{"id":704,"date":"2015-05-12T18:37:44","date_gmt":"2015-05-12T18:37:44","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=704"},"modified":"2024-09-03T12:57:49","modified_gmt":"2024-09-03T12:57:49","slug":"resolucao-comentada-da-unespar-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2015\/exercicios-resolvidos-da-unespar-2015\/resolucao-comentada-da-unespar-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada da UNESPAR 2015"},"content":{"rendered":"

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

01- <\/b><\/span><\/span><\/span>)<\/b><\/span><\/span><\/span> Na <\/b><\/span><\/span><\/span>irradia\u00e7\u00e3o<\/b><\/span><\/span><\/span>, para que a energia t\u00e9rmica se propague \u00e9 necess\u00e1rio <\/b><\/span><\/span><\/span>existir um meio material.<\/b><\/span><\/span><\/span><\/p>\n

Irradia\u00e7\u00e3o ou radia\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/p>\n

\u00c9 o fen\u00f4meno pelo qual a energia emitida pelo Sol chega at\u00e9 a Terra <\/b><\/span><\/span><\/span>atrav\u00e9s do v\u00e1cuo<\/b><\/span><\/span><\/span>. Realiza-se <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

eletromagn\u00e9ticas que s\u00e3o compostas por diversas ondas de freq\u00fc\u00eancias diferentes (raios c\u00f3smicos, raios X, raios ultravioleta, luz vis\u00edvel, raios infravermelhos, microondas, etc.), chamadas radia\u00e7\u00f5es t\u00e9rmicas. Qualquer corpo que possua temperatura superior ao zero absoluto (0K ou -273<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C) emite energia radiante e as que se transformam mais facilmente em calor quando absorvidas pelo receptor s\u00e3o as infravermelhas, denominadas tamb\u00e9m de ondas de calor.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0A <\/b><\/span><\/span><\/span>radia\u00e7\u00e3o <\/b><\/span><\/span><\/span>ocorre tamb\u00e9m no ar, como voc\u00ea pode observar na figura abaixo onde a m\u00e3o est\u00e1 recebendo calor por irradia\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

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

(F )<\/b><\/span><\/span><\/span> A condu\u00e7\u00e3o acontece quando a troca de energia t\u00e9rmica \u00e9 feita <\/b><\/span><\/span><\/span>sem a presen\u00e7a de um meio material.<\/b><\/span><\/span><\/span><\/p>\n

Condu\u00e7\u00e3o t\u00e9rmica<\/b><\/span><\/span><\/span><\/p>\n

\u00a0O calor \u00e9 conduzido de um ponto a outro do corpo (meio material) <\/b><\/span><\/span><\/span>sem que haja deslocamento das part\u00edculas.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Explicando microscopicamente o fen\u00f4meno: a regi\u00e3o pr\u00f3xima da chama tem o movimento vibrat\u00f3rio de suas mol\u00e9culas aumentado, adquirindo assim maior energia cin\u00e9tica, que \u00e9 transferida atrav\u00e9s de choques \u00e0s part\u00edculas vizinhas, que tamb\u00e9m aumentam seu movimento vibrat\u00f3rio. Atrav\u00e9s desse transporte de energia, toda a barra \u00e9 aquecida.<\/b><\/span><\/span><\/span><\/p>\n

(V )<\/b><\/span><\/span><\/span> A <\/b><\/span><\/span><\/span>convec\u00e7\u00e3o <\/b><\/span><\/span><\/span>\u00e9 o processo de transmiss\u00e3o de calor em que a energia t\u00e9rmica se propaga em meios fluidos (l\u00edquidos e gases) atrav\u00e9s de correntes de convec\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

Convec\u00e7\u00e3o t\u00e9rmica<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Trata-se da transfer\u00eancia de energia t\u00e9rmica (calor) pela mat\u00e9ria em movimento devido \u00e0 diferen\u00e7a de densidades dessa mat\u00e9ria. Essa mat\u00e9ria s\u00f3 pode ser fluida (l\u00edquido, g\u00e1s ou vapor). Observe a figura onde se tem tubos de vidro contendo \u00e1gua l\u00edquida.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

