Beban seberat 400 N digantungkan pada kawat baja sepanjang 2 meter dan luas penampang 25 . Apabila modulus Young baja 20 x N/, besar tetapan kawat tersebut . . . N/m. A. 2,5 x B. 5 x C. 2,5 x D. 4 x E. 4 x
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Dzaky111
Dik : F = 400 N l = 2 m A = 25 mm² = 25 x 10⁻⁶ m³ E = 20 x 10¹° N/m² Dit : K = F/Δl = .....? Jwb: E = (F/A)x(L/ΔL) = (F/ΔL) x (L/A) F/ΔL = EA/L = (20x10¹° x 25 x 10⁻⁶)/2 = 25 x 10⁵ N/m = 2,5 x 10⁶ N/m
l = 2 m
A = 25 mm²
= 25 x 10⁻⁶ m³
E = 20 x 10¹° N/m²
Dit : K = F/Δl = .....?
Jwb: E = (F/A)x(L/ΔL)
= (F/ΔL) x (L/A)
F/ΔL = EA/L
= (20x10¹° x 25 x 10⁻⁶)/2
= 25 x 10⁵ N/m
= 2,5 x 10⁶ N/m