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8. Two 40 ft long wires made of differing materials are supported from the ceiling of a testing laboratory. Wire (1) is made of material H and has a diameter of 3/8 in. Wire (2) is made of material K and has a diameter of 3/16 in. When a load of 225 lb is applied to its lower end, wire (1) stretches 0.10 in. When the same 225 lb load is applied to the lower end of wire (2), wire (2) stretches 0.25 in. Compare materials H and K. Which material has the greater modulus of elasticity? Which is the stiffer material?

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Answer:

Material K has a modulus of elasticity E=3.389× 10¹¹ Pa

Material H has a modulus of elasticity E=1.009 × 10⁹ Pa

Material K has higher value of modulus of elasticity than material H

Material K is stiffer.

Explanation:

Wire 1 material H

Length=L = 40 ft =12.192 m

Diameter= 3/8 in = 0.009525 m

Area= A= πr²,where r=0.009525/2 =0.004763

A=3.142*0.004763² =0.00007126 m²

Force, F= 225 lb=  225*4.45 =1001.25 N

Change in length =Δ L= 0.10 in = 0.00254

To find modulus of elasticity apply'

E=F*L/A*ΔL

E=1001.25*12.192/(0.004763*0.00254)

E= 1009027923.58 Pa

E=1.009 × 10⁹ Pa

For Wire 2 material K

Length=L= 40 ft =12.192 m

Diameter = 3/16 in = 0.1875 in = 0.004763 m

Area= πr² = 3.142 * (0.004763/2)² = 0.00000567154 m²

Force, F= 225 lb=  225*4.45 =1001.25 N

Change in length =Δ L= 0.25 in =0.00635 m

To find modulus of elasticity apply'

E=F*L/A*ΔL

E= (1001.25*12.192)/(0.00000567154 * 0.00635 )

E=338955422575 Pa

E=3.389× 10¹¹ Pa

Material  K has a greater modulus of elasticity

The material with higher value of E is stiffer than that with low value of E.The stiffer material is K.

The material that has the greater modulus of elasticity and the material that is stiffer is; Material K in both cases

What is the Elastic Modulus?

Wire 1 of material H;

Length; L = 40 ft = 12.192 m

Diameter; d = ³/₈ in = 0.009525 m

Area; A = πd²/4

A = π * 0.009525²/4

A = 0.00007126 m²

Force; F = 225 lb = 1001.25 N

Change in length; ΔL= 0.10 in = 0.00254 m

Formula for modulus of elasticity is;

E = (F × L)/(A × ΔL)

E = (1001.25 × 12.192/(0.004763 × 0.00254)

E = 1.009 × 10⁹ Pa

Wire 2 of material K;

Length; L= 40 ft = 12.192 m

Diameter; d = 3/16 in = 0.004763 m

Area; A = πd²/4

A =  π * 0.004763²/4

A = 0.00000567154 m²

Force; F = 225 lb = 1001.25 N

Change in length; Δ L = 0.25 in = 0.00635 m

Formula for the modulus of elasticity is;

E = (F × L)/(A × ΔL)

E = (1001.25 × 12.192/(0.00000567154 × 0.00635)

E = 3.389 × 10¹¹ Pa

Thus, Material  K has a greater modulus of elasticity

The greater the modulus, the stiffer the material and as such  material K is stiffer.

Read more about Modulus of Elasticity at; https://brainly.com/question/14468674

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