{"id":3026,"date":"2017-05-30T02:38:43","date_gmt":"2017-05-30T02:38:43","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3026"},"modified":"2024-09-03T16:58:54","modified_gmt":"2024-09-03T16:58:54","slug":"resolucao-comentada-das-questoes-de-fisica-da-uerr-rr-2017","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2017\/uerr-2017\/resolucao-comentada-das-questoes-de-fisica-da-uerr-rr-2017\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da UERR-RR-2017"},"content":{"rendered":"

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

UERR-RR-2017<\/b><\/span><\/span><\/span><\/p>\n

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

Espelhos planos angulares<\/b><\/span><\/span><\/span><\/p>\n

Sendo\u00a0<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span><\/span>n\u00famero de imagens<\/b><\/span><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span><\/span>\u03b1<\/b><\/span><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span><\/span>\u00e2ngulo entre os dois espelhos<\/b><\/span><\/span><\/span>, a <\/b><\/span><\/span><\/span>rela\u00e7\u00e3o entre eles \u00e9 fornecida pela express\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como o <\/b><\/span><\/span><\/span>n\u00famero de imagens que foram formadas <\/b><\/span><\/span><\/span>equivale <\/b><\/span><\/span><\/span>1\/4 de vezes do valor do \u00e2ngulo <\/b><\/span><\/span><\/span>formado entre os dois espelhos \"\" <\/b><\/span><\/span><\/span>n = \"\"<\/b><\/span><\/span><\/span>\/4 \"\"<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span>= 4n \"\"<\/b><\/span><\/span><\/span> <\/b>n = 360\/\"\"<\/b><\/span><\/span><\/span> <\/b>– <\/b><\/span><\/span><\/span>1 \"\" n = \"\" \u2013 1 \"\" n = \"\" \u2013 1 \"\"<\/b><\/span><\/span><\/span><\/p>\n

n<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + n \u2013 90 = 0 \"\" \u2206 = 1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 360 = 361 \"\" \"\" = 19 \"\"<\/b><\/span><\/span><\/span> <\/b>n = ( -1 \"\"\/2<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>n <\/b><\/span><\/span><\/span>= \"\" <\/b><\/span><\/span><\/span>= 9 imagens \"\" <\/b><\/span><\/span><\/span>a outra resposta \u00e9 negativa e desprezada.\"\"<\/b><\/span><\/span><\/span>=<\/b><\/span><\/span><\/span> 4n = 4.<\/b><\/span><\/span><\/span>9 = 36<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/b><\/span><\/span><\/span>Veja quest\u00e3o 01.<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

No caso do exerc\u00edcio \"\"<\/b><\/span><\/span><\/span> <\/b>R = \"\"<\/b><\/span><\/span><\/span> <\/b>= \"\" = . . . = \"\" <\/b><\/span><\/span><\/span>= 0,25 \u2126<\/b><\/span><\/span><\/span><\/p>\n

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

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

Significado f\u00edsico do \u00edndice de refra\u00e7\u00e3o de um dado meio<\/b><\/span><\/span><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span><\/span><\/span>\u00edndice de refra\u00e7\u00e3o de um meio\u00a0<\/b><\/span><\/span><\/span><\/span>representa a\u00a0<\/b><\/span><\/span><\/span><\/span>raz\u00e3o (compara\u00e7\u00e3o)\u00a0<\/b><\/span><\/span><\/span><\/span>entre a<\/b><\/span><\/span><\/span><\/span>velocidade da luz no v\u00e1cuo e \u00a0a velocidade da luz no meio, ou seja:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

V = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>n = \"\"<\/b><\/span><\/span><\/span> \"\" n = \"\" \"\" <\/b><\/span><\/span><\/span>n = \"\"<\/b><\/span><\/span><\/span><\/p>\n

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

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

Carnot demonstrou que:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\" = 1 – \"\"<\/b><\/span><\/span><\/span> \"\" \"\"<\/b><\/span><\/span> = 1 – \"\"<\/b><\/span><\/span><\/span> \"\" \"\"<\/b><\/span><\/span><\/span> = 1 – \"\"<\/b><\/span><\/span><\/span> = (16 \u2013 5)\/16 = \"\" \"\" \"\"<\/b><\/span><\/span><\/span> = 0,6875×100% = 68,75%.<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/b><\/span><\/span><\/span>P = \"\" = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>P = \"\" \"\" <\/b><\/span><\/span><\/span>P = 3000 W = 3 kW.<\/b><\/span><\/span><\/span><\/p>\n

