{"id":2908,"date":"2017-04-20T05:37:57","date_gmt":"2017-04-20T05:37:57","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2908"},"modified":"2024-09-03T17:01:55","modified_gmt":"2024-09-03T17:01:55","slug":"resolucao-comentada-das-questoes-de-fisica-da-ufes-es-017","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/universidades-2017\/ufes-2017\/resolucao-comentada-das-questoes-de-fisica-da-ufes-es-017\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da UFES \u2013 ES \u2013 017"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da <\/span><\/span><\/span><\/p>\n

UFES \u2013 ES \u2013 017<\/span><\/span><\/span><\/p>\n

01- <\/span><\/span><\/span>(A) Pelo gr\u00e1fico, <\/b><\/span><\/span><\/span>enquanto o <\/b><\/span><\/span><\/span>tempo t <\/b><\/span><\/span><\/span>(considerando medido em minutos) <\/b><\/span><\/span><\/span>varia de 0 a 6min (\u2206t <\/b><\/span><\/span><\/span>= 6 <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u2013 0 = 6 minx60 <\/b><\/span><\/span><\/span>= 360 s<\/b><\/span><\/span><\/span>) a <\/b><\/span><\/span><\/span>temperatura varia de 20 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C para 50<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C <\/b><\/span><\/span><\/span>(<\/b><\/span><\/span><\/span>\u2206T <\/b><\/span><\/span><\/span>= 50 \u2013 20 = <\/b><\/span><\/span><\/span>30 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span>).<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da <\/b><\/span><\/span><\/span>quantidade de calor<\/b><\/span><\/span><\/span> (energia t\u00e9rmica) <\/b><\/span><\/span><\/span>W<\/b><\/span><\/span><\/span> <\/b>absorvida pela \u00e1gua <\/b><\/span><\/span><\/span>quando sua <\/b><\/span><\/span><\/span>temperatura varia <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>\u2206T = 30 <\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C, <\/b><\/span><\/span><\/span>sendo <\/b><\/span><\/span><\/span>dados m = 4 kg e c = 4,2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> J.kg<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>C<\/b><\/span><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/span>,utilizando a <\/b><\/span><\/span><\/span>equa\u00e7\u00e3o fundamental da calorimetria<\/b><\/span><\/span><\/span> \"\" <\/span>W = Q = m.c. \u2206T = 4.4,2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.30 \"\"<\/b><\/span><\/span><\/span> <\/b>W <\/b><\/span><\/span><\/span>= 504.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> <\/b>= 5,04.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> J.<\/b><\/span><\/span><\/span><\/p>\n

Essa <\/b><\/span><\/span><\/span>energia (W = 5,04.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> J) <\/b><\/span><\/span><\/span>foi <\/b><\/span><\/span><\/span>totalmente fornecida pelo resistor <\/b><\/span><\/span><\/span>de <\/b><\/span><\/span><\/span>pot\u00eancia P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> <\/b>no <\/b><\/span><\/span><\/span>intervalo<\/b><\/span><\/span><\/span> de <\/b><\/span><\/span><\/span>tempo de \u2206t = 360 s \"\"<\/b><\/span><\/span><\/span> <\/b>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> <\/b>= \"\" \"\" P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 0,014.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> W \"\" <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 1400 W ou 1,4 kW.<\/b><\/span><\/span><\/span><\/p>\n

B) <\/b><\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>pot\u00eancia de P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 1400 W <\/b><\/span><\/span><\/span>\u00e9 a mesma da <\/b><\/span><\/span><\/span>bateria ideal <\/b><\/span><\/span><\/span>(de resist\u00eancia interna nula) e de <\/b><\/span><\/span><\/span>fem U <\/b><\/span><\/span><\/span>fornecida<\/b><\/span><\/span><\/span> <\/b>ao <\/b><\/span><\/span><\/span>resistor de resist\u00eancia R = 3,5 \u2126 \"\" P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span>1400 = \"\" \"\" <\/b><\/span><\/span><\/span>U = \"\" = \"\" \"\" <\/b><\/span><\/span><\/span>U = 70 V.<\/b><\/span><\/span><\/span><\/p>\n

