{"id":3546,"date":"2018-01-19T12:23:16","date_gmt":"2018-01-19T12:23:16","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3546"},"modified":"2024-09-04T12:36:25","modified_gmt":"2024-09-04T12:36:25","slug":"resolucao-comentada-das-questoes-de-fisica-do-insper-sp-2018","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2018\/insper-2018\/resolucao-comentada-das-questoes-de-fisica-do-insper-sp-2018\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA do INSPER \u2013 SP \u2013 2018"},"content":{"rendered":"

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

INSPER \u2013 SP \u2013 2018<\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

For\u00e7a de atrito est\u00e1tico<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

O <\/b><\/span><\/span><\/span>que faz com que se diferenciem <\/b><\/span><\/span><\/span>e n\u00e3o <\/b><\/span><\/span><\/span>interfiram uma na outra <\/b><\/span><\/span><\/span>\u00e9 <\/b><\/span><\/span><\/span>a frequ\u00eancia na qual operam e seu\u00a0comprimento de onda.<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\" Antes da troca \"\"<\/b><\/span><\/span><\/span> <\/b>energia consumida <\/b><\/span><\/span><\/span>pelas <\/b><\/span><\/span><\/span>duas l\u00e2mpadas fluorescentes <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>20 W cada <\/b><\/span><\/span><\/span>acesas <\/b><\/span><\/span><\/span>durante \u2206t = 4 h <\/b><\/span><\/span><\/span>por noite \"\" <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 2.P. \u2206t <\/b><\/span><\/span><\/span>= 2.20.4 <\/b><\/span><\/span><\/span>= 160 Wh <\/b><\/span><\/span><\/span>+ dispositivos de <\/b><\/span><\/span><\/span>200 W ligados por 4 h = 800 Wh.<\/b><\/span><\/span><\/span><\/p>\n

W<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span> = 160 + 800 \"\" <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span> = 960 Wh.<\/b><\/span><\/span><\/span><\/p>\n

\"\" Depois da troca \"\"<\/b><\/span><\/span><\/span> <\/b>energia consumida <\/b><\/span><\/span><\/span>pela <\/b><\/span><\/span><\/span>duas l\u00e2mpadas de LED <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>9 W cada <\/b><\/span><\/span><\/span>acesas<\/b><\/span><\/span><\/span> durante \u2206t = 6 h <\/b><\/span><\/span><\/span>por noite \"\" <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 2.P. \u2206t <\/b><\/span><\/span><\/span>= 2.9.6 <\/b><\/span><\/span><\/span>= 108 Wh <\/b><\/span><\/span><\/span>+ dispositivos de <\/b><\/span><\/span><\/span>200 W ligados por 5 h = 1000 Wh.<\/b><\/span><\/span><\/span><\/p>\n

W<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span> = 108 + 1000 \"\" <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span> = 1108 Wh.<\/b><\/span><\/span><\/span><\/p>\n

Diferen\u00e7a \"\" W\u2019 = W<\/b><\/span><\/span><\/span>depois<\/b><\/span><\/span><\/sub><\/span> \u2013 W<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/sub><\/span> <\/b>= 1108 \u2013 960 \"\" <\/b><\/span><\/span><\/span>W\u2019 = 148 Wh.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

O perfil do pneu \u00e9 o percentual entre a altura e a largura da se\u00e7\u00e3o do pneu em mil\u00edmetros, ou seja, os pneus tem aproximadamente o mesmo perfil.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

