{"id":2847,"date":"2017-04-12T04:47:36","date_gmt":"2017-04-12T04:47:36","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2847"},"modified":"2024-09-03T17:39:59","modified_gmt":"2024-09-03T17:39:59","slug":"resolucao-comentada-das-questoes-de-fisica-da-unimontes-mg-2017","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2017\/unimontes-2017\/resolucao-comentada-das-questoes-de-fisica-da-unimontes-mg-2017\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da UNIMONTES \u2013 MG \u2013 2017"},"content":{"rendered":"

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

UNIMONTES \u2013 MG \u2013 2017<\/span><\/span><\/span><\/p>\n

PAES \u2013 1<\/span><\/span><\/span>a<\/span><\/span><\/sup><\/span> etapa<\/span><\/span><\/span><\/p>\n

01- <\/span><\/span><\/span>Quando o <\/b><\/span><\/span><\/span>sinal fechado 1 abre os carros aceleram<\/b><\/span><\/span><\/span> com <\/b><\/span><\/span><\/span>a = 5,5 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>at\u00e9 atingirem a <\/b><\/span><\/span><\/span>velocidade de V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = 11 m\/s cujo tempo decorrido para que isso ocorra deve ser <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + a.t \"\"<\/b><\/span><\/span><\/span> 11 = 0 + 5,5.t \"\" t = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>t = 2 s.<\/b><\/span><\/span><\/span><\/p>\n

D<\/b><\/span><\/span><\/span>12<\/b><\/span><\/span><\/sub><\/span> \"\" dist\u00e2ncia \u2206S <\/b><\/span><\/span><\/span>que eles <\/b><\/span><\/span><\/span>percorreram nesse tempo \"\"<\/b><\/span><\/span><\/span> <\/b>\u2206S = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t + \"\"<\/b><\/span><\/span><\/span> <\/b>= 0.t + \"\" \"\" <\/b><\/span><\/span><\/span>\u2206S = 11 m.<\/b><\/span><\/span><\/span><\/p>\n

Como o <\/b><\/span><\/span><\/span>sinal 2 deve abrir quando eles estiverem a 25 m d<\/b><\/span><\/span><\/span>ele a <\/b><\/span><\/span><\/span>dist\u00e2ncia <\/b><\/span><\/span><\/span>a ser percorrida com <\/b><\/span><\/span><\/span>velocidade constante de V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = 11 m\/s <\/b><\/span><\/span><\/span>ser\u00e1 \"\" <\/b><\/span><\/span><\/span>\u2206S<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> <\/b>= (300 \u2013 (11 + 25)) <\/b><\/span><\/span><\/span>= 264 m \"\"<\/b><\/span><\/span><\/span> V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" 11 = \"\" \"\" <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 24 s.<\/b><\/span><\/span><\/span><\/p>\n

Ent\u00e3o, o <\/b><\/span><\/span><\/span>tempo decorrido desde a abertura do sinal 1 foi de t\u2019 <\/b><\/span><\/span><\/span>= 24 + 2 <\/b><\/span><\/span><\/span>= 26 s.<\/b><\/span><\/span><\/span><\/p>\n

D<\/b><\/span><\/span><\/span>12<\/b><\/span><\/span><\/sub><\/span> \"\" dist\u00e2ncia \u2206S <\/b><\/span><\/span><\/span>que eles <\/b><\/span><\/span><\/span>percorreram nesse tempo \"\"<\/b><\/span><\/span><\/span> <\/b>\u2206S = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t + \"\"<\/b><\/span><\/span><\/span> <\/b>= 0.t + \"\" \"\" <\/b><\/span><\/span><\/span>\u2206S = 11 m.<\/b><\/span><\/span><\/span><\/p>\n

Como o <\/b><\/span><\/span><\/span>sinal deve abrir quando eles estiverem a 25 m d<\/b><\/span><\/span><\/span>ele a <\/b><\/span><\/span><\/span>dist\u00e2ncia <\/b><\/span><\/span><\/span>a ser percorrida com <\/b><\/span><\/span><\/span>velocidade constante de V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = 11 m\/s <\/b><\/span><\/span><\/span>ser\u00e1 \"\" <\/b><\/span><\/span><\/span>\u2206S<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> <\/b>= (300 \u2013 (11 + 25)) <\/b><\/span><\/span><\/span>= 264 m \"\"<\/b><\/span><\/span><\/span> V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" 11 = \"\" \"\" <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 24 s.<\/b><\/span><\/span><\/span><\/p>\n

