{"id":814,"date":"2015-05-14T15:07:27","date_gmt":"2015-05-14T15:07:27","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=814"},"modified":"2024-09-03T13:20:20","modified_gmt":"2024-09-03T13:20:20","slug":"resolucao-comentada-ifce-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-ifce-2015\/resolucao-comentada-ifce-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada IFCE 2015"},"content":{"rendered":"

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

01- <\/b><\/span><\/span><\/span>Quando <\/b><\/span><\/span>h = 30m e t= 2s<\/b><\/span><\/span><\/span> e ela encontra-se pela <\/b><\/span><\/span>primeira vez (na subida) a 30 m<\/b><\/span><\/span><\/span> do solo, lan\u00e7ada com <\/b><\/span><\/span>velocidade inicial V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>, <\/b><\/span><\/span><\/span>num local onde<\/b><\/span><\/span> g=10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \"\" <\/b><\/span><\/span>h = h<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span> \"\" 30 = 0 + V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.2<\/b><\/span><\/span> <\/b> – 10.2<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 \"\" 30 + 20 = 2V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub> \"\" <\/b><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 25 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

Velocidade <\/b><\/span><\/span><\/span>com que ela passa pela <\/b><\/span><\/span>altura de 30 m pela primeira, com V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 25 m\/s e t = 2 s<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

V = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \u2013 gt =<\/b><\/span><\/span><\/span> 25 \u2013 10.2 \"\" <\/b><\/span><\/span><\/span>V <\/b><\/span><\/span><\/span>= 25 \u2013 10.2 <\/b><\/span><\/span><\/span>= 5 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

Se ela passa pela altura de 30m na <\/b><\/span><\/span><\/span>subida com velocidade de V= + 5m\/s, na descida ela passar\u00e1<\/b><\/span><\/span><\/span> <\/b><\/p>\n

\"\"<\/p>\n

pela mesma altura<\/b><\/span><\/span><\/span> (30m) com <\/b><\/span><\/span><\/span>velocidade de V = – 5 m\/s<\/b><\/span><\/span><\/span>, no <\/b><\/span><\/span><\/span>instante t\u2019<\/b><\/span><\/span><\/span> dado por \"\" <\/b><\/span><\/span><\/span>V = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> \u2013 10t\u2019<\/b><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span><\/p>\n

-5 = 25 \u2013 10t\u2019 \"\" 10t\u2019<\/b><\/span><\/span><\/span> <\/b>= 30 \"\" <\/b><\/span><\/span><\/span>t\u2019 = 3 s.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/b><\/span><\/span><\/span>C\u00e1lculo da <\/b><\/span><\/span>densidade linear \u03bc<\/b><\/span><\/span><\/span> da corda sendo dados m=10 g = 10.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup> <\/b>= 10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> kg<\/b><\/span><\/span><\/span> e <\/b><\/span><\/span>L <\/b><\/span><\/span><\/span>= 26cm = <\/b><\/span><\/span><\/p>\n

26.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span> \"\" \u03bc = m\/L = 10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\/26.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup> \"\" <\/b><\/span><\/span>\u03bc = 1\/26 kg\/m.<\/b><\/span><\/span><\/span><\/p>\n

Velocidade <\/b><\/span><\/span><\/span>V <\/b><\/span><\/span><\/span>de propaga\u00e7\u00e3o da onda na corda sendo dados <\/b><\/span><\/span><\/span>f = 260 Hz<\/b><\/span><\/span><\/span> e <\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span> = 2 cm <\/b><\/span><\/span><\/span>=2.10<\/b><\/span><\/span><\/span>-2 <\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>V =<\/b><\/span><\/span><\/span> <\/b>\u03bb.f<\/b><\/span><\/span><\/span> = 2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.260 \"\" V = 520.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> \"\" <\/b><\/span><\/span><\/span>V = 5,2 m\/s.<\/b><\/span><\/span><\/span><\/p>\n

V = \u221a(T\/\u03bc)<\/b><\/span><\/span><\/span> \"\" 5,2 = \u221a(T\/(1\/26) = \u221a(26T) \"\" (5,2)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> = [\u221a(26T)]<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup> \"\" 27,04 = 26T \"\" <\/b><\/span><\/span>T <\/b><\/span><\/span><\/span>= 27,04\/26 = <\/b><\/span><\/span>1,04 N.<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

04- <\/b><\/span><\/span><\/span>Veja <\/b><\/span><\/span>tabela <\/b><\/span><\/span><\/span>abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Colis\u00e3o inel\u00e1stica<\/b><\/span><\/span><\/span><\/p>\n

