{"id":3491,"date":"2018-01-19T02:02:12","date_gmt":"2018-01-19T02:02:12","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3491"},"modified":"2024-09-04T12:49:03","modified_gmt":"2024-09-04T12:49:03","slug":"resolucao-comentada-das-questoes-de-fisica-da-unifesp-2018","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2018\/unifesp-2018\/resolucao-comentada-das-questoes-de-fisica-da-unifesp-2018\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da UNIFESP-2018"},"content":{"rendered":"

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

UNIFESP-2018<\/b><\/span><\/span><\/span><\/p>\n

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

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

Trata-se de um lan\u00e7amento horizontal<\/b><\/span><\/span><\/span><\/p>\n

Se voc\u00ea n\u00e3o domina a teoria ela est\u00e1 a seguir:<\/b><\/span><\/span><\/span><\/p>\n

Decomp\u00f5e-se o movimento em\u00a0<\/b><\/span><\/span>duas parcelas:\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Segundo o eixo X\u00a0\"\"<\/b><\/span><\/span>\u00a0 trata-se de um\u00a0<\/b><\/span><\/span>movimento horizontal uniforme com velocidade constante de<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>intensidade\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>, que \u00e9 a\u00a0<\/b><\/span><\/span>velocidade de lan\u00e7amento<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>S = S<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>\u00a0+ V.t <\/b><\/span><\/span>\u00a0\"\"\u00a0X= 0 + V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>.t\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span>X = V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>.t<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Segundo o eixo Y<\/b><\/span><\/span>\u00a0\"\"\u00a0 trata-se de um\u00a0<\/b><\/span><\/span>movimento uniformemente variado com <\/b><\/span><\/span>velocidade inicial\u00a0V<\/b><\/span><\/span>oy <\/b><\/span><\/sub><\/span>= 0,<\/b><\/span><\/span>\u00a0ou seja, \u00e9 uma\u00a0<\/b><\/span><\/span>queda livre com o corpo abandonado da origem<\/b><\/span><\/span>,\u00a0sujeito\u00a0<\/b><\/span><\/span>apenas \u00e0 acelera\u00e7\u00e3o da<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>gravidade, de intensidade\u00a0g, dire\u00e7\u00e3o vertical e sentido para baixo<\/b><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

b) <\/b><\/span><\/span>Agora o <\/b><\/span><\/span>proj\u00e9til deve sair do canh\u00e3o com velocidade V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span> (pedida) <\/b><\/span><\/span>e deve <\/b><\/span><\/span>atingir o avi\u00e3o com velocidade final vertical nula (V<\/b><\/span><\/span>y<\/b><\/span><\/sub><\/span> = 0) para que o tempo para atingir o avi\u00e3o seja m\u00ednimo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Trata-se de um lan\u00e7amento obl\u00edquo, cuja teoria est\u00e1 fornecida a seguir:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Segundo a vertical (eixo y)<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Segundo a horizontal (eixo x)<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

b) Veja<\/b><\/span><\/span><\/span> na <\/b><\/span><\/span><\/span>figura abaixo <\/b><\/span><\/span><\/span>as <\/b><\/span><\/span><\/span>for\u00e7as <\/b><\/span><\/span><\/span>que <\/b><\/span><\/span><\/span>agem<\/b><\/span><\/span><\/span> <\/b>sobre<\/b><\/span><\/span><\/span> o <\/b><\/span><\/span><\/span>sistema <\/b><\/span><\/span><\/span>que se <\/b><\/span><\/span><\/span>encontra em equil\u00edbrio (for\u00e7a resultante nula).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.g <\/b><\/span><\/span><\/span>= 10.10 \"\" <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 100 N \"\"<\/b><\/span><\/span><\/span> <\/b>P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.g <\/b><\/span><\/span><\/span>= 8.10 \"\" <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = 80 N.<\/b><\/span><\/span><\/span><\/p>\n

Empuxo de Arquimedes<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

b) S\u00e3o dados:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

a) <\/b><\/span><\/span><\/span>Como o <\/b><\/span><\/span><\/span>objeto est\u00e1 a 6 m do espelho plano <\/b><\/span><\/span><\/span>e nesse espelho a <\/b><\/span><\/span><\/span>imagem \u00e9 sim\u00e9trica ao objeto <\/b><\/span><\/span><\/span>(<\/b><\/span><\/span><\/span>dist\u00e2ncia do objeto ao espelho <\/b><\/span><\/span><\/span>plano \u00e9 a <\/b><\/span><\/span><\/span>mesma que a da imagem ao espelho <\/b><\/span><\/span><\/span>plano), a <\/b><\/span><\/span><\/span>imagem I<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> <\/b>estar\u00e1 <\/b><\/span><\/span><\/span>6m \u00e0 esquerda <\/b><\/span><\/span><\/span>da superf\u00edcie refletora <\/b><\/span><\/span><\/span>do espelho plano <\/b><\/span><\/span><\/span>(veja figura).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Assim, a <\/b><\/span><\/span><\/span>dist\u00e2ncia d entre I<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> e I<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span> <\/b>ser\u00e1 \"\" d = 6 + 6 + 4 + 2 = 18 m \"\" <\/b><\/span><\/span><\/span>d = 18 m.<\/b><\/span><\/span><\/span><\/p>\n

