A 3.5-kg object is supported by an aluminum wire of length 1.8 m and diameter 2.0...

A 3.5-kg object is supported by an aluminum wire of length 1.8 m and diameter 2.0 mm . How much will the wire stretch?

Solutions

Expert Solution

he weight of the object is W = mg

oungs modulus of aluminum wire is Y = 6.9x10¹⁰ N/m²

Youngs modulus of aluminum wire is Y = (F / A) / (ΔL / L)

= FL / AΔL

= mgL / AΔL

A= pi*r^2 = pi*( 2/2 *10^-3)^2 = 3.14159265e-6 m²

ΔL = mgL / AY

= 3.5*9.8*1.8 / ( 3.14159265e-6 * 6.9x10^10)

= 0.0002848181506 m

=0.28481*10^-3 m

= 0.28481 mm answer

Goodluck for exam Comment in case any doubt, will reply for sure..

genius_generous answered 2 months ago

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