{"id":982,"date":"2015-06-29T01:40:48","date_gmt":"2015-06-29T01:40:48","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=982"},"modified":"2024-09-02T17:18:13","modified_gmt":"2024-09-02T17:18:13","slug":"resolucao-comentada-da-escs-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-escs-2015\/resolucao-comentada-da-escs-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada da ESCS 2015"},"content":{"rendered":"

RESOLU\u00c7\u00c3O DAS QUEST\u00d5ES DA MEDICINA ESCS-DF-2015<\/b><\/span><\/span><\/span><\/p>\n

01- <\/b><\/span><\/span><\/span>Leia as informa\u00e7\u00f5es abaixo e lembre-se de que o princ\u00edpio de funcionamento \u00e9 o mesmo tanto para submarinos quanto para navios:<\/b><\/span><\/span><\/span><\/p>\n

Os submarinos <\/b><\/span><\/span>possuem tanques de ar<\/b><\/span><\/span><\/span> (c\u00e2maras de flutua\u00e7\u00e3o) que s\u00e3o instaladas entre as estruturas interna e externa do casco, com paredes bem refor\u00e7adas para resistirem \u00e0 press\u00e3o. O <\/b><\/span><\/span>sobe e desce<\/b><\/span><\/span><\/span> <\/b>do submarino<\/b><\/span><\/span><\/span> depende da <\/b><\/span><\/span>quantidade de \u00e1gua nesses tanques, que podem aumentar ou diminuir devido ao ar comprimido.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n


\nQuando est\u00e1 na <\/b><\/span><\/span>superf\u00edcie<\/b><\/span><\/span><\/span>, um submarino funciona exatamente como um <\/b><\/span><\/span>navio<\/b><\/span><\/span><\/span>. Como <\/b><\/span><\/span>seu peso<\/b><\/span><\/span><\/span> \u00e9 <\/b><\/span><\/span>igual ao da massa l\u00edquida deslocada (empuxo), ele flutua.<\/b><\/span><\/span><\/span> Para<\/b><\/span><\/span>imergir,<\/b><\/span><\/span><\/span> \u00e9 preciso torn\u00e1-lo <\/b><\/span><\/span>mais pesado<\/b><\/span><\/span><\/span> e para isso deve-se <\/b><\/span><\/span>completar com a \u00e1gua do mar os tanques de ar (c\u00e2maras de flutua\u00e7\u00e3o).<\/b><\/span><\/span><\/span> Quando totalmente submerso, o empuxo sobre ele \u00e9 sempre o mesmo independente da profundidade e, para subir <\/b><\/span><\/span>o ar comprimido expulsa a \u00e1gua, ele fica mais leve e a \u00e1gua do mar, devido ao empuxo empurra-o para cima e para descer deixa-se entrar \u00e1gua, diminuindo a press\u00e3o do ar, ele fica mais pesado que o empuxo, e desce.<\/b><\/span><\/span><\/span>Lembre-se de que, quando totalmente submerso o empuxo \u00e9 o<\/b><\/span><\/span><\/span> mesmo, quem varia \u00e9 o peso do submarino.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

02- <\/span><\/span><\/span>Timbre<\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>timbre <\/b><\/span><\/span><\/span>\u00e9 uma qualidade sonora que permite <\/b><\/span><\/span><\/span>distinguir dois sons de mesma altura (<\/b><\/span><\/span><\/span>mesma<\/b><\/span><\/span><\/span> <\/b>freq\u00fc\u00eancia<\/b><\/span><\/span><\/span>) e mesma intensidade, emitidos por instrumentos diferentes tocando a mesma nota musical ou acorde.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\"\"<\/p>\n

Assim, conseguimos <\/b><\/span><\/span><\/span>distinguir se a mesma nota musical \u00e9 emitida por um viol\u00e3o ou por um piano<\/b><\/span><\/span><\/span> pelo timbre, pois ele <\/b><\/span><\/span><\/span>difere <\/b><\/span><\/span><\/span>nos dois instrumentos e nos fornece<\/b><\/span><\/span><\/span>sensa\u00e7\u00f5es sonoras diferentes,<\/b><\/span><\/span><\/span> devido \u00e0s <\/b><\/span><\/span><\/span>diferentes composi\u00e7\u00f5es de harm\u00f4nicos gerados por cada instrumento.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/span><\/span><\/span>A. Correta<\/b><\/span><\/span><\/span> \"\" A corrente m\u00e1xima \u00e9 a de <\/b><\/span><\/span>curto circuito<\/b><\/span><\/span><\/span> e fornecida por <\/b><\/span><\/span>i<\/b><\/span><\/span><\/span>cc<\/b><\/span><\/span><\/sub><\/span> = E\/r <\/b><\/span><\/span><\/span>= 12\/0,2 <\/b><\/span><\/span>= 60 A<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

