A shearing of 50N is applied to an aluminum rod with a length of 10m a cross sectional area of 1.0×10-5 and a shear modulus of 2.5×1010 as result the rod is sheared through a distance

0.002 m or 2 mm

Explanation:

Given that:

Force, F = 50N

Area = 1 * 10^-5

Length, L = 10m

Shear modulus, = 2.5 * 10^10

Using the relation ;

D = (50 ÷ 1*10^-5) ÷ (2.5 * 10^10 ÷ 10)

D = 5000000 ÷ 2.5 * 10^9

D = 5 * 10^6 ÷ 2.5 * 10^9

D = (5/2.5) * 10^(6-9)

D = 2 * 10^-3

D = 0.002 m

1m = 1000 mm

0.002m = (1000 * 0.002) = 2 mm

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explanation:

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