{"id":736,"date":"2015-05-12T19:56:13","date_gmt":"2015-05-12T19:56:13","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=736"},"modified":"2024-09-03T13:22:48","modified_gmt":"2024-09-03T13:22:48","slug":"resolucao-comentada-uscs-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-uscs-2015\/resolucao-comentada-uscs-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada USCS 2015"},"content":{"rendered":"

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

01- <\/span><\/span><\/span>Dados: <\/b><\/span><\/span><\/span> V = <\/b><\/span><\/span><\/span>300000 km\/s<\/b><\/span><\/span><\/span> = 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> m\/s \"\" \u2206S <\/b><\/span><\/span><\/span>= 2×56.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span> km<\/b><\/span><\/span><\/span> = 112.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span> m <\/b><\/span><\/span><\/span>(ida e volta)<\/b><\/span><\/span><\/span> \"\" \u2206t = ?.<\/b><\/span><\/span><\/span><\/p>\n

V =<\/b><\/span><\/span><\/span> <\/b>\u2206S\/\u2206t \"\" <\/b><\/span><\/span><\/span>3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> = 112.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\/\u2206t \"\" <\/b><\/span><\/span><\/span>\u2206t<\/b><\/span><\/span><\/span> = 112.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>\/3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> = 37.10<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sup><\/span> = 370s\/60 <\/b><\/span><\/span><\/span>= 6,1 min.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/span><\/span><\/span>Como <\/b><\/span><\/span><\/span>n\u00e3o deve ocorrer deslizamento<\/b><\/span><\/span><\/span> entre os blocos A e B, eles devem se <\/b><\/span><\/span><\/span>mover juntos<\/b><\/span><\/span><\/span> como um <\/b><\/span><\/span><\/span>\u00fanico bloco de massa m<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> = 5 = 10 = 15 kg.<\/b><\/span><\/span><\/span><\/p>\n

Colocando as for\u00e7a que agem sobre o <\/b><\/span><\/span><\/span>sistema m<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span> = 15 kg e m<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span> = 5 kg, na dire\u00e7\u00e3o do movimento:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Bloco C<\/b><\/span><\/span><\/span> \"\" desce com acelera\u00e7\u00e3o <\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span> \"\" F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> \u2013 m<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>.a \"\" <\/b><\/span><\/span><\/span>50 \u2013 T = 5.a (I)<\/b><\/span><\/span><\/span><\/p>\n

Bloco AB \"\"<\/b><\/span><\/span><\/span> move-se <\/b><\/span><\/span><\/span>sem atrito para a direita<\/b><\/span><\/span><\/span> com acelera\u00e7\u00e3o <\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span> \"\" F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>.a \"\" <\/b><\/span><\/span><\/span>T = 15.a (II)<\/b><\/span><\/span><\/span><\/p>\n

(II) em (I)<\/b><\/span><\/span><\/span> \"\" 50 \u2013 15.a = 5.a \"\" 20.a = 50 \"\" <\/b><\/span><\/span><\/span>a = 2,5 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