Efeito Joule<\/b><\/span><\/span><\/span><\/p>\n

Efeito Joule<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0 num\u00a0<\/b><\/span><\/span><\/span>condutor met\u00e1lico<\/b><\/span><\/span><\/span>, os\u00a0<\/b><\/span><\/span><\/span>el\u00e9trons livres da corrente el\u00e9trica,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>durante suas<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>movimenta\u00e7\u00f5es,<\/b><\/span><\/span><\/span>\u00a0sofrem continuamente\u00a0<\/b><\/span><\/span><\/span>colis\u00f5es <\/b><\/span><\/span><\/span>com os\u00a0<\/b><\/span><\/span><\/span>\u00e1tomos da rede cristalina desse condutor,<\/b><\/span><\/span><\/span>\u00a0transferindo a eles\u00a0<\/b><\/span><\/span><\/span>parte\u00a0<\/b><\/span><\/span><\/span>de sua\u00a0<\/b><\/span><\/span><\/span>energia cin\u00e9tica e, como resultado, os \u00e1tomos do condutor, <\/b><\/span><\/span><\/span>como um todo, passam a\u00a0<\/b><\/span><\/span><\/span>vibrar com uma energia maior.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Esse\u00a0<\/b><\/span><\/span><\/span>aumento do \u201cn\u00edvel de vibra\u00e7\u00e3o\u201d dos \u00e1tomos<\/b><\/span><\/span><\/span>\u00a0do condutor\u00a0<\/b><\/span><\/span><\/span>provocam um aumento<\/b><\/span><\/span><\/span>\u00a0de sua\u00a0<\/b><\/span><\/span><\/span>temperatura,<\/b><\/span><\/span><\/span>\u00a0fazendo-o\u00a0<\/b><\/span><\/span><\/span>liberar energia t\u00e9rmica (calor).<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Esse fen\u00f4meno \u00e9 denominado efeito Joule<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

Sendo a <\/span><\/span>pot\u00eancia da P da fonte (usina geradora) constante, <\/span><\/span>pela express\u00e3o <\/span><\/span>P = i.U, <\/span><\/span>quanto <\/span><\/span>maior a tens\u00e3o U <\/span><\/span>no gerador, <\/span><\/span>menor ser\u00e1 a corrente <\/span><\/span>(menor perda por efeito Joule) <\/span><\/span>na linha <\/span><\/span>capaz de transmitir uma determinada pot\u00eancia ao consumidor.<\/span><\/span><\/p>\n

Essa <\/b><\/span><\/span><\/span>perda de pot\u00eancia P <\/b><\/span><\/span><\/span>(e consequentemente de energia el\u00e9trica) por <\/b><\/span><\/span><\/span>efeito Joule <\/b><\/span><\/span><\/span>\u00e9 fornecida pela <\/b><\/span><\/span><\/span>express\u00e3o P = R.i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>onde <\/b><\/span><\/span><\/span>P \u00e9 a pot\u00eancia da fonte constante, R a resist\u00eancia equivalente da linha <\/b><\/span><\/span><\/span>de transmiss\u00e3o e <\/b><\/span><\/span><\/span>i a intensidade da corrente el\u00e9trica na mesma. <\/b><\/span><\/span><\/span>Assim, como <\/b><\/span><\/span><\/span>conv\u00e9m P dissipada menor, voc\u00ea deve diminuir R e i.<\/b><\/span><\/span><\/span><\/p>\n

Para <\/b><\/span><\/span><\/span>diminuir<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>resist\u00eancia equivalente R <\/b><\/span><\/span><\/span>da <\/b><\/span><\/span><\/span>linha de transmiss\u00e3o, <\/b><\/span><\/span><\/span>pela <\/b><\/span><\/span><\/span>segunda lei de Ohm<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

fornecida acima, mantendo L e S constantes, <\/b><\/span><\/span><\/span>voc\u00ea deve <\/b><\/span><\/span><\/span>diminuir o valor da resistividade <\/b><\/span><\/span><\/span>(R e \"\" s\u00e3o diretamente proporcionais)<\/b><\/span><\/span><\/span><\/p>\n

R- D<\/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 do INSPER \u2013 SP \u2013 2018 01- For\u00e7a de atrito est\u00e1tico 02- 03- 04- O que faz com que se diferenciem e n\u00e3o interfiram uma na outra \u00e9 a frequ\u00eancia na qual operam e seu\u00a0comprimento de onda. R- C 05- Antes da troca energia consumida pelas duas l\u00e2mpadas fluorescentes de 20 W cada acesas<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3544,"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-3546","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3546","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=3546"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3546\/revisions"}],"predecessor-version":[{"id":3548,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3546\/revisions\/3548"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3544"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3546"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}