{"id":678,"date":"2015-05-12T17:39:34","date_gmt":"2015-05-12T17:39:34","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=678"},"modified":"2024-09-02T17:42:45","modified_gmt":"2024-09-02T17:42:45","slug":"resolucao-comentada-da-uenp-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-uenp-2015\/resolucao-comentada-da-uenp-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada da UENP 2015"},"content":{"rendered":"

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

01- <\/b><\/span><\/span><\/span>I. Falsa<\/b><\/span><\/span><\/span> \"\" No c\u00e1lculo da <\/b><\/span><\/span><\/span>for\u00e7a resultante<\/b><\/span><\/span><\/span> sobre um corpo <\/b><\/span><\/span><\/span>todas as caracter\u00edsticas vetoriais (intensidade, dire\u00e7\u00e3o e sentido) influem.<\/b><\/span><\/span><\/span><\/p>\n

II. Correta \"\" <\/b><\/span><\/span><\/span>conseq\u00fc\u00eancia do <\/b><\/span><\/span><\/span>princ\u00edpio da in\u00e9rcia.<\/b><\/span><\/span><\/span><\/p>\n

III. Falsa \"\" <\/b><\/span><\/span><\/span>s\u00e3o massa e acera\u00e7\u00e3o<\/b><\/span><\/span><\/span> \"\" segunda lei de Newton \"\" F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = m.a.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/b><\/span><\/span><\/span>C\u00e1lculo da <\/b><\/span><\/span><\/span>resist\u00eancia<\/b><\/span><\/span><\/span> do resistor <\/b><\/span><\/span><\/span>equivalente<\/b><\/span><\/span><\/span>, curto-circuitando o amper\u00edmetro e retirando o volt\u00edmetro, pois s\u00e3o <\/b><\/span><\/span><\/span>ideais:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = U\/i<\/b><\/span><\/span><\/span> \"\" 7 = 70\/i \"\" <\/b><\/span><\/span><\/span>i <\/b><\/span><\/span><\/span>= 7\/70 = <\/b><\/span><\/span><\/span>10 A.<\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>amper\u00edmetro<\/b><\/span><\/span><\/span> marca i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> tal que i<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> =U\/R = 30\/12 <\/b><\/span><\/span><\/span>= 2,5 A (figura abaixo).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe na figura abaixo que o <\/b><\/span><\/span><\/span>volt\u00edmetro<\/b><\/span><\/span><\/span> indica <\/b><\/span><\/span><\/span>U <\/b><\/span><\/span><\/span>= R.i = 6.3,75 <\/b><\/span><\/span><\/span>= 22,5 V.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

03- <\/b><\/span><\/span><\/span>P = W\/\u2206t<\/b><\/span><\/span><\/span> \"\" 30.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> = W\/5 \"\" <\/b><\/span><\/span><\/span>W =<\/b><\/span><\/span><\/span> 150.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> J = <\/b><\/span><\/span><\/span>150 kJ.<\/b><\/span><\/span><\/span><\/p>\n

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

04- <\/b><\/span><\/span><\/span>A falsa \u00e9 a III<\/b><\/span><\/span><\/span> \"\" O <\/b><\/span><\/span><\/span>tempo medido<\/b><\/span><\/span><\/span> pelo observador de <\/b><\/span><\/span><\/span>dentro de nave que se move com velocidade pr\u00f3xima \u00e0 da luz<\/b><\/span><\/span><\/span> \u00e9 numericamente <\/b><\/span><\/span><\/span>menor que o tempo medido por um observador na Terra<\/b><\/span><\/span><\/span>.<\/span><\/span><\/span><\/p>\n

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

05- <\/b><\/span><\/span><\/span>Trecho submarino<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>\u2206S<\/b><\/span><\/span><\/span> = 40000 + 400 <\/b><\/span><\/span><\/span>= 40400 m<\/b><\/span><\/span><\/span> \"\" V = 300\/3,6 m\/s \"\" \u2206t = ? \"\" V = \u2206S\/\u2206t \"\" <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

