{"id":769,"date":"2015-05-13T15:17:10","date_gmt":"2015-05-13T15:17:10","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=769"},"modified":"2024-09-02T17:31:00","modified_gmt":"2024-09-02T17:31:00","slug":"resolucao-comentada-da-puc-pr-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-puc-pr-2015\/resolucao-comentada-da-puc-pr-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada da PUC PR 2015"},"content":{"rendered":"

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

01- <\/b><\/span><\/span><\/span>P = U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span> = 120<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/14,4 = 14400\/14,4 \"\" <\/b><\/span><\/span>P = 1000W = 1 kW.<\/b><\/span><\/span><\/span><\/p>\n

P = W\/\u2206t<\/b><\/span><\/span><\/span><\/span> \"\" 1 = W\/3×30 \"\" <\/b><\/span><\/span><\/span><\/span>W = 90 kWh.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Custo \"\" <\/b><\/span><\/span><\/span>C$<\/b><\/span><\/span><\/span><\/span> = 90×0,25 <\/b><\/span><\/span><\/span>= 22,50.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

02- <\/b><\/span><\/span><\/span>Pelo <\/b><\/span><\/span>gr\u00e1fico<\/b><\/span><\/span><\/span>, quando <\/b><\/span><\/span>m=0,4<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>P = 15 cm<\/b><\/span><\/span><\/span> \"\" m = – P\u2019\/P \"\" 0,4 = – P\u2019\/15 \"\" <\/b><\/span><\/span>P\u2019 = – 6 cm.<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da dist\u00e2ncia focal f dessa lente \"\" <\/b><\/span><\/span><\/span>1\/f = 1\/P + 1\/P\u2019<\/b><\/span><\/span><\/span> \"\" 1\/f = 1\/15 + 1\/(-6) \"\" 1\/f = (2 \u2013 5)\/30 \"\" 1\/f = – 3\/30 \"\" <\/b><\/span><\/span><\/span>f = – 10 cm (f < 0, lente divergente).<\/b><\/span><\/span><\/span><\/p>\n

Se o objeto estiver a P = 6cm da lente de f= – 10 cm, a dist\u00e2ncia P\u2019 da imagem \u00e0 lente ser\u00e1 \"\" <\/b><\/span><\/span><\/span>1\/f = 1\/P + 1\/P\u2019<\/b><\/span><\/span><\/span> \"\" 1\/(-10) = 1\/6 + 1\/P\u2019 \"\" – 1\/10 \u2013 1\/6 = 1\/P\u2019 \"\" (- 3 \u2013 5)\/30 = 1\/P\u2019 \"\" <\/b><\/span><\/span><\/span>P\u2019 <\/b><\/span><\/span><\/span>= – 30\/8 <\/b><\/span><\/span><\/span>= – 3,75 cm<\/b><\/span><\/span><\/span><\/p>\n

(<\/b><\/span><\/span><\/span>P\u2019 < 0<\/b><\/span><\/span><\/span>, imagem <\/b><\/span><\/span><\/span>virtual<\/b><\/span><\/span><\/span> e toda imagem <\/b><\/span><\/span><\/span>virtual \u00e9 direita<\/b><\/span><\/span><\/span>).<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>imagem<\/b><\/span><\/span><\/span> fornecida por lente <\/b><\/span><\/span><\/span>divergente<\/b><\/span><\/span><\/span> \u00e9 sempre <\/b><\/span><\/span><\/span>menor <\/b><\/span><\/span><\/span>que o objeto.<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/b><\/span><\/span><\/span>Varia\u00e7\u00e3o de velocidade<\/b><\/span><\/span><\/span> da bola:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

sen60<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup><\/span> = 9\/\u2206V<\/b><\/span><\/span><\/span><\/span> \"\" \u221a3\/2 = 9\/\u2206V \"\" <\/b><\/span><\/span><\/span>\u2206V<\/b><\/span><\/span><\/span><\/span> = 18\u221a3\/3 = 18×1,7\/3 <\/b><\/span><\/span><\/span>= 10,2 m\/s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Varia\u00e7\u00e3o da quantidade de movimento \"\" <\/b><\/span><\/span>\u2206Q = m.\u2206V <\/b><\/span><\/span><\/span>= 0,4.10,2 <\/b><\/span><\/span><\/span>= 4,08 kgm\/s.<\/b><\/span><\/span><\/span><\/p>\n

