{"id":892,"date":"2015-06-28T23:51:59","date_gmt":"2015-06-28T23:51:59","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=892"},"modified":"2024-09-03T12:53:33","modified_gmt":"2024-09-03T12:53:33","slug":"resolucao-comentada-da-uncisal-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-uncisal-2015\/resolucao-comentada-da-uncisal-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada da UNCISAL 2015"},"content":{"rendered":"

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

01- <\/span><\/span><\/span>Q <\/b><\/span><\/span><\/span>= 7000 mAh = 7.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup> <\/b>= 7 Ah<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>i = 2 A<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>i = Q\/t<\/b><\/span><\/span><\/span> \"\" 2 = 7\/t \"\" t = 7\/2 \"\" <\/b><\/span><\/span>t = 3,5 h.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/span><\/span><\/span>Determinando a <\/b><\/span><\/span>energia (W) dissipada<\/b><\/span><\/span><\/span> quando a velocidade do carro <\/b><\/span><\/span>varia de V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span> = 30 m\/s at\u00e9 V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span><\/b><\/p>\n

\"\"<\/p>\n

= 0<\/b><\/span><\/span><\/span>,pelo teorema da <\/b><\/span><\/span>varia\u00e7\u00e3o de energia cin\u00e9tica<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>W = \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> = mV<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2 \u2013 mV<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2 = 800.0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 -800.30<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 = 0 – 36000 \"\" <\/b><\/span><\/span>W = – 360000 J<\/b><\/span><\/span><\/span> (o sinal negativo significa que perdeu energia) \"\" em m\u00f3dulo <\/b><\/span><\/span>W = 360000J.<\/b><\/span><\/span><\/span><\/p>\n

Dessa energia, <\/b><\/span><\/span>40%<\/b><\/span><\/span><\/span> \u00e9 armazenada sob forma de energia el\u00e9trica \"\" <\/b><\/span><\/span>W\u2019<\/b><\/span><\/span><\/span> = 0,4×360000 <\/b><\/span><\/span>= 144000 J.<\/b><\/span><\/span><\/span><\/p>\n

Essa energia <\/b><\/span><\/span>W\u2019 = 144000 J<\/b><\/span><\/span><\/span> \u00e9 utilizada para manter acesa uma l\u00e2mpada de <\/b><\/span><\/span>P = 20 W<\/b><\/span><\/span><\/span> durante um intervalo de tempo <\/b><\/span><\/span>\u2206t,<\/b><\/span><\/span><\/span> tal que \"\" <\/b><\/span><\/span>P = W\u2019\/\u2206<\/b><\/span><\/span><\/span>t \"\" 20 = 144000\/\u2206t \"\" <\/b><\/span><\/span>\u2206t<\/b><\/span><\/span><\/span> = 144000\/20 = 7200 s\/60 =<\/b><\/span><\/span><\/p>\n

120 min.<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/span><\/span><\/span>Observe na <\/b><\/span><\/span>figura abaixo<\/b><\/span><\/span><\/span> que, se voc\u00ea <\/b><\/span><\/span>projetar cada for\u00e7a de tra\u00e7\u00e3o na vertical, voc\u00ea obter\u00e1 T\u2019 =<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Tsen\u03b8<\/b><\/span><\/span><\/span> e a <\/b><\/span><\/span>soma dessas duas for\u00e7a deve equilibrar a for\u00e7a peso<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>2Tsen\u03b8 = P<\/b><\/span><\/span><\/span> \"\" 2Tsen\u03b8 = Mg \"\" <\/b><\/span><\/span>T = Mg\/2sen\u03b8.<\/b><\/span><\/span><\/span><\/p>\n

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

04- <\/span><\/span><\/span>Obs. Uma <\/b><\/span><\/span>imagem virtual pode ser projetada sim<\/b><\/span><\/span><\/span>, veja a ilustra\u00e7\u00e3o do enunciado.<\/b><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>I. Correta<\/b><\/span><\/span><\/span> \"\" se a <\/b><\/span><\/span>temperatura \u00e9 constante<\/b><\/span><\/span><\/span>, trata-se de uma <\/b><\/span><\/span>isot\u00e9rmica<\/b><\/span><\/span><\/span> de equa\u00e7\u00e3o <\/b><\/span><\/span>P.V = constante <\/b><\/span><\/span><\/span>e, nela, voc\u00ea observa que a <\/b><\/span><\/span>press\u00e3o P do g\u00e1s confinado no bal\u00e3o \u00e9 inversamente proporcional ao volume do mesmo.<\/b><\/span><\/span><\/span><\/p>\n

Ent\u00e3o, se quando o <\/b><\/span><\/span>bal\u00e3o sobe, sua temperatura diminui<\/b><\/span><\/span><\/span> o que faz que seu <\/b><\/span><\/span>volume aumente<\/b><\/span><\/span><\/span> tal que, dependente da altura o bal\u00e3o pode explodir.<\/b><\/span><\/span><\/p>\n

II. Correta<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>Baixando drasticamente a temperatura<\/b><\/span><\/span><\/span> no interior de um ve\u00edculo fechado, a <\/b><\/span><\/span>fun\u00e7\u00e3o dos<\/b><\/span><\/span><\/span> <\/b>\u201c<\/b><\/span><\/span><\/span>air bags<\/b><\/i><\/span><\/span><\/span>\u201d<\/b><\/span><\/span><\/span> pode ficar <\/b><\/span><\/span>comprometida<\/b><\/span><\/span><\/span>, pois o <\/b><\/span><\/span>volume de g\u00e1s em seu interior diminui, diminuindo suas dimens\u00f5es .<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/span><\/span><\/span>O <\/b><\/span><\/span>travamento das rodas<\/b><\/span><\/span><\/span> faz o <\/b><\/span><\/span>atrito est\u00e1tico respons\u00e1vel pela for\u00e7a centr\u00edpeta parar de atuar, <\/b><\/span><\/span><\/span>transformando-se em <\/b><\/span><\/span><\/span>atrito din\u00e2mico que \u00e9 menor que o est\u00e1tico,<\/b><\/span><\/span><\/span>facilitando a derrapagem e <\/b><\/span><\/span>impedindo a realiza\u00e7\u00e3o da curva.<\/b><\/span><\/span><\/span><\/p>\n

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

Voltar para os Exerc\u00edcios<\/a><\/span><\/h2>\n","protected":false},"excerpt":{"rendered":"

RESOLU\u00c7\u00d5ES 01- Q = 7000 mAh = 7.10310-3 = 7 Ah i = 2 A i = Q\/t 2 = 7\/t t = 7\/2 t = 3,5 h. R- E 02- Determinando a energia (W) dissipada quando a velocidade do carro varia de Vi = 30 m\/s at\u00e9 Vf = 0,pelo teorema da varia\u00e7\u00e3o de energia cin\u00e9tica W = \u2206Ec<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":890,"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-892","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/892","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=892"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/892\/revisions"}],"predecessor-version":[{"id":894,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/892\/revisions\/894"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/890"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=892"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}