{"id":664,"date":"2015-05-12T17:01:21","date_gmt":"2015-05-12T17:01:21","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=664"},"modified":"2024-09-03T13:21:54","modified_gmt":"2024-09-03T13:21:54","slug":"resolucao-da-ucs-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-ucs-2015\/resolucao-da-ucs-2015\/","title":{"rendered":"Resolucao da UCS 2015"},"content":{"rendered":"

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

01-\u00a0Transformando\u00a0<\/b><\/span><\/span>30Psi<\/b><\/span><\/span><\/span>\u00a0em\u00a0<\/b><\/span><\/span>Pa\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>=30×6,895.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=206,91.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=2,07.10<\/b><\/span><\/span><\/span>5\u00a0<\/b><\/span><\/span><\/sup><\/span>Pa.<\/b><\/span><\/span><\/span><\/p>\n

1 Pa = 1 N\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

02-\u00a0Dados\u00a0<\/b><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>t=3s — \u2206S=40,5cm=0,405m — V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0 (partiu do repouso) — a=?\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>\u2206S=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>2\u00a0<\/b><\/span><\/span><\/span>—<\/b><\/span><\/span><\/span><\/p>\n

0,405 = 0.3 + a.3<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 — a=0,405\/4,5 —\u00a0<\/b><\/span><\/span><\/span>a=0,09m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

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

03-\u00a0Energia potencial gravitacional\u00a0<\/b><\/span><\/span>que ele adquire ao pular a\u00a0<\/b><\/span><\/span><\/span>altura h=40cm=0,40m\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=m.g.h=\u00a0<\/b><\/span><\/span><\/span>80.10.0,4<\/b><\/span><\/span><\/span>=320J.<\/b><\/span><\/span><\/span><\/p>\n

Cada\u00a0<\/b><\/span><\/span><\/span>100g de almo\u00e7o\u00a0<\/b><\/span><\/span><\/span>corresponde<\/b><\/span><\/span><\/span>\u00a0a 100cal\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>1g<\/b><\/span><\/span><\/span>\u00a0corresponde a\u00a0<\/b><\/span><\/span><\/span>1 cal\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>regra de tr\u00eas<\/b><\/span><\/span>\u00a0— 1cl \u2013 4,2j — m cal \u2013 320J —\u00a0<\/b><\/span><\/span><\/span>m=<\/b><\/span><\/span>320\/4,2=<\/b><\/span><\/span><\/span>76,2g.<\/b><\/span><\/span><\/p>\n

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

04-\u00a01 vaga-lume\u00a0<\/b><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>=0,5 J\/s\u00a0<\/b><\/span><\/span><\/span>= 0,5W.<\/b><\/span><\/span><\/span><\/p>\n

Ilumina\u00e7\u00e3o artificial\u00a0<\/b><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=300W.<\/b><\/span><\/span><\/span><\/p>\n

n=<\/b><\/span><\/span><\/span>300W\/0,5W=<\/b><\/span><\/span><\/span>600 vaga-lumes<\/b><\/span><\/span><\/p>\n

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

05-\u00a0Pot\u00eancia mec\u00e2nica da pancada\u00a0<\/b><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=0,5J\/0,5s<\/b><\/span><\/span><\/span>=1W\u00a0<\/b><\/span><\/span><\/span>— ela \u00e9 transformada em el\u00e9trica —\u00a0<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=1W\u00a0<\/b><\/span><\/span><\/span>—<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=i.U\u00a0<\/b><\/span><\/span><\/span>— 1=4.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.U — U=1\/4.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>=0,25.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>V —\u00a0<\/b><\/span><\/span><\/span>U=250V.<\/b><\/span><\/span><\/p>\n

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

06-\u00a0<\/b><\/span><\/span>E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=1200.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 1000.20<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2= 60000 + 200000=260000<\/b><\/span><\/span><\/span>=260.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>J.<\/b><\/span><\/span><\/p>\n