C) Como a <\/b><\/span><\/span><\/span>taxa de aquecimento \u00e9 constante<\/b><\/span><\/span><\/span> voc\u00ea pode utilizar uma <\/b><\/span><\/span><\/span>regra de tr\u00eas:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

02- <\/span><\/span><\/span>A) Veja a equa\u00e7\u00e3o de Gauss abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sendo a <\/b><\/span><\/span><\/span>lente a mesma, <\/b><\/span><\/span><\/span>a <\/b><\/span><\/span><\/span>dist\u00e2ncia focal f tamb\u00e9m \u00e9 a mesma <\/b><\/span><\/span><\/span>e a <\/b><\/span><\/span><\/span>amplia\u00e7\u00e3o A <\/b><\/span><\/span><\/span>(aumento linear transversal) \u00e9 fornecida pela <\/b><\/span><\/span><\/span>express\u00e3o A = – \"\" <\/b><\/span><\/span><\/span>que mostra que <\/b><\/span><\/span><\/span>A depende apenas das dist\u00e2ncias do objeto e da imagem \u00e0 lente.<\/b><\/span><\/span><\/span><\/p>\n

Se a <\/b><\/span><\/span><\/span>amplia\u00e7\u00e3o fosse a mesma nas duas posi\u00e7\u00f5es, P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> seria igual \u00e0 P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u2019, P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> seria igual \u00e0 P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u2019 <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>\u2502A\u2502= \u25021\u2502(objeto e imagem teriam o mesmo tamanho)<\/b><\/span><\/span><\/span>. <\/b><\/span><\/span><\/span><\/p>\n

Observe na <\/b><\/span><\/span><\/span>figura <\/b><\/span><\/span><\/span>que, com o <\/b><\/span><\/span><\/span>objeto mais pr\u00f3ximo da lente P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u2019 > P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> e <\/b><\/span><\/span><\/span>\u2502A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u2502> \u25021\u2502(tamanho aumentado).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Da mesma maneira<\/b><\/span><\/span><\/span>, com o objeto mais afastado da lente <\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u2019 < P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> e \u2502A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u2502< \u25021\u2502(tamanho diminu\u00eddo).<\/b><\/span><\/span><\/span><\/p>\n

B) <\/b><\/span><\/span><\/span>Est\u00e1 sendo <\/b><\/span><\/span><\/span>pedida <\/b><\/span><\/span><\/span>a dist\u00e2ncia entre <\/b><\/span><\/span><\/span>o objeto e a lente<\/b><\/span><\/span><\/span> na situa\u00e7\u00e3o em que est\u00e3o mais pr\u00f3ximos entre si (posi\u00e7\u00e3o P), ou seja, est\u00e1 pedindo P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Nessas condi\u00e7\u00f5es a <\/b><\/span><\/span><\/span>dist\u00e2ncia do objeto \u00e0 lente ser\u00e1 P<\/b><\/span><\/span><\/span>1 <\/b><\/span><\/span><\/sub><\/span>e da <\/b><\/span><\/span><\/span>imagem \u00e0 lente P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u2019 = L \u2013 P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 10 \u2013 P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> <\/b>e na<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

outra posi\u00e7\u00e3o<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>dist\u00e2ncia do objeto \u00e0 lente ser\u00e1 P<\/b><\/span><\/span><\/span>2 <\/b><\/span><\/span><\/sub><\/span>e <\/b><\/span><\/span><\/span>da imagem \u00e0 lente P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u2019 = L \u2013 P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 10 \u2013 P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Observe ainda, na <\/b><\/span><\/span><\/span>figura acima, que P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> \u2013 P<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 6 (I). <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

