To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105 mm−2 . Suppose that all the dislocations in 1000 mm3 (1 cm3 ) were somehow removed and linked end to end. How far (in miles) would this chain extend? Now suppose that the density is increased to 109 mm−2 by cold working. What would be the chain length of dislocations in 1000 mm3 of the material?

A) At Rd₁ = 10^(5) mm^(-2), chain length = 62.14 miles

B) At Rd₂ = 10^(9) mm^(-2), chain length = 621,372.74 miles

Explanation:

From the question, let's call dislocation density Rd and thus

Rd₁ = 10^(5) mm^(-2)

Rd₂ = 10^(9) mm^(-2)

And

Volume (V) = 1000 mm³

Now, the formula for chain length dislocation is given as;

L = Rd x V

Thus;

At, Rd = Rd₁ = 10^(5) mm^(-2), we have;

L1 = (10^(5) mm^(-2)) x (1000mm³) = 10×10^(7) mm = 100000

At Rd = Rd₂ = 10^(9) mm^(-2), we have;

L2 = (10^(9)mm^(-2))(1000mm³) = 10^(12) mm = 10^(9)m

Now, let's convert both lengths from metre to miles:

1 mile = 1609.34m

Thus;

L1 = 100000 / 1609. 34m = 62. 14 miles

Similarly,

L2 = 10^(9)m / 1609.34 m = 621,372.74 miles.

f=ma newton's second law 7x8=56n

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2