{"id":2059,"date":"2016-03-27T01:56:43","date_gmt":"2016-03-27T01:56:43","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2059"},"modified":"2024-08-21T14:51:28","modified_gmt":"2024-08-21T14:51:28","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-dilatacao-linear-superficial-e-volumetrica-dos-solidos-e-dilatacao-dos-liquidos","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/fisica-termica\/dilatometria\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-dilatacao-linear-superficial-e-volumetrica-dos-solidos-e-dilatacao-dos-liquidos\/","title":{"rendered":"Dilata\u00e7\u00e3o linear, superficial e volum\u00e9trica dos s\u00f3lidos e dilata\u00e7\u00e3o dos l\u00edquidos."},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span><\/p>\n

dilata\u00e7\u00e3o linear, superficial e volum\u00e9trica dos s\u00f3lidos e dilata\u00e7\u00e3o dos l\u00edquidos<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como A est\u00e1 fixo, as dimens\u00f5es do sistema devem aumentar, pois est\u00e1 sendo aquecido e o ponto B deve afastar-se de A (veja a figura)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0R- B<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quando a temperatura aumenta (\u03b8 > 27 \u00baC) o alum\u00ednio se dilata mais e ela se curva para cima e quando a temperatura diminui (\u03b8 < 27 \u00baC) o alum\u00ednio se dilata menos e ela se curva para baixo\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

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

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo a barra de mesmo material (mesmo coeficiente de dilata\u00e7\u00e3o) e como sofre a mesma varia\u00e7\u00e3o de temperatura, todos os seus pontos sofrer\u00e3o a mesma dilata\u00e7\u00e3o e ela n\u00e3o ser\u00e1 deformada\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span> <\/b>R- D\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

05-\u00a0<\/b><\/span><\/span>A dilata\u00e7\u00e3o do alum\u00ednio \u00e9 o dobro da dilata\u00e7\u00e3o do a\u00e7o\u00a0 —\u00a0 veja figura<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que as l\u00e2minas est\u00e3o se encurvando para dentro, ent\u00e3o \u03b1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0> \u03b1<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e, quanto mais baixo estiver o contato (mais apertado o parafuso), menor ser\u00e1 a dilata\u00e7\u00e3o das l\u00e2minas e consequentemente menor ser\u00e1 a temperatura\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span> <\/b>R- B\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00c1gua fria contrai o interno e \u00e1gua quente dilata o externo\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo \u03b1<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0(azul) maior que \u03b1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0(vermelho), com o aquecimento, a l\u00e2mina azul se dilata mais que a vermelha e a espiral se fecha (contrai) fazendo o ponteiro girar no sentido hor\u00e1rio\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Para que a ponte permane\u00e7a sempre na horizontal, os dois pilares devem sofrer a mesma dilata\u00e7\u00e3o para a mesma varia\u00e7\u00e3o de temperatura\u00a0 —\u00a0 \u0394L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>= \u0394L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 L<\/b><\/span><\/span>01<\/b><\/span><\/span><\/sub>.\u03b1<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.\u0394t= L<\/b><\/span><\/span>02<\/b><\/span><\/span><\/sub>.\u03b1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.\u0394t\u00a0 —\u00a0 40.18.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=30.\u03b1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03b1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=24.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u0394L=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1.\u0394t=100.1,2.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>.20\u00a0 —\u00a0 \u0394L=2,4.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m=2,4 cm\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u0394L=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1.\u0394t=12.1,1.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>.40 = 528.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>=52.800.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0 \u00a0\u0394L=0,528 cm\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>c\u00e1lculo do coeficiente de dilata\u00e7\u00e3o da haste\u00a0 —\u00a0 \u0394L=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1.\u0394t\u00a0 —\u00a0 2.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>.\u03b1.1\u00a0 —\u00a0 \u03b1=2.10<\/b><\/span><\/span>-5o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u0394L=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1.\u0394t=<\/b><\/span><\/span><\/p>\n

800.2.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>.20\u00a0 ===\u00a0 \u0394L=32.000.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>=0,32mm\u00a0 —\u00a0 R-<\/b><\/span><\/span>\u00a0B<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u0394L = L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0. \u03b1 . \u0394\u03b8\u00a0 —\u00a0 \u0394L = 30 . (11 . 10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>) . (40 \u2013 10) = 99 . 10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0m\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u0394L=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1.\u0394t\u00a0 —\u00a0 15=1.000.\u03b1.(500 \u2013 0)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- \u03b1=30.10<\/b><\/span><\/span>-6 o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u0394L=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1.\u0394t\u00a0 —\u00a0 \u0394L=10.11.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>(2,7 \u2013 30)\u00a0 —\u00a0 \u0394L=110.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.(-27,3)\u00a0 —\u00a0 \u0394L= – 0,00303\u00a0 —\u00a0 L=10,00 \u2013 0,00303\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>L=9,99697cm<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a\u00e7o de 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C a 100<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 L<\/b><\/span><\/span>a\u00e7o<\/b><\/span><\/span><\/sub>=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>(1 + \u03b1.\u0394t)=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.12.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.100)\u00a0 —\u00a0 L<\/b><\/span><\/span>a\u00e7o<\/b><\/span><\/span><\/sub>=1,0012L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 vidro de 0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C a 100<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 L<\/b><\/span><\/span>vidro<\/b><\/span><\/span><\/sub>=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>(1 + \u03b1.\u0394t)=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.8.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.100)\u00a0 —\u00a0 L<\/b><\/span><\/span>a\u00e7o<\/b><\/span><\/span><\/sub>=1,0008L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 L<\/b><\/span><\/span>a\u00e7o<\/b><\/span><\/span><\/sub>\u00a0\u2013 L<\/b><\/span><\/span>vidro<\/b><\/span><\/span><\/sub>=0,1\u00a0 —\u00a0 1,0012L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 1,0008L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0,1\u00a0 —\u00a0 L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0,1\/0.0004 = 0,25.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=250 cm\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A reta A tem maior inclina\u00e7\u00e3o\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como o coeficiente de dilata\u00e7\u00e3o linear do lat\u00e3o \u00e9 maior que o do a\u00e7o, o lat\u00e3o se dilata mais quando aquecido e se contrai mais<\/b><\/span><\/span><\/p>\n

quando resfriado\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03b1=\u0394L\/L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u0394t=0,24\/2.200 \u00a0—\u03b1=0,24\/400\u00a0 —\u00a0 \u03b1=0,0006\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Para que o balan\u00e7o permane\u00e7a sempre na horizontal, independente da temperatura, eles devem, para qualquer varia\u00e7\u00e3o de temperatura, sofrer sempre a mesma dilata\u00e7\u00e3o \u0394L\u00a0 —\u00a0 \u0394L<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=\u0394L<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 L<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.1,5.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>.\u0394t=72.2.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>.\u0394t\u00a0 —\u00a0 L<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=144.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\/1,5.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

