Here, as the charge is uniformly distributed in the sphere, we will consider s as an area of the sphere which is, s=4πr2 and r is radius of the gaussian surface shown in the figure above. From this, it can be seen that
Answer:
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A. Ocean winds bring rain and fog and often bring warm water that keeps the climate mild
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Answer:
A_f= 15,769 m²
Explanation:
Este es un ejercicio de dilatación térmica,
ΔA = (2α) A₀ ΔT
el arrea de recipiente
A₀ = L A
A₀ = 3,5 4,5
A₀= 15,75 m²
el coeficiente de dilatación térmica es alfa = 16,6 10⁻⁶ C⁻¹
calculemos
ΔA = 2 16,6 10⁻⁶ 15,75 ( 40 -4)
ΔA = 522,9 (36) 10⁻⁶
A= 1,88 10⁻² m2
el cambio de volumen es
ΔA = A_f – A₀
A_f = A₀ + ΔA
A_f= 15,75 +1,88 10⁻²
A_f= 15,769 m²