Ent\u00e3o o <\/b><\/span><\/span><\/span>tempo decorrido desde a abertura do sinal 1 foi de t\u2019 <\/b><\/span><\/span><\/span>= 24 + 2 <\/b><\/span><\/span><\/span>= 26 s.<\/b><\/span><\/span><\/span><\/p>\n

Quando o <\/b><\/span><\/span><\/span>sinal fechado 2 abre os carros aceleram<\/b><\/span><\/span><\/span> com <\/b><\/span><\/span><\/span>a = 5,5 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>at\u00e9 atingirem a <\/b><\/span><\/span><\/span>velocidade de V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = 11 m\/s cujo tempo decorrido para que isso ocorra deve ser <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + a.t \"\"<\/b><\/span><\/span><\/span> 11 = 0 + 5,5.t \"\" t = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>t = 2 s.<\/b><\/span><\/span><\/span><\/p>\n

D<\/b><\/span><\/span><\/span>23<\/b><\/span><\/span><\/sub><\/span> \"\" dist\u00e2ncia \u2206S <\/b><\/span><\/span><\/span>que eles <\/b><\/span><\/span><\/span>percorreram nesse tempo \"\"<\/b><\/span><\/span><\/span> <\/b>\u2206S = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t + \"\"<\/b><\/span><\/span><\/span> <\/b>= 0.t + \"\" \"\" <\/b><\/span><\/span><\/span>\u2206S = 11 m.<\/b><\/span><\/span><\/span><\/p>\n

Como o <\/b><\/span><\/span><\/span>sinal 3 deve abrir quando eles estiverem a 25 m <\/b><\/span><\/span><\/span>dele a <\/b><\/span><\/span><\/span>dist\u00e2ncia <\/b><\/span><\/span><\/span>a ser percorrida com <\/b><\/span><\/span><\/span>velocidade constante de V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = 11 m\/s <\/b><\/span><\/span><\/span>ser\u00e1 \"\" <\/b><\/span><\/span><\/span>\u2206S<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> <\/b>= (110 \u2013 (11 + 25)) <\/b><\/span><\/span><\/span>= 74 m \"\"<\/b><\/span><\/span><\/span> V<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" 11 = \"\" \"\" <\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 6,7 s.<\/b><\/span><\/span><\/span><\/p>\n

Ent\u00e3o, o <\/b><\/span><\/span><\/span>tempo decorrido desde a abertura do sinal 1 foi de t\u2019\u2019 = (6,7 + 2) = 8,7 s.<\/b><\/span><\/span><\/span><\/p>\n

Portanto, o <\/b><\/span><\/span><\/span>tempo total decorrido desde a abertura do sinal 1 foi de t\u2019 <\/b><\/span><\/span><\/span>= 26 + 8,7 <\/b><\/span><\/span><\/span>= 34,7 s, supondo que o tempo de parada no farol 2 foi de 1,3 s.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/span><\/span><\/span>Na demonstra\u00e7\u00e3o abaixo, substituir o carro pela moto.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"<\/p>\n

PAES \u2013 2<\/span><\/span><\/span>a<\/span><\/span><\/sup><\/span> etapa<\/span><\/span><\/span><\/p>\n

01- <\/span><\/span><\/span>C\u00e1lculo da <\/b><\/span><\/span><\/span>velocidade V<\/b><\/span><\/span><\/span> com que a <\/b><\/span><\/span><\/span>esfera de massa m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 2 kg <\/b><\/span><\/span><\/span>chega <\/b><\/span><\/span><\/span>at\u00e9 a caixa <\/b><\/span><\/span><\/span>pelo <\/b><\/span><\/span><\/span>teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica:<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/span><\/span><\/span>Dados:<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span> <\/b>= 1 g\/cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> <\/b>= 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> \"\" V<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span> <\/b>= 3 L = <\/b><\/span><\/span><\/span>3.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> kg \"\" c<\/b><\/span><\/span><\/span>\u00e1gua <\/b><\/span><\/span><\/sub><\/span>=4200 J\/kgK \"\" \u2206\"\" <\/b><\/span><\/span><\/span>=10 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C = <\/b><\/span><\/span><\/span><\/p>\n