L<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=96cm\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quanto maior o coeficiente de dilata\u00e7\u00e3o mais o corpo se dilata quando aquecido e mais se contrais quando resfriado.<\/b><\/span><\/span><\/p>\n

(01) A se dilata mais que B\u00a0 —\u00a0\u00a0 Correta<\/b><\/span><\/span><\/p>\n

(02) Falsa\u00a0 —\u00a0 veja (01)<\/b><\/span><\/span><\/p>\n

(04) A folga diminuir\u00e1\u00a0 —\u00a0 Falsa<\/b><\/span><\/span><\/p>\n

(08) Possuem diferentes L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Falsa<\/b><\/span><\/span><\/p>\n

(16) Apenas a placa se dilatar\u00e1\u00a0 —\u00a0 Correta<\/b><\/span><\/span><\/p>\n

(32) Apenas a placa se dilatar\u00e1\u00a0 —\u00a0 Correta<\/b><\/span><\/span><\/p>\n

R- (01 + 16 + 32)=49\u00a0<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span> <\/b>R- A<\/b><\/span><\/span>\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como um metal se dilata quando se aquece a estrutura met\u00e1lica do lado direito do pr\u00e9dio passa a ter um comprimento maior do que a estrutura met\u00e1lica em seu lado esquerdo devido ao aquecimento provocado pelo inc\u00eandio que ocorreu no lado direito. Para que a altura do pr\u00e9dio medida em seu lado direito fique maior do que a medida pelo lado esquerdo, o pr\u00e9dio entortar\u00e1 necessariamente para o lado esquerdo, como indicado na figura 2.<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) \u03b1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=\u0394L<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/L<\/b><\/span><\/span>oA<\/b><\/span><\/span><\/sub>\u0394t<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=0,0022\/1,0022.100\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u03b1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=22.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u03b1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=\u0394L<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/B\u0394t<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=0,0011\/1,0011.100\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03b1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=11.10<\/b><\/span><\/span>-6o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/p>\n

b) \u03b1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/\u03b1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=22.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\/11.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u03b1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/\u03b1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=2<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Para que o circuito seja ligado com menor aumento de temperatura voc\u00ea deve escolher o que se dilata muito mais que o cobre, que no caso \u00e9 o alum\u00ednio\u00a0 —\u00a0 R- B<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Esfriou o conjunto, o alum\u00ednio tem maior coeficiente de dilata\u00e7\u00e3o e se contrai mais que o ferro.<\/b><\/span><\/span><\/p>\n

b) \u0394L<\/b><\/span><\/span>Al<\/b><\/span><\/span><\/sub>=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1<\/b><\/span><\/span>Al<\/b><\/span><\/span><\/sub>.\u0394t\u00a0 —\u00a0 \u0394L<\/b><\/span><\/span>Fe<\/b><\/span><\/span><\/sub>=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1<\/b><\/span><\/span>Fe<\/b><\/span><\/span><\/sub>.\u0394t\u00a0 —\u00a0 \u0394L<\/b><\/span><\/span>Al<\/b><\/span><\/span><\/sub>\/ \u0394L<\/b><\/span><\/span>Fe<\/b><\/span><\/span><\/sub>=24.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\/11.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u22482,2\u00a0 —<\/b><\/span><\/span>\u00a0 R- 2,2<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Deve-se mergulhar a tampa do frasco na \u00e1gua quente. O zinco ir\u00e1 dilatar mais que o\u00a0 vidro,\u00a0\u00a0soltando-se do gargalo.<\/b><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u0394V=0,006V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u00a0\u0394V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b3.\u0394t\u00a0 —\u00a0 0,006V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o.<\/b><\/span><\/span><\/sub>\u03b3.200\u00a0 —\u00a0 \u03b3=3.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u03b1\/1=\u03b3\/3\u00a0 —\u00a0 \u03b1\/1=3.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\/3\u00a0 —\u00a0 \u03b1=1,0.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

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

30-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a massa \u00e9 a mesma para qualquer temperatura\u00a0 —\u00a0 d=m\/V\u00a0 —\u00a0 volume inicial\u00a0 —\u00a0 10=m\/V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=m\/10\u00a0 —\u00a0 volume final\u00a0 —\u00a0 32<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>F=0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 10,03=m\/V\u00a0 —\u00a0 V=m\/10,03\u00a0 —32<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>F=0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.(1 + \u03b3(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>))\u00a0 —\u00a0 m\/10,03=(m\/10).(1 +\u00a0<\/b><\/span><\/span>\u03b3<\/b><\/span><\/span><\/span>(0 \u2013 100)\u00a0 —\u00a0 1\/10.03=1\/10 \u2013 100<\/b><\/span><\/span>\u03b3<\/b><\/span><\/span><\/span>\/10\u00a0 —\u00a0 1003<\/b><\/span><\/span>\u03b3<\/b><\/span><\/span><\/span>=0,03\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03b3<\/b><\/span><\/span><\/span>=2,99.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>=3.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03b1<\/b><\/span><\/span><\/span>\/1=<\/b><\/span><\/span>\u03b3<\/b><\/span><\/span><\/span>\/3\u00a0 —<\/b><\/span><\/span>\u03b1<\/b><\/span><\/span><\/span>\/1=3.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\/3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03b1<\/b><\/span><\/span><\/span>=1,0.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Tempo que a chapa demora para atingir a eleva\u00e7\u00e3o\u00a0 —\u00a0 \u0394S=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t + gt<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 11.045=0.t + 10.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 t=\u221a2.209\u00a0 —\u00a0 t=47s\u00a0 —\u00a0 at\u00e9 abrir o p\u00e1ra-quedas ele demorou 5 minutos=5.60=300s\u00a0 —\u00a0 \u0394t=300 \u2013 47=253s\u00a0 —\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>\u0394t=253s<\/b><\/span><\/span><\/p>\n