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

07-\u00a0<\/b><\/span><\/span>O\u00a0<\/b><\/span><\/span><\/span>calor recebido\u00a0<\/b><\/span><\/span>por um corpo pode fazer com que sua temperatura aumente ou que ele\u00a0<\/b><\/span><\/span><\/span>sofra uma mudan\u00e7a de estado<\/b><\/span><\/span>. No\u00a0<\/b><\/span><\/span><\/span>segundo caso voc\u00ea tem uma quantidade de calor latente,\u00a0ou seja, calor latente se refere \u00e0 quantidade de calor que um corpo recebe apenas para mudar de estado, o que ocorre \u00e0 temperatura constante.<\/b><\/span><\/span><\/p>\n

A quantidade de\u00a0<\/b><\/span><\/span><\/span>calor latente (Q) cedida ou retirada de uma subst\u00e2ncia para que ela mude de estado o que sempre ocorre \u00e0 temperatura constante\u00a0<\/b><\/span><\/span><\/span>\u00e9 diretamente proporcional \u00e0 sua massa (m), e depende das caracter\u00edsticas da subst\u00e2ncia (L). Matematicamente:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

08-\u00a0<\/b><\/span><\/span>Refra\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span><\/span>\u2013 Fen\u00f4meno que ocorre quando a\u00a0<\/b><\/span><\/span><\/span><\/span>luz passa de um meio para outro<\/b><\/span><\/span><\/span>, variando sua velocidade de propaga\u00e7\u00e3o e\u00a0<\/b><\/span><\/span><\/span><\/span>sofrendo desvio.<\/b><\/span><\/span><\/span><\/p>\n

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

09-\u00a0m<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.c<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub>.(t \u2013 60) + (100 + 100 + 30) = 0\u00a0<\/b><\/span><\/span>— 100t \u2013 6000 + 230 = 0 —\u00a0<\/b><\/span><\/span><\/span>t=<\/b><\/span><\/span><\/span>5770\/100=<\/b><\/span><\/span><\/span>57,7<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C.<\/b><\/span><\/span><\/span><\/p>\n

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

10-\u00a0Altura do som\u00a0<\/b><\/span><\/span>est\u00e1 relacionada com sua\u00a0<\/b><\/span><\/span><\/span>freq\u00fc\u00eancia, ou seja, a altura (tom) \u00e9 a qualidade do som que permite ao ouvido distinguir um som grave, de baixa frequ\u00eancia, de um som agudo, de alta freq\u00fc\u00eancia.\u00a0 \u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0O som\u00a0<\/b><\/span><\/span><\/span>mais grave aud\u00edvel\u00a0<\/b><\/span><\/span>por um ouvido humano \u00e9 de\u00a0<\/b><\/span><\/span><\/span>aproximadamente 20 Hz\u00a0<\/b><\/span><\/span>e o\u00a0<\/b><\/span><\/span><\/span>mais agudo\u00a0<\/b><\/span><\/span>\u00e9 de\u00a0<\/b><\/span><\/span><\/span>aproximadamente\u00a020 000 Hz.<\/b><\/span><\/span><\/p>\n

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

 <\/p>\n

Voltar para os Exercicios<\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

RESOLU\u00c7\u00d5ES 01-\u00a0Transformando\u00a030Psi\u00a0em\u00a0Pa\u00a0—\u00a0P=30×6,895.103=206,91.103=2,07.105\u00a0Pa. 1 Pa = 1 N\/m2. R- B 02-\u00a0Dados\u00a0—\u00a0t=3s — \u2206S=40,5cm=0,405m — Vo=0 (partiu do repouso) — a=?\u00a0—\u00a0\u2206S=Vo.t + at22\u00a0— 0,405 = 0.3 + a.32\/2 — a=0,405\/4,5 —\u00a0a=0,09m\/s2. R- C 03-\u00a0Energia potencial gravitacional\u00a0que ele adquire ao pular a\u00a0altura h=40cm=0,40m\u00a0—\u00a0Ep=m.g.h=\u00a080.10.0,4=320J. Cada\u00a0100g de almo\u00e7o\u00a0corresponde\u00a0a 100cal\u00a0—\u00a01g\u00a0corresponde a\u00a01 cal\u00a0—\u00a0regra de tr\u00eas\u00a0— 1cl \u2013 4,2j — m cal \u2013 320J —\u00a0m=320\/4,2=76,2g. R- C 04-\u00a01<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":662,"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-664","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/664","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=664"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/664\/revisions"}],"predecessor-version":[{"id":666,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/664\/revisions\/666"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/662"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=664"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}