{"id":3021,"date":"2017-05-30T02:36:36","date_gmt":"2017-05-30T02:36:36","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3021"},"modified":"2024-09-03T17:05:41","modified_gmt":"2024-09-03T17:05:41","slug":"resolucao-comentada-das-questoes-de-fisica-da-ufrr-rr-017","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2017\/ufrr-2017\/resolucao-comentada-das-questoes-de-fisica-da-ufrr-rr-017\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da UFRR-RR-017"},"content":{"rendered":"

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

UFRR-RR-017<\/b><\/span><\/span><\/span><\/p>\n

01- \"\"<\/b><\/span><\/span><\/span> A <\/b><\/span><\/span><\/span>velocidade escalar m\u00e9dia independe da dire\u00e7\u00e3o e sentido do deslocamento <\/b><\/span><\/span><\/span>(num carro seria a indica\u00e7\u00e3o do od\u00f4metro), sendo ent\u00e3o, a <\/b><\/span><\/span><\/span>soma dos m\u00f3dulos das dist\u00e2ncias percorridas dividido pelo tempo gasto para percorr\u00ea-las.<\/b><\/span><\/span><\/span><\/p>\n

Velocidade escalar m\u00e9dia na chegada \"\"<\/b><\/span><\/span><\/span> <\/b>V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span>= \"\" \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = 7 km\/h<\/b><\/span><\/span><\/span>\/3,6 <\/b><\/span><\/span><\/span>= 1,94 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

Velocidade escalar m\u00e9dia na metade do percurso \"\"<\/b><\/span><\/span><\/span> <\/b>V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span>= \"\" \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = 7 km\/h<\/b><\/span><\/span><\/span>\/3,6 <\/b><\/span><\/span><\/span>= 1,94 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Velocidade m\u00e9dia<\/b><\/span><\/span><\/span> (velocidade vetorial m\u00e9dia) na <\/b><\/span><\/span><\/span>metade do caminho \"\"<\/b><\/span><\/span><\/span> o <\/b><\/span><\/span><\/span>deslocamento vetorial <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

tem m\u00f3dulo d<\/b><\/span><\/span><\/span>t <\/b><\/span><\/span><\/sub><\/span> que<\/b><\/span><\/span><\/span>, por <\/b><\/span><\/span><\/span>Pit\u00e1goras<\/b><\/span><\/span><\/span> vale <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span> = 5 km <\/b><\/span><\/span><\/span>demorando <\/b><\/span><\/span><\/span>t = 1 h \"\"<\/b><\/span><\/span><\/span> <\/b>V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = d<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span>\/t <\/b><\/span><\/span><\/span>= 5\/1 <\/b><\/span><\/span><\/span>= 5km\/h\/<\/b><\/span><\/span><\/span>3,6 = 1,388 = <\/b><\/span><\/span><\/span>1,39 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/b><\/span><\/span><\/span>S\u00e3o dados: m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = m; m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = 2m; V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/2 \"\" V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 2V<\/b><\/span><\/span><\/span>B; <\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>final<\/b><\/span><\/span><\/sub><\/span> = 0 e d<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> = d<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span> = d.<\/b><\/span><\/span><\/span><\/p>\n

Bloco A<\/b><\/span><\/span><\/span> \"\" Ele <\/b><\/span><\/span><\/span>penetra no plano com atrito <\/b><\/span><\/span><\/span>com <\/b><\/span><\/span><\/span>velocidade inicial V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>, <\/b><\/span><\/span><\/span>adquire <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o<\/b><\/span><\/span><\/span> de retardamento <\/b><\/span><\/span><\/span>\u2013 a<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> <\/b>sujeito apenas \u00e0 <\/b><\/span><\/span><\/span>for\u00e7a de atrito F<\/b><\/span><\/span><\/span>atA<\/b><\/span><\/span><\/sub><\/span> <\/b>que \u00e9 a <\/b><\/span><\/span><\/span>for\u00e7a resultante sobre ele, <\/b><\/span><\/span><\/span>atuando at\u00e9 que <\/b><\/span><\/span><\/span>ele pare (V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = 0) <\/b><\/span><\/span><\/span>ap\u00f3s <\/b><\/span><\/span><\/span>percorrer uma dist\u00e2ncia d.<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>atA<\/b><\/span><\/span><\/sub><\/span> =\u00b5<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.N<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = \u00b5<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.g = \u00b5<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.m.g \"\" <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>atA<\/b><\/span><\/span><\/sub><\/span> = = \u00b5<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.m.g. <\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>RA<\/b><\/span><\/span><\/sub><\/span> = ma<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>F<\/b><\/span><\/span><\/span>atA<\/b><\/span><\/span><\/sub><\/span> = = \u00b5<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.m.g \"\"<\/b><\/span><\/span><\/span> ma<\/b><\/span><\/span><\/span>A =<\/b><\/span><\/span><\/sub><\/span>= \u00b5<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.m.g \"\" <\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = <\/b><\/span><\/span><\/span>=<\/b><\/span><\/span><\/sub><\/span>= \u00b5<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.g.<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o de Torricelli \"\" V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> = V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> \u2013 2.a<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.d \"\"<\/b><\/span><\/span><\/span> 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> – 2. \u00b5<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.g.d \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 2. \u00b5<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.g.d, mas V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 2V<\/b><\/span><\/span><\/span>B <\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>(2V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 2. \u00b5<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.g.d \"\" 4V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 2. \u00b5<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.g.d (I)<\/b><\/span><\/span><\/span><\/p>\n