b) \u0394S=S<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b2.\u0394t=500.4,8.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>.(40 – -55)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u0394S=2,28cm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O orif\u00edcio se dilata na mesma propor\u00e7\u00e3o que a chapa, ou seja, de 4%\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1.\u0394t=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>.25\u00a0 —\u00a0 \u0394L=25.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u0394d=0,02d<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u0394d=d<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1.\u0394t\u00a0 —\u00a0 0,02.d<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=d<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1.100\u00a0 —\u00a0 \u03b1=2.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/p>\n

espessura\u00a0 —\u00a0 \u0394L=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u03b1\u0394t\u00a0 —\u00a0 \u0394L\/L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=\u03b1\u0394t=2.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>.100\u00a0 —\u00a0 \u0394L\/L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=2.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>=0,2%<\/b><\/span><\/span><\/p>\n

\u00e1rea\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>S=S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/span>\u03b2<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>S\/S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>=2\u03b1\u0394t=2.2.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/span><\/sup>.100\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>S\/S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>=4.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/span><\/sup>=0,4%<\/b><\/span><\/span><\/span><\/p>\n

volume\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/span>\u03b3<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>V\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>=<\/b><\/span><\/span><\/span>\u03b3<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>t =3.2.10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/span><\/sup>.100\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>V\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>=6.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/span><\/sup>=0,6%\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

35-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) \u0394L=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1.\u0394T\u00a0 — \u03b1 = \u0394L\/(L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u0394T)\u00a0 —\u00a0 metal I\u00a0 —\u00a0 \u03b1<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>=300.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\/30\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03b1<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>=1,0.10<\/b><\/span><\/span>-5o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 metal II\u00a0 —\u00a0 \u03b1<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>=600.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\/30\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03b1<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>=2,0.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\"\"<\/p>\n

b) A l\u00e2mina II deve estar na parte superior que deve se dilatar mais para que o dispositivo se encurve para baixo, pois ela tem maior coeficiente de dilata\u00e7\u00e3o.<\/b><\/span><\/span><\/p>\n

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

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u0394L=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1.\u0394T=2.22.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.110=484.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>m\u00a0 —\u00a0 R=1 + 0,00484=1,00484m\u00a0 —\u00a0 S=\u03c0R<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=3,1416.(1,00484)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S=3,172 m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

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

38-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) DL = a.L0.DT\u00a0\"\"\u00a0DL = 11.10-6.50.(45 \u2013 15) = 16500.10-6 = 0,0165 m = 1,65 cm<\/b><\/span><\/span><\/p>\n

b) \u0394V = g.V0.DT\u00a0\"\"\u00a0DV = 9,6.10-4.20000.(10-35) = – 4800000.10-4 = – 480 litros<\/b><\/span><\/span><\/p>\n

c) \u0394L = a.L0.DT\u00a0\"\"\u00a0DL = 12.10-6.1.50 = 600.10-6 = 0,0006 m = 0,06 cm = 0,6 mm<\/b><\/span><\/span><\/p>\n

39-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u0394V = V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0(\u03b3<\/b><\/span><\/span>Hg<\/b><\/span><\/span><\/sub>) DT \u00a0—\u00a0 \u00a0DV = V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0(3\u03b1<\/b><\/span><\/span>Hg<\/b><\/span><\/span><\/sub>) DT\u00a0 —\u00a0 \u0394T=\u0394V\/3\u03b1<\/b><\/span><\/span>Hg<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=A<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.h\/3\u03b1<\/b><\/span><\/span>Hg<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u00a0\u0394T=10<\/b><\/span><\/span>-7<\/b><\/span><\/span><\/sup>.6.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\/(3.40.<\/b><\/span><\/span><\/p>\n

10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>)=6.10<\/b><\/span><\/span>-9<\/b><\/span><\/span><\/sup>\/120.10<\/b><\/span><\/span>-11<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u0394T=5<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 \u0394T=T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0\u2013 T<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 5=T<\/b><\/span><\/span>f\u00ad\u00a0<\/b><\/span><\/span><\/sub>\u2013 12\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0T<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=17<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

40-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) O recipiente B porque se dilata menos.<\/b><\/span><\/span><\/p>\n

b) O n\u00edvel do l\u00edquido continuaria sendo o mesmo nos dois recipientes.<\/b><\/span><\/span><\/p>\n

41-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo a densidade inversamente proporcionsal ao volume\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

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

43-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe no gr\u00e1fico que, se voc\u00ea diminuir a temperatura do sistema, o volume do recipiente fica menor que o da \u00e1gua e ela transborda e que se voc\u00ea aumentar a temperatura do sistema, o volume da \u00e1gua fica maior qie o do recipiente, e ela transborda\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

44-<\/b><\/span><\/span><\/span> <\/b>R- C\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

