{"id":2913,"date":"2017-04-20T05:43:00","date_gmt":"2017-04-20T05:43:00","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2913"},"modified":"2024-09-03T17:29:32","modified_gmt":"2024-09-03T17:29:32","slug":"resolucao-comentada-das-questoes-de-fisica-do-curso-de-medicina-da-unesc-es-2017","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2017\/unesc-2017\/resolucao-comentada-das-questoes-de-fisica-do-curso-de-medicina-da-unesc-es-2017\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA do curso de Medicina da UNESC-ES-2017"},"content":{"rendered":"

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

Medicina da UNESC-ES-2017<\/span><\/span><\/span><\/p>\n

01-<\/span><\/span><\/span> Primeira etapa<\/b><\/span><\/span><\/span> \"\" com <\/b><\/span><\/span><\/span>velocidade de 30 km\/h <\/b><\/span><\/span><\/span>ele <\/b><\/span><\/span><\/span>percorre \u2206S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 30 km em \u2206t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 1 h.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Segunda etapa<\/b><\/span><\/span><\/span> \"\" com <\/b><\/span><\/span><\/span>velocidade de 60 km\/h <\/b><\/span><\/span><\/span>ele <\/b><\/span><\/span><\/span>percorre \u2206S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 30 km em \u2206t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 1\/2 h.<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>mtotal<\/b><\/span><\/span><\/sub><\/span> = (\u2206S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> + \u2206S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>)\/(\u2206t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> + \u2206t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>) = <\/b><\/span><\/span><\/span>(30 + 30)\/1,5) \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>mtotal<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" V<\/b><\/span><\/span><\/span>mtotal<\/b><\/span><\/span><\/sub><\/span> = 40 km\/h.<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

Trem <\/b><\/span><\/span><\/span>(caminh\u00e3o) de <\/b><\/span><\/span><\/span>comprimento (x) <\/b><\/span><\/span><\/span>atravessando um <\/b><\/span><\/span><\/span>t\u00fanel <\/b><\/span><\/span><\/span>(ponte) de <\/b><\/span><\/span><\/span>comprimento ( c ):<\/b><\/span><\/span><\/span> observe que <\/b><\/span><\/span><\/span>para atravessar completamente o t\u00fanel <\/b><\/span><\/span><\/span>(ponte), um <\/b><\/span><\/span><\/span>ponto P fixo em qualquer parte do trem <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

(caminh\u00e3o) deve <\/b><\/span><\/span><\/span>percorrer uma dist\u00e2ncia \u0394S = x + c<\/b><\/span><\/span><\/span>, <\/b><\/span><\/span>com <\/b><\/span><\/span><\/span>velocidade V num intervalo de tempo \u0394t \u2014 V= (x +c)\/\u0394t. <\/b><\/span><\/span><\/span><\/p>\n

X = 400 m \"\" V = 20 m\/s \"\" c = 1800 m \"\" \u2206t = ?<\/b><\/span><\/span><\/span><\/p>\n

V = \"\" \"\" <\/b><\/span><\/span><\/span>20 = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>20<\/b><\/span><\/span><\/span>\u2206t = 2200 \"\" \u2206t = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>\u2206t = 110 s.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Teorema do Impulso<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>\u201c\u00a0O impulso da for\u00e7a resultante de um sistema de for\u00e7as que age sobre um corpo \u00e9 igual \u00e0 varia\u00e7\u00e3o da quantidade de movimento do corpo<\/b><\/span><\/span><\/span>\u201d<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0<\/span>Unidades no SI<\/b><\/span><\/span><\/span><\/p>\n

Impulso (I)\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0newton vezes segundo (N.s)<\/b><\/span><\/span><\/span><\/p>\n

Quantidade de movimento (Q)<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0quilograma vezes metro por segundo (kg.m\/s)<\/b><\/span><\/span><\/span><\/p>\n

