A shear force of 400 N is applied to one face of an aluminum cube with sides of 30 cm. What is the resulting relative displacement? (the shear modulus for aluminum is 2.5x1010N/m2)

5.3 × 10^(-8) m

Explanation:

We are given;

Shear force; F = 400 N

Length of cube; L_o = 30 cm = 0.3 m

Shear modulus; S = 2.5 × 10^(10) N/m²

Now,the resulting relative displacement can be gotten from the formula;

F = A × S × Δx/L_o

Where Δx is resulting relative displacement

A is area.

Area of cube = (L_o)² = 0.3² = 0.09

Thus, making Δx the subject, we have;

Δx = (F × L_o)/(A × S)

Plugging in the relevant values;

Δx = (400 × 0.3)/(0.09 × 2.5 × 10^(10))

Δx = 5.3 × 10^(-8) m

a.   mass number from atomic number