B. Falsa<\/b><\/span><\/span><\/span> \"\" a <\/b><\/span><\/span>resist\u00eancia interna<\/b><\/span><\/span><\/span> da pilha \u00e9 <\/b><\/span><\/span>constante<\/b><\/span><\/span><\/span>, <\/b><\/span><\/span>independente da corrente el\u00e9trica<\/b><\/span><\/span><\/span> que a est\u00e1 pecorrendo.<\/b><\/span><\/span><\/p>\n

C. Falsa<\/b><\/span><\/span><\/span> \"\" se a <\/b><\/span><\/span>bateria<\/b><\/span><\/span><\/span> possui <\/b><\/span><\/span>resist\u00eancia interna nula,<\/b><\/span><\/span><\/span> ela \u00e9 <\/b><\/span><\/span>ideal<\/b><\/span><\/span><\/span> e <\/b><\/span><\/span>fornece sempre a mesma ddp<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

(tens\u00e3o) que \u00e9 constante V.<\/b><\/span><\/span><\/span><\/p>\n

D. Falsa<\/b><\/span><\/span><\/span> \"\" mesmo com a <\/b><\/span><\/span>pilha carregada<\/b><\/span><\/span><\/span>, segurando os terminais da pilha com os <\/b><\/span><\/span>dedos secos,<\/b><\/span><\/span><\/span> <\/b>cuja pele tem resist\u00eancia relativamente alta \u00e0 passagem de corrente el\u00e9trica, voc\u00ea n\u00e3o levar\u00e1 choque.<\/b><\/span><\/span><\/span><\/p>\n

Mesmo com os <\/b><\/span><\/span>dedos molhados<\/b><\/span><\/span><\/span>, voc\u00ea <\/b><\/span><\/span>n\u00e3o levar\u00e1 choque<\/b><\/span><\/span><\/span> porque a diferen\u00e7a de potencial de uma pilha \u00e9 muito baixa para que <\/b><\/span><\/span>haja uma corrente que se possa perceber.<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

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

04- \"\"<\/span><\/span><\/span><\/p>\n

For\u00e7a magn\u00e9tica sobre um fio condutor retil\u00edneo de comprimento <\/b><\/span><\/span><\/span>\u2113 percorrido por uma corrente i e imerso num campo magn\u00e9tico de intensidade B.<\/b><\/span><\/span><\/span>
\n<\/b><\/span><\/span>A dire\u00e7\u00e3o e sentido de\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0\u00e9 fornecida pela <\/b><\/span><\/span>regra da m\u00e3o esquerda<\/b><\/span><\/span><\/span> (veja figura) onde o <\/b><\/span><\/span>dedo m\u00e9dio<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Indica a dire\u00e7\u00e3o e o sentido da<\/b><\/span><\/span><\/span> <\/b>corrente el\u00e9trica i, o indicador, do campo magn\u00e9tico B e o polegar, da for\u00e7a magn\u00e9tica F<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

No caso do exerc\u00edcio, onde <\/b><\/span><\/span>B e i s\u00e3o perpendiculares<\/b><\/span><\/span><\/span> a intensidade da <\/b><\/span><\/span>for\u00e7a magn\u00e9tica<\/b><\/span><\/span><\/span> fica F<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub> = B.i.\u2113.<\/b><\/span><\/span><\/p>\n

sen90<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup> = B.i.\u2113.1 \"\" <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span> = B.i.\u2113.<\/b><\/span><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>figura abaixo<\/b><\/span><\/span><\/span> mostra a <\/b><\/span><\/span><\/span>dire\u00e7\u00e3o e o sentido<\/b><\/span><\/span><\/span> da <\/b><\/span><\/span><\/span>for\u00e7a magn\u00e9tica<\/b><\/span><\/span><\/span> sobre <\/b><\/span><\/span><\/span>cada trecho do fio utilizando a regra da m\u00e3o esquerda:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Veja na <\/b><\/span><\/span>primeira figura<\/b><\/span><\/span><\/span> que as <\/b><\/span><\/span>duas for\u00e7as de intensidades 4BIL se anulam<\/b><\/span><\/span><\/span> e o <\/b><\/span><\/span>mesmo acontece<\/b><\/span><\/span><\/span> com as <\/b><\/span><\/span>outras duas de intensidades BIL.<\/b><\/span><\/span><\/span><\/p>\n

