{"id":3047,"date":"2017-05-30T03:06:42","date_gmt":"2017-05-30T03:06:42","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3047"},"modified":"2024-09-03T16:52:13","modified_gmt":"2024-09-03T16:52:13","slug":"resolucao-comentada-das-questoes-de-fisica-da-ueg-go-2017","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2017\/ueg-2017\/resolucao-comentada-das-questoes-de-fisica-da-ueg-go-2017\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da UEG-GO-2017"},"content":{"rendered":"

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

UEG-GO-2017<\/b><\/span><\/span><\/span><\/p>\n

01- <\/b><\/span><\/span><\/span>Sendo a <\/b><\/span><\/span><\/span>part\u00edcula com carga q positiva <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>for\u00e7a el\u00e9trica sobre ela tem a mesma dire\u00e7\u00e3o e sentido do campo el\u00e9trico <\/b><\/span><\/span><\/span>(vertical e para baixo).<\/b><\/span><\/span><\/span><\/p>\n

<\/a>Para determinar a <\/b><\/span><\/span><\/span>dire\u00e7\u00e3o e sentido da for\u00e7a magn\u00e9tica <\/b><\/span><\/span><\/span>voc\u00ea deve utilizar a <\/b><\/span><\/span><\/span>regra da m\u00e3o esquerda <\/b><\/span><\/span><\/span>com o <\/b><\/span><\/span><\/span>campo magn\u00e9tico entrando na folha e a velocidade horizontal e para a direita <\/b><\/span><\/span><\/span>(veja fisicaevestibular.com.br \u2013 eletricidade \u2013 eletromagnetismo- for\u00e7a magn\u00e9tica sobre cargas em movimento).<\/b><\/span><\/span><\/span><\/p>\n

Pela <\/b><\/span><\/span><\/span>regra da m\u00e3o esquerda a for\u00e7a magn\u00e9tica ter\u00e1 dire\u00e7\u00e3o vertical e sentido para cima.<\/b><\/span><\/span><\/span><\/p>\n

Assim, para que a <\/b><\/span><\/span><\/span>carga n\u00e3o sofra desvio na vertical a for\u00e7a el\u00e9trica <\/b><\/span><\/span><\/span>(vertical e para baixo) e a <\/b><\/span><\/span><\/span>magn\u00e9tica <\/b><\/span><\/span><\/span>(vertical e para cima)<\/b><\/span><\/span><\/span> devem se anular, ou seja, possu\u00edrem o mesmo m\u00f3dulo.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/b><\/span><\/span><\/span>\u2206L = <\/b><\/span><\/span><\/span>8 mm <\/b><\/span><\/span><\/span>= 8.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> <\/b>m \"\" <\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 5m \"\" \"\"<\/b><\/span><\/span><\/span> = ? \"\" \"\"<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 0 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C \"\" \"\" = 100<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C.<\/b><\/span><\/span><\/span><\/p>\n

\u2206L = L<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\"\"<\/b><\/span><\/span><\/span> \u2013 \"\"<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) \"\"<\/b><\/span><\/span><\/span> 8.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> = 5.\"\"(100 \u2013 0) \"\" \"\"= \"\" <\/b><\/span><\/span><\/span>= 1,6.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span> <\/b>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>.<\/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 UEG-GO-2017 01- Sendo a part\u00edcula com carga q positiva a for\u00e7a el\u00e9trica sobre ela tem a mesma dire\u00e7\u00e3o e sentido do campo el\u00e9trico (vertical e para baixo). Para determinar a dire\u00e7\u00e3o e sentido da for\u00e7a magn\u00e9tica voc\u00ea deve utilizar a regra da m\u00e3o esquerda com o campo magn\u00e9tico entrando na folha e a<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3045,"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-3047","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3047","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=3047"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3047\/revisions"}],"predecessor-version":[{"id":3048,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3047\/revisions\/3048"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3045"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3047"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}