A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the center of mass of the bent wire?

the center of mass is 7.07 cm apart from the bend

Explanation:

the centre of mass of a wire of length L is L/2 (assuming uniform density). Then initially the x coordinate of the centre of mass is

x₁ = L/2 = 20 cm / 2 = 10 cm

when the wire is bent in a right angle the coordinates of the new centre of mass will be

x₂ = L₂/2

y₂ = L₂/2

where L₂ is the length of the horizontal piece and vertical piece. Then L₂=L/2

x₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm

y₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm

x₂=y₂=X

locating the bend in the origin (0,0) the distance to the centre of mass is

d = √ (x₂²+y₂²) = √ (2X²) = √2*X=√2*5cm = 7.07 cm

d = 7.07 cm