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A force of 5000 n is applied outwardly to each end of a 5.0-m long rod with a radius of 34.0 mm and a young's modulus of 125 x 108 n/m2. the elongation of the rod is:

Respuesta :

carlosego
Por definicion tenemos que
 (F/A) = E(∆/0)
 Sustituyendo los valores tenemos y despejando ∆:
 ∆ = (F/(πr2 × E))*0
 (5000×5)/(3.14×(34×10^−2)^2×(125×10^8))
 5.5×10^−6 m
 
asifejaz20

Answer:

The elongation will be 5.51 × 10^-4 m.

Explanation:

Force = F = 5000 N

Length = l = 5.0 m

Area = A = πr^2 = (3.14)(34 × 10^-3)^2 = 3.63 × 10^-3 m^2

Young’s Modulus = E = 125 × 10^8 N/m2

We know that:

E = (F/A)/(∆l/l)

∆l = (F l)/AE

∆l=(5000 ×5)/(3.63 × 10^(-3))( 125 × 10^8)

∆l  = 5.51 × 10^-4 m

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