A \u00e1gua em A, aquecida, expande-se, fica menos densa, mais leve e sobe o trecho AB. A \u00e1gua mais fria, mais densa e mais pesada do trecho CD desce e ainda empurra para cima a \u00e1gua quente do ramo AB. Obt\u00e9m-se ent\u00e3o uma circula\u00e7\u00e3o de \u00e1gua na qual a transfer\u00eancia de calor \u00e9 feita\u00a0 atrav\u00e9s da mat\u00e9ria (no caso, \u00e1gua) e que recebe o nome de<\/b><\/span><\/span><\/span>corrente de convec\u00e7\u00e3o<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

( F)<\/b><\/span><\/span><\/span> A forma como ocorre a refrigera\u00e7\u00e3o de uma geladeira \u00e9 um exemplo de <\/b><\/span><\/span><\/span>condu\u00e7\u00e3o<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

O congelador deve ficar na parte superior da geladeira pois o ar frio mais denso (mais pesado) desce <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

e o ar quente menos denso (mais leve) sobe, originando no interior da geladeira, correntes de convec\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

( V)<\/b><\/span><\/span><\/span> Sentimos o calor do Sol devido as radia\u00e7\u00f5es eletromagn\u00e9ticas que ele emite e que atravessam o v\u00e1cuo existente entre o Sol e a Terra, isto \u00e9 um exemplo de Irradia\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

Veja a primeira. <\/b><\/span><\/span><\/span><\/p>\n

a)<\/b><\/span><\/span><\/span> F V F F V; <\/b><\/span><\/span><\/span>b)<\/b><\/span><\/span><\/span> F F V F V; <\/b><\/span><\/span><\/span>c)<\/b><\/span><\/span><\/span> V F V F V; <\/b><\/span><\/span><\/span>d) <\/b><\/span><\/span><\/span>F F V V V; <\/b><\/span><\/span><\/span>e)<\/b><\/span><\/span><\/span> V V F V F.<\/b><\/span><\/span><\/span><\/p>\n

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

02- <\/b><\/span><\/span><\/span>P = i.U<\/b><\/span><\/span><\/span> \"\" 80 = i.220 \"\" <\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span> = 80\/220 <\/b><\/span><\/span><\/span>= 0,36 A.<\/b><\/span><\/span><\/span><\/p>\n

P = W\/t<\/b><\/span><\/span><\/span> \"\" 80 = W\/5 \"\" <\/b><\/span><\/span><\/span>W <\/b><\/span><\/span><\/span>= 5.80 = 400 Wh <\/b><\/span><\/span><\/span>= 0,4 kWh.<\/b><\/span><\/span><\/span><\/p>\n

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

03- <\/b><\/span><\/span><\/span>F = K.q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 9.10<\/b><\/span><\/span><\/span>9<\/b><\/span><\/span><\/sup><\/span>.2.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>.6.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 108.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>\/4 = 27.10<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span> N \"\" F<\/b><\/span><\/span><\/span>el\u00e9trica<\/b><\/span><\/span><\/sub><\/span> = 2,7 N.<\/b><\/span><\/span><\/span><\/p>\n

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

04- <\/b><\/span><\/span><\/span>Veja na figura abaixo a sequ\u00eancia do c\u00e1lculo da resist\u00eancia do resistor equivalente <\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/sub><\/span> = 175\u2126<\/b><\/span><\/span><\/span> <\/b><\/p>\n

\"\"<\/p>\n

\"\" R<\/b><\/span><\/span><\/span><\/span>eq<\/b><\/span><\/span><\/span><\/sub><\/span> = U\/i<\/b><\/span><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span><\/sub><\/span> \"\" 175 = 20\/i<\/b><\/span><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span><\/span>i<\/b><\/span><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span><\/sub><\/span> = 0,114 A<\/b><\/span><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/span>t<\/b><\/span><\/span><\/span><\/sub><\/span> <\/b>= i.U = 0,114.20 <\/b><\/span><\/span><\/span><\/span>= 2,28 W.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