45-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Correta\u00a0 —\u00a0 elas possuem a mesma densidade\u00a0 —\u00a0 est\u00e3o em equil\u00edbrio est\u00e1tico\u00a0 —\u00a0 o peso da esfera \u00e9 igual ao empuxo.<\/b><\/span><\/span><\/p>\n

b) Correta\u00a0 —\u00a0 sofrem dilata\u00e7\u00e3o volum\u00e9trica e a esfera se dilata mais que o \u00f3leo, deslocando mais l\u00edquido eaumentando o empuxo.<\/b><\/span><\/span><\/p>\n

c) Correta<\/b><\/span><\/span><\/p>\n

d) Falsa\u00a0 —\u00a0 a esfera se contrai mais que o \u00f3leo diesel, seu volume diminui mais e sua sensidade aumenta ficando maior que a do \u00f3leo diesel.<\/b><\/span><\/span><\/p>\n

e) Correta<\/b><\/span><\/span><\/p>\n

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

46-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u0394L<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>= \u0394L<\/b><\/span><\/span>aparente(tyransbordado)<\/b><\/span><\/span><\/sub>\u00a0+ \u0394L<\/b><\/span><\/span>frasco<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u0394L<\/b><\/span><\/span>aparente(tyransbordado)<\/b><\/span><\/span><\/sub>= \u0394L<\/b><\/span><\/span>l\u00edquido<\/b><\/span><\/span><\/sub>\u00a0\u2013 \u0394L<\/b><\/span><\/span>recipiente\u00ad<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

47-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>01. Falso\u00a0 —\u00a0 como a temperatura aumentou o volume do tanque tamb\u00e9m aumentou<\/b><\/span><\/span><\/p>\n

02. Correto\u00a0 —\u00a0 a temperatura aumentou<\/b><\/span><\/span><\/p>\n

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

08. Falsa\u00a0 —\u00a0 o tanque tamb\u00e9m se dilatou<\/b><\/span><\/span><\/p>\n

16. Falsa\u00a0 —\u00a0 foi maior, pois parte da gasolina vazou<\/b><\/span><\/span><\/p>\n

R- (02 + 04)=06<\/b><\/span><\/span><\/p>\n

48-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) \u0394V<\/b><\/span><\/span>gas<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b3<\/b><\/span><\/span>gas<\/b><\/span><\/span><\/sub>.\u0394t=15.000.9,6.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.(30 \u2013 15)\u00a0 —<\/b><\/span><\/span>\u00a0\u0394V<\/b><\/span><\/span>gas<\/b><\/span><\/span><\/sub>=216 L<\/b><\/span><\/span><\/p>\n

b)\u00a0 d=0,8g\/cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=0,8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>g\/L\u00a0 —\u00a0 d=m\/V\u00a0 — 0,8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=m\/20.000\u00a0 —\u00a0 m=16.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>g\u00a0 —\u00a0 Q=m.c.<\/b><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>t=16.10<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sup>.0,6.15\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q= 1,44.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>\u00a0cal<\/b><\/span><\/span><\/p>\n

49-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u2206V = \u03b3 .V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u2206T\u00a0 —\u00a0 \u2206V = 2.10\u00ad<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.(S.20).4\u00a0 —\u00a0 S.\u2206h = 160.S.10\u00ad<\/b><\/span><\/span>-4\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0 \u2206h = 16.10\u00ad<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>m = 1,6 cm\u00a0 —\u00a0\u00a0\u00a0<\/b><\/span><\/span>\u2206h=1,6cm<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

50-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>\u2206V<\/b><\/span><\/span>ta\u00e7a<\/b><\/span><\/span><\/sub>\u00a0= \u03b3 .V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u2206T=120.2,3.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.18=0,49cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>al<\/b><\/span><\/span><\/sub>=120,49cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u00a0\u00a0\u2206V<\/b><\/span><\/span>glic<\/b><\/span><\/span><\/sub>\u00a0= \u03b3 .V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u2206T=5,1.<\/b><\/span><\/span><\/p>\n

10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.119.18=1,092cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>glic<\/b><\/span><\/span><\/sub>=120,092cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 a glicerina n\u00e3o transbordar\u00e1 pois a ta\u00e7a passar\u00e1 a ter um volume de 120,49 cent\u00edmetros c\u00fabicos, enquanto que o volume total da glicerina passar\u00e1 a ser de 120,092 cent\u00edmetros c\u00fabicos. Esta diferen\u00e7a 120,49 – 120,092 = 0,398 cent\u00edmetros c\u00fabicos \u00e9 quanto ainda se poderia preencher de glicerina, na temperatura final.\u00a0<\/b><\/span><\/span><\/p>\n

51-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u2206V = \u03b3 .V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u2206T=4.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>.1,1.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.20\u00a0 —\u00a0 \u0394V=880L\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

52-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Analisando o gr\u00e1fico, notamos que o volume espec\u00edfico diminui de 0 \u00b0C at\u00e9 4\u00b0C, aumentando a partir dessa temperatura.<\/b><\/span><\/span><\/p>\n

Aproximando os valores lidos no gr\u00e1fico, constatamos uma redu\u00e7\u00e3o de 1,00015 cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/g para 1,00000 cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/g de 0 \u00b0C a 4 \u00b0C, ou seja, de 0,00015 cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\/g. Isso representa uma redu\u00e7\u00e3o percentual de 0,015%, o que \u00e9 menos que 0,04 %\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

53-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quando o recipiente estiver completamente cheio de l\u00edquido, eles dever\u00e3o ter o mesmo volume\u00a0 —\u00a0 \u0394V<\/b><\/span><\/span>V<\/b><\/span><\/span><\/sub>= \u0394V<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 100(1 + 10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>(t \u2013 10) = 99.(1 + 2.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>)(t \u2013 10)\u00a0 —\u00a0 t=19,6.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\/18,6.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0t\u224810,5<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

54-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O volume de glicerina que extravasou corresponde \u00e0 dilata\u00e7\u00e3o aparente\u00a0 —\u00a0 \u00a0\u0394V<\/b><\/span><\/span>ap<\/b><\/span><\/span><\/sub>=Vo.\u03b3<\/b><\/span><\/span>ap<\/b><\/span><\/span><\/sub>.\u0394t\u00a0 —\u00a0 352.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>=10.\u03b3<\/b><\/span><\/span>ap<\/b><\/span><\/span><\/sub>.(90 \u2013 10)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u03b3<\/b><\/span><\/span>ap<\/b><\/span><\/span><\/sub>=352.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\/8.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u03b3<\/b><\/span><\/span>ap<\/b><\/span><\/span><\/sub>=44.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u03b3<\/b><\/span><\/span>g<\/b><\/span><\/span><\/sub>= \u03b3<\/b><\/span><\/span>ap<\/b><\/span><\/span><\/sub>+ \u03b3<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 5,0.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>=44.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0+ \u03b3<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span><\/sup>\u00a0\u00a0—\u00a0 \u03b3<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=5,0.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0\u2013 4,4.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u03b3<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=6,0.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u03b1<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=6.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\/3\u00a0 —\u00a0 \u03b1<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=2,0.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

