{"id":3567,"date":"2018-02-05T02:47:00","date_gmt":"2018-02-05T02:47:00","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3567"},"modified":"2024-09-04T12:28:46","modified_gmt":"2024-09-04T12:28:46","slug":"resolucao-comentada-das-questoes-de-fisica-da-escola-preparatoria-de-cadetes-do-exercito-espcex-20172018","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2018\/espcex-aman-20172018\/resolucao-comentada-das-questoes-de-fisica-da-escola-preparatoria-de-cadetes-do-exercito-espcex-20172018\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da Escola Preparat\u00f3ria de Cadetes do Ex\u00e9rcito (EsPCEx) \u2013 2017\/2018"},"content":{"rendered":"

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

Escola Preparat\u00f3ria de Cadetes do Ex\u00e9rcito (EsPCEx) \u2013 2017\/2018.<\/b><\/span><\/span><\/span><\/p>\n

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

Marcando os pontos com mesmo potencial<\/b><\/span><\/span><\/span> (curto circuito), <\/b><\/span><\/span><\/span>refazendo o circuito, tirando o volt\u00edmetro <\/b><\/span><\/span><\/span>porque \u00e9 ideal e possui resist\u00eancia interna infinita e <\/b><\/span><\/span><\/span>colocando as correntes lembrando <\/b><\/span><\/span><\/span>que o dispositivo de <\/b><\/span><\/span><\/span>8V \u00e9 o gerador<\/b><\/span><\/span><\/span> (corrente i sai do polo positivo) e <\/b><\/span><\/span><\/span>o de 4 V \u00e9 o receptor <\/b><\/span><\/span><\/span>(corrente i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> dai pelo polo negativo).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe que, como o <\/b><\/span><\/span><\/span>volt\u00edmetro que indica a tens\u00e3o U est\u00e1 em paralelo com os tr\u00eas dispositivos todos eles possuem a mesma ddp U.<\/b><\/span><\/span><\/span><\/p>\n

Para o gerador \"\" <\/b><\/span><\/span><\/span>equa\u00e7\u00e3o do gerador \"\"<\/b><\/span><\/span><\/span> <\/b>U = E \u2013 r.i \"\" U = 8 \u2013 2.i (I)<\/b><\/span><\/span><\/span><\/p>\n

Para o receptor \"\" <\/b><\/span><\/span><\/span>equa\u00e7\u00e3o do receptor \"\" U = E\u2019 + r\u2019.i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> \"\" U = 4 + 4.i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> (II)<\/b><\/span><\/span><\/span><\/p>\n

Para o resistor \"\" <\/b><\/span><\/span><\/span>lei de Ohmn \"\" U = R.i<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> \"\" U = 4.i<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sub><\/span>(III)<\/b><\/span><\/span><\/span><\/p>\n

I = i<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> + i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> (IV)<\/b><\/span><\/span><\/span><\/p>\n

Um <\/b><\/span><\/span><\/span>bloco A de massa 100 kg sobe, <\/b><\/span><\/span><\/span>em <\/b><\/span><\/span><\/span>movimento retil\u00edneo uniforme, <\/b><\/span><\/span><\/span>um\"\" <\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

Plano inclinado com atrito<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

C\u00e1lculo\u00a0<\/b><\/span><\/span>da intensidade de PP\u00a0e PN<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>utilizando o\u00a0<\/b><\/span><\/span>tri\u00e2ngulo ret\u00e2ngulo\u00a0<\/b><\/span><\/span>da figura abaixo:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0<\/span><\/p>\n

\"\"<\/p>\n

Se n\u00e3o existisse atrito,<\/b><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span>bloco descria<\/b><\/span><\/span>\u00a0com\u00a0<\/b><\/span><\/span>acelera\u00e7\u00e3o de intensidade\u00a0a<\/b><\/span><\/span>,<\/b><\/span><\/span>tal que\u00a0<\/b><\/span><\/span>FR\u00a0= m.a<\/b><\/span><\/span>\u00a0\u00a0\"\"\u00a0\u00a0\u00a0PP\u00a0= m.a\u00a0\"\"\u00a0\u00a0\u00a0<\/b><\/span><\/span>m,g.sen\u03b1 = m.a<\/b><\/span><\/span>\u00a0\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span>a = g.sen\u03b1 <\/b><\/span><\/span>(acelera\u00e7\u00e3o de descida do bloco de massa m).<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

