(\beta = 3.4 \times 10^-5 , \text°C^-1), (\Delta A = 0.00272 , \textm^2), (A_f = 0.50272 , \textm^2).
$$\beta = 2\alpha = 2 \times (17 \times 10^-6) = 34 \times 10^-6 , ^\circ\textC^-1$$ dilatacion superficial ejercicios resueltos
A steel ring has an internal area of (12 , \textcm^2) at (15 , \text°C). If the ring is heated to (215 , \text°C), what is its new internal area? ((\alpha_\textsteel = 1.2 \times 10^-5 , \text°C^-1)) (\beta = 3
Ejercicio 1: Cálculo del área final de una plancha de cobre Una plancha de cobre mide . Determine su superficie final a . Considere 1. Calcular el área inicial y el coeficiente superficial Primero, obtenemos el área inicial ( ) y calculamos duplicando el coeficiente lineal: ((\alpha_\textsteel = 1
A2 = A1 + ΔA = 2 m^2 + 0,00132 m^2 = 2,00132 m^2
Ambos se expanden: