{"id":588,"date":"2015-03-18T18:03:29","date_gmt":"2015-03-18T18:03:29","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=588"},"modified":"2024-09-02T17:33:35","modified_gmt":"2024-09-02T17:33:35","slug":"resolucao-comentada-da-puc-mg-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-puc-mg-2015\/resolucao-comentada-da-puc-mg-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada da PUC-MG 2015"},"content":{"rendered":"

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

01- <\/b><\/span><\/span><\/span>Ano-luz<\/b><\/span><\/span><\/span> \u00e9 uma unidade de medida utilizada em astronomia e corresponde \u00e0 <\/b><\/span><\/span><\/span>dist\u00e2ncia percorrida pela<\/b><\/span><\/span><\/span> <\/b>luz em um ano, no v\u00e1cuo<\/b><\/span><\/span><\/span>, ou seja, se a luz <\/b><\/span><\/span><\/span>dista 1 bilh\u00e3o de anos-luz da Terra, ela demora 1 bilh\u00e3o de anos at\u00e9 chegar \u00e0 Terra.<\/b><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span>
\n<\/b><\/span><\/span><\/span>02- <\/span><\/span><\/span>O <\/b><\/span><\/span><\/span>calor recebido<\/b><\/span><\/span><\/span> por um corpo pode fazer com que sua temperatura aumente ou <\/b><\/span><\/span><\/span>que ele sofra uma<\/b><\/span><\/span><\/span> <\/b>mudan\u00e7a de estado<\/b><\/span><\/span><\/span>. No segundo caso voc\u00ea tem uma<\/b><\/span><\/span><\/span>quantidade de calor latente<\/b><\/span><\/span><\/span>,\u00a0ou seja, calor latente se refere \u00e0 <\/b><\/span><\/span><\/span>quantidade de calor que um corpo recebe apenas para mudar de estado, o que ocorre \u00e0 temperatura constante, no caso, transformando-se em vapor \u00e0 temperatura constante de 100<\/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

03- <\/span><\/span><\/span>Observe na express\u00e3o <\/b><\/span><\/span><\/span>P = U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span> que, se voc\u00ea <\/b><\/span><\/span><\/span>n\u00e3o alterar a resist\u00eancia R do chuveiro<\/b><\/span><\/span><\/span>, que \u00e9 o caso do exerc\u00edcio, a <\/b><\/span><\/span><\/span>pot\u00eancia P \u00e9 diretamente proporcional ao quadrado da voltagem <\/b><\/span><\/span><\/span>U, o que significa que, se voc\u00ea <\/b><\/span><\/span><\/span>reduzir U \u00e0 metade, a pot\u00eancia do chuveiro ficar\u00e1 2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 4 vezes menor.<\/b><\/span><\/span><\/span><\/p>\n

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

04- <\/span><\/span><\/span>Se a <\/b><\/span><\/span><\/span>velocidade de queda \u00e9 constante<\/b><\/span><\/span><\/span>, ele dever\u00e1 <\/b><\/span><\/span><\/span>percorrer dist\u00e2ncias iguais em intervalos de tempo<\/b><\/span><\/span><\/span> <\/b>iguais.<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>Enunciado do princ\u00edpio de Arquimedes:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u201cTodo corpo total ou parcialmente<\/b><\/span><\/span><\/span><\/span> mergulhado <\/b><\/span><\/span><\/span><\/span>num l\u00edquido em equil\u00edbrio<\/b><\/span><\/span><\/span><\/span>, recebe <\/b><\/span><\/span><\/span><\/span>uma for\u00e7a de<\/b><\/span><\/span><\/span><\/span> <\/b>dire\u00e7\u00e3o vertical e sentido\u00a0 para cima<\/b><\/span><\/span><\/span><\/span> denominada <\/b><\/span><\/span><\/span><\/span>de Empuxo, cuja intensidade\u00a0 \u00e9 igual ao peso do volume de l\u00edquido deslocado\u201c<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