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

Constru\u00e7\u00e3o geom\u00e9trica de imagens nos espelhos esf\u00e9ricos convexos<\/b><\/span><\/span><\/span><\/span><\/p>\n

Espelho convexo\u00a0\"\"\u00a0Para qualquer localiza\u00e7\u00e3o do objeto<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Independentemente da posi\u00e7\u00e3o do objeto, a imagem ter\u00e1 sempre:<\/b><\/span><\/span><\/span><\/p>\n

Natureza<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>virtual<\/b><\/span><\/span><\/span><\/p>\n

Localiza\u00e7\u00e3o<\/b><\/span><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>atr\u00e1s do espelho e entre V e F\u00a0<\/b><\/span><\/span><\/span>e observe nas\u00a0<\/b><\/span><\/span><\/span>figuras acima <\/b><\/span><\/span><\/span>que\u00a0<\/b><\/span><\/span><\/span>\u00e0 medida que o objeto se aproxima do espelho, a imagem tamb\u00e9m se aproxima e aumenta de tamanho, mas est\u00e1 sempre entre V e F.<\/b><\/span><\/span><\/span><\/p>\n

Tamanho\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>menor que o do objeto<\/b><\/span><\/span><\/span><\/p>\n

Orienta\u00e7\u00e3o\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>direita em rela\u00e7\u00e3o ao objeto<\/b><\/span><\/span><\/span><\/p>\n

Utilidades:<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Os\u00a0<\/b><\/span><\/span><\/span>espelhos convexos\u00a0<\/b><\/span><\/span><\/span>s\u00e3o empregados como\u00a0<\/b><\/span><\/span><\/span>retrovisores em ve\u00edculos, cabines de seguran\u00e7a, elevadores, etc.<\/b><\/span><\/span><\/span><\/p>\n

Sua vantagem sobre o espelho plano<\/b><\/span><\/span><\/span>, nesse particular, \u00e9 ter\u00a0<\/b><\/span><\/span><\/span>maior campo visual.\u00a0<\/b><\/span><\/span><\/span>T\u00eam, entretanto, o\u00a0<\/b><\/span><\/span><\/span>inconveniente de n\u00e3o darem no\u00e7\u00e3o da dist\u00e2ncia.<\/b><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

Equil\u00edbrio na horizontal \"\" N<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = T.<\/b><\/span><\/span><\/span><\/p>\n

Equil\u00edbrio na vertical \"\" P = N.<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo do <\/b><\/span><\/span><\/span>momento de cada for\u00e7a <\/b><\/span><\/span><\/span>que \u00e9 o valor da <\/b><\/span><\/span><\/span>for\u00e7a multiplicada <\/b><\/span><\/span><\/span>pela <\/b><\/span><\/span><\/span>dist\u00e2ncia do prolongamento da for\u00e7a at\u00e9 o polo 0 (colocado em B).<\/b><\/span><\/span><\/span><\/p>\n

Com o <\/b><\/span><\/span><\/span>polo em B <\/b><\/span><\/span><\/span>vamos <\/b><\/span><\/span><\/span>calcular o momento de cada for\u00e7a <\/b><\/span><\/span><\/span>estabelecento o <\/b><\/span><\/span><\/span>sentido hor\u00e1rio d rota\u00e7\u00e3o como positivo:<\/b><\/span><\/span><\/span><\/p>\n

M<\/b><\/span><\/span><\/span>N1<\/b><\/span><\/span><\/sub><\/span> = – N<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.AM = – T.1,5\"\"<\/b><\/span><\/span><\/span><\/p>\n

