{"id":2775,"date":"2017-03-15T04:51:47","date_gmt":"2017-03-15T04:51:47","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2775"},"modified":"2024-09-03T14:29:07","modified_gmt":"2024-09-03T14:29:07","slug":"resolucao-comentada-das-questoes-de-fisica-do-vestibular-da-fgv-rj2017","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2017\/fgv-rj-2017\/resolucao-comentada-das-questoes-de-fisica-do-vestibular-da-fgv-rj2017\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00edsica do vestibular da FGV-RJ\/2017"},"content":{"rendered":"

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

FGV-RJ\/2017<\/span><\/span><\/span><\/p>\n

01- <\/span><\/span><\/span>Rela\u00e7\u00e3o entre pot\u00eancia m\u00e9dia (P<\/b><\/span><\/span>m<\/b><\/span><\/sub><\/span>) e velocidade m\u00e9dia (V<\/b><\/span><\/span>m<\/b><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Considere uma<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>for\u00e7a constante de intensidade F<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>realizando um<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>trabalho W <\/b><\/span><\/span>num<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>deslocamento d,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>de modo que a<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>dire\u00e7\u00e3o de\u00a0F\u00a0forme um \u00e2ngulo\u00a0\u03b1<\/b><\/span><\/span>\u00a0com a<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>dire\u00e7\u00e3o do deslocamento.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>pot\u00eancia instant\u00e2nea P<\/b><\/span><\/span>\u00a0\u00e9 obtida substituindo-se a<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>velocidade m\u00e9dia V<\/b><\/span><\/span>m<\/b><\/span><\/sub><\/span>\u00a0pela instant\u00e2nea V<\/b><\/span><\/span>\u00a0 \"\"\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= F.V \"\" <\/b><\/span><\/span>24200 = F.10 \"\" F = \"\" \"\" <\/b><\/span><\/span>F = 2420 N.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Segunda lei de Newton \"\" F = m.a \"\" <\/b><\/span><\/span>2420 = 1100.a \"\"<\/b><\/span><\/span> a = \"\" \"\" a = 2,2 m\/s<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- C\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

02- <\/span><\/span><\/span>\u00c9 uma <\/b><\/span><\/span><\/span>transforma\u00e7\u00e3o isob\u00e1rica <\/b><\/span><\/span><\/span>(press\u00e3o constante) de equa\u00e7\u00e3o <\/b><\/span><\/span><\/span>P.V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>\/300 = V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\/400 \"\"<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = 4V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>\/3.<\/b><\/span><\/span><\/span><\/p>\n

Para que ele <\/b><\/span><\/span><\/span>comece a subir seu volume aumentou <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>\u2206V <\/b><\/span><\/span><\/span>= 4V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>\/3 \u2013 V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>\/3 <\/b><\/span><\/span><\/span>e, nesse caso seu peso P vale <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span> = m.g = constante <\/b><\/span><\/span><\/span>= d<\/b><\/span><\/span><\/span>arbal\u00e3o<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.g <\/b><\/span><\/span><\/span>(constante, antes e depois) e o <\/b><\/span><\/span><\/span>empuxo (peso do volume de ar externo deslocado)<\/b><\/span><\/span><\/span> sobre ele vale <\/b><\/span><\/span><\/span>E <\/b><\/span><\/span><\/span>= d<\/b><\/span><\/span><\/span>arfora<\/b><\/span><\/span><\/sub><\/span>. \u2206V.g <\/b><\/span><\/span><\/span>= 1,3.(V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>\/3).g.<\/b><\/span><\/span><\/span><\/p>\n

Na <\/b><\/span><\/span><\/span>situa\u00e7\u00e3o final <\/b><\/span><\/span><\/span>ele est\u00e1 na <\/b><\/span><\/span><\/span>imin\u00eancia de subir <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>P = E \"\"<\/b><\/span><\/span><\/span> <\/b>d<\/b><\/span><\/span><\/span>arbal\u00e3o<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>.g = 1,3.(V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>\/3).g \"\"<\/b><\/span><\/span><\/span> <\/b>d<\/b><\/span><\/span><\/span>arbal\u00e3o<\/b><\/span><\/span><\/sub><\/span> <\/b>= 1,3\/3 <\/b><\/span><\/span><\/span>= 0,43 kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/span><\/span><\/span>Aplicando a <\/b><\/span><\/span><\/span>express\u00e3o fornecida<\/b><\/span><\/span><\/span> para o c\u00e1lculo da <\/b><\/span><\/span><\/span>porcentagem de energia luminosa refletida <\/b><\/span><\/span><\/span>em <\/b><\/span><\/span><\/span>uma das superf\u00edcies de separa\u00e7\u00e3o entre os dois meios<\/b><\/span><\/span><\/span> (ar e vidro) \"\" <\/b><\/span><\/span><\/span>R = 100.\"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>R = 100<\/b><\/span><\/span><\/span>.\"\" \"\"<\/b><\/span><\/span><\/span> R =100×0,04 \"\" <\/b><\/span><\/span><\/span>R = 4%.<\/b><\/span><\/span><\/span><\/p>\n

