{"id":564,"date":"2015-03-05T18:00:56","date_gmt":"2015-03-05T18:00:56","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=564"},"modified":"2024-09-02T17:26:34","modified_gmt":"2024-09-02T17:26:34","slug":"resolucao-comentada-da-ifmg-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-ifmg-2015\/resolucao-comentada-da-ifmg-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada da IFMG 2015"},"content":{"rendered":"

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

01- <\/span><\/span><\/span>C\u00e1lculo da <\/b><\/span><\/span><\/span>acelera\u00e7\u00e3<\/b><\/span><\/span><\/span>o de retardamento— <\/b><\/span><\/span><\/span>V = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> + at<\/b><\/span><\/span><\/span>— 0 = 1 + a.2 — <\/b><\/span><\/span><\/span>a = – 1\/2 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> — em m\u00f3dulo — <\/b><\/span><\/span><\/span>a = 0,5 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>for\u00e7a resultante <\/b><\/span><\/span><\/span>sobre ele \u00e9 a <\/b><\/span><\/span><\/span>pr\u00f3pria for\u00e7a de atrito F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>— <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>R <\/b><\/span><\/span><\/sub><\/span>= F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span>= ma<\/b><\/span><\/span><\/span>— \u00b5N = ma — — \u00b5P = ma — \u00b5mg = ma — — \u00b5g =a — \u00b5 =a\/g = 0,5\/10 <\/b><\/span><\/span><\/span>— \u00b5 = 0,05.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/span><\/span><\/span>\u2206L = L<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u03b1. \u2206t <\/b><\/span><\/span><\/span>= 15.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>.(65 \u2013 25) = 600.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span> — <\/b><\/span><\/span><\/span>\u2206L = 0,006 m <\/b><\/span><\/span><\/span>— L = 15,0 + 0,006 — L = 15,006 m.<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>(3)<\/b><\/span><\/span><\/sub><\/span> = P<\/b><\/span><\/span><\/span>atm<\/b><\/span><\/span><\/sub><\/span> + <\/b><\/span><\/span><\/span>\u03c1<\/b><\/span><\/span><\/span>gasolina<\/b><\/span><\/span><\/sub><\/span>.g.5 + \u03c1<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.g.6 <\/b><\/span><\/span><\/span>— <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>(3)<\/b><\/span><\/span><\/sub><\/span> = 10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> + <\/b><\/span><\/span><\/span>8.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.10.5 + 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10.6 — <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>(3)<\/b><\/span><\/span><\/sub><\/span> = 10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> + 0,4.10<\/b><\/span><\/span><\/span>5 <\/b><\/span><\/span><\/sup><\/span>+ 0,6.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> — <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>(3) <\/b><\/span><\/span><\/sub><\/span>= 2,0.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> Pa.<\/b><\/span><\/span><\/span><\/p>\n

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

04- <\/span><\/span><\/span>O <\/b><\/span><\/span><\/span>centro de gravidade<\/b><\/span><\/span><\/span> da garota variou de <\/b><\/span><\/span><\/span>0,8m para 4m.<\/b><\/span><\/span><\/span><\/p>\n

\u2206E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> =m.g.\u2206h<\/b><\/span><\/span><\/span> = 50.10.(4 \u2013 0,8) — <\/b><\/span><\/span><\/span>\u2206E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span> =500.3,2 = 1600J = <\/b><\/span><\/span><\/span>1,6 kJ.<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>A imagem, para <\/b><\/span><\/span><\/span>qualquer posi\u00e7\u00e3o do objeto<\/b><\/span><\/span><\/span>, fornecida por um <\/b><\/span><\/span><\/span>espelho esf\u00e9rico convexo<\/b><\/span><\/span><\/span>, \u00e9 sempre <\/b><\/span><\/span><\/span>virtual, direita e menor que o objeto<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/span><\/span><\/span>P = i.U<\/b><\/span><\/span><\/span> — 4500 = 220.i — <\/b><\/span><\/span><\/span>i <\/b><\/span><\/span><\/span>= 4500\/220 <\/b><\/span><\/span><\/span>= 20,45 A<\/b><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

Voltar para os Exerc\u00edcios<\/span><\/a><\/h3>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

RESOLU\u00c7\u00d5ES 01- C\u00e1lculo da acelera\u00e7\u00e3o de retardamento— V = Vo + at— 0 = 1 + a.2 — a = – 1\/2 m\/s2 — em m\u00f3dulo — a = 0,5 m\/s2. A for\u00e7a resultante sobre ele \u00e9 a pr\u00f3pria for\u00e7a de atrito Fat— FR = Fat= ma— \u00b5N = ma — — \u00b5P = ma — \u00b5mg = ma —<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":562,"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-564","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/564","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=564"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/564\/revisions"}],"predecessor-version":[{"id":565,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/564\/revisions\/565"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/562"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=564"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}