{"id":835,"date":"2015-05-14T17:44:02","date_gmt":"2015-05-14T17:44:02","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=835"},"modified":"2024-09-03T13:07:15","modified_gmt":"2024-09-03T13:07:15","slug":"resolucao-comentada-da-unifor-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-unifor-2015\/resolucao-comentada-da-unifor-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada da UNIFOR 2015"},"content":{"rendered":"

RESOLU\u00c7\u00d5ES<\/span><\/span><\/span><\/p>\n

01- <\/span><\/span><\/span>Observe, atrav\u00e9s dos raios not\u00e1veis as caracter\u00edsticas da imagem pedida<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

Objeto entre o foco F e o v\u00e9rtice V ou entre o foco F e o espelho<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A <\/b><\/span><\/span>imagem ter\u00e1:<\/b><\/span><\/span><\/span><\/p>\n

Natureza <\/b><\/span><\/span><\/span>\u2013 <\/b><\/span><\/span>virtual<\/b><\/span><\/span><\/span> (obtida na interse\u00e7\u00e3o do prolongamento dos raios de luz ) \u2013 n\u00e3o pode ser projetada, fotografada, etc.<\/b><\/span><\/span><\/p>\n

Localiza\u00e7\u00e3o \u2013 atr\u00e1s do espelho<\/b><\/span><\/span><\/span><\/p>\n

Tamanho \u2013 maior<\/b><\/span><\/span><\/span> que o do <\/b><\/span><\/span>objeto<\/b><\/span><\/span><\/span><\/p>\n

Orienta\u00e7\u00e3o<\/b><\/span><\/span><\/span> \u2013 <\/b><\/span><\/span>direita<\/b><\/span><\/span><\/span> em <\/b><\/span><\/span>rela\u00e7\u00e3o ao objeto<\/b><\/span><\/span><\/span><\/p>\n

Utilidades:<\/b><\/span><\/span><\/span> s\u00e3o empregados com freq\u00fc\u00eancia quando se deseja obter uma imagem <\/b><\/span><\/span>virtual e ampliada<\/b><\/span><\/span><\/span> <\/b>de um objeto<\/b><\/span><\/span><\/span>, como \u00e9 o caso dos espelhos de barbear, toalete, <\/b><\/span><\/span>de dentista<\/b><\/span><\/span><\/span>, espelho de otorrinolaringologia, etc.<\/b><\/span><\/span><\/p>\n

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

02- <\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>\u00fanica que n\u00e3o satisfaz \u00e9<\/b><\/span><\/span><\/span> \u201cavenidas paralelas de tr\u00e2nsito) o que <\/b><\/span><\/span><\/span>exclui a (IV).<\/b><\/span><\/span><\/span><\/p>\n

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

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

Carga el\u00e9trica\u00a0<\/b><\/span><\/span><\/span>q<\/b><\/span><\/span><\/span>\u00a0lan\u00e7ada com velocidade\u00a0\"\"\u00a0lan\u00e7ada perpendicularmente <\/b><\/span><\/span><\/span>\u00e0s linhas de indu\u00e7\u00e3o de um <\/b><\/span><\/span>campo magn\u00e9tico uniforme<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0\u00a0—\u00a0 observe que, neste caso o <\/b><\/span><\/span>\u00e2ngulo entre\u00a0\"\"e\u00a0\"\" \u00e9 90<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0e que <\/b><\/span><\/span>sen90<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=1.<\/b><\/span><\/span><\/span><\/p>\n

