{"id":2855,"date":"2017-04-12T04:54:07","date_gmt":"2017-04-12T04:54:07","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2855"},"modified":"2024-09-03T17:02:27","modified_gmt":"2024-09-03T17:02:27","slug":"resolucao-comentada-das-questoes-de-fisica-da-ufjf-mg-2017","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2017\/ufjf-2017\/resolucao-comentada-das-questoes-de-fisica-da-ufjf-mg-2017\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da UFJF \u2013 MG \u2013 2017"},"content":{"rendered":"

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

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

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

\"\"<\/p>\n

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

02- <\/span><\/span><\/span>As <\/b><\/span><\/span><\/span>alternativas (A) e (E) est\u00e3o erradas,<\/b><\/span><\/span><\/span> pois a <\/b><\/span><\/span><\/span>temperatura de fus\u00e3o ou de vaporiza\u00e7\u00e3o n\u00e3o <\/b><\/span><\/span><\/span>depende da <\/b><\/span><\/span><\/span>massa do corpo, apenas do material que o constitui.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>(C) tamb\u00e9m est\u00e1 errada, <\/b><\/span><\/span><\/span>pois a <\/b><\/span><\/span><\/span>subst\u00e2ncia n\u00e3o \u00e9 a \u00e1gua que sofre fus\u00e3o a 0<\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/sup><\/span>C.<\/b><\/span><\/span><\/span><\/p>\n

Na <\/b><\/span><\/span><\/span>fus\u00e3o desse corpo de massa m = 20 g que ocorre a 50 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C <\/b><\/span><\/span><\/span>voc\u00ea deve usar a express\u00e3o o <\/b><\/span><\/span><\/span>calor latente fornecida por Q = m.L<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> (200 \u2013 100) = 20L<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span> \"\" L<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span> = 5 cal\/g.<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/span><\/span><\/span>L\u00edquido A \"\" valores obtidos pelo gr\u00e1fico \"\"<\/b><\/span><\/span><\/span> <\/b>\u2206P<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.g.h<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> (2 \u2013 1).10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> = \"\"<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.10.4 \"\" \"\"<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> = 0,025.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> \"\" \"\"<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = 2,5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span><\/p>\n

L\u00edquido B \"\" valores obtidos pelo gr\u00e1fico \"\"<\/b><\/span><\/span><\/span> <\/b>\u2206P<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.g.h<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> (3 \u2013 1).10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> = \"\"<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.10.4 \"\" \"\"<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> = 0,05.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> \"\" \"\"<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = 5,0.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span><\/p>\n

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

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

\"\"<\/p>\n

05- <\/span><\/span><\/span>Como o <\/b><\/span><\/span><\/span>n\u00edvel da \u00e1gua no reservat\u00f3rio diminui \u00e0 raz\u00e3o de 4 mm por minuto <\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>volume<\/b><\/span><\/span><\/span> que desce para a mangueira em <\/b><\/span><\/span><\/span>cada minuto ser\u00e1 \u2206V = 1mx1mx4mm <\/b><\/span><\/span><\/span>= 10dmx10dmx4.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>dm <\/b><\/span><\/span><\/span>= 4 dm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>\u2206V = 4 L.<\/b><\/span><\/span><\/span><\/p>\n

A vaz\u00e3o Q ser\u00e1 \"\"<\/b><\/span><\/span><\/span> <\/b>Q = \"\"<\/b><\/span><\/span><\/span> = \"\" \"\" Q = \"\" \"\" <\/b><\/span><\/span><\/span>Q = ( \"\" L\/s.<\/b><\/span><\/span><\/span><\/p>\n