55-<\/b><\/span><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span><\/p>\n

56-<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) T=2\u03c0\u221a(L\/g)\u00a0 —\u00a0 L=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1.\u0394t\u00a0 —\u00a0 T=2\u03c0\u221a( L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1.\u0394t\/g)<\/b><\/span><\/span><\/p>\n

57<\/b><\/span><\/span><\/span>–\u00a0<\/b><\/span><\/span><\/span>\u0394V<\/b><\/span><\/span>ap<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>. \u03b3<\/b><\/span><\/span>ap<\/b><\/span><\/span><\/sub>.\u0394t\u00a0 —\u00a0 25=500.\u03b3<\/b><\/span><\/span>ap<\/b><\/span><\/span><\/sub>.50\u00a0 —\u00a0 \u03b3<\/b><\/span><\/span>ap<\/b><\/span><\/span><\/sub>=25\/25.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u03b3<\/b><\/span><\/span>ap<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u03b3<\/b><\/span><\/span>l\u00edq<\/b><\/span><\/span><\/sub>= \u03b3<\/b><\/span><\/span>V<\/b><\/span><\/span><\/sub>\u00a0= \u03b3<\/b><\/span><\/span>ap<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u03b3<\/b><\/span><\/span>l\u00edq<\/b><\/span><\/span><\/sub>=0,001 + 0,00001\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u03b3<\/b><\/span><\/span>l\u00edq<\/b><\/span><\/span><\/sub>=0,00101\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup><\/p>\n

58-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo de \u2113<\/b><\/span><\/span>0A<\/b><\/span><\/span><\/sub>\u00a0e de \u2113<\/b><\/span><\/span>0B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u2113<\/b><\/span><\/span>0A<\/b><\/span><\/span><\/sub>. \u2113<\/b><\/span><\/span>0B<\/b><\/span><\/span><\/sub>=75\u00a0 —\u00a0 3\u2113<\/b><\/span><\/span>0B<\/b><\/span><\/span><\/sub>. \u2113<\/b><\/span><\/span>0B<\/b><\/span><\/span><\/sub>=75\u00a0 —\u00a0 \u2113<\/b><\/span><\/span>0B<\/b><\/span><\/span><\/sub>=5cm e \u2113<\/b><\/span><\/span>0A<\/b><\/span><\/span><\/sub>=15cm\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

Ap\u00f3s o aquecimento, os comprimentos finais das hastes s\u00e3o iguais, pois a figura transforma-se em um quadrado\u00a0 —\u00a0 \u2113<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>= \u2113<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u2113= \u2113<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sub>(1 + \u03b1\u0394t)\u00a0 — \u00a015.(1 + \u03b1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u0394t) = 5.(1 + \u03b1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u0394t)\u00a0\u00a0 —\u00a0 a rela\u00e7\u00e3o entre os coeficientes de dilata\u00e7\u00e3o das hastes equivale a 9 e a varia\u00e7\u00e3o da temperatura a 300 (320\u00ba \u2013 20\u00ba)\u00a0 —\u00a0 logo\u00a0 —\u00a0 15.(1 + 300.\u03b1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\/9) = 5.(1 + 300\u03b1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 15 + 500\u03b1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=5 + 155\u03b1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u03b1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>= 1,0.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/p>\n

59-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A express\u00e3o matem\u00e1tica que fornece o di\u00e2metro do aro ap\u00f3s aquecido ou resfriado\u00a0 vale\u00a0 —\u00a0\u00a0 d = d<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>(1+a.\u0394T) \u00a0—\u00a0 ao aquecer de 1\u00baC o aro da rede no lado do time visitante eles provocaram uma dilata\u00e7\u00e3o em seu di\u00e2metro modificando-o para \u00a0\u00a0—\u00a0 d = 230,1(1+4,8.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.(21\u201320)) = 230,21mm \u00a0\u00a0—\u00a0 o que facilita a marca\u00e7\u00e3o de pontos pelo time local, j\u00e1 que o di\u00e2metro do aro foi aumentado.<\/b><\/span><\/span><\/p>\n

Por outro lado ao resfriar o aro da cesta em seu lado eles provocaram uma contra\u00e7\u00e3o deste, reduzindo-o para \u00a0\u00a0—\u00a0 d = 230,1(1+4,8.10<\/b><\/span><\/span>-4<\/b><\/span><\/span><\/sup>.(19 \u2013 20)) = 229,99mm \u00a0\u00a0—\u00a0 assim, o time visitante n\u00e3o conseguir\u00e1 marcar pontos, uma vez que o aro tem di\u00e2metro menor do que o da bola.<\/b><\/span><\/span><\/p>\n

As atitudes dos torcedores facilitaram as realiza\u00e7\u00f5es de pontos para o time local e impossibilitaram a marca\u00e7\u00e3o de pontos pelo time advers\u00e1rio.<\/b><\/span><\/span><\/p>\n

O t\u00e9cnico do time visitante est\u00e1 reclamando dessas atitudes dos anfitri\u00f5es em utilizar conhecimentos cient\u00edficos para fraudar o resultado da partida.<\/b><\/span><\/span><\/p>\n

60-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe que o valor do coeficiente de dilata\u00e7\u00e3o do alum\u00ednio \u00e9 o dobro do coeficiente de dilata\u00e7\u00e3o do concreto e, como elas tem as mesmas dimens\u00f5es iniciais e sofrem a mesma varia\u00e7\u00e3o de temperatura, a chapa de alum\u00ednio se dilata dias vezes mais do que a de concreto\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