Na figura abaixo uma carga positiva\u00a0<\/b><\/span><\/span>q\u00a0<\/b><\/span><\/span><\/span>penetra com <\/b><\/span><\/span>velocidade \"\"\u00a0no ponto A<\/b><\/span><\/span><\/span> numa regi\u00e3o em que existe um campo <\/b><\/span><\/span>magn\u00e9tico uniforme\u00a0\"\"\u00a0penetrando na folha.<\/b><\/span><\/span><\/span> Observe que\u00a0\"\"e\u00a0\"\" s\u00e3o <\/b><\/span><\/span>perpendiculares<\/b><\/span><\/span><\/span> e, como a velocidade\u00a0\"\"\u00a0\u00e9 <\/b><\/span><\/span>sempre tangente \u00e0 trajet\u00f3ria em cada ponto<\/b><\/span><\/span><\/span>, a for\u00e7a magn\u00e9tica\"\", obtida pela <\/b><\/span><\/span>regra da m\u00e3o esquerda<\/b><\/span><\/span><\/span> e indicada na figura \u00e9 sempre dirigida para o <\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0centro de uma circunfer\u00eancia de raio R<\/b><\/span><\/span><\/span>. Assim, a <\/b><\/span><\/span>carga\u00a0<\/b><\/span><\/span><\/span>q<\/b><\/span><\/span><\/span>\u00a0realizar\u00e1 um <\/b><\/span><\/span>movimento circular uniforme com velocidade de intensidade constante.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A <\/b><\/span><\/span>express\u00e3o matem\u00e1tica<\/b><\/span><\/span><\/span> dessa for\u00e7a magn\u00e9tica <\/b><\/span><\/span>\u00e9 F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=q.V.B.sen\u03b8<\/b><\/span><\/span><\/span>=q.V.B.1\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=q.V.B<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 lembrando que a <\/b><\/span><\/span>for\u00e7a magn\u00e9tica<\/b><\/span><\/span><\/span> \"\",\u00a0respons\u00e1vel pelo <\/b><\/span><\/span>movimento circular \u00e9 a for\u00e7a resultante centr\u00edpeta <\/b><\/span><\/span><\/span>de intensidade <\/b><\/span><\/span>F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 q.V.B=m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span>\u00a0 —<\/b><\/span><\/span><\/p>\n

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

 <\/p>\n

centr\u00edpeta <\/b><\/span><\/span><\/span>de intensidade <\/b><\/span><\/span>F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 q.V.B=m.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span>\u00a0 — <\/b><\/span><\/span>R=m.V\/q.B<\/b><\/span><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/p>\n

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

04- <\/span><\/span><\/span>For\u00e7a de <\/b><\/span><\/span>tra\u00e7\u00e3o resultante T<\/b><\/span><\/span><\/span>, vertical e para baixo de <\/b><\/span><\/span>intensidade<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Lei dos cossenos<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = T\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + T\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 2.T\u2019.T\u2019.cos120<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> = 2T\u2019<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> + 2.T\u2019<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.(-1\/2) \"\" T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> = 2T\u2019<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> \u2013 T\u2019<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> = T\u2019<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> \"\" <\/b><\/span><\/span>T = T\u2019= 6N<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/p>\n

Colocando <\/b><\/span><\/span>todas as for\u00e7as que agem sobre o corpo<\/b><\/span><\/span><\/span>, que s\u00e3o:<\/b><\/span><\/span><\/p>\n

Tra\u00e7\u00e3o \"\" <\/b><\/span><\/span>T = 6 N.<\/b><\/span><\/span><\/span><\/p>\n

Peso \"\" <\/b><\/span><\/span>P <\/b><\/span><\/span><\/span>= mg<\/b><\/span><\/span><\/span> = 0,1.10 <\/b><\/span><\/span>= 1N <\/b><\/span><\/span><\/span>(vertical e para baixo).<\/b><\/span><\/span><\/span><\/p>\n

Empuxo \"\" <\/b><\/span><\/span>E =<\/b><\/span><\/span><\/span> d<\/b><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>corpo<\/b><\/span><\/span><\/sub>.g = d<\/b><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub>.2.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup> <\/b>.10 = <\/b><\/span><\/span>d<\/b><\/span><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span>2.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> N <\/b><\/span><\/span><\/span>(vertical e para cima)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

