Question

determinar a area final da chapa a 87(graus) celsíus sabendo sua que é de 80cm(quadrado) a 0(graus)celsíus e o coeficiente de dilatacão é 48.10 elevado(a-6 graus)celsíus elevado(a-1)

Answer 1

Legenda:

$\rightarrow \Delta A\:=\text{Variacao da Area}\\\rightarrow A\:=\text{Area final}\\\rightarrow A\:=\text{Area inicial}\\\rightarrow \beta \:=\text{Coeficiente de dilacao superficial}\\\rightarrow \Delta \theta\:=\text{Variacao da temperatura}$
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$\Delta A=A_o\cdot \beta \cdot \Delta \theta \\\Delta A=A_o\cdot \beta \:\cdot \Delta \theta \:\\A_f-A_o=A_o\cdot \beta  \cdot \Delta \theta \\A_f=\left(A_o\cdot \beta  \cdot \Delta \theta \right)+A_o\\A_f=A_o\cdot \left(1+\beta  \:\cdot \:\Delta \theta \:\right)\\A_f=80\cdot \left(1+\left(48\cdot 10^{-6}\right)\:\cdot \:\left(87-0\right)\:\right)\\A_f=80\cdot \left(1+\left(48\cdot 10^{-6}\cdot 87\right)\right)\\A_f=80\cdot \left(1+\left(4176\cdot 10^{-6}\right)\right)\\A_f=80\cdot \:\left(1+\left(4,176\cdot \:10^{-3}\right)\right)\\A_f=80+334,08\cdot 10^{-3}\\A_f=80+0,33408\\\boxed{\bold{A_f=80,33\:cm^2}}$