03- <\/span><\/span><\/span><\/p>\n

A) <\/b><\/span><\/span><\/span>Colocando o <\/b><\/span><\/span><\/span>polo <\/b><\/span><\/span><\/span>(eixo de rota\u00e7\u00e3o) no <\/b><\/span><\/span><\/span>ponto O<\/b><\/span><\/span><\/span>, e estabelecendo o <\/b><\/span><\/span><\/span>sentido de tend\u00eancia<\/b><\/span><\/span><\/span> de <\/b><\/span><\/span><\/span>rota\u00e7\u00e3o em torno de O como positivo para o hor\u00e1rio e negativo para o anti hor\u00e1rio, <\/b><\/span><\/span><\/span>vamos <\/b><\/span><\/span><\/span>calcular<\/b><\/span><\/span><\/span> o <\/b><\/span><\/span><\/span>momento (torque) de cada for\u00e7a em rela\u00e7\u00e3o a O,<\/b><\/span><\/span><\/span> sendo <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> e d<\/b><\/span><\/span><\/span>m <\/b><\/span><\/span><\/sub><\/span>as dist\u00e2ncias entre o ponto de aplica\u00e7\u00e3o das for\u00e7as O at\u00e9 as for\u00e7as F<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span> e F<\/b><\/span><\/span><\/span>m:<\/b><\/span><\/span><\/sub><\/span> <\/b><\/p>\n

M<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span><\/span> P.d = P.0 <\/b><\/span><\/span><\/span><\/span>= 0<\/b><\/span><\/span><\/span><\/span><\/p>\n

M<\/b><\/span><\/span><\/span><\/span>FP<\/b><\/span><\/span><\/span><\/sub><\/span> <\/b>= + F<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span> <\/b><\/p>\n

M<\/b><\/span><\/span><\/span><\/span>Fm<\/b><\/span><\/span><\/span><\/sub><\/span> = – F<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span> <\/b><\/p>\n

A <\/b><\/span><\/span><\/span><\/span>condi\u00e7\u00e3o de equil\u00edbrio de rota\u00e7\u00e3o \u00e9 de que a soma dos momentos (torques) de cada for\u00e7a seja nul<\/b><\/span><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\" 0 + F<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span><\/span>P <\/b><\/span><\/span><\/span><\/sub><\/span>\u2013 F<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span> = <\/b><\/span><\/span><\/span><\/span>0 \"\" F<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span><\/span>P <\/b><\/span><\/span><\/span><\/sub><\/span>= F<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span> = F<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

B) O momento (torque) exercido pelo prego sobre o martelo (e vice versa) \u00e9 fornecido e tem intensidade M<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span> = F<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span> = 30 Nm<\/b><\/span><\/span><\/span><\/span> com <\/b><\/span><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span> = 0,2 m, <\/b><\/span><\/span><\/span><\/span>que substituidos em <\/b><\/span><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span> = F<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>\/d<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span> <\/b>fornrcem<\/b><\/span><\/span><\/span><\/span> \"\"<\/b><\/span><\/span><\/span><\/span> <\/b>F<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span> = 150 N<\/b><\/span><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

C) <\/b><\/span><\/span><\/span><\/span>Aqu\u00ed o <\/b><\/span><\/span><\/span><\/span>peso<\/b><\/span><\/span><\/span><\/span> de intensidade <\/b><\/span><\/span><\/span><\/span>P <\/b><\/span><\/span><\/span><\/span>participa e sua <\/b><\/span><\/span><\/span><\/span>dist\u00e2ncia ao polo \u00e9 d<\/b><\/span><\/span><\/span><\/span>, e as <\/b><\/span><\/span><\/span><\/span>tr\u00eas for\u00e7as <\/b><\/span><\/span><\/span><\/span>que influem na <\/b><\/span><\/span><\/span><\/span>rota\u00e7\u00e3o <\/b><\/span><\/span><\/span><\/span>tem <\/b><\/span><\/span><\/span><\/span>intensidades P, F<\/b><\/span><\/span><\/span><\/span>p<\/b><\/span><\/span><\/span><\/sub><\/span> e F<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>, <\/b><\/span><\/span><\/span><\/span>cujos <\/b><\/span><\/span><\/span><\/span>momentos (torques) <\/b><\/span><\/span><\/span><\/span>em rela\u00e7\u00e3o \u00e0 O, <\/b><\/span><\/span><\/span><\/span>estabelecendo o <\/b><\/span><\/span><\/span>sentido de tend\u00eancia<\/b><\/span><\/span><\/span> de <\/b><\/span><\/span><\/span>rota\u00e7\u00e3o em torno de O como positivo para o hor\u00e1rio e negativo para o anti hor\u00e1rio, <\/b><\/span><\/span><\/span>ser\u00e3o:<\/b><\/span><\/span><\/span><\/span><\/p>\n