No equil\u00edbrio<\/b><\/span><\/span><\/span> \"\" F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub> = 0 \"\" <\/b><\/span><\/span>T + P = E<\/b><\/span><\/span><\/span> \"\" 6 + 1 = 2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.d<\/b><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub> \"\" <\/b><\/span><\/span>d<\/b><\/span><\/span><\/span>\u00f3leo<\/b><\/span><\/span><\/sub><\/span> = 7\/2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup> <\/b>= 3,5.10<\/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><\/p>\n

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

05- <\/span><\/span><\/span>C\u00e1lculo do <\/b><\/span><\/span>n\u00famero n de recargas<\/b><\/span><\/span><\/span> sabendo que <\/b><\/span><\/span>percorre 70km<\/b><\/span><\/span><\/span> e que <\/b><\/span><\/span>cada recarga <\/b><\/span><\/span><\/span>ocorre a <\/b><\/span><\/span>cada 14km<\/b><\/span><\/span><\/span> <\/b>percorrido<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>n <\/b><\/span><\/span><\/span>= 70km\/14km <\/b><\/span><\/span>= 5 recargas.<\/b><\/span><\/span><\/span><\/p>\n

Observe na figura abaixo que <\/b><\/span><\/span>na realidade s\u00e3o 4 recargas: <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Tempo total de recargas<\/b><\/span><\/span><\/span> se <\/b><\/span><\/span>cada uma demora 2h<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>\u2206t<\/b><\/span><\/span><\/span>r\u00ad<\/b><\/span><\/span><\/sub><\/span> <\/b>=n.2 = 4.2 <\/b><\/span><\/span>= 8 h.<\/b><\/span><\/span><\/span><\/p>\n

Tempo total da viajem sem as recargas<\/b><\/span><\/span><\/span>, percorrendo <\/b><\/span><\/span><\/span>70km<\/b><\/span><\/span><\/span> com velocidade de <\/b><\/span><\/span><\/span>20 km\/h<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>V = \u2206S\/\u2206t<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span> \"\"20 = 70\/\u2206t<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>\u2206t<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span> = 70\/20 <\/b><\/span><\/span><\/span>= 3,5 h.<\/b><\/span><\/span><\/span><\/p>\n

Tempo total<\/b><\/span><\/span><\/span> da viajem com as recargas \"\" <\/b><\/span><\/span>\u2206t<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span> = <\/b><\/span><\/span><\/span>8 + 3,5<\/b><\/span><\/span><\/span> = 11,5 h.<\/b><\/span><\/span><\/span><\/p>\n

Tempo pedido<\/b><\/span><\/span><\/span> sabendo que ele <\/b><\/span><\/span><\/span>partiu \u00e0s 5 h<\/b><\/span><\/span><\/span> e que a <\/b><\/span><\/span><\/span>viajem demorou 11,5h<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>t <\/b><\/span><\/span><\/span>= 5 + 11,5 <\/b><\/span><\/span><\/span>= 16,5 h.<\/b><\/span><\/span><\/span><\/p>\n

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

<\/h2>\n

Voltar para os exerc\u00edcios.<\/span><\/a><\/h2>\n","protected":false},"excerpt":{"rendered":"

RESOLU\u00c7\u00d5ES 01- Observe, atrav\u00e9s dos raios not\u00e1veis as caracter\u00edsticas da imagem pedida: Objeto entre o foco F e o v\u00e9rtice V ou entre o foco F e o espelho A imagem ter\u00e1: Natureza \u2013 virtual (obtida na interse\u00e7\u00e3o do prolongamento dos raios de luz ) \u2013 n\u00e3o pode ser projetada, fotografada, etc. Localiza\u00e7\u00e3o \u2013 atr\u00e1s do espelho Tamanho \u2013 maior<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":833,"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-835","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/835","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=835"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/835\/revisions"}],"predecessor-version":[{"id":836,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/835\/revisions\/836"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/833"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=835"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}