Como, <\/b><\/span><\/span><\/span>pelo enunciado, <\/b><\/span><\/span><\/span>entre a <\/b><\/span><\/span><\/span>tela e o vidro h\u00e1 uma delgada camada de ar <\/b><\/span><\/span><\/span>s\u00e3o <\/b><\/span><\/span><\/span>duas reflex\u00f5es<\/b><\/span><\/span><\/span>, ent\u00e3o,<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R <\/b><\/span><\/span><\/span>= 2×4 <\/b><\/span><\/span><\/span>= 8%. <\/b><\/span><\/span><\/span><\/p>\n

Como <\/b><\/span><\/span><\/span>incide <\/b><\/span><\/span><\/span>no quadro <\/b><\/span><\/span><\/span>60 W,<\/b><\/span><\/span><\/span> <\/b>8% dessa quantia ser\u00e3o refletidas \"\"<\/b><\/span><\/span><\/span> <\/b>R <\/b><\/span><\/span><\/span>= 0,08×60 <\/b><\/span><\/span><\/span>= 4,8 W.<\/b><\/span><\/span><\/span><\/p>\n

Assim, <\/b><\/span><\/span><\/span>chega \u00e0 tela <\/b><\/span><\/span><\/span>a pot\u00eancia pedida de <\/b><\/span><\/span><\/span>P <\/b><\/span><\/span><\/span>= 60 \u2013 4,8 <\/b><\/span><\/span><\/span>= 55,2 W.<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

S\u00e3o dados: f = <\/b><\/span><\/span><\/span>640 Hz \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 0 \"\" <\/b><\/span><\/span><\/span>V = <\/b><\/span><\/span><\/span>340 m\/s \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> <\/b>= 20 m\/s \"\" <\/b><\/span><\/span><\/span> <\/b> <\/b>f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> <\/b>= ?.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>fonte<\/b><\/span><\/span><\/span> se <\/b><\/span><\/span><\/span>aproxima do observador \"\"<\/b><\/span><\/span><\/span> <\/b>f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> = 640. \"\"<\/b><\/span><\/span><\/span> = 640. \"\" \"\" <\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> <\/b>= 640.1,0625 <\/b><\/span><\/span><\/span>= 680 Hz.<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>Os <\/b><\/span><\/span><\/span>aparelhos est\u00e3o associados em paralelo, <\/b><\/span><\/span><\/span>pois est\u00e3o instalados em <\/b><\/span><\/span><\/span>uma rede, <\/b><\/span><\/span><\/span>de forma a <\/b><\/span><\/span><\/span>funcionarem com sua tens\u00e3o nominal<\/b><\/span><\/span><\/span> e, uma das <\/b><\/span><\/span><\/span>caracter\u00edsticas da associa\u00e7\u00e3o paralelo <\/b><\/span><\/span><\/span>\u00e9 que a <\/b><\/span><\/span><\/span>tens\u00e3o U = 220 V \u00e9 a mesma para cada um dos aparelhos.<\/b><\/span><\/span><\/span><\/p>\n

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

Corrente i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> na torneira el\u00e9trica \"\" P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.U \"\" <\/b><\/span><\/span><\/span>3300 = i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.220 \"\" <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> = <\/b><\/span><\/span><\/span>15 A<\/b><\/span><\/span><\/span><\/p>\n

Corrente i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> no chuveiro el\u00e9trico \"\" P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.U \"\" <\/b><\/span><\/span><\/span>4400 = i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.220 \"\" <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> <\/b>= 20 A<\/b><\/span><\/span><\/span><\/p>\n

Corrente total i que atravessa o disjuntor<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>i = i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> + i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> <\/b>= 15 + 20 \"\" <\/b><\/span><\/span><\/span>i = 35 A<\/b><\/span><\/span><\/span><\/p>\n

Como <\/b><\/span><\/span><\/span>o enunciado pede <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>corrente m\u00e1xima<\/b><\/span><\/span><\/span> <\/b>no disjuntor para proteger esse circuito<\/b><\/span><\/span><\/span>, o valor m\u00e1ximo mais pr\u00f3ximo de 35 A \u00e9 <\/b><\/span><\/span><\/span>40 A.<\/b><\/span><\/span><\/span><\/p>\n

R- B<\/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 FGV-RJ\/2017 01- Rela\u00e7\u00e3o entre pot\u00eancia m\u00e9dia (Pm) e velocidade m\u00e9dia (Vm) Considere uma\u00a0for\u00e7a constante de intensidade F\u00a0realizando um\u00a0trabalho W num\u00a0deslocamento d,\u00a0de modo que a\u00a0dire\u00e7\u00e3o de\u00a0F\u00a0forme um \u00e2ngulo\u00a0\u03b1\u00a0com a\u00a0dire\u00e7\u00e3o do deslocamento. A\u00a0pot\u00eancia instant\u00e2nea P\u00a0\u00e9 obtida substituindo-se a\u00a0velocidade m\u00e9dia Vm\u00a0pela instant\u00e2nea V\u00a0 \u00a0 P\u00a0= F.V 24200 = F.10 F = F = 2420<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2773,"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-2775","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2775","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=2775"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2775\/revisions"}],"predecessor-version":[{"id":2777,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2775\/revisions\/2777"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2773"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}