{"id":2739,"date":"2017-02-07T04:07:43","date_gmt":"2017-02-07T04:07:43","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2739"},"modified":"2024-09-03T14:29:40","modified_gmt":"2024-09-03T14:29:40","slug":"resolucao-comentada-das-questoes-de-fisica-do-vestibular-da-faculdade-de-medicina-do-abc-fmabc-2017","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2017\/fmabc-2017\/resolucao-comentada-das-questoes-de-fisica-do-vestibular-da-faculdade-de-medicina-do-abc-fmabc-2017\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00edsica do vestibular da Faculdade de Medicina do ABC (FMABC) \u2013 2017"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00edsica do vestibular da <\/span><\/span><\/span><\/p>\n

Faculdade de Medicina do ABC (FMABC) \u2013 2017<\/span><\/span><\/span><\/p>\n

<\/a><\/p>\n

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

\"\"<\/p>\n

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

Como <\/b><\/span><\/span><\/span>o empuxo<\/b><\/span><\/span><\/span> (for\u00e7a vertical e para cima) \u00e9 o <\/b><\/span><\/span><\/span>peso do volume do fluido deslocado<\/b><\/span><\/span><\/span>, o <\/b><\/span><\/span><\/span>volume de ar<\/b><\/span><\/span><\/span> deslocado pela esfera <\/b><\/span><\/span><\/span>tem peso desprez\u00edvel, <\/b><\/span><\/span><\/span>ent\u00e3o vamos calcular <\/b><\/span><\/span><\/span>apenas o empuxo sobre a esfera apenas quando est\u00e1 totalmente imersa na \u00e1gua.<\/b><\/span><\/span><\/span><\/p>\n

A intensidade do empuxo sobre a esfera na \u00e1gua \u00e9 fornecida por:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como a esfera est\u00e1 <\/b><\/span><\/span><\/span>totalmente imersa <\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>volume de \u00e1gu<\/b><\/span><\/span><\/span>a que ela desloca \u00e9 pr\u00f3prio <\/b><\/span><\/span><\/span>volume do corpo.<\/b><\/span><\/span><\/span><\/p>\n

Intensidade do empuxo \"\" E = d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>esfera<\/b><\/span><\/span><\/sub><\/span>.g <\/b><\/span><\/span><\/span>= 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.5.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>.10 \"\" <\/b><\/span><\/span><\/span>E = 5.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> N.<\/b><\/span><\/span><\/span><\/p>\n

As <\/b><\/span><\/span><\/span>intensidades das for\u00e7as <\/b><\/span><\/span><\/span>que agem <\/b><\/span><\/span><\/span>sobre a esfera totalmente imersa <\/b><\/span><\/span><\/span>na \u00e1gua s\u00e3o seu peso <\/b><\/span><\/span><\/span>P = 4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> N <\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/span> o empuxo E = 5.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> N.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

Intensidade <\/b><\/span><\/span><\/span>da <\/b><\/span><\/span><\/span>for\u00e7a resultante <\/b><\/span><\/span><\/span>sobre a esfera <\/b><\/span><\/span><\/span>na descida \"\" F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = P \u2013 E <\/b><\/span><\/span><\/span>= 4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> \u2013 0,5.10<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/sup><\/span>2<\/b><\/span><\/span><\/sup><\/span> \"\" F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = 3,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> N\"\"<\/b><\/span><\/span><\/span><\/p>\n

Aplicando a <\/b><\/span><\/span><\/span>segunda lei de Newton <\/b><\/span><\/span><\/span>voc\u00ea pode achar a <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o de descidada esfera \"\" F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = m.a \"\" <\/b><\/span><\/span><\/span>3,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> = 4.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.a<\/b><\/span><\/span><\/span> \"\" a = \"\" \"\" a = <\/b><\/span><\/span><\/span>0,875.10<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sup><\/span> \"\"<\/b><\/span><\/span><\/span> a = 8,75 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Agora vamos <\/b><\/span><\/span><\/span>calcular o tempo t <\/b><\/span><\/span><\/span>que a esfera demora para descer <\/b><\/span><\/span><\/span>S = 17,5 m <\/b><\/span><\/span><\/span>partindo do <\/b><\/span><\/span><\/span>repouso V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 0, <\/b><\/span><\/span><\/span>com <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o a = 8,75 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>e <\/b><\/span><\/span><\/span>supondo S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 0<\/b><\/span><\/span><\/span>, utilizando a <\/b><\/span><\/span><\/span>equa\u00e7\u00e3o dos espa\u00e7os do MUV <\/b><\/span><\/span><\/span>fornecida a seguir:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