61-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O f\u00f3sforo acender\u00e1 primeiro no cubo que atingir a temperatura final de igni\u00e7\u00e3o mais rapidamente, ou seja, aquele que possuir maior condutividade t\u00e9rmica. Neste caso \u00e9 o cubo feito de ouro, o cubo A. Quanto ao tamanho da aresta do cubo ela dilatar\u00e1 conforme seu coeficiente de dilata\u00e7\u00e3o, que \u00e9 maior no chumbo. Assim a aresta do cubo A ser\u00e1 menor que a do cubo B\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

62-\u00a0<\/b><\/span><\/span><\/span>O cobre deve se dilatar mais e deve ter maior coeficiente de dilata\u00e7\u00e3o\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

63-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados: Ao = 1 m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 106 mm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>; DA = 0,36 mm<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 e\u00a0 Vo = 1 dm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0= 10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0mm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 DA = Ao.2 a DT\u00a0 —\u00a0\u00a0 0,36 = 10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a02 a DT\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

DT = 0,36\/2.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>=0,18\/10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 DV = Vo3aDT\u00a0 — \u0394V=10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.3.0,18\/10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0 DV= 0,54 mm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

64-\u00a0<\/b><\/span><\/span>Veja o esquema abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Nas figuras acima\u00a0 —\u00a0 \u2113: lado inicial do quadrado\u00a0 —\u00a0\u00a0 \u2113\u2019: lado do quadrado depois do aquecimento\u00a0 —\u00a0\u00a0 L: comprimento da corda\u00a0 —\u00a0 h: dist\u00e2ncia \u00a0OB\u00a0 —\u00a0 na Fig 1, no tri\u00e2ngulo ABO, aplicando o teorema de Pit\u00e1goras\u00a0 —\u00a0 h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ \u2113\u2019<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/4= \u2113<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/4\u00a0 —\u00a0 h<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>= \u2113<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/4 – \u2113\u2019<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/4\u00a0 —\u00a0 h=1\/2.\u221a (\u2113<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0– \u2113\u2019<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>) (I)\u00a0 —\u00a0 na Fig 2, como o quadro est\u00e1 em equil\u00edbrio, a resultante das for\u00e7as \u00e9 nula\u00a0 —\u00a0\u00a0 assim\u00a0 —\u00a0\u00a0 2 Fy = P \u00a0—\u00a0 \u00a02 Fy = m g \u00a0—\u00a0 F<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=mg\/2 (II)\u00a0\u00a0 —\u00a0 o tri\u00e2ngulo ABO da Fig 1 \u00e9 semelhante ao tri\u00e2ngulo das for\u00e7as na Fig 3\u00a0 —\u00a0\u00a0 ent\u00e3o\u00a0 —\u00a0 F<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>\/h=F\/ \u2113\/2\u00a0 —\u00a0 substituindo nessa express\u00e3o as equa\u00e7\u00f5es (I) e (II)\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

65-\u00a0<\/b><\/span><\/span>A express\u00e3o da dilata\u00e7\u00e3o superficial \u00e9: A = A<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0(1+ b DT). Como As \u00e1reas finais ter\u00e3o que ser iguais\u00a0 —\u00a0 A<\/b><\/span><\/span>eixo<\/b><\/span><\/span><\/sub>\u00a0= A<\/b><\/span><\/span>orif<\/b><\/span><\/span><\/sub>\u00a0\u00de 1,02 A<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0[(1 + 5\u00b410-5) (-50)] = Ao (1 + 5\u00b410-5) DT\u00a0 —\u00a0 1,02 \u2013 2,55\u00b410-3 = 1 + 5\u00b410-5 DT\u00a0 —\u00a0 \u0394T=0.02 \u2013 2,55.10-3.5.10-6\u00a0\u00a0 —\u00a0 \u0394T=349<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

66-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados: R = 6.400 km = 6,4\u00b410<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0m; L = 6,4 m; b =\u00a0\"\"\u00b0C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>; A<\/b><\/span><\/span>agua<\/b><\/span><\/span><\/sub>\u00a0= 75%A<\/b><\/span><\/span>Terra<\/b><\/span><\/span><\/sub>\u00a0=\u00a0\"\".<\/b><\/span><\/span><\/p>\n

Da figura dada: sen<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>q =\u00a0\"\"\u00a0\u00de r = R<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>sen<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>q<\/b><\/span><\/span><\/p>\n

O comprimento da base da \u00e1rea de avan\u00e7o do oceano (DA) \u00e9 b = 2<\/b><\/span><\/span>\u00a0p\u00a0<\/b><\/span><\/span>r e a altura \u00e9 L. Assim:<\/b><\/span><\/span><\/p>\n

DA = (2<\/b><\/span><\/span>\u00a0p\u00a0<\/b><\/span><\/span>r)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>L = (2<\/b><\/span><\/span>\u00a0p\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>sen<\/b><\/span><\/span>\u00a0q)\u00a0<\/b><\/span><\/span>L. Mas:<\/b><\/span><\/span><\/p>\n

DA = A<\/b><\/span><\/span>agua\u00a0<\/b><\/span><\/span><\/sub>b<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>DT. Igualando essas duas express\u00f5es:<\/b><\/span><\/span><\/p>\n

(2<\/b><\/span><\/span>\u00a0p\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>sen<\/b><\/span><\/span>\u00a0q)\u00a0<\/b><\/span><\/span>L =\u00a0\"\"b<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>DT. fazendo os cancelamentos e isolando DT, vem:<\/b><\/span><\/span><\/p>\n

DT =\u00a0\"\". Substituindo os valores dados, temos:<\/b><\/span><\/span><\/p>\n

DT =\u00a0\"\"\u00de DT =\u00a0\"\".<\/b><\/span><\/span><\/p>\n

DT = 4,3<\/b><\/span><\/span>\u00a0\u00b4\u00a0<\/b><\/span><\/span>10<\/b><\/span><\/span>\u20133<\/b><\/span><\/span><\/sup>\u00a0\u00b0C. \u00a0<\/b><\/span><\/span><\/p>\n

