Auniform 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

TL;DR: 3.535 cm

Explanation:

Xcm = ΣxMoments/ΣMasses = (10*0 + 10*5)/(10+10) = 50/20 = 2.5 cm

by symmetry,

Ycm = 2.5 cm

The distance D from the point Xcm,Ycm to the origin is D = √(2.5²+2.5²) = 3.535 cm