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

07- <\/b><\/span><\/span><\/span>For\u00e7a resultante \"\" F = m.a<\/b><\/span><\/span><\/span> = 5.1,5 \"\" <\/b><\/span><\/span><\/span>F = 7,5 N.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>decomposi\u00e7\u00e3o dessa for\u00e7a <\/b><\/span><\/span><\/span>nos <\/b><\/span><\/span><\/span>semieixos x e y <\/b><\/span><\/span><\/span>est\u00e1 na <\/b><\/span><\/span><\/span>figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

F<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span> = Fcos30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> = <\/b><\/span><\/span><\/span>7,5.\"\"<\/b><\/span><\/span><\/span> = 7,5×0,86 \"\" <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span> = 6,45 N<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span> = Fsen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> <\/b>= 7,5.\"\"<\/b><\/span><\/span><\/span> = 7,5×0,86 \"\" <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span> = 3,75 N<\/b><\/span><\/span><\/span><\/p>\n

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

08- <\/b><\/span><\/span><\/span><\/p>\n

Energia potencial gravitacional<\/b><\/span><\/span><\/span><\/p>\n

Considere um\u00a0<\/b><\/span><\/span><\/span>corpo de massa\u00a0m<\/b><\/span><\/span><\/span>\u00a0que se encontra num\u00a0<\/b><\/span><\/span><\/span>ponto P<\/b><\/span><\/span><\/span>\u00a0que est\u00e1 a uma determinada\u00a0<\/b><\/span><\/span><\/span>altura\u00a0h\u00a0de outro ponto Q.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

onde:<\/b><\/span><\/span><\/span><\/p>\n

Ep\u00a0<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span><\/span>energia potencial gravitacional<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>no SI, medida em joules\u00a0(J)<\/b><\/span><\/span><\/span><\/p>\n

m\u00a0<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span><\/span>massa<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>no SI, medida em quilograma\u00a0(kg)<\/b><\/span><\/span><\/span><\/p>\n

g<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o da gravidade local<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>no SI, medida em\u00a0m\/s2\u00a0ou\u00a0N\/kg<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = m.g.h <\/b><\/span><\/span><\/span>= 307.9,8.1,9 <\/b><\/span><\/span><\/span>= 5716,34 J \"\"<\/b><\/span><\/span><\/span> <\/b>E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = 5,71634 kJ.<\/b><\/span><\/span><\/span><\/p>\n

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

09- <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

W = q.(V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> \u2013 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span> = 1,2.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>.(10 \u2013 3) <\/b><\/span><\/span><\/span>= 8,4.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span> J = 8,4 \"\"<\/b><\/span><\/span><\/span>J.<\/b><\/span><\/span><\/span><\/p>\n

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

10- <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> <\/b>= 9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.7.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\/5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> <\/b>= 12,6.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> V<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> <\/b>= 9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.7.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\/59.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> <\/b>= 7.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> V<\/b><\/span><\/span><\/span><\/p>\n

U<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> = (V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> \u2013 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>) <\/b><\/span><\/span><\/span>= (12,6 -7).10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> <\/b>= 5,6.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> V.<\/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\u00cdSICA da UERR-RR-2017 01- Espelhos planos angulares Sendo\u00a0n\u00a0o\u00a0n\u00famero de imagens\u00a0e\u00a0\u03b1\u00a0o\u00a0\u00e2ngulo entre os dois espelhos, a rela\u00e7\u00e3o entre eles \u00e9 fornecida pela express\u00e3o: Como o n\u00famero de imagens que foram formadas equivale 1\/4 de vezes do valor do \u00e2ngulo formado entre os dois espelhos n = \/4 = 4n n = 360\/ – 1 n =<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3024,"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-3026","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3026","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=3026"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3026\/revisions"}],"predecessor-version":[{"id":3029,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3026\/revisions\/3029"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3024"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3026"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}