67-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span>cat<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span>= A\"\"\u00b0C<\/b><\/span><\/span>\u20131<\/b><\/span><\/span><\/sup>\u00a0e\u00a0<\/b><\/span><\/span>a<\/b><\/span><\/span>hip<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span>=\u00a0\"\"\u00b0C<\/b><\/span><\/span>\u20131<\/b><\/span><\/span><\/sup>.<\/b><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>como o tri\u00e2ngulo, no in\u00edcio, \u00e9 ret\u00e2ngulo e is\u00f3sceles, os catetos possuem inicialmente o mesmo comprimento,\u00a0<\/b><\/span><\/span>L<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/i><\/span><\/span><\/sub>\u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span>o comprimento da hipotenusa,\u00a0<\/b><\/span><\/span>a<\/b><\/span><\/span>, \u00e9 calculado pelo teorema de Pit\u00e1goras\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

a<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>=\u00a0\"\"\u00a0\u00a0\u00de\u00a0 a =\u00a0\"\".\u00a0\u00a0\u00a0<\/b><\/span><\/span>(I)<\/b><\/span><\/span>\u00a0 —\u00a0 para que o tri\u00e2ngulo se torne equil\u00e1tero, de lado\u00a0<\/b><\/span><\/span>L<\/b><\/span><\/span>, temos\u00a0 —\u00a0 a(1 + a<\/b><\/span><\/span>hip<\/b><\/span><\/span><\/sub>\u00a0DT) = L<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sub>(1 + a<\/b><\/span><\/span>cat<\/b><\/span><\/span><\/sub>\u00a0DT)\u00a0 —\u00a0 substituindo os dados\u00a0 e a express\u00e3o\u00a0<\/b><\/span><\/span>(I)<\/b><\/span><\/span>, vem\u00a0 —\u00a0\u00a0\"\"\"\"= L<\/b><\/span><\/span>0\u00a0<\/b><\/span><\/span><\/sub>(1 + A\u00a0\"\"DT) \u00de\u00a0\"\"+ A<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>DT = 1 + A\"\"DT\u00a0 \u00a0—\u00a0 \u00a0\u00a0\u00a0(\"\"\u2013 1)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>A<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>DT =\u00a0\"\"\u00a0\u2013 1\u00a0 \u00a0—\u00a0\u00a0<\/b><\/span><\/span>DT =\u00a0\"\"\u00b0C<\/b><\/span><\/span>\u20131<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

68-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u2113<\/b><\/span><\/span>oA<\/b><\/span><\/span><\/sub>=0,99\u2113<\/b><\/span><\/span>oB<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 \u2113<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=\u2113<\/b><\/span><\/span>oA<\/b><\/span><\/span><\/sub>(1 + \u03b1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.\u2206t)\u00a0 —\u00a0 \u2113<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=0,99\u2113<\/b><\/span><\/span>oB<\/b><\/span><\/span><\/sub>(1 + \u03b1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.\u2206t)\u00a0 —\u00a0 \u2113<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=\u2113<\/b><\/span><\/span>oB<\/b><\/span><\/span><\/sub>(1 + \u03b1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.\u2206t) —\u00a0 quando o comprimento for o mesmo\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 \u2113<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0= \u2113<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0,99\u2113<\/b><\/span><\/span>oB<\/b><\/span><\/span><\/sub>(1 + \u03b1<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.\u2206t) = \u2113<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=\u2113<\/b><\/span><\/span>oB<\/b><\/span><\/span><\/sub>(1 + \u03b1<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.\u2206t)\u00a0 —\u00a0 0,99\u2113<\/b><\/span><\/span>oB<\/b><\/span><\/span><\/sub>(1 + 10.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>.\u2206t) = \u2113<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=\u2113<\/b><\/span><\/span>oB<\/b><\/span><\/span><\/sub>(1 + 9,1.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>.\u2206t)\u00a0 —\u00a0 \u2206t=1.250<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —<\/b><\/span><\/span><\/p>\n

\u2206t=t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1.250=t \u2013 20\u00a0 —\u00a0 t=1.270<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

69-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Novo comprimento \u2113 do p\u00eandulo quando a temperatura subir de 10<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 \u00a0\u2113 = \u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0(1 + \u03b1(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 \u2113 = 10(1 + 10.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.10\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u2113=1,0001m\u00a0 —\u00a0 c\u00e1lculo do per\u00edodo T que pelo enunciado vale T= \u221a(4\u2113)\u00a0 —\u00a0 T= \u221a(4.1,0001)\u00a0 —\u00a0 T= 2.\u221a1,0001 s\u00a0 —\u00a0 para meio ciclo de oscila\u00e7\u00e3o o per\u00edodo ser\u00e1 T=\u221a1,0001 s\u00a0 —\u00a0 se, para cada meia oscila\u00e7\u00e3o o per\u00edodo aumenta de T=\u221a1,0001 s, o rel\u00f3gio ir\u00e1 atrasar\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

70-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo do coeficiente de dilata\u00e7\u00e3o linear do material\u00a0 —\u00a0 L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=50m\u00a0 —\u00a0 L=50 + 50×0.02\/100\u00a0 —\u00a0 L=50,01m\u00a0 — \u2206L= L \u2013 L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=L<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sub>\u03b1\u2206\u03b8\u00a0 —\u00a0 0,01=50.\u03b1.20\u00a0 —\u00a0 \u03b1=10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\/10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u03b1=10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 o espa\u00e7o (d) entre as barras \u00e9 preenchido pelas duas metades das dilata\u00e7\u00f5es (\u2206L) de cada barra\u00a0 —\u00a0 d=2. \u2206L\/2=2.(0,01)\/2\u00a0 —\u00a0 d=0,01=10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>cm\u00a0 —\u00a0 \u2206L=d=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u03b1\u2206\u03b8\u00a0 —\u00a0 10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>=50.10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>.\u2206\u03b8\u00a0 —<\/b><\/span><\/span><\/p>\n

\u2206\u03b8=10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\/50×10<\/b><\/span><\/span>-5<\/b><\/span><\/span><\/sup>=1.000\/50\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u2206\u03b8=20\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span><\/p>\n

