Question

ME AJUDEM,É PRA AMANHÃ !

Answer 1

Dilatação linear:

$\Delta l=l_{0}\cdot\alpha\cdot\Delta\theta$

Δl ---> Variação do comprimento (l - l₀)
l₀ ----> Comprimento inicial
α ----> Coeficiente de dilatação linear
Δθ --> Variação da temperatura
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1)

$l_{0}=1200~m\\\Delta l=l-l_{0}=l-1200\\\alpha=4\cdot10^{-6}~\ºC^{-1}\\\Delta\theta=150\º-0\º=150\º$

Achando o comprimento:

$\Delta l=l_{0}\cdot\alpha\cdot\Delta\theta\\l-1200=1200\cdot4\cdot10^{-6}\cdot150\\l-1200=12\cdot4\cdot15\cdot10^{-6}\cdot10^{3}\\l-1200=720\cdot10^{-3}\\l-1200=0,72\\l=0,72+1200\\\\\boxed{\boxed{l=1200,72~m}}$

2)

$l_{0}=38000~cm\\\Delta l=l-l_{0}=l-38000\\\alpha=12\cdot10^{-6}~\ºC^{-1}\\\Delta\theta=90\ºC-(-10\ºC)=100\ºC$

Substituindo:

$l-38000=38000\cdot12\cdot10^{-6}\cdot100\\l-38000=38\cdot10^{5}\cdot12\cdot10^{-6}\\l-38000=456\cdot10^{-1}\\l-38000=45,6\\\\\boxed{\boxed{l=38045,6~cm}}$

3)
$l_{0}=150~m\\\alpha=23\cdot10^{-6}~\ºC^{-1}\\\Delta\theta=100\ºC-20\ºC=80\ºC$

Achando o aumento no comprimento:

$\Delta l=150\cdot23\cdot10^{-6}\cdot80\\\Delta l=12\cdot10^{3}\cdot23\cdot10^{-6}\\\Delta l=276\cdot10^{-3}\\\Delta l=0,276~m$

b)

Achando o comprimento final:

$\Delta l=0,276~m\\l-l_{0}=0,276\\l-150=0,276\\l=0,276+150\\\\\boxed{\boxed{l=150,276~m}}$