05- <\/b><\/span><\/span><\/span>Energia mec\u00e2nica em A<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh<\/b><\/span><\/span><\/span> = 23.2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 23.10.2,65 = 46 + 609,5 \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = 655,5J.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Energia mec\u00e2nica em B<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> = mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + mgh<\/b><\/span><\/span><\/span> = 23.V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 23.10.0,4 = 11,5V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 92 \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> = 11,5V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 92.<\/b><\/span><\/span><\/span> Como <\/b><\/span><\/span><\/span>n\u00e3o existe atrito<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span>mA<\/b><\/span><\/span><\/sub><\/span> = E<\/b><\/span><\/span><\/span>mB<\/b><\/span><\/span><\/sub><\/span> \"\" 655,5 = 11,5V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 92 \"\" V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 563,5\/11,5 \"\" V = \u221a49 \"\" <\/b><\/span><\/span><\/span>V = 7m\/s.<\/b><\/span><\/span><\/span><\/p>\n

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

06- <\/b><\/span><\/span><\/span>f=R\/2 = 20\/2 = 10 cm \"\" P = 15 cm \"\" o = 5 cm.<\/b><\/span><\/span><\/span><\/p>\n

1\/f = 1\/P + 1\/P\u2019 \"\" 1\/10 = 1\/15 + 1\/P\u2019 \"\" (3 \u2013 2)\/30 = 1\/P\u2019 \"\" <\/b><\/span><\/span><\/span>P\u2019 = 30 cm (dist\u00e2ncia da imagem ao v\u00e9rtice do espelho e \u00e9 real, pois P\u2019 > 0).<\/b><\/span><\/span><\/span><\/p>\n

i\/o = – P\u2019\/P \"\" i\/5 = – 30\/15 \"\" <\/b><\/span><\/span><\/span>i = – 10 cm (a imagem tem 10cm de altura, dobro da altura do objeto, e \u00e9 invertida, pois i < 0)<\/b><\/span><\/span><\/span><\/p>\n

2. Correta<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

4. Falsa.<\/b><\/span><\/span><\/span><\/p>\n

8. Falsa<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

16. Correta.<\/b><\/span><\/span><\/span><\/p>\n

32. Correta.<\/b><\/span><\/span><\/span><\/p>\n

64. Falsa, o centro de curvatura est\u00e1 a 20 cm do v\u00e9rtice do espelho<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span><\/p>\n

Soma (2 + 16 + 32) = 50<\/b><\/span><\/span><\/span><\/p>\n

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

07- <\/b><\/span><\/span><\/span>Menor densidade<\/b><\/span><\/span><\/span>, mais leve, fica mais em <\/b><\/span><\/span><\/span>cima<\/b><\/span><\/span><\/span>, (<\/b><\/span><\/span><\/span>podendo flutuar).<\/b><\/span><\/span><\/span><\/p>\n

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

08- <\/b><\/span><\/span><\/span>Trata-se do teorema do <\/b><\/span><\/span><\/span>Impulso:<\/b><\/span><\/span><\/span> “O <\/b><\/span><\/span><\/span>impulso da for\u00e7a resultante<\/b><\/span><\/span><\/span> de um sistema de for\u00e7as que age sobre um corpo \u00e9 <\/b><\/span><\/span><\/span>igual \u00e0 varia\u00e7\u00e3o da quantidade<\/b><\/span><\/span><\/span> de movimento do corpo” e da quantidade de movimento: <\/b><\/span><\/span><\/span>princ\u00edpio da conserva\u00e7\u00e3o da quantidade de movimento<\/b><\/span><\/span><\/span>:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

09- <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span> = d\/t \"\" 80 = d\/(45\/60) \"\" d = 45×80\/60 = 60 km.<\/b><\/span><\/span><\/span><\/p>\n

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

<\/h2>\n

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

RESOLU\u00c7\u00d5ES 01- ) Na irradia\u00e7\u00e3o, para que a energia t\u00e9rmica se propague \u00e9 necess\u00e1rio existir um meio material. Irradia\u00e7\u00e3o ou radia\u00e7\u00e3o \u00c9 o fen\u00f4meno pelo qual a energia emitida pelo Sol chega at\u00e9 a Terra atrav\u00e9s do v\u00e1cuo. Realiza-se eletromagn\u00e9ticas que s\u00e3o compostas por diversas ondas de freq\u00fc\u00eancias diferentes (raios c\u00f3smicos, raios X, raios ultravioleta, luz vis\u00edvel, raios infravermelhos, microondas,<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":702,"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-704","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/704","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=704"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/704\/revisions"}],"predecessor-version":[{"id":705,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/704\/revisions\/705"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/702"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=704"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}