M<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span> = – P.d<\/b><\/span><\/span><\/span><\/span><\/p>\n

M<\/b><\/span><\/span><\/span><\/span>Fm<\/b><\/span><\/span><\/span><\/sub><\/span> = – F<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span><\/span>m <\/b><\/span><\/span><\/span><\/sub><\/span>= <\/b><\/span><\/span><\/span><\/span>– F<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>.20d <\/b><\/span><\/span><\/span><\/span>= – 20dF<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span> <\/b><\/p>\n

M<\/b><\/span><\/span><\/span><\/span>FP<\/b><\/span><\/span><\/span><\/sub><\/span> <\/b>= + F<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>.d<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span> = <\/b><\/span><\/span><\/span><\/span>+ F<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>.4d <\/b><\/span><\/span><\/span><\/span>= + 4dF<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span><\/p>\n

No <\/b><\/span><\/span><\/span><\/span>equil\u00edbrio de rota\u00e7\u00e3o a soma desses momentos <\/b><\/span><\/span><\/span><\/span>deve ser <\/b><\/span><\/span><\/span><\/span>nula<\/b><\/span><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span><\/span>– P.d – 20dF<\/b><\/span><\/span><\/span><\/span>m <\/b><\/span><\/span><\/span><\/sub><\/span>+ 4dF<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span> = 0 \"\" <\/b><\/span><\/span><\/span><\/span>20F<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span> = P \u2013 4<\/b><\/span><\/span><\/span><\/span>FP \"\" <\/b><\/span><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span> =<\/b><\/span><\/span><\/span><\/span> (P \u2013 4F<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>\/20 = \"\"<\/b><\/span><\/span><\/span><\/span> – \"\" <\/b><\/span><\/span><\/span><\/span><\/p>\n

04- <\/span><\/span><\/span>A) <\/b><\/span><\/span><\/span>A <\/b><\/span><\/span><\/span>energia interna de certa massa de um g\u00e1s<\/b><\/span><\/span><\/span> <\/b>perfeito<\/b><\/span><\/span><\/span> \u00e9 <\/b><\/span><\/span><\/span>fun\u00e7\u00e3o exclusiva da temperatura <\/b><\/span><\/span><\/span>desse g\u00e1s<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>\u00a0<\/span>Aumento de temperatura\u00a0<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>(\u0394T > 0) \u00a0\"\"\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>aumento de energia interna\u00a0<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>(\u0394U > 0).<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Diminui\u00e7\u00e3o de temperatura\u00a0<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>(\u0394T < 0)\u00a0<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>diminui\u00e7\u00e3o de energia interna\u00a0<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>(\u0394U<0).<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Temperatura constante<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0(\u0394T = 0)\u00a0<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>energia interna constante\u00a0<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>(\u0394U = 0).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Vamos ent\u00e3o <\/b><\/span><\/span>calcular a varia\u00e7\u00e3o da energia interna em cada transforma\u00e7\u00e3o:<\/b><\/span><\/span><\/span><\/span><\/p>\n

I: E \"\"\"\"\"\"<\/b><\/span><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" 1.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>.0,6\/T<\/b><\/span><\/span>E<\/b><\/span><\/sub><\/span> = 3.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>.0,2\/T<\/b><\/span><\/span>F<\/b><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span>6.10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span> \/ T<\/b><\/span><\/span>E<\/b><\/span><\/sub><\/span> = 6.10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span>\/T<\/b><\/span><\/span>F<\/b><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span> T<\/b><\/span><\/span>E<\/b><\/span><\/sub><\/span> = T<\/b><\/span><\/span>F<\/b><\/span><\/sub><\/span> \"\" energia interna constante.<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

