{"id":259,"date":"2014-12-13T20:06:54","date_gmt":"2014-12-13T20:06:54","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=259"},"modified":"2024-09-02T16:55:20","modified_gmt":"2024-09-02T16:55:20","slug":"resolucao-comentada-da-uerj-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-da-uerj-2015\/resolucao-comentada-da-uerj-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada da UERJ 2015"},"content":{"rendered":"

Resolu\u00e7\u00f5es<\/b><\/span><\/span><\/span><\/p>\n

01- <\/b><\/span><\/span><\/span>Observe que s\u00e3o desprezados os atritos e o <\/b><\/span><\/span><\/span>helic\u00f3ptero<\/b><\/span><\/span><\/span> est\u00e1 em <\/b><\/span><\/span><\/span>movimento retil\u00edneo <\/b><\/span><\/span><\/span>e, como <\/b><\/span><\/span><\/span>cada pacote atinge o solo em um ponto exatamente embaixo do helic\u00f3ptero<\/b><\/span><\/span><\/span>, ele (helic\u00f3ptero) est\u00e1 com <\/b><\/span><\/span><\/span>velocidade constante (acelera\u00e7\u00e3o nula).<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

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

05- <\/b><\/span><\/span><\/span>A fun\u00e7\u00e3o de um <\/b><\/span><\/span><\/span>transformador <\/b><\/span><\/span><\/span>\u00e9 aumentar ou diminuir a diferen\u00e7a de potencial\u00a0 (voltagem).\u00a0Trata-se<\/b><\/span><\/span><\/span><\/p>\n

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

de um \u00a0dispositivo de corrente alternada que opera <\/b><\/span><\/span><\/span>baseado nos princ\u00edpios eletromagn\u00e9ticos da Lei de Faraday<\/b><\/span><\/span><\/span> e da Lei de Lenz.<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

15- <\/b><\/span><\/span><\/span>Na <\/b><\/span><\/span><\/span>figura abaixo<\/b><\/span><\/span><\/span> uma carga <\/b><\/span><\/span><\/span>positiva\u00a0<\/b><\/span><\/span><\/span>q<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>penetra com <\/b><\/span><\/span><\/span>velocidade\"\"<\/b><\/span><\/span><\/span>\u00a0no ponto A numa regi\u00e3o em que existe <\/b><\/span><\/span><\/span>um campo magn\u00e9tico uniforme\u00a0\"\"\u00a0penetrando na folha<\/b><\/span><\/span><\/span>. Observe que\"\"\u00a0e\u00a0\"\" s\u00e3o <\/b><\/span><\/span><\/span>perpendiculare<\/b><\/span><\/span><\/span>s e, como a velocidade\"\"\u00a0\u00e9 sempre tangente \u00e0 trajet\u00f3ria em cada ponto, a <\/b><\/span><\/span><\/span>for\u00e7a magn\u00e9tica\u00a0\"\"<\/b><\/span><\/span><\/span>, obtida pela regra da m\u00e3o esquerda e indicada na figura \u00e9 sempre dirigida para o<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

A <\/b><\/span><\/span><\/span>express\u00e3o matem\u00e1tica dessa for\u00e7a magn\u00e9tica \u00e9 F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=q.V.B.sen\u03b8=q.V.B.1<\/b><\/span><\/span><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0F<\/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><\/span>for\u00e7a magn\u00e9tica F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>,\u00a0<\/b><\/span><\/span><\/span> respons\u00e1vel pelo movimento circular \u00e9 a <\/b><\/span><\/span><\/span>for\u00e7a resultante<\/b><\/span><\/span><\/span> <\/b>centr\u00edpeta<\/b><\/span><\/span><\/span> de intensidade <\/b><\/span><\/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<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 <\/b><\/span><\/span><\/span>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\u00a0 — <\/b><\/span><\/span><\/span>R=m.V\/q.B\u00a0<\/b><\/span><\/span><\/span>(express\u00e3o matem\u00e1tica pedida).<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Confira os exerc\u00edcios<\/a> <\/span>
\n<\/span><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00f5es 01- Observe que s\u00e3o desprezados os atritos e o helic\u00f3ptero est\u00e1 em movimento retil\u00edneo e, como cada pacote atinge o solo em um ponto exatamente embaixo do helic\u00f3ptero, ele (helic\u00f3ptero) est\u00e1 com velocidade constante (acelera\u00e7\u00e3o nula). R- C 02- 03- 04- R- D 05- A fun\u00e7\u00e3o de um transformador \u00e9 aumentar ou diminuir a diferen\u00e7a de potencial\u00a0 (voltagem).\u00a0Trata-se de<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":257,"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-259","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/259","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=259"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/259\/revisions"}],"predecessor-version":[{"id":10993,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/259\/revisions\/10993"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/257"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=259"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}