{"id":637,"date":"2015-04-12T14:31:48","date_gmt":"2015-04-12T14:31:48","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=637"},"modified":"2024-09-02T16:51:23","modified_gmt":"2024-09-02T16:51:23","slug":"resolucao-comentada-da-puc-rj-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-da-puc-rj-2015\/resolucao-comentada-da-puc-rj-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada da PUC RJ 2015"},"content":{"rendered":"

RESOLU\u00c7\u00d5ES<\/b><\/span><\/span><\/span><\/p>\n

01- <\/b><\/span><\/span><\/span>Em toda <\/b><\/span><\/span><\/span>colis\u00e3o inel\u00e1stica<\/b><\/span><\/span><\/span> eles se movem <\/b><\/span><\/span><\/span>juntos ap\u00f3s o choque<\/b><\/span><\/span><\/span>, com a mesma velocidade <\/b><\/span><\/span><\/span>V\u2019:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Quantidade de movimento<\/b><\/span><\/span><\/span> do sistema <\/b><\/span><\/span>antes<\/b><\/span><\/span><\/span> da colis\u00e3o \"\" Q<\/b><\/span><\/span>sa<\/b><\/span><\/span><\/sub> =10.5 + 15.0 \"\" <\/b><\/span><\/span>Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = 50kgm\/s.<\/b><\/span><\/span><\/span><\/p>\n

Quantidade de movimento<\/b><\/span><\/span><\/span> do sistema <\/b><\/span><\/span>depois<\/b><\/span><\/span><\/span> da colis\u00e3o \"\" Q<\/b><\/span><\/span>sd<\/b><\/span><\/span><\/sub> =10.V\u2019 + 15.V\u2019 \"\" <\/b><\/span><\/span>Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = 25V\u2019.<\/b><\/span><\/span><\/span><\/p>\n

Princ\u00edpio da <\/b><\/span><\/span>conserva\u00e7\u00e3o da quantidade de movimento<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = Q<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span>d<\/b><\/span><\/span><\/sub> \"\" 50 = 25V\u2019 \"\" <\/b><\/span><\/span>V\u2019 = 2m\/s.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/b><\/span><\/span><\/span>R=10cm=0,1m \"\" m=20g=0,02kg \"\" V=2,0m\/s \"\" g=10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \"\" T=? \"\" <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span><\/b><\/p>\n

\"\"<\/p>\n

T \u2013 P = mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span> \"\" T \u2013 mg = mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R \"\" T \u2013 0,02.10 = 0,02.2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/0,1 \"\" T = 0,2 + 0,8 \"\" <\/b><\/span><\/span><\/span>T=1,0N<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3o gravitacional<\/b><\/span><\/span><\/span> nesse planeta, em <\/b><\/span><\/span><\/span>m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>,<\/b><\/span><\/span><\/span> \u00e9:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

h = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t + gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span> \"\" 12 = 0.2 + g.2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \"\" g = 12\/2 \"\" <\/b><\/span><\/span><\/span>g = 6m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

04- <\/span><\/span><\/span><\/span>(<\/b><\/span><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub><\/span> + Q<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub><\/span> = 0 \"\" m<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub><\/span>.c.(t \u2013 25) + m<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub><\/span>.c.(t \u2013 15) = 0<\/b><\/span><\/span><\/span><\/span> \"\" 0,1t \u2013 2,5 + 0,15t \u2013 2,25 = 0 \"\" t = 4,75\/0,25<\/b><\/span><\/span><\/span><\/span><\/p>\n

t = 19<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C.<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>P = U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span> \"\" 40 = 14400\/R \"\" <\/b><\/span><\/span><\/span>R = 360\u2126<\/b><\/span><\/span><\/span><\/b><\/p>\n

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

06- <\/span><\/span><\/span>n = c\/V<\/b><\/span><\/span><\/span> \"\" n = 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>\/1,2.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> \"\" <\/b><\/span><\/span><\/span>n = 2,5<\/b><\/span><\/span><\/span><\/b><\/p>\n