\u00a0\"\"<\/span><\/p>\n

S\u00e3o dados: m <\/b><\/span><\/span><\/span>= 100 g <\/b><\/span><\/span><\/span>= 0,1 kg;<\/b><\/span><\/span><\/span> <\/b>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = V = <\/b><\/span><\/span><\/span>20 m\/s e <\/b><\/span><\/span><\/span>F = <\/b><\/span><\/span><\/span>100 N.<\/b><\/span><\/span><\/span><\/p>\n

F.\u2206t = 2mV \"\" <\/b><\/span><\/span><\/span>100.\u2206t = 2.0,1.20 \"\"<\/b><\/span><\/span><\/span> \u2206t = \"\"<\/b><\/span><\/span><\/span> <\/b>= 0,04 s. <\/b><\/span><\/span><\/span><\/p>\n

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

04- <\/span><\/span><\/span>Volume da piscina<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>V = S.h <\/b><\/span><\/span><\/span>= 40.1 <\/b><\/span><\/span><\/span>= 40 m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Considerando <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span> <\/b>= 1 g\/cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> <\/b>= 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> = \"\" \"\" <\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span> = 4.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> kg. <\/b><\/span><\/span><\/span><\/p>\n

Q = W = m.c.(\"\" \u2013 \"\"<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) <\/b><\/span><\/span><\/span>com <\/b><\/span><\/span><\/span>c =<\/b><\/span><\/span><\/span> 4,18 J\/g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C <\/b><\/span><\/span><\/span>= 4,18.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> J\/kg<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C \"\" <\/b><\/span><\/span><\/span>W = m.c.(\"\" \u2013 \"\"<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) <\/b><\/span><\/span><\/span>= 4.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.4,18.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.2<\/b><\/span><\/span><\/span><\/p>\n

W = 33,44.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span> J.<\/b><\/span><\/span><\/span><\/p>\n

Se <\/b><\/span><\/span><\/span>1 m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> de \u00e1gua absorve P = 834 W, 40 m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> absorver\u00e3o P =<\/b><\/span><\/span><\/span> 40.834 <\/b><\/span><\/span><\/span>= 33.360 W.<\/b><\/span><\/span><\/span><\/p>\n

P = \"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> 33.360 = 33,44.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>\/t \"\" t = 33,44.10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>\/33.360 \"\" <\/b><\/span><\/span><\/span>t = 10.000 s.<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>Colocando a <\/b><\/span><\/span><\/span>origem da trajet\u00f3ria (S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 0) no ponto de partida <\/b><\/span><\/span><\/span>dos dois trens e <\/b><\/span><\/span><\/span>deduzindo<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>equa\u00e7\u00e3o hor\u00e1ria de cada um \"\"<\/b><\/span><\/span><\/span> <\/b>S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.t <\/b><\/span><\/span><\/span>= 0 + 300t \"\" <\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 300t \"\"<\/b><\/span><\/span><\/span> <\/b>S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.t <\/b><\/span><\/span><\/span>= 0 + 250t \"\" <\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 250t.<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span><\/p>\n

Quando eles se <\/b><\/span><\/span><\/span>encontram a 10 km um do outro<\/b><\/span><\/span><\/span>, com <\/b><\/span><\/span><\/span>1 na frente voc\u00ea ter\u00e1 \"\"<\/b><\/span><\/span><\/span> <\/b>S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> \u2013 S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 10 km \"\"<\/b><\/span><\/span><\/span> 300t \u2013 250t = 10 \"\" 50t = 10 \"\" t = \"\"<\/b><\/span><\/span><\/span> = \"\" x 60 = \"\" \"\" <\/b><\/span><\/span><\/span>t = 12 minutos.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Apoio horizontal m\u00f3vel<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>(peso aparente, elevadores)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

No caso do <\/b><\/span><\/span><\/span>exerc\u00edcio, <\/b><\/span><\/span><\/span>como o <\/b><\/span><\/span><\/span>elevador sobe acelerando \"\"<\/b><\/span><\/span><\/span> <\/b>N \u2013 P = m.a \"\"<\/b><\/span><\/span><\/span> N = mg + ma = 95.10 + 95.0,5 \"\" N = 950 + 47,5 \"\" <\/b><\/span><\/span><\/span>N = 997,5 N. <\/b><\/span><\/span><\/span><\/p>\n