For\u00e7as <\/b><\/span><\/span><\/span>que agem sobre os <\/b><\/span><\/span><\/span>blocos A e B<\/b><\/span><\/span><\/span>, que se <\/b><\/span><\/span><\/span>movem juntos com acelera\u00e7\u00e3o a = 2,5 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>, devido \u00e0 <\/b><\/span><\/span><\/span>for\u00e7a de atrito F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span> que eles trocam entre si:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sobre o <\/b><\/span><\/span><\/span>bloco A,<\/b><\/span><\/span><\/span> que se desloca para a direita com acelera\u00e7\u00e3o <\/b><\/span><\/span><\/span>a = 2,5 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>, a <\/b><\/span><\/span><\/span>for\u00e7a resultante<\/b><\/span><\/span><\/span>, na dire\u00e7\u00e3o do movimento, \u00e9 a <\/b><\/span><\/span><\/span>for\u00e7a de atrito<\/b><\/span><\/span><\/span> \"\" F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span> = F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span> = m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.a \"\" F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span> = 5.2,5 \"\" <\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>at<\/b><\/span><\/span><\/sub><\/span> = 12,5 N.<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/span><\/span><\/span>Energia mec\u00e2nica no ponto P<\/b><\/span><\/span><\/span>, mais alto do plano inclinado onde <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> = 0 e h = 20m<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>cP<\/b><\/span><\/span><\/sub><\/span> + E<\/b><\/span><\/span><\/span>pP<\/b><\/span><\/span><\/sub><\/span> \"\" E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span> = mV<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh = 100.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 100.10.20 \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span> = 20000 J<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Energia mec\u00e2nica no ponto Q<\/b><\/span><\/span><\/span>, mais baixo do plano inclinado onde <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span> = 10 m\/s e h = 0<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>cQ<\/b><\/span><\/span><\/sub><\/span> + E<\/b><\/span><\/span><\/span>pQ<\/b><\/span><\/span><\/sub><\/span> \"\" E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span> = mV<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh = 100.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 100.10.0 \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span> = 5000 J<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>energia dissipada pelo atrito<\/b><\/span><\/span><\/span> corresponde \u00e0 <\/b><\/span><\/span><\/span>diferen\u00e7<\/b><\/span><\/span><\/span>a entre as energias totais inicial e final \"\" \u2206E =<\/b><\/span><\/span><\/span><\/p>\n

E<\/b><\/span><\/span><\/span>mP<\/b><\/span><\/span><\/sub><\/span> \u2013 E<\/b><\/span><\/span><\/span>mQ<\/b><\/span><\/span><\/sub><\/span> = 20000 \u2013 5000 \"\" \u2206E = 15000J = 15kJ.<\/b><\/span><\/span><\/span><\/p>\n

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

04- <\/span><\/span><\/span>Massa do navio \"\" <\/b><\/span><\/span><\/span>m =<\/b><\/span><\/span><\/span> 90 ton = <\/b><\/span><\/span><\/span>90.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>kg<\/b><\/span><\/span><\/span> \"\" peso do navio \"\" <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span> = m.g = 90.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10 <\/b><\/span><\/span><\/span>= 9.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> N.<\/b><\/span><\/span><\/span><\/p>\n

O <\/b><\/span><\/span><\/span>empuxo<\/b><\/span><\/span><\/span> (for\u00e7a vertical e para cima) sobre o navio \u00e9 fornecido por <\/b><\/span><\/span><\/span>E = d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/sub><\/span>.g<\/b><\/span><\/span><\/span> = 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/sub><\/span>.10 \"\" <\/b><\/span><\/span><\/span>E = 10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Como o navio est\u00e1 <\/b><\/span><\/span><\/span>flutuando,<\/b><\/span><\/span><\/span> o peso deve anular o empuxo \"\" <\/b><\/span><\/span><\/span>P = E<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>9.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> = 10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/sub><\/span> \"\" V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

9.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\/10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>imerso<\/b><\/span><\/span><\/sub><\/span> = 9.10<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sup><\/span> = 90 m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>Como o <\/b><\/span><\/span><\/span>furo (raio R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>) \u00e9 menor<\/b><\/span><\/span><\/span> que o <\/b><\/span><\/span><\/span>disco de alum\u00ednio (raio R),<\/b><\/span><\/span><\/span> voc\u00ea deve <\/b><\/span><\/span><\/span>diminuir R (resfriando <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

o disco de alum\u00ednio)<\/b><\/span><\/span><\/span> at\u00e9 que ele se encaixe no anel de cobre (ficando com raio R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>).<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/span><\/span><\/span>Intensidade<\/b><\/span><\/span><\/span><\/p>\n