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

07- <\/span><\/span><\/span>a)<\/b><\/span><\/span><\/span> Se o processo ocorre \u00e0 <\/b><\/span><\/span><\/span>temperatura constante<\/b><\/span><\/span><\/span> trata-se de uma transforma\u00e7\u00e3o <\/b><\/span><\/span><\/span>isot\u00e9rmica <\/b><\/span><\/span><\/span>e, aplicando a equa\u00e7\u00e3o dos gases ideais \"\" <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/T<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = P.V\/T \"\"P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = P.V<\/b><\/span><\/span><\/span> \"\" volume inicial de ar no tubo cil\u00edndrico de \u00e1rea da base S \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \"\" volume final de ar no tubo cil\u00edndrico \"\" <\/b><\/span><\/span><\/span>V =(4\/5)V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \"\" P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 1 atm \"\" P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = P.V \"\" 1.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = P.(4\/5)V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \"\" P = 5\/4 \"\" <\/b><\/span><\/span><\/span>P = 1,25 atm.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Mas, uma das conseq\u00fc\u00eancias do princ\u00edpio de Stevin \u00e9 que <\/b><\/span><\/span>todos os pontos do mesmo n\u00edvel<\/b><\/span><\/span><\/span> <\/b>horizontal suportam a mesma press\u00e3o<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>P<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span><\/span> 1,35.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup> = 10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup> + 10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>H \"\" 1,35.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup> \u2013 1,0.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup> = 10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>H \"\" H = 0,35.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>\/10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup> \"\" H = 0,35.10 \"\" <\/b><\/span><\/span>H = 3,5m<\/b><\/span><\/span><\/span><\/p>\n

08- <\/span><\/span><\/span>a)<\/b><\/span><\/span><\/span> Para se obter uma corrente el\u00e9trica <\/b><\/span><\/span>de i=0,8 A sob tens\u00e3o de U = 12V<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span>resist\u00eancia equivalente<\/b><\/span><\/span><\/span> deve valer \"\" R<\/b><\/span><\/span>eq<\/b><\/span><\/span><\/sub> = U\/i = 12\/0,8 \"\" <\/b><\/span><\/span>R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = 15 \u2126<\/b><\/span><\/span><\/span><\/b><\/p>\n

Para obter <\/b><\/span><\/span><\/span>essa resist\u00eancia equivalente<\/b><\/span><\/span><\/span> voc\u00ea pode montar o circuito abaixo onde <\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 10\u2126 e R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 20\u2126<\/b><\/span><\/span><\/span> <\/b>s\u00e3o associados em s\u00e9rie, fornecendo R<\/b><\/span><\/span><\/span>eq1<\/b><\/span><\/span><\/sub><\/span> = 30\u2126 e esse conjunto associado em paralelo com R<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

30\u2126 fornecendo R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = 30.30\/(30 + 30)=900\/60=15\u2126 (figura acima).<\/b><\/span><\/span><\/span><\/p>\n

b) e c)<\/b><\/span><\/span><\/span> Veja na express\u00e3o <\/b><\/span><\/span><\/span>R = U\/i que i \u00e9 inversamente proporcional \u00e0 R<\/b><\/span><\/span><\/span> e, assim, a <\/b><\/span><\/span><\/span>m\u00e1xima corrente<\/b><\/span><\/span><\/span> obtida ocorre quando a <\/b><\/span><\/span><\/span>resist\u00eancia equivalente \u00e9 a menor poss\u00edvel<\/b><\/span><\/span><\/span>, o que s\u00f3 pode ocorrer quando os<\/b><\/span><\/span><\/span><\/p>\n

\"\"4 resistores estiverem associados em paralelo<\/b><\/span><\/span><\/span>, cujo <\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> <\/b>ser\u00e1 \"\" 1\/R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span>= 1\/10 + 1\/20 + 1\/30 + 1\/40 \"\" <\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = 4,8\u2126<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Nesse caso, a <\/b><\/span><\/span><\/span>corrente el\u00e9trica total ser\u00e1 m\u00ednima<\/b><\/span><\/span><\/span> e de valor R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> =U\/i \"\" 4,8 = 12\/i \"\" <\/b><\/span><\/span><\/span>i = 2,5 A.<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os Exerc\u00edcios<\/a>\u00a0<\/span><\/h3>\n","protected":false},"excerpt":{"rendered":"

RESOLU\u00c7\u00d5ES 01- Em toda colis\u00e3o inel\u00e1stica eles se movem juntos ap\u00f3s o choque, com a mesma velocidade V\u2019: Quantidade de movimento do sistema antes da colis\u00e3o Qsa =10.5 + 15.0 Qsa = 50kgm\/s. Quantidade de movimento do sistema depois da colis\u00e3o Qsd =10.V\u2019 + 15.V\u2019 Qsa = 25V\u2019. Princ\u00edpio da conserva\u00e7\u00e3o da quantidade de movimento Qsa = Qsd 50 =<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":635,"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-637","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/637","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=637"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/637\/revisions"}],"predecessor-version":[{"id":639,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/637\/revisions\/639"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/635"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=637"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}