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Suppose that all the dislocations in 1600 mm3 of crystal were somehow removed and linked end to end. Given 1 m = 0.0006214 mile, how far (in miles) would this chain extend for dislocation densities of (a) 10^4 mm-2 (undeformed metal)? (b) 10^10 mm-2 (cold-worked metal)?

Respuesta :

Answer:

(a) 9.924 mile

(b) 0.0994 mile

Explanation:

We have given dislocation volume [tex]V=1600mm^3

(a) Density of the dislocation d = [tex]10^4mm^{-2}[/tex]

We know that dislocation density is given by

[tex]d=\frac{length\ of\ dislocation}{volume}[/tex]

So [tex]{length\ of\ dislocation}={volume}\times density[/tex]

[tex]{length\ of\ dislocation}=1600\times 10^4mm=16000m =16000\times 0006214=9.924mile[/tex]

(b) [tex]{length\ of\ dislocation}=1600\times 10^2mm=160m =160\times 0006214=0.0994mile[/tex]

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