Na <\/b><\/span><\/span>figura da direita<\/b><\/span><\/span><\/span> as <\/b><\/span><\/span>tr\u00eas for\u00e7as<\/b><\/span><\/span><\/span> que restaram fornecem uma <\/b><\/span><\/span>for\u00e7a resultante<\/b><\/span><\/span><\/span>, pedida no exerc\u00edcio, de intensidade <\/b><\/span><\/span>F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> <\/b>= 2BiL + BiL + BiL <\/b><\/span><\/span>= 4BiL<\/b><\/span><\/span><\/span>, de dire\u00e7\u00e3o <\/b><\/span><\/span>vertical e sentido para cima<\/b><\/span><\/span><\/span> (sentido positivo do eixo y).<\/b><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>Sendo a <\/b><\/span><\/span><\/span>densidade da \u00e1gua 1g\/cm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> = 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> = 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>kg\/10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>dm<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> <\/b>= 1kg\/L<\/b><\/span><\/span><\/span>, ent\u00e3o 1 volume de <\/b><\/span><\/span><\/span>1L tem massa de 1kg.<\/b><\/span><\/span><\/span><\/p>\n

m = 2kg = 2000 g \"\" c = 1 cal\/g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C \"\" \u2206<\/b><\/span><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span><\/span> <\/b>= 45 \u2013 40 <\/b><\/span><\/span><\/span>= 5<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span> \"\" equa\u00e7\u00e3o fundamental da calorimetria <\/b><\/span><\/span><\/span>Q =<\/b><\/span><\/span><\/span> <\/b>m.c.\u2206<\/b><\/span><\/span><\/span>\u03b8<\/b><\/span><\/span><\/span><\/span> = 2000.1.5 \"\" <\/b><\/span><\/span><\/span>Q = 10000 cal.<\/b><\/span><\/span><\/span><\/p>\n

1 cal = 4,2 J \"\" Q = W = <\/b><\/span><\/span><\/span><\/span>10000×4,2<\/b><\/span><\/span><\/span> = 42000 J.<\/b><\/span><\/span><\/span><\/span><\/p>\n

P = W\/\u2206t \"\" P <\/b><\/span><\/span><\/span><\/span>= 42000\/50×60 = 42000\/3000<\/b><\/span><\/span><\/span><\/span> = 14 W.<\/b><\/span><\/span><\/span><\/span><\/p>\n

P = U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R \"\" <\/b><\/span><\/span><\/span>14 = 36<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R<\/b><\/span><\/span> \"\" R <\/b><\/span><\/span><\/span>= 1296\/14<\/b><\/span><\/span><\/span> = 92,6 \u2126.<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/span><\/span><\/span>A. Falsa \"\" g<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span> =GM<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> <\/b>= G.6.10<\/b><\/span><\/span><\/span>24<\/b><\/span><\/span><\/sup><\/span>\/(150.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 6<\/b><\/span><\/span><\/span>.10<\/b><\/span><\/span>24<\/b><\/span><\/sup><\/span>G\/22500.10<\/b><\/span><\/span>12<\/b><\/span><\/sup><\/span> <\/b>= 6<\/b><\/span><\/span><\/span>.10<\/b><\/span><\/span>24<\/b><\/span><\/sup><\/span>G\/2,25.10<\/b><\/span><\/span>16<\/b><\/span><\/sup><\/span> \"\" g<\/b><\/span><\/span>T<\/b><\/span><\/sub><\/span> = 2,7 .10<\/b><\/span><\/span>8<\/b><\/span><\/sup><\/span>G.g<\/b><\/span><\/span>M<\/b><\/span><\/sub><\/span> = <\/b><\/span><\/span>GM<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> = <\/b><\/span><\/span>G.6,5.10<\/b><\/span>23<\/b><\/span><\/sup>\/(230.10<\/b><\/span>6<\/b><\/span><\/sup>)<\/b><\/span>2<\/b><\/span><\/sup> = G.6,5.10<\/b><\/span>23<\/b><\/span><\/sup>\/52900.10<\/b><\/span>12<\/b><\/span><\/sup> = G.6,5.10<\/b><\/span>23<\/b><\/span><\/sup>\/5,29.10<\/b><\/span>16<\/b><\/span><\/sup> \"\" <\/b><\/span>g<\/b><\/span><\/span>M<\/b><\/span><\/sub><\/span> = 1,23.10<\/b><\/span><\/span>7<\/b><\/span><\/sup><\/span>G.<\/b><\/span><\/span><\/p>\n