06- <\/span><\/span><\/span>Se a luz se propaga com <\/b><\/span><\/span><\/span>velocidade constante<\/b><\/span><\/span><\/span>, pela equa\u00e7\u00e3o fundamental da ondulat\u00f3ria, <\/b><\/span><\/span><\/span>V = \u03bbf, se V \u00e9 a mesma, o comprimento de onda \u03bb deve ser inversamente proporcional \u00e0 freq\u00fc\u00eancia f.<\/b><\/span><\/span><\/span><\/p>\n

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

07- <\/span><\/span><\/span>Trabalho da for\u00e7a resultante como varia\u00e7\u00e3o de energia cin\u00e9tica:<\/b><\/span><\/span><\/span><\/p>\n

W<\/b><\/span><\/span><\/span>F1<\/b><\/span><\/span><\/sub><\/span> = 70 \u2013 20 <\/b><\/span><\/span><\/span>= 50 J<\/b><\/span><\/span><\/span><\/p>\n

W<\/b><\/span><\/span><\/span>F2<\/b><\/span><\/span><\/sub><\/span> <\/b>= 160 \u2013100 <\/b><\/span><\/span><\/span>= 60 J<\/b><\/span><\/span><\/span><\/p>\n

W<\/b><\/span><\/span><\/span>F3<\/b><\/span><\/span><\/sub><\/span> <\/b>= 220 – 160 <\/b><\/span><\/span><\/span>= 60 J<\/b><\/span><\/span><\/span><\/p>\n

Mas, <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>FR<\/b><\/span><\/span><\/sub><\/span> = F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span> e sendo a <\/b><\/span><\/span><\/span>dist\u00e2ncia d a mesma a for\u00e7a resultante F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> \u00e9 diretamente proporcional ao W<\/b><\/span><\/span><\/span> e, assim, <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> < F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = F<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

08- <\/span><\/span><\/span>Se voc\u00ea <\/b><\/span><\/span><\/span>retirar o ar do interior da esfera<\/b><\/span><\/span><\/span> ela fica <\/b><\/span><\/span><\/span>menos densa, mais leve<\/b><\/span><\/span><\/span>, e subir\u00e1 <\/b><\/span><\/span><\/span>diminuindo seu volume imerso.<\/b><\/span><\/span><\/span><\/p>\n

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

09- <\/b><\/span><\/span><\/span>Press\u00e3o no fundo do dep\u00f3sito<\/b><\/span><\/span><\/span> que faz a <\/b><\/span><\/span><\/span>v\u00e1lvula abrir<\/b><\/span><\/span><\/span> quando a \u00e1gua est\u00e1 na <\/b><\/span><\/span><\/span>altura de 4m acima dela:<\/b><\/span><\/span><\/span><\/p>\n

Teorema de Stevin<\/b><\/span><\/span><\/span> — <\/b><\/span><\/span><\/span>P =<\/b><\/span><\/span><\/span> <\/b>\u03c1.g.h = 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10.4 <\/b><\/span><\/span><\/span>—<\/b><\/span><\/span><\/span> P = 4.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> (Pa).<\/b><\/span><\/span><\/span><\/p>\n

For\u00e7a necess\u00e1ria<\/b><\/span><\/span><\/span> para <\/b><\/span><\/span><\/span>abrir a v\u00e1lvula<\/b><\/span><\/span><\/span> de <\/b><\/span><\/span><\/span>\u00e1rea S <\/b><\/span><\/span><\/span>= 120cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>= 120.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span> m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> quando <\/b><\/span><\/span><\/span>sob press\u00e3o de P = 4.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

P = F\/S<\/b><\/span><\/span><\/span> — 4.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> = F\/120.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span> — F = 4.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.120.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span> — <\/b><\/span><\/span><\/span>F = 480 N<\/b><\/span><\/span><\/span><\/p>\n

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

10- <\/span><\/span><\/span>\u00c9 fornecido que a <\/b><\/span><\/span><\/span>pot\u00eancia solar m\u00e9dia<\/b><\/span><\/span><\/span> recebida nos Estados Unidos, considerando-se <\/b><\/span><\/span><\/span>dia e noite, ver\u00e3o e inverno<\/b><\/span><\/span><\/span>, \u00e9 <\/b><\/span><\/span><\/span>cerca de 15% da pot\u00eancia incidente nas partes altas da atmosfera que \u00e9 de 1400 W\/m<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sup><\/span>— <\/b><\/span><\/span><\/span>pot\u00eancia recebida nos Estados Unidos<\/b><\/span><\/span><\/span> — P\u00ad<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span> = 0,15×1400 = 210 W\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Regra de tr\u00eas — 210 W \u2013 1 m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> — 4200 W \u2013 S m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> — S = 4200\/210 — S = 20 m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

