[tex]4) 1,2075.10^{-6}\ m\\[/tex]
[tex]5) \beta = 8.10^{-6}\ ^0C^{-1}\\ \\\alpha = 4.10^{-6}\ ^ 0C^{-1}[/tex]
Considerando-se que o raio da chapa seja [tex]10 \\[/tex] cm a sua área inicial será:
[tex]A = \pi.r^2\\A = 3,14.10^2\\A = 3,14.100\\A = 314\ cm^2[/tex]
[tex]\beta = 2.\alpha \\\beta = 2.12.10^{-6}\ ^0C^{-1}\\ \beta = 24.10^{-6}\ ^0C^{-1}\\ \Delta\beta = Ao.\beta.\Delta\ T\\\Delta\beta = 314.24.10^{-6}.200 \\ \Delta\beta = 1,5072.10^{-6}\ m \\\\5) \Delta\beta = Ao.\beta.\Delta\ T\\0,004 = 5.\beta.(100 - 0)\\0,004 = 500\beta\\\beta = 0,004/ 500\\\beta = 0,000008\\\beta = 8.10^{-6}\ ^0C^{-1}\ Coeficiente\ de\ dilatacao\ superficial\\\alpha = \beta / 2\\\alpha = 8.10^{-6 / 2}\\\alpha = 4.10^{-6}\ ^0C^{-1} Coeficiente\ de\ dilatacao\ linear\\[/tex]
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