S = S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t + \"\"<\/b><\/span><\/span><\/span> \"\" 17,5 = 0 + 0.t <\/b><\/span><\/span><\/span>+ \"\"<\/b><\/span><\/span><\/span> \"\" 17,5 = 4,375.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \"\" t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 4 \"\" <\/b><\/span><\/span><\/span>t = 2 s. <\/b><\/span><\/span><\/span><\/p>\n

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

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

Propriedade do gr\u00e1fico for\u00e7axdeslocamento<\/b><\/span><\/span><\/span><\/p>\n

Em\u00a0<\/b><\/span><\/span><\/span>todo gr\u00e1fico da for\u00e7a\u00a0F\u00a0em fun\u00e7\u00e3o do deslocamento\u00a0d, o\u00a0trabalho realizado pela for\u00e7a F \u00e9<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>numericamente\u00a0 igual \u00e0 \u00e1rea<\/b><\/span><\/span><\/span>\u00a0compreendida entre a <\/b><\/span><\/span><\/span>reta representativa (linha cheia) e o eixo do deslocamento.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Se, ao final desse <\/b><\/span><\/span><\/span>deslocamento de d= 50 m <\/b><\/span><\/span><\/span>a intensidade dessa <\/b><\/span><\/span><\/span>for\u00e7a F (suposta resultante) fosse mantida constante (F = 310 N), o trabalho <\/b><\/span><\/span><\/span>realizado por ela seria \"\"<\/b><\/span><\/span><\/span> W = F.d <\/b><\/span><\/span><\/span>= 310.50 \"\"<\/b><\/span><\/span><\/span><\/p>\n

W = 15 500 J.<\/b><\/span><\/span><\/span><\/p>\n

Supondo que essa for\u00e7a F seja a for\u00e7a resultante voc\u00ea pode aplicar o teorema do Trabalho como varia\u00e7\u00e3o da energia cin\u00e9tica:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

W<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span> \u2013 E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span> \u2013 mV<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 <\/b><\/span><\/span><\/span>= mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u2013 m0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span> \u2013 0 \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span> = 15 500 J. <\/b><\/span><\/span><\/span><\/p>\n

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

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

Transformando 77 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>F em <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C \"\" \"\"<\/b><\/span><\/span><\/span> <\/b>=<\/b><\/span><\/span><\/span> \"\" \"\" \"\" <\/b><\/span><\/span><\/span>=<\/b><\/span><\/span><\/span> \"\" \"\" \"\" <\/b><\/span><\/span><\/span>=<\/b><\/span><\/span><\/span> \"\" \"\" \"\"<\/b><\/span><\/span><\/span> = 5 \"\" C = 25 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>circuito se fecha <\/b><\/span><\/span><\/span>quando as <\/b><\/span><\/span><\/span>l\u00e2minas, por dilata\u00e7\u00e3o se encostam e, <\/b><\/span><\/span><\/span>isso ocorre quando,<\/b><\/span><\/span><\/span> <\/b>pelo.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

enunciado, <\/b><\/span><\/span><\/span>elas que est\u00e3o separadas por uma dist\u00e2ncia equivalente a um mil\u00e9simo do comprimento da l\u00e2mina L<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> se deslocarem essa dist\u00e2ncia, ou seja. \u2206L = L<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/1000 = L<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> \"\" \u2206L = 10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> L<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>. <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/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\u00edsica do vestibular da Faculdade de Medicina do ABC (FMABC) \u2013 2017 01- 02- Como o empuxo (for\u00e7a vertical e para cima) \u00e9 o peso do volume do fluido deslocado, o volume de ar deslocado pela esfera tem peso desprez\u00edvel, ent\u00e3o vamos calcular apenas o empuxo sobre a esfera apenas quando est\u00e1 totalmente imersa na<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2737,"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-2739","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2739","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=2739"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2739\/revisions"}],"predecessor-version":[{"id":2740,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2739\/revisions\/2740"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2737"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2739"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}