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

11- <\/span><\/span><\/span>W = P.\u2206t =<\/b><\/span><\/span><\/span> 1400×100 = 140000 J\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> — regra de tr\u00eas — <\/b><\/span><\/span><\/span>140000 j \u2013 1 m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> — W (J) \u2013 1km<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> (10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span><\/p>\n

W =<\/b><\/span><\/span><\/span> 14.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>x10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span> = 14.10<\/b><\/span><\/span><\/span>10<\/b><\/span><\/span><\/sup><\/span> J <\/b><\/span><\/span><\/span>= 1,4.10<\/b><\/span><\/span><\/span>11<\/b><\/span><\/span><\/sup><\/span> J.<\/b><\/span><\/span><\/span><\/p>\n

R \u2013 B<\/b><\/span><\/span><\/span><\/p>\n

12- <\/b><\/span><\/span><\/span>C\u00e1lculo do <\/b><\/span><\/span><\/span>per\u00edodo T<\/b><\/span><\/span><\/span>, tempo que <\/b><\/span><\/span><\/span>as rodas de raio R<\/b><\/span><\/span><\/span> = 0,5\/2 <\/b><\/span><\/span><\/span>= 0,25m<\/b><\/span><\/span><\/span> demoram ao <\/b><\/span><\/span><\/span>efetuar uma volta completa, com velocidade angular de W = 40 rad\/s — W = 2\u03c0\/T<\/b><\/span><\/span><\/span>— <\/b><\/span><\/span><\/span><\/b>40 = 2\u03c0\/T — <\/b><\/span><\/span><\/span>T <\/b><\/span><\/span><\/span>= 2\u03c0\/40 <\/b><\/span><\/span><\/span>= \u03c0\/20 s.<\/b><\/span><\/span><\/span> C\u00e1lculo do <\/b><\/span><\/span><\/span>n\u00famero de voltas completas em 1 min = 60 s<\/b><\/span><\/span><\/span> — Regra de tr\u00eas — <\/b><\/span><\/span><\/span>1 volta – \u03c0\/20 s — n<\/b><\/span><\/span><\/span> <\/b>voltas \u2013 60s<\/b><\/span><\/span><\/span> — n.(\u03c0\/20) = 1.60 — <\/b><\/span><\/span><\/span>n = 1200\/\u03c0 voltas.<\/b><\/span><\/span><\/span><\/p>\n

Se, em <\/b><\/span><\/span><\/span>cada volta completa da roda<\/b><\/span><\/span><\/span> o carro percorre — <\/b><\/span><\/span><\/span>\u2206S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 2\u03c0R<\/b><\/span><\/span><\/span> = 2.\u03c0.0,25 — <\/b><\/span><\/span><\/span>\u2206S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 0,5\u03c0 m<\/b><\/span><\/span><\/span>, em <\/b><\/span><\/span><\/span>n = 1200\/\u03c0 voltas ele percorrer\u00e1 \u2206S =<\/b><\/span><\/span><\/span> 0,5\u03c0x1200\/\u03c0 <\/b><\/span><\/span><\/span>= 600 m.<\/b><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

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

\u00a0<\/span><\/h3>\n","protected":false},"excerpt":{"rendered":"

RESOLU\u00c7\u00d5ES 01- Ano-luz \u00e9 uma unidade de medida utilizada em astronomia e corresponde \u00e0 dist\u00e2ncia percorrida pela luz em um ano, no v\u00e1cuo, ou seja, se a luz dista 1 bilh\u00e3o de anos-luz da Terra, ela demora 1 bilh\u00e3o de anos at\u00e9 chegar \u00e0 Terra. R- C 02- O calor recebido por um corpo pode fazer com que sua temperatura<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":586,"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-588","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/588","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=588"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/588\/revisions"}],"predecessor-version":[{"id":590,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/588\/revisions\/590"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/586"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=588"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}