Question

resolva a seguinte questão:

Answer 1

2024 mm = 2,024 m

Lf = Li.[1 + coeficiente linear.(Tf - Ti)]

$2,024 = 2.[1 + 1,6.10^{ - 5}.(Tf - 20)] \\ \frac{2,024}{2} = 1 + 1,6.10^{ - 5}.(Tf - 20) \\ 1,012 = 1 + 1,6.10^{ - 5}.(Tf - 20) \\ 1,012 - 1= 1,6.10^{ - 5}.(Tf - 20) \\ 0,012= 1,6.10^{ - 5}.(Tf - 20) \\ 12. {10}^{ - 3} = 1,6.10^{ - 5}.(Tf - 20) \\ \frac{12.{10}^{ - 3} }{1,6.10^{ - 5}} = Tf - 20 \\ 7,5. {10}^{2} = Tf - 20 \\ 750 = Tf - 20 \\ 750 - 20 = Tf \\ 730 = Tf$

Alternativa (A), 730 °C.