Uma chapa de alumínio tem um furo central de 100cm de raio, estando numa temperatura de 22ºC. Sabendo-se que o coeficiente linear do alumínio equivale a 20.10^-6ºC, a nova área do furo, quando a chapa for aquecida até 122ºC, será equivalente a qual valor em metros? (Considere pi = 3)
Resposta certa: 3,012 m²
Me ajudem no cálculo.
Soluções para a tarefa
Respondido por
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Dilatação superficial (2 dimensões)
![Informa\c{c}\~oes:~\begin{Bmatrix}raio~do~furo~na~chapa~de~alum.=100cm\\coef.~linear~do~alum\'inio~(\alpha)=20\cdot10^{-6}\\varia\c{c}\~ao~de~temperatura~(\Delta\theta)=100^\circ C\\\'area~do~circulo=\pi r^2\\considerar~\pi=3\end. Informa\c{c}\~oes:~\begin{Bmatrix}raio~do~furo~na~chapa~de~alum.=100cm\\coef.~linear~do~alum\'inio~(\alpha)=20\cdot10^{-6}\\varia\c{c}\~ao~de~temperatura~(\Delta\theta)=100^\circ C\\\'area~do~circulo=\pi r^2\\considerar~\pi=3\end.](https://tex.z-dn.net/?f=Informa%5Cc%7Bc%7D%5C%7Eoes%3A%7E%5Cbegin%7BBmatrix%7Draio%7Edo%7Efuro%7Ena%7Echapa%7Ede%7Ealum.%3D100cm%5C%5Ccoef.%7Elinear%7Edo%7Ealum%5C%27inio%7E%28%5Calpha%29%3D20%5Ccdot10%5E%7B-6%7D%5C%5Cvaria%5Cc%7Bc%7D%5C%7Eao%7Ede%7Etemperatura%7E%28%5CDelta%5Ctheta%29%3D100%5E%5Ccirc+C%5C%5C%5C%27area%7Edo%7Ecirculo%3D%5Cpi+r%5E2%5C%5Cconsiderar%7E%5Cpi%3D3%5Cend.)
Vamos calcular primeiro a área inicial do furo na chapa
![\begin{array}{l}\'area~do~c\'irculo=\pi r^2\\\\=3\cdot100^2\\\\=3\cdot10.000\\\\=30.000~cm^2~\Leftrightarrow~3m^2\\\\\\f\'ormula:~\fbox{$A=A_0+A_0\cdot\beta\cdot\Delta\theta $}~~~~(\beta=2\cdot\alpha)\end{array} \begin{array}{l}\'area~do~c\'irculo=\pi r^2\\\\=3\cdot100^2\\\\=3\cdot10.000\\\\=30.000~cm^2~\Leftrightarrow~3m^2\\\\\\f\'ormula:~\fbox{$A=A_0+A_0\cdot\beta\cdot\Delta\theta $}~~~~(\beta=2\cdot\alpha)\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%5C%27area%7Edo%7Ec%5C%27irculo%3D%5Cpi+r%5E2%5C%5C%5C%5C%3D3%5Ccdot100%5E2%5C%5C%5C%5C%3D3%5Ccdot10.000%5C%5C%5C%5C%3D30.000%7Ecm%5E2%7E%5CLeftrightarrow%7E3m%5E2%5C%5C%5C%5C%5C%5Cf%5C%27ormula%3A%7E%5Cfbox%7B%24A%3DA_0%2BA_0%5Ccdot%5Cbeta%5Ccdot%5CDelta%5Ctheta+%24%7D%7E%7E%7E%7E%28%5Cbeta%3D2%5Ccdot%5Calpha%29%5Cend%7Barray%7D)
Substituindo...
![\begin{array}{l}A=3+3\cdot40\cdot10^{-6}\cdot100\\\\A=3+12.000\cdot10^{-6}\\\\A=3+12.000\cdot\dfrac{1}{10^6}\\\\A=3+12.000\cdot\dfrac{1}{1000000}\\\\A=3+\dfrac{12.000}{1000000}\\\\A=3+\dfrac{12}{1000}\\\\A=3+0.012\\\\A=3.012m^2~~\leftarrow~~resposta\end{array} \begin{array}{l}A=3+3\cdot40\cdot10^{-6}\cdot100\\\\A=3+12.000\cdot10^{-6}\\\\A=3+12.000\cdot\dfrac{1}{10^6}\\\\A=3+12.000\cdot\dfrac{1}{1000000}\\\\A=3+\dfrac{12.000}{1000000}\\\\A=3+\dfrac{12}{1000}\\\\A=3+0.012\\\\A=3.012m^2~~\leftarrow~~resposta\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7DA%3D3%2B3%5Ccdot40%5Ccdot10%5E%7B-6%7D%5Ccdot100%5C%5C%5C%5CA%3D3%2B12.000%5Ccdot10%5E%7B-6%7D%5C%5C%5C%5CA%3D3%2B12.000%5Ccdot%5Cdfrac%7B1%7D%7B10%5E6%7D%5C%5C%5C%5CA%3D3%2B12.000%5Ccdot%5Cdfrac%7B1%7D%7B1000000%7D%5C%5C%5C%5CA%3D3%2B%5Cdfrac%7B12.000%7D%7B1000000%7D%5C%5C%5C%5CA%3D3%2B%5Cdfrac%7B12%7D%7B1000%7D%5C%5C%5C%5CA%3D3%2B0.012%5C%5C%5C%5CA%3D3.012m%5E2%7E%7E%5Cleftarrow%7E%7Eresposta%5Cend%7Barray%7D)
Vamos calcular primeiro a área inicial do furo na chapa
Substituindo...
viniciushenrique406:
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Respondido por
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A = Ao.(1 + β.ΔӨ)
A = Л.r²..(1 + β.ΔӨ)
A = Л.1².(1 + 44.10-6.110)
A = Л.(1 + 0,00484)
A = Л.(1,00484)
A = 3,155 em valor aproximado.
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