{"id":646,"date":"2015-04-12T14:43:52","date_gmt":"2015-04-12T14:43:52","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=646"},"modified":"2024-09-02T17:02:28","modified_gmt":"2024-09-02T17:02:28","slug":"resolucao-comentada-da-unifeso-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-da-unifeso-2015\/resolucao-comentada-da-unifeso-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada da UNIFESO – 2015"},"content":{"rendered":"

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

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

\"\"<\/p>\n

02- <\/b><\/span><\/span><\/span>Observe na <\/b><\/span><\/span><\/span>figura abaixo<\/b><\/span><\/span><\/span> que, pelo <\/b><\/span><\/span><\/span>princ\u00edpio da reversibilidade<\/b><\/span><\/span><\/span> (caminho inverso igual), se os <\/b><\/span><\/span><\/span>raios de luz incidirem da direita para a esquerda paralelamente ao eixo principal<\/b><\/span><\/span><\/span> eles se refratar\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

divergindo<\/b><\/span><\/span><\/span> de modo que os <\/b><\/span><\/span><\/span>prolongamentos dos raios refratados devem passar pelo foco da lente L<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>, que \u00e9 de f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 10cm.<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/b><\/span><\/span><\/span>C\u00e1lculo da <\/b><\/span><\/span><\/span>resist\u00eancia equivalente e da fem do geradorcom a chave C aberta:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Essa fem \u03b5 = 15R \u00e9 <\/b><\/span><\/span><\/span>constante<\/b><\/span><\/span><\/span>, pois o gerador \u00e9 o mesmo:<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>sequ\u00eancia <\/b><\/span><\/span><\/span>de figuras abaixo mostra o <\/b><\/span><\/span><\/span>c\u00e1lculo da resist\u00eancia do resistor equivalente<\/b><\/span><\/span><\/span> com a chave <\/b><\/span><\/span><\/span>C fechada<\/b><\/span><\/span><\/span> e retirando o amper\u00edmetro ideal (resist\u00eancia interna nula):<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Observe <\/b><\/span><\/span><\/span>atentamente na figura abaixo<\/b><\/span><\/span><\/span> que a corrente el\u00e9trica <\/b><\/span><\/span><\/span>indicada pelo amper\u00edmetro \u00e9 i<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 3 A.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

04- <\/b><\/span><\/span><\/span>Com <\/b><\/span><\/span><\/span>velocidade de 3cm\/s, em 2s o ponto P do pulso percorreu 2×3=6cm<\/b><\/span><\/span><\/span> e estar\u00e1 na <\/b><\/span><\/span><\/span>posi\u00e7\u00e3o <\/b><\/span><\/span><\/span>indicada a fnigura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

05-<\/span><\/span><\/span> Devido ao <\/b><\/span><\/span>teorema de Stevin, P=d<\/b><\/span><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub><\/span>.g.h<\/b><\/span><\/span><\/span>, para um mesmo l\u00edquido, <\/b><\/span><\/span>quanto maior a profundidade,maior ser\u00e1 a press\u00e3o<\/b><\/span><\/span><\/span>. Observe na figura abaixo, onde a esfera est\u00e1 imersa, que as <\/b><\/span><\/span>for\u00e7as exercidas<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0pelo l\u00edquido sobre ela s\u00e3o maiores nas \u00e1reas inferiores (onde h \u00e9 maior) que nas superiores (onde h\u00e9 menor).<\/b><\/span><\/span><\/span> A <\/b><\/span><\/span>for\u00e7a resultante de todas essas for\u00e7as est\u00e1 dirigida para cima<\/b><\/span><\/span><\/span>, e corresponde ao <\/b><\/span><\/span>empuxo <\/b><\/span><\/span><\/span>que age sobre essa esfera. Nas <\/b><\/span><\/span>laterais essas for\u00e7as se anulam,<\/b><\/span><\/span><\/span><\/p>\n

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

06- \"\"<\/span><\/span><\/span>For\u00e7as<\/b><\/span><\/span><\/span> que agem sobre o <\/b><\/span><\/span><\/span>sistema em equil\u00edbrio:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe que <\/b><\/span><\/span><\/span>as for\u00e7as que o bloco A troca com o plano inclinado s\u00e3o F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span> = 5N e P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span> = 12 N.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

O7- <\/span><\/span><\/span>Sendo que <\/b><\/span><\/span><\/span>n\u00e3o existe atrito<\/b><\/span><\/span><\/span>, na parte <\/b><\/span><\/span><\/span>superior <\/b><\/span><\/span><\/span>da haste a <\/b><\/span><\/span><\/span>\u00fanica for\u00e7a existente \u00e9 a normal N que \u00e9 afor\u00e7a pedida F, ou seja,<\/b><\/span><\/span><\/span> N = F.<\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>peso P = mg est\u00e1 no centro geom\u00e9trico<\/b><\/span><\/span><\/span> da haste, pois a mesma \u00e9 <\/b><\/span><\/span><\/span>homog\u00eanea <\/b><\/span><\/span><\/span>e, na <\/b><\/span><\/span><\/span>parte inferior<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

existem as duas componentes N<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span> e N<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Colocando o <\/b><\/span><\/span><\/span>polo 0<\/b><\/span><\/span><\/span> (eixo de rota\u00e7\u00e3o) na <\/b><\/span><\/span><\/span>parte inferior esquerda da haste<\/b><\/span><\/span><\/span>, vamos calcular o <\/b><\/span><\/span><\/span>momento de cada for\u00e7a em rela\u00e7\u00e3o a ele<\/b><\/span><\/span><\/span>, estabelecendo o <\/b><\/span><\/span><\/span>sentido hor\u00e1rio de rota\u00e7\u00e3o como positivo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

 <\/p>\n

Voltar para os Exerc\u00edcios<\/span><\/a><\/h3>\n

\u00a0<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

RESOLU\u00c7\u00d5ES 01- 02- Observe na figura abaixo que, pelo princ\u00edpio da reversibilidade (caminho inverso igual), se os raios de luz incidirem da direita para a esquerda paralelamente ao eixo principal eles se refratar\u00e3o divergindo de modo que os prolongamentos dos raios refratados devem passar pelo foco da lente L1, que \u00e9 de f1 = 10cm. R- B 03- C\u00e1lculo da<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":644,"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-646","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/646","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=646"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/646\/revisions"}],"predecessor-version":[{"id":648,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/646\/revisions\/648"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/644"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=646"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}