\u00c8 a <\/b><\/span><\/span><\/span>caracter\u00edstica do som<\/b><\/span><\/span><\/span> que nos permite distinguir um <\/b><\/span><\/span><\/span>som forte de um som fraco<\/b><\/span><\/span><\/span> e est\u00e1 relacionada com a <\/b><\/span><\/span><\/span>energia transportada pela onda<\/b><\/span><\/span><\/span> que, <\/b><\/span><\/span><\/span>quanto mais perto da fonte mais forte ser\u00e1 o som e mais afastado da fonte, mais fraco.<\/b><\/span><\/span><\/span><\/p>\n

Dessa maneira, o som emitido, por exemplo, por uma britadeira,\u00a0 ser\u00e1 mais forte para uma pessoa <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

perto da mesma e mais fraco para uma pessoa afastada da mesma.<\/b><\/span><\/span><\/span><\/p>\n

A <\/b><\/span><\/span><\/span>freq\u00fc\u00eancia do som<\/b><\/span><\/span><\/span>, independente do meio onde est\u00e1 se propagando, \u00e9 a <\/b><\/span><\/span><\/span>mesma que a da fonte, que no caso \u00e9 constante.<\/b><\/span><\/span><\/span><\/p>\n

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

07- <\/span><\/span><\/span>Como est\u00e3o associados em <\/b><\/span><\/span><\/span>paralelo, a tens\u00e3o U = 100 V \u00e9 a mesma<\/b><\/span><\/span><\/span> para todos os aparelhos:<\/b><\/span><\/span><\/span><\/p>\n

L\u00e2mpada<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>P = U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span> = 100<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/100 <\/b><\/span><\/span><\/span>= 100 W<\/b><\/span><\/span><\/span> \"\" energia el\u00e9trica \"\" <\/b><\/span><\/span><\/span>W = P.\u2206<\/b><\/span><\/span><\/span>t = 100.5 = 500 Wh <\/b><\/span><\/span><\/span>= 0,5 kWh.<\/b><\/span><\/span><\/span><\/p>\n

Torneira el\u00e9trica<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>P = U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span> = 100<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/50 <\/b><\/span><\/span><\/span>= 200 W<\/b><\/span><\/span><\/span> \"\" energia el\u00e9trica \"\" <\/b><\/span><\/span><\/span>W = P.\u2206<\/b><\/span><\/span><\/span>t = 200.0,5 = 100 Wh <\/b><\/span><\/span><\/span>= 0,1 kWh.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Chuveiro el\u00e9trico<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>P = U<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R<\/b><\/span><\/span><\/span> = 100<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/10 <\/b><\/span><\/span><\/span>= 1000 W<\/b><\/span><\/span><\/span> \"\" energia el\u00e9trica \"\" <\/b><\/span><\/span><\/span>W = P.\u2206<\/b><\/span><\/span><\/span>t = 1000.(45\/60) = 750 Wh <\/b><\/span><\/span><\/span>= 0,75 kWh.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

Energia el\u00e9trica total<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span><\/sub><\/span> <\/b>=0,5 + 0,1 + 0,75 <\/b><\/span><\/span><\/span>= 1,35 kWh.<\/b><\/span><\/span><\/span><\/p>\n

Custo<\/b><\/span><\/span><\/span> \"\" C$ = 1,35×20 <\/b><\/span><\/span><\/span>= 27.<\/b><\/span><\/span><\/span><\/p>\n

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

<\/h2>\n

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

RESOLU\u00c7\u00d5ES 01- Dados: V = 300000 km\/s = 3.108 m\/s \u2206S = 2×56.106 km = 112.109 m (ida e volta) \u2206t = ?. V = \u2206S\/\u2206t 3.108 = 112.109\/\u2206t \u2206t = 112.109\/3.108 = 37.101 = 370s\/60 = 6,1 min. R- A 02- Como n\u00e3o deve ocorrer deslizamento entre os blocos A e B, eles devem se mover juntos como um<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":734,"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-736","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/736","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=736"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/736\/revisions"}],"predecessor-version":[{"id":737,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/736\/revisions\/737"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/734"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=736"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}