>(B)<\/b><\/span><\/span><\/span> que a <\/b><\/span><\/span>for\u00e7a que provoca a descida da crian\u00e7a<\/b><\/span><\/span><\/span> \u00e9 \"\" <\/b><\/span><\/span>P<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = Psen\u03b1 \"\" <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> = mgsen\u03b1 =<\/b><\/span><\/span><\/span> <\/b>30.10.0,5<\/b><\/span><\/span><\/span> \"\" P<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> = 150 N.<\/b><\/span><\/span><\/span><\/p>\n

Essa for\u00e7a \u00e9 a <\/b><\/span><\/span><\/span>resultante<\/b><\/span><\/span><\/span> na dire\u00e7\u00e3o do movimento<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>aplicando a <\/b><\/span><\/span><\/span>segunda lei de Newton \"\"<\/b><\/span><\/span><\/span><\/p>\n

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06- <\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span><\/span> os <\/b><\/span><\/span>m\u00f3dulos E<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>, E<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> e E<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span> <\/b>do <\/b><\/span><\/span>campo el\u00e9trico<\/b><\/span><\/span><\/span> nos pontos <\/b><\/span><\/span>A, B e C, respectivamente;<\/b><\/span><\/span><\/span><\/p>\n

S\u00e3o fornecidos<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span>diferen\u00e7a de potencial entre as placas<\/b><\/span><\/span><\/span> V = 300V e a <\/b><\/span><\/span><\/span>dist\u00e2ncia entre as placas<\/b><\/span><\/span><\/span> d=5 mm = 5.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> m.<\/b><\/span><\/span><\/span><\/p>\n

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 <\/p>\n

Voltar para os Exerc\u00edcios<\/span><\/a><\/h3>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

RESOLU\u00c7\u00d5ES 01- a) b) Veja figura abaixo: 02- a) Do gr\u00e1fico fornecido: c) \u00c9 pedida a temperatura TM sendo fornecido que, nessa temperatura, a corrente el\u00e9trica \u00e9 i = 5 A. Como V \u00e9 constante e calculado em (b), de valor V = 120 V voc\u00ea pode aplicar a primeira lei de Ohm: R = V \/ i = 120<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":426,"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-428","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/428","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=428"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/428\/revisions"}],"predecessor-version":[{"id":434,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/428\/revisions\/434"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/426"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=428"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}