Essa <\/b><\/span><\/span><\/span>vaz\u00e3o<\/b><\/span><\/span><\/span> <\/b>Q =<\/b><\/span><\/span><\/span> <\/b>(\"\"10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\/s <\/b><\/span><\/span><\/span>que sai <\/b><\/span><\/span><\/span>pela extremidade da mangueira de \u00e1rea S = 1 cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 1.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span> m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \u00e9 fornecida pela express\u00e3o Q = S.V (demonstra\u00e7\u00e3o em fisicaevestibular.com.br, mec\u00e2nica, hidrodin\u00e2mica, equa\u00e7\u00e3o da continuidade), onde V \u00e9 a velocidade de sa\u00edda da \u00e1gua (\"\").10<\/b><\/span><\/span><\/span>-3\u00a0<\/b><\/span><\/span><\/sup><\/span> = 10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>.V \"\" <\/b><\/span><\/span><\/span>V = \"\"<\/b><\/span><\/span><\/span> <\/b>= \"\" <\/b><\/span><\/span><\/span>V = \"\" m\/s<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/span><\/span><\/span>a) Em (A), <\/b><\/span><\/span><\/span>para que o <\/b><\/span><\/span><\/span>halter suba com acelera\u00e7\u00e3o de 1 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>, <\/b><\/span><\/span><\/span>pela <\/b><\/span><\/span><\/span>segunda lei de Newton <\/b><\/span><\/span><\/span>deve surgir uma <\/b><\/span><\/span><\/span>for\u00e7a resultante vertical e para cima de modo que F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = m.a <\/b><\/span><\/span><\/span>e, para que isso ocorra a <\/b><\/span><\/span><\/span>intensidade da for\u00e7a F deve ser maior que a intensidade da for\u00e7a peso. Isso ocorre na op\u00e7\u00e3o 1 (A).<\/b><\/span><\/span><\/span><\/p>\n

Em (B), <\/b><\/span><\/span><\/span>com o <\/b><\/span><\/span><\/span>halter em repouso <\/b><\/span><\/span><\/span>(equil\u00edbrio est\u00e1tico) a <\/b><\/span><\/span><\/span>for\u00e7a resultante deve ser nula <\/b><\/span><\/span><\/span>e, para que <\/b><\/span><\/span><\/span>isso ocorra, as for\u00e7as P e F devem ter a mesma intensidade. Isso ocorre na op\u00e7\u00e3o 1 (B).<\/b><\/span><\/span><\/span><\/p>\n

Observa\u00e7\u00e3o: <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>op\u00e7\u00e3o 3 est\u00e1 errada <\/b><\/span><\/span><\/span>porque <\/b><\/span><\/span><\/span>tanto em A quanto em B o peso deve ter a mesma intensidade (dimens\u00e3o, tamanho). <\/b><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span>Na situa\u00e7\u00e3o (A)<\/b><\/span><\/span><\/span> <\/b>onde o <\/b><\/span><\/span><\/span>halter est\u00e1 subindo com acelera\u00e7\u00e3o de a = 1m\/s<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sup><\/span>a <\/b><\/span><\/span><\/span>for\u00e7a resultante F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> <\/b>vale <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = ma = 216.1 \"\" F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = 216 N <\/b><\/span><\/span><\/span>e o <\/b><\/span><\/span><\/span>peso P tem intensidade P = m.g<\/b><\/span><\/span><\/span> = 216.10 \"\" <\/b><\/span><\/span><\/span>P = 2160 N. <\/b><\/span><\/span><\/span><\/p>\n

Como ele <\/b><\/span><\/span><\/span>sobe F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = F \u2013 P \"\"<\/b><\/span><\/span><\/span> 216 = F \u2013 2160 \"\" <\/b><\/span><\/span><\/span>F = 2376 N.<\/b><\/span><\/span><\/span><\/p>\n

Na situa\u00e7\u00e3o B,<\/b><\/span><\/span><\/span> <\/b>como o <\/b><\/span><\/span><\/span>halter est\u00e1 em repouso \"\" F = P = 2160 N.<\/b><\/span><\/span><\/span><\/p>\n

