{"id":3497,"date":"2018-01-19T02:11:30","date_gmt":"2018-01-19T02:11:30","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3497"},"modified":"2024-09-04T12:30:10","modified_gmt":"2024-09-04T12:30:10","slug":"resolucao-comentadada-das-questoes-de-fisica-da-famerp-2018","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2018\/famerp\/resolucao-comentadada-das-questoes-de-fisica-da-famerp-2018\/","title":{"rendered":"Resolu\u00e7\u00e3o comentadada das quest\u00f5es de F\u00cdSICA da FAMERP- 2018"},"content":{"rendered":"

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

FAMERP- 2018<\/b><\/span><\/span><\/span><\/p>\n

01- <\/b><\/span><\/span><\/span>Em <\/b><\/span><\/span><\/span>rela\u00e7\u00e3o \u00e0 pessoa, <\/b><\/span><\/span><\/span>como o vag\u00e3o est\u00e1 em <\/b><\/span><\/span><\/span>trajet\u00f3ria retil\u00ednea e em MRU<\/b><\/span><\/span><\/span> com <\/b><\/span><\/span><\/span>velocidade constante, <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>trajet\u00f3ria <\/b><\/span><\/span><\/span>da pequena esfera <\/b><\/span><\/span><\/span>ser\u00e1 vertical na subida e na descida,<\/b><\/span><\/span><\/span> <\/b>retornando \u00e0 sua m\u00e3o <\/b><\/span><\/span><\/span>(lan\u00e7amento vertical para cima).<\/b><\/span><\/span><\/span><\/p>\n

Mas, em <\/b><\/span><\/span><\/span>rela\u00e7\u00e3o ao solo <\/b><\/span><\/span><\/span>voc\u00ea <\/b><\/span><\/span><\/span>observar\u00e1 dois movimentos,<\/b><\/span><\/span><\/span> um na <\/b><\/span><\/span><\/span>horizontal<\/b><\/span><\/span><\/span> (devido \u00e0 velocidade do vag\u00e3o, constante e para a direita) e <\/b><\/span><\/span><\/span>outro vertical, <\/b><\/span><\/span><\/span>devido ao lan\u00e7amento para cima.<\/b><\/span><\/span><\/span><\/p>\n

Assim, um <\/b><\/span><\/span><\/span>observador no solo observar\u00e1 um movimento parab\u00f3lico, <\/b><\/span><\/span><\/span>ou seja, um <\/b><\/span><\/span><\/span>lan\u00e7amento obl\u00edquo cujas caracter\u00edsticas s\u00e3o fornecidas a seguir:<\/b><\/span><\/span><\/span><\/p>\n

Movimento parcial na dire\u00e7\u00e3o vertical (eixo Y)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Movimento parcial na dire\u00e7\u00e3o horizontal (eixo X)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Sendo o\u00a0<\/b><\/span><\/span>movimento apenas na horizontal,<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>proje\u00e7\u00e3o do vetor acelera\u00e7\u00e3o da gravidade (\"\") no eixo X \u00e9 nula, ou seja, na dire\u00e7\u00e3o horizontal do lan\u00e7amento obl\u00edquo n\u00e3o existe acelera\u00e7\u00e3o\u00a0<\/b><\/span><\/span>e, nesse caso o\u00a0<\/b><\/span><\/span>movimento \u00e9 <\/b><\/span><\/span>uniforme (MU)<\/b><\/span><\/span>\u00a0com velocidade constante\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>ox<\/b><\/span><\/sub><\/span>\u00a0= V<\/b><\/span><\/span>o<\/b><\/span><\/sub><\/span>cos\u03b1.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

O\u00a0<\/b><\/span><\/span>vetor velocidade\u00a0\"\"\u00a0\u00e9 sempre tangente \u00e0 trajet\u00f3ria em cada ponto e sempre decomposto em\u00a0\"\"e\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe que\u00a0no <\/b><\/span><\/span>ponto mais alto da trajet\u00f3ria a velocidade vetorial\u00a0\"\"\u00a0<\/b><\/span><\/span>n\u00e3o \u00e9 nula<\/b><\/span><\/span>, tem intensidade <\/b><\/span><\/span>m\u00ednima e \u00e9 igual \u00e0 componente horizontal, ou seja,\u00a0\"\".<\/b><\/span><\/span><\/span><\/span><\/p>\n