M<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> = + P.BN = 20.\"\"<\/b><\/span><\/span><\/span>.cos60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span> <\/b>= 20.\"\"<\/b><\/span><\/span><\/span>. \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> = 15 N.m<\/b><\/span><\/span><\/span><\/p>\n

No <\/b><\/span><\/span><\/span>equil\u00edbrio de rota\u00e7\u00e3o <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>soma dos momentos de cada for\u00e7a <\/b><\/span><\/span><\/span>em <\/b><\/span><\/span><\/span>rela\u00e7\u00e3o ao polo deve ser nula \"\"<\/b><\/span><\/span><\/span><\/p>\n

– T.1,5\"\"<\/b><\/span><\/span><\/span> + 15 = 0 \"\"<\/b><\/span><\/span><\/span> T = \"\"<\/b><\/span><\/span><\/span> <\/b>=<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>=<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>T = \"\" N.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Como as <\/b><\/span><\/span><\/span>for\u00e7as externas s\u00e3o desprezadas <\/b><\/span><\/span><\/span>voc\u00ea pode aplicar <\/b><\/span><\/span><\/span>o princ\u00edpio da conserva\u00e7\u00e3o da quantidade de movimento .<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

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

Sendo o <\/b><\/span><\/span><\/span>movimento na horizontal <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>altura n\u00e3o varia <\/b><\/span><\/span><\/span>(n\u00e3o existe varia\u00e7\u00e3o de energia potencial gravitacional). Ent\u00e3o, a <\/b><\/span><\/span><\/span>varia\u00e7\u00e3o \u00e9 apenas de energia cin\u00e9tica.<\/b><\/span><\/span><\/span><\/p>\n

Teorema da energia cin\u00e9tica:<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u201c O\u00a0<\/b><\/span><\/span><\/span>trabalho da resultante<\/b><\/span><\/span><\/span>\u00a0de\u00a0<\/b><\/span><\/span><\/span>todas as for\u00e7as<\/b><\/span><\/span><\/span>\u00a0que agem sobre um corpo\u00a0<\/b><\/span><\/span><\/span>\u00e9 igual \u00e0 varia\u00e7\u00e3o da energia cin\u00e9tica sofrida pelo corpo \u201c<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Assim, o\u00a0<\/b><\/span><\/span><\/span>trabalho da for\u00e7a resultante<\/b><\/span><\/span><\/span>\u00a0sobre o carro de\u00a0<\/b><\/span><\/span><\/span>massa m<\/b><\/span><\/span><\/span>\u00a0da figura acima no seu\u00a0<\/b><\/span><\/span><\/span>deslocamento<\/b><\/span><\/span><\/span>\u00a0de\u00a0<\/b><\/span><\/span><\/span>A para B<\/b><\/span><\/span><\/span>\u00a0\u00e9 fornecido por:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

No <\/b><\/span><\/span><\/span>caso do exerc\u00edcio <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>for\u00e7a resultante sobre o bloco \u00e9 a for\u00e7a de atrito <\/b><\/span><\/span><\/span>j\u00e1 que a normal e o peso se anulam (n\u00e3o existe movimento vertical), ent\u00e3o voc\u00ea pode aplicar o <\/b><\/span><\/span><\/span>teorema da energia cin\u00e9tica.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

Veja abaixo as duas condi\u00e7\u00f5es para que a for\u00e7a magn\u00e9tica sobre a carga seja nula.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

<\/a><\/p>\n

\"\"<\/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 Escola Preparat\u00f3ria de Cadetes do Ex\u00e9rcito (EsPCEx) \u2013 2017\/2018. 01- Marcando os pontos com mesmo potencial (curto circuito), refazendo o circuito, tirando o volt\u00edmetro porque \u00e9 ideal e possui resist\u00eancia interna infinita e colocando as correntes lembrando que o dispositivo de 8V \u00e9 o gerador (corrente i sai do polo positivo) e o<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3565,"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-3567","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3567","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=3567"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3567\/revisions"}],"predecessor-version":[{"id":3568,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3567\/revisions\/3568"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3565"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3567"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}