07- <\/span><\/span><\/span>a) <\/b><\/span><\/span><\/span>A energia cin\u00e9tica inicial da bola branca de massa m = 300 g = 0,3 kg \u00e9 nula E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span> = 0 (estava inicialmente parada, V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = 0). <\/b><\/span><\/span><\/span><\/p>\n

Depois sua velocidade passou a ser V = 50 cm\/s = 0,5 m\/s e ent\u00e3o sua energia cin\u00e9tica ser\u00e1 <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span> = m.\"\"<\/b><\/span><\/span><\/span> \"\" E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span> = 0,3.\"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span> 0,15×0,25 <\/b><\/span><\/span><\/span>= 0,0375 J.<\/b><\/span><\/span><\/span><\/p>\n

\u2206E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>cf<\/b><\/span><\/span><\/sub><\/span> \u2013 E<\/b><\/span><\/span><\/span>ci<\/b><\/span><\/span><\/sub><\/span> = <\/b><\/span><\/span><\/span>0,0375 \u2013 0 \"\" <\/b><\/span><\/span><\/span>\u2206E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = <\/b><\/span><\/span><\/span>0,0375 J.<\/b><\/span><\/span><\/span><\/p>\n

b) <\/b><\/span><\/span><\/span>C\u00e1lculo da <\/b><\/span><\/span><\/span>quantidade de movimento do sistema antes da colis\u00e3o <\/b><\/span><\/span><\/span>entre as duas bolas \"\" <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>b.<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>b <\/b><\/span><\/span><\/sub><\/span>+ m<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sub><\/span> <\/b>= 0,3.0,5 + 0.2.0 \"\" <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = 0,15 kgm\/s.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

C\u00e1lculo da <\/b><\/span><\/span><\/span>quantidade de movimento do sistema depois da colis\u00e3o <\/b><\/span><\/span><\/span>entre as duas bolas com <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span> = 0,1 m\/s e V<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span> <\/b>pedido \"\"<\/b><\/span><\/span><\/span> <\/b>Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>b.<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>b <\/b><\/span><\/span><\/sub><\/span>+ m<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sub><\/span> <\/b>= 0,3.0,1 + 0,2V<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u2019 \"\" <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span> = 0,03 + 0,2V<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u2019.<\/b><\/span><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> 0,15 = 0,03 + 0,2V<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u2019 \"\" V<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span> <\/b>= \"\" \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span> <\/b>= 0,6 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

c) A energia mec\u00e2nica inicial <\/b><\/span><\/span><\/span>foi acha<\/b><\/span><\/span><\/span>da em a) e vale E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span> = 0,0375 J.<\/b><\/span><\/span><\/span><\/p>\n

Energia mec\u00e2nica final \"\"<\/b><\/span><\/span><\/span> E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span> = 0,3.\"\" + 0,2.\"\" = 0,15×0,01 + 0,2×0,18 \"\" E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span> = 0,0015 + 0,036 \"\"<\/b><\/span><\/span><\/span> E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span> = 0,0375 J.<\/b><\/span><\/span><\/span><\/p>\n

Como E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span> = 0,0375 J, a <\/b><\/span><\/span><\/span>energia mec\u00e2nica se conserva e o choque \u00e9 perfeitamente el\u00e1stico.<\/b><\/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 UFJF \u2013 MG \u2013 2017 01- R- D 02- As alternativas (A) e (E) est\u00e3o erradas, pois a temperatura de fus\u00e3o ou de vaporiza\u00e7\u00e3o n\u00e3o depende da massa do corpo, apenas do material que o constitui. A (C) tamb\u00e9m est\u00e1 errada, pois a subst\u00e2ncia n\u00e3o \u00e9 a \u00e1gua que sofre fus\u00e3o a 0o<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2853,"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-2855","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2855","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=2855"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2855\/revisions"}],"predecessor-version":[{"id":2858,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2855\/revisions\/2858"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2853"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2855"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}