II:<\/b><\/span><\/span><\/span> <\/b>F \"\"<\/b><\/span><\/span> <\/b>3.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>.0,2\/T<\/b><\/span><\/span>F<\/b><\/span><\/sub><\/span> = 6.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>.0,6\/T<\/b><\/span><\/span>H <\/b><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span>6.10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span>\/T<\/b><\/span><\/span>E<\/b><\/span><\/sub><\/span> = 6.10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span>\/T<\/b><\/span><\/span>F<\/b><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span> 6.10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span>T<\/b><\/span><\/span>G<\/b><\/span><\/sub><\/span> = 12.10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span>T<\/b><\/span><\/span>F<\/b><\/span><\/sub><\/span> \"\" T<\/b><\/span><\/span>F<\/b><\/span><\/sub><\/span> > T<\/b><\/span><\/span>G <\/b><\/span><\/sub><\/span> \"\" T<\/b><\/span><\/span>G<\/b><\/span><\/sub><\/span> = 2T<\/b><\/span><\/span>F<\/b><\/span><\/sub><\/span> \"\" energia interna n\u00e3o constante.<\/b><\/span><\/span><\/p>\n

III:<\/b><\/span><\/span><\/span> <\/b>G \"\"<\/b><\/span><\/span> <\/b>6.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>.0,2\/T<\/b><\/span><\/span>G<\/b><\/span><\/sub><\/span> = 4.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>.0,6\/T<\/b><\/span><\/span>H <\/b><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span>12.10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span>\/T<\/b><\/span><\/span>G<\/b><\/span><\/sub><\/span> = 24.10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span>\/T<\/b><\/span><\/span>H <\/b><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span> 24.10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span>T<\/b><\/span><\/span>G<\/b><\/span><\/sub><\/span> = 12.10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span>T<\/b><\/span><\/span>H <\/b><\/span><\/sub><\/span> \"\" T<\/b><\/span><\/span>H<\/b><\/span><\/sub><\/span> = 2T<\/b><\/span><\/span>H<\/b><\/span><\/sub><\/span> \"\" T<\/b><\/span><\/span>H<\/b><\/span><\/sub><\/span> \"\" T<\/b><\/span><\/span>G<\/b><\/span><\/sub><\/span> \"\" energia interna n\u00e3o constante.<\/b><\/span><\/span><\/p>\n

IV: <\/b><\/span><\/span><\/span>H \"\"\"\"\"\"<\/b><\/span><\/span> <\/b>4.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>.0,2\/T<\/b><\/span><\/span>H <\/b><\/span><\/sub><\/span> = 1.10<\/b><\/span><\/span>5<\/b><\/span><\/sup><\/span>.0,6\/T<\/b><\/span><\/span>E <\/b><\/span><\/sub><\/span> \"\" <\/b><\/span><\/span>8.10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span>\/T<\/b><\/span><\/span>H<\/b><\/span><\/sub><\/span> = 6.10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span>\/T<\/b><\/span><\/span>E <\/b><\/span><\/sub><\/span> \"\"<\/b><\/span><\/span> 8.10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span>T<\/b><\/span><\/span>E<\/b><\/span><\/sub><\/span> = 6.10<\/b><\/span><\/span>4<\/b><\/span><\/sup><\/span>T<\/b><\/span><\/span>H <\/b><\/span><\/sub><\/span> \"\" T<\/b><\/span><\/span>E<\/b><\/span><\/sub><\/span> \"\" T<\/b><\/span><\/span>H<\/b><\/span><\/sub><\/span> \"\" energia interna n\u00e3o constante.<\/b><\/span><\/span><\/p>\n

R- A energia interna \u2206U \u00e9 constante apenas em I.<\/b><\/span><\/span><\/p>\n

B) <\/b><\/span><\/span>Em <\/b><\/span><\/span>I n\u00e3o ocorre varia\u00e7\u00e3o <\/b><\/span><\/span>de <\/b><\/span><\/span>energia interna.<\/b><\/span><\/span><\/p>\n

Em <\/b><\/span><\/span>II e III houve aumento de temperatura <\/b><\/span><\/span>e, consequentemente <\/b><\/span><\/span>aumento de energia interna.<\/b><\/span><\/span><\/p>\n

Em <\/b><\/span><\/span>IV <\/b><\/span><\/span>houve <\/b><\/span><\/span>diminui\u00e7\u00e3o de temperatura <\/b><\/span><\/span>e, consequentemente <\/b><\/span><\/span>diminui\u00e7\u00e3o de energia interna.<\/b><\/span><\/span><\/p>\n

O aumento de enet=rgia interna ocorreu apenas em II e III e, como a energia interna \u00e9 diretamente proporcional \u00e0 temperatura absoluta, de A:<\/b><\/span><\/span><\/p>\n

