{"id":3842,"date":"2018-04-15T21:30:09","date_gmt":"2018-04-15T21:30:09","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3842"},"modified":"2024-09-04T12:24:15","modified_gmt":"2024-09-04T12:24:15","slug":"resolucao-comentada-das-questoes-de-fisica-da-cederj-rj-2018","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2018\/cederj-2018\/resolucao-comentada-das-questoes-de-fisica-da-cederj-rj-2018\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da CEDERJ-RJ \u2013 2018"},"content":{"rendered":"

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

CEDERJ-RJ \u2013 2018<\/span><\/span><\/span><\/p>\n

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

Se o\u00a0<\/b><\/span><\/span><\/span><\/span>corpo imerso<\/b><\/span><\/span><\/span><\/span>\u00a0recebe do\u00a0<\/b><\/span><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/span><\/span>\u00a0uma\u00a0<\/b><\/span><\/span><\/span><\/span>for\u00e7a vertical e para cima (Empuxo) \"\"<\/b><\/span><\/span><\/span><\/span>, pelo princ\u00edpio da<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

a\u00e7\u00e3o e rea\u00e7\u00e3o\u00a0o\u00a0<\/b><\/span><\/span><\/span><\/span>corpo reage sobre o l\u00edquido\u00a0com for\u00e7a de mesma intensidade (Empuxo), mesma dire\u00e7\u00e3o\u00a0(vertical) e sentido contr\u00e1rio (para baixo) – \"\".<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

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

Equa\u00e7\u00e3o geral dos gases perfeitos<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

As <\/b><\/span><\/span><\/span>for\u00e7as el\u00e9tricas trocadas entre A e B s\u00e3o de repuls\u00e3o<\/b><\/span><\/span><\/span> (q<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> e q<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> se repelem <\/b><\/span><\/span><\/span>\u201cmesmo sinal\u201d<\/b><\/span><\/span><\/span>) e possuem a <\/b><\/span><\/span><\/span>mesma intensidade (m\u00f3dulo) pelo princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Como as setas representam as acelera\u00e7\u00f5es, a intensidade (m\u00f3dulo) de B \u00e9 maior que a de A \"\" a<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> > a<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Segunda lei de Newton \"\" F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>A <\/b><\/span><\/span><\/sub><\/span>e F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>B.<\/b><\/span><\/span><\/sub><\/span>a<\/b><\/span><\/span><\/span>B \"\"<\/b><\/span><\/span><\/sub><\/span> <\/b>F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> \"\" m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.a<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>B.<\/b><\/span><\/span><\/sub><\/span>a<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> como <\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> > a<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>, m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> deve ser menor que m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> \"\" m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> < m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Intensidade Sonora<\/b><\/span><\/span><\/span><\/p>\n

\u00c8 a\u00a0<\/b><\/span><\/span><\/span>caracter\u00edstica do som\u00a0<\/b><\/span><\/span><\/span>que nos permite\u00a0<\/b><\/span><\/span><\/span>distinguir um som forte de um som fraco\u00a0<\/b><\/span><\/span><\/span>e est\u00e1\u00a0<\/b><\/span><\/span><\/span>relacionada<\/b><\/span><\/span><\/span>\u00a0com a\u00a0<\/b><\/span><\/span><\/span>energia transportada pela onda\u00a0<\/b><\/span><\/span><\/span>que, <\/b><\/span><\/span><\/span>quanto mais perto da fonte mais forte ser\u00e1 o som e mais afastado da fonte, mais fraco.<\/b><\/span><\/span><\/span><\/p>\n

Dessa maneira, o\u00a0<\/b><\/span><\/span><\/span>som emitido<\/b><\/span><\/span><\/span>, por exemplo, por\u00a0<\/b><\/span><\/span><\/span>uma britadeira<\/b><\/span><\/span><\/span>,\u00a0\u00a0<\/b><\/span><\/span><\/span>ser\u00e1 mais forte para uma pessoa <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

perto da<\/b><\/span><\/span><\/span> mesma e mais fraco para uma pessoa afastada da mesma.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

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

A <\/b><\/span><\/span><\/span>tens\u00e3o nos terminais dos dois resistores paralelos \u00e9 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> <\/b>(caracter\u00edstica da associa\u00e7\u00e3o paralelo) e <\/b><\/span><\/span><\/span>nos terminais dos outros dois tamb\u00e9m \u00e9 V<\/b><\/span><\/span><\/span>1.<\/b><\/span><\/sub><\/span><\/p>\n

Assim, a <\/b><\/span><\/span>for\u00e7a eletromotriz do gerador \u00e9 V<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span> <\/b>tal que <\/b><\/span><\/span>V<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span> = 3 V<\/b><\/span><\/span>1.<\/b><\/span><\/sub><\/span><\/p>\n

R – C<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

No caso do exerc\u00edcio, o <\/b><\/span><\/span><\/span>raio de luz ao passar do meio 1 para o meio 2 ele se aproxima da normal, <\/b><\/span><\/span><\/span>ent\u00e3o <\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> > n<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Quando o <\/b><\/span><\/span><\/span>raio de luz passa do meio 2 para o meio 3 ele se afasta da normal,<\/b><\/span><\/span><\/span> ent\u00e3o <\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> > n<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

n<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> > n<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> porque em 3 ele afasta-se mais da normal do que em 1.<\/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 CEDERJ-RJ \u2013 2018 01- Se o\u00a0corpo imerso\u00a0recebe do\u00a0l\u00edquido\u00a0uma\u00a0for\u00e7a vertical e para cima (Empuxo) , pelo princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o\u00a0o\u00a0corpo reage sobre o l\u00edquido\u00a0com for\u00e7a de mesma intensidade (Empuxo), mesma dire\u00e7\u00e3o\u00a0(vertical) e sentido contr\u00e1rio (para baixo) – . R- A 02- R- D 03- Equa\u00e7\u00e3o geral dos gases perfeitos 04- As for\u00e7as el\u00e9tricas<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3840,"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-3842","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3842","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=3842"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3842\/revisions"}],"predecessor-version":[{"id":11061,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3842\/revisions\/11061"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3840"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3842"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}