Ent\u00e3o, no <\/b><\/span><\/span>caso do exerc\u00edcio <\/b><\/span><\/span>em que o <\/b><\/span><\/span>movimento horizontal <\/b><\/span><\/span>tem <\/b><\/span><\/span>velocidade constante de V = 4,0 m\/s, <\/b><\/span><\/span>no <\/b><\/span><\/span>ponto mais alto da trajet\u00f3ria, <\/b><\/span><\/span>j\u00e1 que <\/b><\/span><\/span>V<\/b><\/span><\/span>y<\/b><\/span><\/sub><\/span> = 0 <\/b><\/span><\/span>a <\/b><\/span><\/span>velocidade ser\u00e1 V<\/b><\/span><\/span>x<\/b><\/span><\/sub><\/span> = V = <\/b><\/span><\/span>4,0 m\/s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

pessoa arremessa uma pequena esfera verticalmente para cima, <\/b><\/span><\/span><\/span>com <\/b><\/span><\/span><\/span>velocidade 3,0 m\/s em rela\u00e7\u00e3o ao vag\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

Como a pessoa arremessa uma pequena esfera verticalmente para cima, com velocidade 3,0 m\/s em rela\u00e7\u00e3o ao vag\u00e3o, e n\u00e3o existe atrito, ela retornar\u00e1 \u00e0 sua m\u00e3o com velocidade vertical, de V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span> – 3,0 m\/s e, em m\u00f3dulo V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span> = 3 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

Nesse ponto, como em qualquer outro ponto horizontal sua velocidade \u00e9 V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span> = 4,0 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

02- <\/b><\/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

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

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

\"\"<\/p>\n

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

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

\"\"<\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

06- <\/b><\/span><\/span><\/span>Em <\/b><\/span><\/span><\/span>todo gr\u00e1fico press\u00e3oxvolume <\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>trabalho realizado <\/b><\/span><\/span><\/span>pela for\u00e7a exercida pelo g\u00e1s <\/b><\/span><\/span><\/span>\u00e9 numericamente igual \u00e0 \u00e1rea compreendida entre a reta representativa <\/b><\/span><\/span><\/span>da varia\u00e7\u00e3o sofrida pelo g\u00e1s nesse <\/b><\/span><\/span><\/span>trecho e o eixo do volume.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Assim, o <\/b><\/span><\/span><\/span>maior trabalho pedido <\/b><\/span><\/span><\/span>corresponde <\/b><\/span><\/span><\/span>\u00e0 maior \u00e1rea <\/b><\/span><\/span><\/span>que est\u00e1 <\/b><\/span><\/span><\/span>representada no trecho 1-2.<\/b><\/span><\/span><\/span><\/p>\n

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

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

Equa\u00e7\u00e3o dos pontos conjugados ou equa\u00e7\u00e3o de Gauss para espelhos esf\u00e9ricos<\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Veja na figura abaixo os valores de P e de P fornecidos:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

S\u00e3o dados,<\/b><\/span><\/span><\/span> <\/b>V = 3,0.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> m\/s,<\/b><\/span><\/span><\/span> <\/b>\u0001<\/b><\/span><\/span><\/span> =<\/b><\/span><\/span><\/span> 2,0 cm = 0,02 m <\/b><\/span><\/span><\/span>= 2,0.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> m e pedido f = ?.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

Campo El\u00e9trico Uniforme<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

10- <\/b><\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>sequ\u00eancia abaixo<\/b><\/span><\/span><\/span> mostra a <\/b><\/span><\/span><\/span>representa\u00e7\u00e3o simb\u00f3lica <\/b><\/span><\/span><\/span>do circuito e o <\/b><\/span><\/span><\/span>c\u00e1lculo da resist\u00eancia equivalente R<\/b><\/span><\/span><\/span>eq <\/b><\/span><\/span><\/sub><\/span>= 5,8 + r <\/b><\/span><\/span><\/span>(soma porque (r e 5,8 \u2126) est\u00e3o associados em s\u00e9rie).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/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 comentadada das quest\u00f5es de F\u00cdSICA da FAMERP- 2018 01- Em rela\u00e7\u00e3o \u00e0 pessoa, como o vag\u00e3o est\u00e1 em trajet\u00f3ria retil\u00ednea e em MRU com velocidade constante, a trajet\u00f3ria da pequena esfera ser\u00e1 vertical na subida e na descida, retornando \u00e0 sua m\u00e3o (lan\u00e7amento vertical para cima). Mas, em rela\u00e7\u00e3o ao solo voc\u00ea observar\u00e1 dois movimentos, um na horizontal (devido<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3494,"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-3497","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3497","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=3497"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3497\/revisions"}],"predecessor-version":[{"id":3499,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3497\/revisions\/3499"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3494"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3497"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}