Com o <\/b><\/span><\/span><\/span>elevador em repouso \"\" N = P<\/b><\/span><\/span><\/span> = mg = 95.10 \"\" <\/b><\/span><\/span><\/span>N = 950 N <\/b><\/span><\/span><\/span><\/p>\n

Diferen\u00e7a <\/b><\/span><\/span><\/span>entre <\/b><\/span><\/span><\/span>peso real e peso aparente<\/b><\/span><\/span><\/span> \"\" 997,5 \u2013 950 <\/b><\/span><\/span><\/span>= 47,7 N.<\/b><\/span><\/span><\/span><\/p>\n

Diferen\u00e7a porcentual \"\"<\/b><\/span><\/span><\/span> d = 47,7×100\/997,5 = \"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> d = 4,78 %.<\/b><\/span><\/span><\/span><\/p>\n

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

07- <\/span><\/span><\/span>Equa\u00e7\u00e3o fundamental da ondulat\u00f3ria \"\"<\/b><\/span><\/span><\/span> <\/b>V = <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span>f \"\" 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> = <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>.4,6.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span> \"\" <\/b><\/span><\/span><\/span>\u0001 <\/b><\/span><\/span><\/span>= \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>\u0001 = 0,65.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span> \"\" <\/b><\/span><\/span><\/span>\u0001 = 650.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span> m. <\/b><\/span><\/span><\/span><\/p>\n

R \u2013 A<\/b><\/span><\/span><\/span><\/p>\n

08<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span><\/span> Sendo o <\/b><\/span><\/span><\/span>sistema conservativo a energia mec\u00e2nica total se conserva.<\/b><\/span><\/span><\/span><\/p>\n

Na <\/b><\/span><\/span><\/span>situa\u00e7\u00e3o inicial (n\u00edvel zero de altura h=0)<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>energia potencial gravitacional \u00e9 nula,<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>energia cin\u00e9tica <\/b><\/span><\/span><\/span>tamb\u00e9m <\/b><\/span><\/span><\/span>\u00e9 nula (V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = 0) <\/b><\/span><\/span><\/span>e a <\/b><\/span><\/span><\/span>\u00fanica energia nessa situa\u00e7\u00e3o \u00e9 a energia potencial el\u00e1stica armazenada<\/b><\/span><\/span><\/span> pela <\/b><\/span><\/span><\/span>mola fornecida por E<\/b><\/span><\/span><\/span>pe<\/b><\/span><\/span><\/sub><\/span> =k.\"\" <\/b><\/span><\/span><\/span>= 50.\"\"<\/b><\/span><\/span><\/span> = 25.0,0225 \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span> = Epe = 0,5625 J.<\/b><\/span><\/span><\/span><\/p>\n

Na <\/b><\/span><\/span><\/span>altura m\u00e1xima (h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>) <\/b><\/span><\/span><\/span>onde a <\/b><\/span><\/span><\/span>energia cin\u00e9tica \u00e9 nula (V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = 0)<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>energia armazenada \u00e9 a penas <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>potencial gravitacional de intensidade E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span> =E<\/b><\/span><\/span><\/span>pg<\/b><\/span><\/span><\/sub><\/span> = m.g.h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span> <\/b>= 2.9,8.h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span> = 19,6h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Sistema conservativo \"\" E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>0,5625 = 19,6h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span> <\/b>=\"\" = 0,028m <\/b><\/span><\/span><\/span>= 2,8 cm.<\/b><\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/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 curso de Medicina da UNESC-ES-2017 01- Primeira etapa com velocidade de 30 km\/h ele percorre \u2206S1 = 30 km em \u2206t1 = 1 h. Segunda etapa com velocidade de 60 km\/h ele percorre \u2206S2 = 30 km em \u2206t2 = 1\/2 h. Vmtotal = (\u2206S1 + \u2206S2)\/(\u2206t1 + \u2206t2) = (30 + 30)\/1,5)<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2911,"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-2913","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2913","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=2913"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2913\/revisions"}],"predecessor-version":[{"id":2914,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2913\/revisions\/2914"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2911"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2913"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}