71-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Todas est\u00e3o corretas, com exce\u00e7\u00e3o da 08 que est\u00e1 errada pois, pela express\u00e3o\u00a0 \u2206V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03bb.\u2206\u03b8 voc\u00ea observa que o coeficiente de dilata\u00e7\u00e3o t\u00e9rmica (\u03bb) \u00e9 diretamente proporcional \u00e0 dilata\u00e7\u00e3o (\u2206V)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- (01 + 02 + 04 + 16) = 23<\/b><\/span><\/span><\/p>\n

72<\/b><\/span><\/span><\/span>–\u00a0<\/b><\/span><\/span><\/span>01. Correta\u00a0 —\u00a0 embora esse \u201cespa\u00e7o vazio\u201d deva, depois de algum tempo, ser preenchido com vapor d\u2019\u00e1gua que se formar\u00e1 devido \u00e0 baixa press\u00e3o.<\/b><\/span><\/span><\/p>\n

02. Falsa\u00a0 —\u00a0 \u00e0 temperatura de \u20135 \u00b0C, a \u00e1gua ser\u00e1 congelada, aumentar\u00e1 de volume, estourando a garrafa.<\/b><\/span><\/span><\/p>\n

04. Correta\u00a0 —\u00a0 essa dilata\u00e7\u00e3o observada \u00e9 chamada de dilata\u00e7\u00e3o aparente\u00a0 —\u00a0 a dilata\u00e7\u00e3o real \u00e9 soma da dilata\u00e7\u00e3o aparente com a dilata\u00e7\u00e3o da garrafa.<\/b><\/span><\/span><\/p>\n

08. Falsa\u00a0 —\u00a0 tanto a \u00e1gua como a garrafa aumentam de volume.\u00a0<\/b><\/span><\/span><\/p>\n

R- \u00a0(01 + 04) = 05<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

74-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dilatometria\u00a0– Estuda a dilata\u00e7\u00e3o dos corpos – Quando a temperatura de um s\u00f3lido aumenta, surge um aumento da amplitude das vibra\u00e7\u00f5es at\u00f4micas e da dist\u00e2ncia m\u00e9dia entre os \u00e1tomos e mol\u00e9culas que o constituem e ent\u00e3o eles se<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

dilatam.\u00a0Se a temperatura diminui, ocorre o fen\u00f4meno inverso, ou seja, eles se contraem.<\/b><\/span><\/span><\/p>\n

16-\u00a0<\/b><\/span><\/span>Rela\u00e7\u00e3o entre os coeficientes de dila\u00e7\u00e3o linear (\u03b1), superficial (\u03b2) e volum\u00e9trica (\u03b3)\u00a0 —\u00a0 \u03b1\/1 = \u03b2\/2 = \u03b3\/3\u00a0 —\u00a0 1,5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\/1 =\u03b3\/3\u00a0 —\u00a0 \u03b3 = 4,5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 volume inicial da esfera\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=(4\/3).\u03c0.R<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>=(4\/3).3.(5)<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=500=5.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 \u2206V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>. \u03b3.\u2206\u03b8=5.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.4,5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.(110 \u2013 10)\u00a0 —\u00a0 \u2206V=225cm<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

75-<\/b><\/span><\/span><\/span> <\/b>Quando a temperatura do trilho passar de 10oC para 40oC um comprimento de trilho de lo=25m sofrer\u00e1 uma dilata\u00e7\u00e3o linear de\u00a0 —\u00a0 \u2206L=Lo.\u03b1.(\u03b8 \u2013 \u03b8o)=25.14.10-6.(40 \u2013 10)=10500.10-6\u00a0 —\u00a0 \u2206L=0,01m\u00a0 —\u00a0 o m\u00e1ximo comprimento do trilho a ser colocado deve ser\u00a0 —\u00a0 L=25,00 \u2013 0,01=24,99m\u00a0 —\u00a0\u00a0R- D.\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

76-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Depois que a r\u00e9gua foi resfriada ela se contraiu e o espa\u00e7o entre cada unidade diminuiu fazendo com que a medida do mesmo l\u00e1pis<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

aumente\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

77-<\/b><\/span><\/span><\/span> <\/b>Para qualquer tipo de dilata\u00e7\u00e3o no caso volum\u00e9trica, quanto maior o coeficiente de dilata\u00e7\u00e3o, maior ser\u00e1 o volume dilatado, pois \u2206V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b3.\u2206\u03b8\u00a0 sendo \u03b3 o coeficiente de dilata\u00e7\u00e3o:<\/b><\/span><\/span><\/p>\n

I. Falsa\u00a0 —\u00a0 a tampa deve se dilatar mais, ou seja, possuir maior coeficiente de dilata\u00e7\u00e3o.<\/b><\/span><\/span><\/p>\n

II. Correta\u00a0 —\u00a0 ambos devem se dilatar mais para se desprenderem.<\/b><\/span><\/span><\/p>\n

III. Correta\u00a0 —\u00a0 como a tampa possui maior coeficiente de dilata\u00e7\u00e3o ela se dilata mais pra aumento de temperatura ou se contrai mais para diminui\u00e7\u00e3o de temperatura.<\/b><\/span><\/span><\/p>\n

IV. Correta\u00a0 —\u00a0 quanto maior a dist\u00e2ncia entre os \u00e1tomos maior ser\u00e1 a facilidade para se dilatar ou contrair com a varia\u00e7\u00e3o de temperatura. —\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre dilata\u00e7\u00e3o linear, superficial e volum\u00e9trica dos s\u00f3lidos e dilata\u00e7\u00e3o dos l\u00edquidos. \u00a0 01-\u00a0Como A est\u00e1 fixo, as dimens\u00f5es do sistema devem aumentar, pois est\u00e1 sendo aquecido e o ponto B deve afastar-se de A (veja a figura) \u00a0R- B 02-\u00a0Quando a temperatura aumenta (\u03b8 > 27 \u00baC) o alum\u00ednio se dilata mais e<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2056,"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-2059","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2059","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=2059"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2059\/revisions"}],"predecessor-version":[{"id":10738,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2059\/revisions\/10738"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2056"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2059"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}