II. T<\/b><\/span><\/span>G<\/b><\/span><\/sub><\/span> = 2T<\/b><\/span><\/span>F <\/b><\/span><\/sub><\/span> \"\" \"\"<\/b><\/span><\/span> <\/b>= 2<\/b><\/span><\/span><\/p>\n

III. T<\/b><\/span><\/span>H<\/b><\/span><\/sub><\/span> = 2T<\/b><\/span><\/span>G <\/b><\/span><\/sub><\/span> \"\" \"\"<\/b><\/span><\/span> <\/b>= 2<\/b><\/span><\/span><\/p>\n

C) Veja informa\u00e7\u00e3o abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

No caso doe exerc\u00edcio vamos calcular a \u00e1rea do ciclo lembrando que nos trechos FG e HE o trabalho \u00e9 nulo (transforma\u00e7\u00e3o isovlum\u00e9trica). <\/b><\/span><\/span><\/p>\n

C\u00e1lculo do trabalho W no trecho GH onde ele fornecido pela \u00e1rea do trap\u00e9zio e \u00e9 positivo (sentido hor\u00e1rio):<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

C\u00e1lculo do trabalho W no trecho EF onde ele \u00e9fornecido pela \u00e1rea do trap\u00e9zio e \u00e9 negativo (sentido anti-hor\u00e1rio):<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

W =<\/b><\/span><\/span><\/span> 2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> \u2013 0,8.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> <\/b>= 1,2.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> J<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = \"\" \"\"<\/b><\/span><\/span><\/span> P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = \"\" = 0,02.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span> \"\" P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 2.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span> W \"\" como <\/b><\/span><\/span><\/span>s\u00e3o 50 ciclos \"\"<\/b><\/span><\/span><\/span> P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 50.2.10<\/b><\/span><\/span><\/span>3 <\/b><\/span><\/span><\/sup><\/span> \"\"<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span> = 100 kW.<\/b><\/span><\/span><\/span><\/p>\n

05- <\/span><\/span><\/span>A) S\u00e3o dados: frequ\u00eancia<\/b><\/span><\/span><\/span> dos f\u00f3tons <\/b><\/span><\/span><\/span>f = 1,2.10<\/b><\/span><\/span><\/span>15<\/b><\/span><\/span><\/sup><\/span> Hz <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>velocidade da luz no v\u00e1cuo c = 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> m\/s \"\"<\/b><\/span><\/span><\/span> equa\u00e7\u00e3o fundamental da ondulat\u00f3ria <\/b><\/span><\/span><\/span>V = c = <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span>f \"\" 3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span> = <\/b><\/span><\/span><\/span>\u0001.1,2.10<\/b><\/span><\/span><\/span>15<\/b><\/span><\/span><\/sup><\/span> \"\" \u0001 = \"\" = 2,5.10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span> = 250.10<\/b><\/span><\/span><\/span>-9<\/b><\/span><\/span><\/sup><\/span> m \"\" <\/b><\/span><\/span><\/span>\u0001<\/b><\/span><\/span><\/span> = 250 nm.<\/b><\/span><\/span><\/span><\/p>\n

Energia de cada f\u00f3ton \"\"<\/b><\/span><\/span><\/span> <\/b>W = h.f <\/b><\/span><\/span><\/span>com <\/b><\/span><\/span><\/span>h <\/b><\/span><\/span><\/span>= constante de Planck <\/b><\/span><\/span><\/span>= 6,6.10<\/b><\/span><\/span><\/span>-34<\/b><\/span><\/span><\/sup><\/span> J.s \"\"<\/b><\/span><\/span><\/span> W = 6,6.10<\/b><\/span><\/span><\/span>-34<\/b><\/span><\/span><\/sup><\/span>.1,2. 10<\/b><\/span><\/span><\/span>15<\/b><\/span><\/span><\/sup><\/span> \"\" <\/b><\/span><\/span><\/span>W = 7,92.10<\/b><\/span><\/span><\/span>-19 <\/b><\/span><\/span><\/sup><\/span>J.<\/b><\/span><\/span><\/span><\/p>\n