10 K.<\/b><\/span><\/span><\/span><\/p>\n

massa de \u00e1gua \"\" \"\"<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>\u00e1gua \"\"<\/b><\/span><\/span><\/sub><\/span> 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/sub><\/span>m<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span> =3 kg.<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da <\/b><\/span><\/span><\/span>quantidade de calor <\/b><\/span><\/span><\/span>(energia t\u00e9rmica) que <\/b><\/span><\/span><\/span>deve ser retirada <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span> =3 kg <\/b><\/span><\/span><\/span>para que a <\/b><\/span><\/span><\/span>temperatura diminua de 10 K \"\" W = Q = m.c. \u2206\"\" \"\" <\/b><\/span><\/span><\/span>W = 3x4200x10 \"\"<\/b><\/span><\/span><\/span> W = 126000 J.<\/b><\/span><\/span><\/span><\/p>\n

P = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>\u2206\"\" = \"\"<\/b><\/span><\/span><\/span> \"\" \u2206\"\" = \"\" \"\" <\/b><\/span><\/span><\/span>\u2206\"\" = 315 s.<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/span><\/span><\/span>Quando o corpo de <\/b><\/span><\/span><\/span>massa m = 2 Kg <\/b><\/span><\/span><\/span>e peso <\/b><\/span><\/span><\/span>P = m.g <\/b><\/span><\/span><\/span>= 2.10 <\/b><\/span><\/span><\/span>= 20 N <\/b><\/span><\/span><\/span>\u00e9 <\/b><\/span><\/span><\/span>suspenso pela mola <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>constante el\u00e1stica k<\/b><\/span><\/span><\/span> e esta sofre uma <\/b><\/span><\/span><\/span>deforma\u00e7\u00e3o de x <\/b><\/span><\/span><\/span>= 2,5 cm <\/b><\/span><\/span><\/span>= 2,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> m <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>for\u00e7a el\u00e1stica <\/b><\/span><\/span><\/span>que est\u00e1 <\/b><\/span><\/span><\/span>deformando a mola \u00e9 seu peso \"\" F<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span> = k.x \"\" P = k.x \"\" <\/b><\/span><\/span><\/span>20 = k.2,5.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> \"\" k = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span><\/p>\n

k = 800 N\/m.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>express\u00e3o matem\u00e1tica da frequ\u00eancia angular <\/b><\/span><\/span><\/span>(velocidade angular) <\/b><\/span><\/span><\/span>W \u00e9 fornecida <\/b><\/span><\/span><\/span>por <\/b><\/span><\/span><\/span>W = \"\"<\/b><\/span><\/span><\/span> (a demonstra\u00e7\u00e3o se encontra em fisicaevestibular.com.br, mec\u00e2nica, din\u00e2mica, MHS, sistema massa-mola).<\/b><\/span><\/span><\/span><\/p>\n

W = \"\"<\/b><\/span><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span><\/span>W = \"\"<\/b><\/span><\/span><\/span><\/span> \"\" W = \"\" \"\" <\/b><\/span><\/span><\/span><\/span>W = 20 rad\/s<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

04- <\/span><\/span><\/span>Sendo a <\/b><\/span><\/span><\/span>energia mec\u00e2nica constante <\/b><\/span><\/span><\/span>(considerando o sistema conservativo) ela \u00e9 a <\/b><\/span><\/span><\/span>soma das energias cin\u00e9tica e potencial gravitacional <\/b><\/span><\/span><\/span>em <\/b><\/span><\/span><\/span>cada ponto, <\/b><\/span><\/span><\/span>ou seja, <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> + E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> + mgh. <\/b><\/span><\/span><\/span><\/p>\n

Veja que nos <\/b><\/span><\/span><\/span>pontos A e C o m\u00f3dulo da velocidade \u00e9 nula <\/b><\/span><\/span><\/span>e consequentemente a <\/b><\/span><\/span><\/span>energia cin\u00e9tica tamb\u00e9m \u00e9 nula, sobrando nesses pontos apenas a energia potencial gravitacional <\/b><\/span><\/span><\/span>que \u00e9 <\/b><\/span><\/span><\/span>igual \u00e0 energia mec\u00e2nica que<\/b><\/span><\/span><\/span>, por sua vez, \u00e9 <\/b><\/span><\/span><\/span>constante.<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span><\/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 UNIMONTES \u2013 MG \u2013 2017 PAES \u2013 1a etapa 01- Quando o sinal fechado 1 abre os carros aceleram com a = 5,5 m\/s2 at\u00e9 atingirem a velocidade de Vp = 11 m\/s cujo tempo decorrido para que isso ocorra deve ser de Vp = Vo + a.t 11 = 0 + 5,5.t<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2844,"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-2847","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2847","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=2847"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2847\/revisions"}],"predecessor-version":[{"id":2849,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2847\/revisions\/2849"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2844"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2847"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}