Dilatacion Superficial Ejercicios Resueltos !!link!!
Increases, same (\beta) as the plate.
Si una placa tiene un agujero central, el agujero se dilata exactamente al mismo ritmo que si estuviera relleno del mismo material. ✅ Conclusión
| | Fórmula | Aplicación | | :--- | :--- | :--- | | Dilatación Superficial | $\Delta A = A_0 \cdot \beta \cdot \Delta T$ | Calcular el cambio de área de un sólido. | | | $A_f = A_0 \left[ 1 + \beta (T_f - T_0) \right]$ | Calcular el área final. | | Relación con $\alpha$ | $\beta = 2\alpha$ | Obtener $\beta$ a partir de $\alpha$ (y viceversa). | | | $\Delta A = A_0 \cdot 2\alpha \cdot \Delta T$ | Usar cuando se conoce el coeficiente lineal. | | Cálculo de $\beta$ | $\beta = \frac\Delta AA_0 \cdot \Delta T$ | Determinar el coeficiente de un material desconocido. | dilatacion superficial ejercicios resueltos
Af=A0⋅(1+β⋅ΔT)cap A sub f equals cap A sub 0 center dot open paren 1 plus beta center dot cap delta cap T close paren
Sustituimos los valores:
At (10^\circ C), a hole in a plate has area (50.00 , cm^2). At (110^\circ C), its area is (50.25 , cm^2). Find (\alpha) of the material. (Answer: (\approx 2.5 \times 10^-5 , ^\circ C^-1))
A glass sheet of area (0.2 , \textm^2) at (30 , \text°C) expands to (0.2005 , \textm^2). What is the final temperature? ((\alpha_\textglass = 9 \times 10^-6 , \text°C^-1)) Increases, same (\beta) as the plate
(168.9 , \text°C).
Un orificio circular en una placa de acero tiene un diámetro de 20∘C20 raised to the composed with power C . Si la temperatura sube a 200∘C200 raised to the composed with power C , ¿cuál es el nuevo diámetro del orificio? ( | | | $A_f = A_0 \left[ 1
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ΔT=0.00110.000044=25∘Ccap delta cap T equals 0.0011 over 0.000044 end-fraction equals 25 raised to the composed with power cap C Paso 3: Calcular la temperatura final ( Tfcap T sub f