1 ev = 1,6.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span> J \"\"<\/b><\/span><\/span><\/span> W = 7,92.10<\/b><\/span><\/span><\/span>-19<\/b><\/span><\/span><\/sup><\/span>x\"\" \"\" <\/b><\/span><\/span><\/span>W = 4,95 eV.<\/b><\/span><\/span><\/span><\/p>\n

B) Consultando<\/b><\/span><\/span><\/span> a <\/b><\/span><\/span><\/span>tabela <\/b><\/span><\/span><\/span>voc\u00ea verifica que <\/b><\/span><\/span><\/span>no<\/b><\/span><\/span><\/span> <\/b>tempo m\u00e1ximo de exposi\u00e7\u00e3o \u2206t = 6,00 s o n\u00edvel de irradia\u00e7\u00e3o vale I<\/b><\/span><\/span><\/span>ef<\/b><\/span><\/span><\/sub><\/span> = 4,95.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span> Wcm<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Como, pelo <\/b><\/span><\/span><\/span>enunciado a irradia\u00e7\u00e3o (I<\/b><\/span><\/span><\/span>ef<\/b><\/span><\/span><\/sub><\/span>) \u00e9 definida como pot\u00eancia (P) incidente por unidade de \u00e1rea (S) \"\"<\/b><\/span><\/span><\/span> <\/b>I<\/b><\/span><\/span><\/span>ef<\/b><\/span><\/span><\/sub><\/span> = \"\" = \"\"<\/b><\/span><\/span><\/span> \"\" I<\/b><\/span><\/span><\/span>ef<\/b><\/span><\/span><\/sub><\/span> = \"\" \"\" W = I<\/b><\/span><\/span><\/span>ef<\/b><\/span><\/span><\/sub><\/span>.S.\u2206t = 4,95.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>.1.6 \"\"<\/b><\/span><\/span><\/span> W <\/b><\/span><\/span><\/span>= 29,7.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span> <\/b>= 2,97.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span> J (energia<\/b><\/span><\/span><\/span> <\/b>total<\/b><\/span><\/span><\/span> absorvida em <\/b><\/span><\/span><\/span>cada 1 cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> de \u00e1rea de pele em \u2206t = 6,00 s).<\/b><\/span><\/span><\/span><\/p>\n

Como \u00e9 <\/b><\/span><\/span><\/span>pedido o n\u00famero n de f\u00f3tons, <\/b><\/span><\/span><\/span>cada um <\/b><\/span><\/span><\/span>com energia W<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span> = 7,92.10<\/b><\/span><\/span><\/span>-19 <\/b><\/span><\/span><\/sup><\/span>J, <\/b><\/span><\/span><\/span>voc\u00ea ter\u00e1 <\/b><\/span><\/span><\/span>n = \"\" <\/b><\/span><\/span><\/span>= \"\" \"\" <\/b><\/span><\/span><\/span>n = 3,75.10<\/b><\/span><\/span><\/span>15 <\/b><\/span><\/span><\/sup><\/span>f\u00f3tons.<\/b><\/span><\/span><\/span> <\/b><\/p>\n

C) <\/b><\/span><\/span><\/span>Utilizando <\/b><\/span><\/span><\/span>a mesma equa\u00e7\u00e3o de (B), mas agora com a nova \u00e1rea S = 2 m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> = 2.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span> cm<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> \"\"<\/b><\/span><\/span><\/span> <\/b>W = I<\/b><\/span><\/span><\/span>ef<\/b><\/span><\/span><\/sub><\/span>.S.\u2206t <\/b><\/span><\/span><\/span>= 4,95.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>.2.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>.6 \"\" <\/b><\/span><\/span><\/span>W = 59,4 J<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA da UFES \u2013 ES \u2013 017 01- (A) Pelo gr\u00e1fico, enquanto o tempo t (considerando medido em minutos) varia de 0 a 6min (\u2206t = 6 \u2013 0 = 6 minx60 = 360 s) a temperatura varia de 20 oC para 50oC (\u2206T = 50 \u2013 20 = 30 oC). C\u00e1lculo da quantidade de<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2906,"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-2908","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2908","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=2908"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2908\/revisions"}],"predecessor-version":[{"id":2909,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2908\/revisions\/2909"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2906"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2908"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}