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

Equa\u00e7\u00e3o dos pontos conjugados ou equa\u00e7\u00e3o de Gauss para espelhos esf\u00e9ricos<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Equa\u00e7\u00e3o do aumento linear transversal<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

C\u00e1lculo<\/b><\/span><\/span><\/span> da <\/b><\/span><\/span><\/span>dist\u00e2ncia focal f <\/b><\/span><\/span><\/span>do <\/b><\/span><\/span><\/span>espelho esf\u00e9rico convexo <\/b><\/span><\/span><\/span>sabendo, <\/b><\/span><\/span><\/span>pela figura, <\/b><\/span><\/span><\/span>que a <\/b><\/span><\/span><\/span>dist\u00e2ncia <\/b><\/span><\/span><\/span>do <\/b><\/span><\/span><\/span>objeto ao espelho \u00e9 de P = 4 m <\/b><\/span><\/span><\/span>e a da <\/b><\/span><\/span><\/span>imagem ao espelho \u00e9 P\u2019 = – 2m <\/b><\/span><\/span><\/span>(espelhos esf\u00e9ricos <\/b><\/span><\/span><\/span>convexos <\/b><\/span><\/span><\/span>formam <\/b><\/span><\/span><\/span>sempre imagens virtuais <\/b><\/span><\/span><\/span>que, por conven\u00e7\u00e3o, s\u00e3o <\/b><\/span><\/span><\/span>sempre negativas<\/b><\/span><\/span><\/span>).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

a) <\/b><\/span><\/span><\/span>Trata-se de uma <\/b><\/span><\/span><\/span>aplica\u00e7\u00e3o das duas leis de Ohm<\/b><\/span><\/span><\/span> cujas <\/b><\/span><\/span><\/span>express\u00f5es s\u00e3o fornecidas a seguir:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Segunda lei de Ohm (Resistividade)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

C\u00e1lculo <\/b><\/span><\/span><\/span>da <\/b><\/span><\/span><\/span>resist\u00eancia el\u00e9trica R<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> <\/b>no <\/b><\/span><\/span><\/span>trecho AB percorrido <\/b><\/span><\/span><\/span>pela <\/b><\/span><\/span><\/span>corrente el\u00e9trica <\/b><\/span><\/span><\/span>fornecida de <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> = 6 A <\/b><\/span><\/span><\/span>e sob <\/b><\/span><\/span><\/span>ddp constante <\/b><\/span><\/span><\/span>do gerador de <\/b><\/span><\/span><\/span>U = 12 V<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

b) <\/b><\/span><\/span><\/span>Observe que o <\/b><\/span><\/span><\/span>fio AB de resist\u00eancia el\u00e9trica R<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> = 2 <\/b><\/span><\/span><\/span>\u2126 <\/b><\/span><\/span><\/span>e o <\/b><\/span><\/span><\/span>fio ACB de resist\u00eancia el\u00e9trica R<\/b><\/span><\/span><\/span>ACB<\/b><\/span><\/span><\/sub><\/span> = 10 \u2126 <\/b><\/span><\/span><\/span>est\u00e3o <\/b><\/span><\/span><\/span>associados em paralelo <\/b><\/span><\/span><\/span>e, portanto, <\/b><\/span><\/span><\/span>submetidos \u00e0 mesma ddp (tens\u00e3o, voltagem) U = 12 V.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/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 da UNIFESP-2018 01- a) Trata-se de um lan\u00e7amento horizontal Se voc\u00ea n\u00e3o domina a teoria ela est\u00e1 a seguir: Decomp\u00f5e-se o movimento em\u00a0duas parcelas: \u00a0Segundo o eixo X\u00a0\u00a0 trata-se de um\u00a0movimento horizontal uniforme com velocidade constante de\u00a0intensidade\u00a0Vo, que \u00e9 a\u00a0velocidade de lan\u00e7amento\u00a0\u00a0S = So\u00a0+ V.t \u00a0\u00a0X= 0 + Vo.t\u00a0\u00a0\u00a0X = Vo.t \u00a0Segundo o eixo<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3489,"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-3491","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3491","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=3491"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3491\/revisions"}],"predecessor-version":[{"id":3492,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3491\/revisions\/3492"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3489"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3491"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}