300\/3,6 = 40400\/\u2206t \"\" <\/b><\/span><\/span><\/span>\u2206t<\/b><\/span><\/span><\/span> = 3,6×40400\/300 = 145440\/300 = <\/b><\/span><\/span><\/span>484,8 s\/60 = 8,08 min.<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/b><\/span><\/span><\/span>Corrente el\u00e9trica no <\/b><\/span><\/span><\/span>secador de cabelo (127V,1400W)<\/b><\/span><\/span><\/span> \"\" P = i.U \"\" 1400 = 127.i<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>s<\/b><\/span><\/span><\/sub><\/span> = 11,0 A.<\/b><\/span><\/span><\/span><\/p>\n

Corrente el\u00e9trica no <\/b><\/span><\/span><\/span>chuveiro (127V,4400W)<\/b><\/span><\/span><\/span> \"\" P = i.U \"\" 4400 = 127.i<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span> = 34,6 A.<\/b><\/span><\/span><\/span><\/p>\n

Corrente el\u00e9trica nas 4 l\u00e2mpadas<\/b><\/span><\/span><\/span> (127V,100W)<\/b><\/span><\/span><\/span> \"\" P = i.U \"\" 100 = 127.i<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>l<\/b><\/span><\/span><\/sub><\/span> = 0,78×4 = 3,12 A.<\/b><\/span><\/span><\/span><\/p>\n

Corrente el\u00e9trica no <\/b><\/span><\/span><\/span>aparelho de som (127V,150W)<\/b><\/span><\/span><\/span> \"\" P = i.U \"\" 150 = 127.i<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span> = 1,18 A.<\/b><\/span><\/span><\/span><\/p>\n

I<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span> <\/b>= 11,0 + 34,6 + 3,12 + 1,18 <\/b><\/span><\/span><\/span>= 49,9 A.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>diferen\u00e7a pedida<\/b><\/span><\/span><\/span> vale 49,9 \u2013 20 <\/b><\/span><\/span><\/span>= 29,9 A<\/b><\/span><\/span><\/span><\/p>\n

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

07- <\/b><\/span><\/span><\/span>I. Correta<\/b><\/span><\/span><\/span> \"\" O processo de <\/b><\/span><\/span><\/span>refrigera\u00e7\u00e3o<\/b><\/span><\/span><\/span> realiza trabalho ao <\/b><\/span><\/span><\/span>retirar calor da fonte fria e transferi-lo para a fonte quente<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

II. Correta <\/b><\/span><\/span><\/span>numa <\/b><\/span><\/span><\/span>isob\u00e1rica<\/b><\/span><\/span><\/span> (press\u00e3o constante), o <\/b><\/span><\/span><\/span>trabalho<\/b><\/span><\/span><\/span> realizado \u00e9 fornecido por <\/b><\/span><\/span><\/span>W = P.\u2206V.<\/b><\/span><\/span><\/span><\/p>\n

III. Correta<\/b><\/span><\/span><\/span> \"\" \u03b7 = (1000 \u2013 600)\/1000 = 0,4 = <\/b><\/span><\/span><\/span>40 %.<\/b><\/span><\/span><\/span><\/p>\n

IV. Correta<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span> = P.\u2206V \"\" W = 2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>.2 <\/b><\/span><\/span><\/span>= 4.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> J.<\/b><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

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

RESOLU\u00c7\u00d5ES 01- I. Falsa No c\u00e1lculo da for\u00e7a resultante sobre um corpo todas as caracter\u00edsticas vetoriais (intensidade, dire\u00e7\u00e3o e sentido) influem. II. Correta conseq\u00fc\u00eancia do princ\u00edpio da in\u00e9rcia. III. Falsa s\u00e3o massa e acera\u00e7\u00e3o segunda lei de Newton FR = m.a. R- E 02- C\u00e1lculo da resist\u00eancia do resistor equivalente, curto-circuitando o amper\u00edmetro e retirando o volt\u00edmetro, pois s\u00e3o ideais:<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":676,"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-678","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/678","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=678"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/678\/revisions"}],"predecessor-version":[{"id":679,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/678\/revisions\/679"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/676"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=678"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}