Impulso \"\" <\/b><\/span><\/span>I = F. \u2206t =<\/b><\/span><\/span><\/span> <\/b>\u2206Q \"\" <\/b><\/span><\/span><\/span>F.10.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> = 4,08 \"\" F = 4,08\/10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span> \"\" <\/b><\/span><\/span><\/span>F = 408 N.<\/b><\/span><\/span><\/span><\/p>\n

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

04- <\/b><\/span><\/span><\/span>Dist\u00e2ncia <\/b><\/span><\/span><\/span>percorrida pelo som at\u00e9 chegar ao refletor \"\" V = d\/t \"\" 340 = d\/30 \"\" <\/b><\/span><\/span>d = 10200m = 10,2 km.<\/b><\/span><\/span><\/span><\/p>\n

Dist\u00e2ncia <\/b><\/span><\/span><\/span>percorrida pelo avi\u00e3o nesses 30s \"\" V = d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/t \"\" 150 = d<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/30 \"\" <\/b><\/span><\/span>d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span> 4500 m <\/b><\/span><\/span>= 4,5 km.<\/b><\/span><\/span><\/span><\/p>\n

Dist\u00e2ncia do refletor ao avi\u00e3o<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span>D<\/b><\/span><\/span><\/span> = 10,2 \u2013 4,5 <\/b><\/span><\/span>= 5,7 km.<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/b><\/span><\/span><\/span>Em toda transforma\u00e7\u00e3o <\/b><\/span><\/span>c\u00edclica<\/b><\/span><\/span><\/span> as temperatura final e inicial s\u00e3o iguais e, consequentemente a <\/b><\/span><\/span>varia\u00e7\u00e3o de energia interna \u00e9 nula \u2206U = 0.<\/b><\/span><\/span><\/span><\/p>\n

Ciclo no sentido anti-hor\u00e1rio \"\" W = – 600 J.<\/b><\/span><\/span><\/span><\/p>\n

\u2206U = Q \u2013 W \"\" <\/b><\/span><\/span><\/span>0 = Q \u2013 (- 600) \"\" <\/b><\/span><\/span><\/span>Q = – 600 <\/b><\/span><\/span><\/span>J (calor cedido pelo g\u00e1s que \u00e9 totalmente transferido \u00e0 mistura \u00e1gua gelo).<\/b><\/span><\/span><\/span><\/p>\n

Esse 600 J de calor \u00e9 utilizado para derreter <\/b><\/span><\/span><\/span>uma massa m<\/b><\/span><\/span><\/span> de gelo tal que <\/b><\/span><\/span><\/span>Q = m.L<\/b><\/span><\/span><\/span> \"\" 600 = m.3.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span><\/p>\n

m <\/b><\/span><\/span><\/span>= 6.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/3.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> = 2.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> kg <\/b><\/span><\/span><\/span>= 2 g.<\/b><\/span><\/span><\/span><\/p>\n

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

<\/h2>\n

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

RESOLU\u00c7\u00d5ES 01- P = U2\/R = 1202\/14,4 = 14400\/14,4 P = 1000W = 1 kW. P = W\/\u2206t 1 = W\/3×30 W = 90 kWh. Custo C$ = 90×0,25 = 22,50. R- A 02- Pelo gr\u00e1fico, quando m=0,4 P = 15 cm m = – P\u2019\/P 0,4 = – P\u2019\/15 P\u2019 = – 6 cm. C\u00e1lculo da dist\u00e2ncia focal f<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":767,"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-769","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/769","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=769"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/769\/revisions"}],"predecessor-version":[{"id":770,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/769\/revisions\/770"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/767"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=769"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}