Neste tipo de choque a <\/b><\/span><\/span><\/span>dissipa\u00e7\u00e3o<\/b><\/span><\/span><\/span> de energia \u00e9 <\/b><\/span><\/span><\/span>m\u00e1xima<\/b><\/span><\/span><\/span>, o coeficiente de restitui\u00e7\u00e3o \u00e9 nulo, e, <\/b><\/span><\/span><\/span>ap\u00f3s o choque<\/b><\/span><\/span><\/span>, os corpos <\/b><\/span><\/span><\/span>obrigatoriamente se juntam e se movem unidos com a mesma velocidade.<\/b><\/span><\/span><\/span> Lembre-se de que em <\/b><\/span><\/span><\/span>qualquer tipo de choque a quantidade de movimento sempre se conserva.<\/b><\/span><\/span><\/span><\/p>\n

No caso do <\/b><\/span><\/span>exerc\u00edcio<\/b><\/span><\/span><\/span>, a alternativa correta \u00e9 a <\/b><\/span><\/span>C.<\/b><\/span><\/span><\/span> Justificando:<\/b><\/span><\/span><\/p>\n

C)<\/b><\/span><\/span><\/span> perfeitamente inel\u00e1stica, podemos ter energia cin\u00e9tica final igual a zero, no caso de uma colis\u00e3o frontal entre dois corpos de mesma massa e velocidades iguais em m\u00f3dulo, por\u00e9m sentidos contr\u00e1rios. <\/b><\/span><\/span><\/p>\n

Veja <\/b><\/span><\/span>figura abaixo<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Antes \"\" <\/b><\/span><\/span>Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> =m(+V) + m(-V)<\/b><\/span><\/span><\/span> = + mV \u2013 mV \"\" <\/b><\/span><\/span>Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = 0<\/b><\/span><\/span><\/span><\/p>\n

Depois \"\" <\/b><\/span><\/span>Q<\/b><\/span><\/span><\/span>sa<\/b><\/span><\/span><\/sub><\/span> = Q<\/b><\/span><\/span><\/span>sd<\/b><\/span><\/span><\/sub><\/span> = 0<\/b><\/span><\/span><\/span> \"\" E<\/b><\/span><\/span>cd<\/b><\/span><\/span><\/sub> =2mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 = 2m.0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 \"\" <\/b><\/span><\/span>E<\/b><\/span><\/span><\/span>cd<\/b><\/span><\/span><\/sub><\/span> = 0.<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> e Q<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> cargas <\/b><\/span><\/span>antes <\/b><\/span><\/span><\/span>de colocar o fio e <\/b><\/span><\/span>Q\u2019<\/b><\/span><\/span><\/span>A <\/b><\/span><\/span><\/sub><\/span>e Q\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>, cargas <\/b><\/span><\/span>depois<\/b><\/span><\/span><\/span> de colocar o fio:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Q<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> + Q<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = Q\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> + Q\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> \"\" 20\u03bc + 40\u03bc = Q\u2019<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub> + Q\u2019<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub> \"\" <\/b><\/span><\/span>60\u03bc = Q\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> + Q\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> (I)<\/b><\/span><\/span><\/span><\/p>\n

Q\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> = Q\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> \"\" Q\u2019<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/1.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup> = Q\u2019<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/3.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup> \"\" <\/b><\/span><\/span>Q\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = 3Q\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> (II)<\/b><\/span><\/span><\/span><\/p>\n

(II) em (I)<\/b><\/span><\/span><\/span> \"\" Q\u2019<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub> + 3Q\u2019<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub> = 60 \u03bc \"\" <\/b><\/span><\/span>Q\u2019<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span> <\/b>= 60\/4<\/b><\/span><\/span>=15 \u03bcC<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>Q\u2019<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span> = 45 <\/b><\/span><\/span><\/span>\u03bcC<\/b><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

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

RESOLU\u00c7\u00d5ES 01- Quando h = 30m e t= 2s e ela encontra-se pela primeira vez (na subida) a 30 m do solo, lan\u00e7ada com velocidade inicial Vo, num local onde g=10m\/s2 h = ho + Vot \u2013 gt2\/2 30 = 0 + Vo.2 – 10.22\/2 30 + 20 = 2Vo Vo = 25 m\/s. Velocidade com que ela passa pela<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":812,"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-814","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/814","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=814"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/814\/revisions"}],"predecessor-version":[{"id":815,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/814\/revisions\/815"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/812"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=814"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}