Bloco B<\/b><\/span><\/span><\/span> \"\" Ele <\/b><\/span><\/span><\/span>penetra no plano com atrito <\/b><\/span><\/span><\/span>com <\/b><\/span><\/span><\/span>velocidade inicial V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>, <\/b><\/span><\/span><\/span>adquire <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o<\/b><\/span><\/span><\/span> de retardamento <\/b><\/span><\/span><\/span>\u2013 a<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> <\/b>sujeito apenas \u00e0 <\/b><\/span><\/span><\/span>for\u00e7a de atrito F<\/b><\/span><\/span><\/span>atB<\/b><\/span><\/span><\/sub><\/span> <\/b>que \u00e9 a <\/b><\/span><\/span><\/span>for\u00e7a resultante sobre ele, <\/b><\/span><\/span><\/span>atuando at\u00e9 que <\/b><\/span><\/span><\/span>ele pare (V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = 0) <\/b><\/span><\/span><\/span>ap\u00f3s <\/b><\/span><\/span><\/span>percorrer uma dist\u00e2ncia d.<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>atB<\/b><\/span><\/span><\/sub><\/span> =\u00b5<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.N<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> <\/b>= \u00b5<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.g = \u00b5<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.2m.g \"\" <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>atB<\/b><\/span><\/span><\/sub><\/span> = = \u00b5<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.2m.g. <\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>RB<\/b><\/span><\/span><\/sub><\/span> = ma<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>F<\/b><\/span><\/span><\/span>atB<\/b><\/span><\/span><\/sub><\/span> = = \u00b5<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.2m.g \"\"<\/b><\/span><\/span><\/span> 2ma<\/b><\/span><\/span><\/span>B =<\/b><\/span><\/span><\/sub><\/span>= \u00b5<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.2m.g \"\" <\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = <\/b><\/span><\/span><\/span>=<\/b><\/span><\/span><\/sub><\/span>= \u00b5<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.g.<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o de Torricelli \"\" V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> = V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> \u2013 2.a<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.d \"\"<\/b><\/span><\/span><\/span> 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> – 2. \u00b5<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.g.d \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 2. \u00b5<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.g.d (II) <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

03- <\/b><\/span><\/span><\/span>Luneta astron\u00f4mica\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>utilizada na<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>observa\u00e7\u00e3o de astros ou de objetos distantes.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Consta de<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>2 lentes convergentes, a objetiva e a ocular.\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>Ao contr\u00e1rio do microsc\u00f3pio composto, a<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>dist\u00e2ncia focal da objetiva \u00e9 muito grande<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>(da ordem de metros). A<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>ocular tem pequena dist\u00e2ncia focal.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Devido ao fato de o<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>objeto estar muito afastado, a imagem (i1) formada pela objetiva est\u00e1 praticamente sobre seu foco (f<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span> objetiva) e \u00e9 real e invertida (veja figura).<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"Essa<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>imagem (i1) vai servir como objeto para a ocular<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>resultando numa <\/b><\/span><\/span>imagem final (i2), virtual<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>e<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>invertida em rela\u00e7\u00e3o ao objeto.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

04- <\/b><\/span><\/span><\/span>Isic\u00f3rica (isom\u00e9trica ou isovolum\u00e9trica) \"\"<\/b><\/span><\/span><\/span> <\/b>volume constante \"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> = \"\"<\/b><\/span><\/span><\/span> = constante \"\"<\/b><\/span><\/span><\/span> proporcionais, o <\/b><\/span><\/span><\/span>gr\u00e1fico \u00e9 uma reta inclinada.<\/b><\/span><\/span><\/span><\/p>\n

Isob\u00e1rica \"\"<\/b><\/span><\/span><\/span> <\/b>press\u00e3o constante \"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>=<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> = constante \"\"<\/b><\/span><\/span><\/span> proporcionais, o <\/b><\/span><\/span><\/span>gr\u00e1fico \u00e9 uma reta inclinada.<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/b><\/span><\/span><\/span>Apenas a (II) est\u00e1 errada, <\/b><\/span><\/span><\/span>pois <\/b><\/span><\/span><\/span>polos de mesmo nome se repelem <\/b><\/span><\/span><\/span>e polos de <\/b><\/span><\/span><\/span>nomes opostos se atraem.<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/b><\/span><\/span><\/span>Veja o <\/b><\/span><\/span><\/span>c\u00e1lculo do resistor equivalente na sequ\u00eancia abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> 1,5 = \"\" \"\" <\/b><\/span><\/span><\/span>i = 8,0 A.<\/b><\/span><\/span><\/span><\/p>\n

R- A<\/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\u00cdSICA da UFRR-RR-017 01- A velocidade escalar m\u00e9dia independe da dire\u00e7\u00e3o e sentido do deslocamento (num carro seria a indica\u00e7\u00e3o do od\u00f4metro), sendo ent\u00e3o, a soma dos m\u00f3dulos das dist\u00e2ncias percorridas dividido pelo tempo gasto para percorr\u00ea-las. Velocidade escalar m\u00e9dia na chegada Vm = = Vm = 7 km\/h\/3,6 = 1,94 m\/s. Velocidade escalar m\u00e9dia<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3019,"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-3021","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3021","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=3021"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3021\/revisions"}],"predecessor-version":[{"id":3022,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3021\/revisions\/3022"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3019"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3021"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}