Observe que <\/b><\/span>g<\/b><\/span><\/span>M<\/b><\/span><\/sub><\/span> < g<\/b><\/span><\/span>T<\/b><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/p>\n

B. Correta \"\" <\/b><\/span><\/span><\/span>quanto <\/b><\/span><\/span><\/span>menor a press\u00e3o atmosf\u00e9rica no planeta, menor ser\u00e1 a temperatura de ebuli\u00e7\u00e3o<\/b><\/span><\/span><\/span> <\/b>da \u00e1gua nele.<\/b><\/span><\/span><\/span> Como a press\u00e3o atmosf\u00e9rica em Marte <\/b><\/span><\/span><\/span>(P<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span> = 0,00 atm) \u00e9 menor que a da Terra (1 atm),<\/b><\/span><\/span><\/span> <\/b>em Marte a \u00e1gua ferver\u00e1 a uma temperatura menor que na Terra..<\/b><\/span><\/span><\/span><\/p>\n

C. Falsa \"\" <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>massa <\/b><\/span><\/span><\/span>de um corpo \u00e9 uma <\/b><\/span><\/span><\/span>grandeza invari\u00e1vel<\/b><\/span><\/span><\/span> <\/b>independente do lugar ou do planeta<\/b><\/span><\/span><\/span> <\/b>onde ele esteja.<\/b><\/span><\/span><\/span><\/p>\n

D. Falsa<\/b><\/span><\/span><\/span> \"\" Na <\/b><\/span><\/span>express\u00e3o<\/b><\/span><\/span><\/span> da terceira lei de Kepler, <\/b><\/span><\/span>T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span>3 <\/b><\/span><\/span><\/sup><\/span>= K = constante<\/b><\/span><\/span><\/span>, observamos que <\/b><\/span><\/span>\u00e0 medida que R<\/b><\/span><\/span><\/span> <\/b>aumenta, T tamb\u00e9m aumenta,<\/b><\/span><\/span><\/span> o que significa que<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>quanto<\/b><\/span><\/span>mais afastado o planeta estiver do Sol<\/b><\/span><\/span><\/span> <\/b>maior ser\u00e1 seu ano<\/b><\/span><\/span><\/span> (tempo que demora para dar um volta completa ao redor do Sol). Como Marte est\u00e1 <\/b><\/span><\/span>mais afastado<\/b><\/span><\/span><\/span> do Sol (R<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub> = 230.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup> km) que a Terra (R<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub> = 150.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup> km), <\/b><\/span><\/span>seu ano ser\u00e1 maior que o da Terra.<\/b><\/span><\/span><\/span><\/p>\n

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

07- <\/span><\/span><\/span>A. Falsa<\/b><\/span><\/span><\/span> \"\" o <\/b><\/span><\/span><\/span>trabalho<\/b><\/span><\/span><\/span> para levantar a carga <\/b><\/span><\/span><\/span>com ou sem polias de uma mesma altura h \u00e9 o mesmo<\/b><\/span><\/span><\/span> e vale <\/b><\/span><\/span><\/span>W = Ph = mgh (observe que P e h s\u00e3o os mesmos).<\/b><\/span><\/span><\/span><\/p>\n

B. Correta \"\" W = Ph = 2P.<\/b><\/span><\/span><\/span><\/p>\n

C. Falsa \"\" suportam tens\u00f5es diferentes (veja figura abaixo):<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

D. Falsa \"\" veja figura acima.<\/b><\/span><\/span><\/span><\/p>\n

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

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

RESOLU\u00c7\u00c3O DAS QUEST\u00d5ES DA MEDICINA ESCS-DF-2015 01- Leia as informa\u00e7\u00f5es abaixo e lembre-se de que o princ\u00edpio de funcionamento \u00e9 o mesmo tanto para submarinos quanto para navios: Os submarinos possuem tanques de ar (c\u00e2maras de flutua\u00e7\u00e3o) que s\u00e3o instaladas entre as estruturas interna e externa do casco, com paredes bem refor\u00e7adas para resistirem \u00e0 press\u00e3o. O sobe e desce<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":980,"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-982","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/982","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=982"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/982\/revisions"}],"predecessor-version":[{"id":984,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/982\/revisions\/984"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/980"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=982"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}