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2 years ago

12

A cylindrical wire made of an unknown alloy hangs from a support in the ceiling. You measure the relaxed length of the wire to b

e 16 m long; and the radius of the wire to be 3.5 m. When hang a 5 kg mass from the wire, you measure that it stretches a distance of 4 x 10 m The average bond length between atoms is 2.3 x 10^0 m for th alloy.
Required:
What is the stiffness of a typical interatomic bond in the alloy

1 answer:

IceJOKER [234]2 years ago

8 0

Answer: hello some of your values are wrongly written hence I will resolve your question using the right values

answer:

stiffness = 1.09 * 10^-6 N/m

Explanation:

Given data:

Length ( l ) = 16 m

radius of wire ( r ) = 3.5 m

mass ( m ) = 5kg

<u>Distance stretched ( Δl ) = 4 * 10^-3 m </u> ( right value )

<u>average bond length ( between atoms ) = 2.3 * 10^-10 m </u>( right value)

first step : calculate the area

area ( A ) = πr^2 = π * ( 3.5)^2 = 38.48 m^2

γ = MgL / A Δl

= [ (5 * 9.81 * 16 ) / ( 38.48 * (4.3*10^-3) ) ]

= 784.8 / 0.165 = 4756.36 N/m^2

hence : stiffness = γ * bond length

= 4756.36 * 2.3 * 10^-10 = 1.09 * 10^-6 N/m

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