How Much Does a Brass Wire Stretch Under a 5 kN Load?

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SUMMARY

The discussion focuses on calculating the stretch of a brass wire under a 5 kN load using Young's modulus, which is 9.1 x 1010 Pa. The wire has a length of 2.1 m and a cross-sectional area of 4.6 mm2. The formula applied is Young's Modulus (YM) = (F/A) / (ΔL/L0), leading to a calculated stretch (ΔL) of 2.5 x 10-4 m, which is equivalent to 0.025 cm. The solution emphasizes the importance of unit consistency throughout the calculation.

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Homework Statement


"A brass wire with Young's modulus of 9.1 1010 Pa is 2.1 m long and has a cross-sectional area of 4.6 mm2. If a weight of 5.0 kN is hung from the wire, by how much does it stretch?"

The answer is to be given in cm.


Homework Equations



I used the equation for Young's Modulus, YM = (F/A) / (\DeltaL/L0)

The Attempt at a Solution



I looked at my given info first.
YM = 9.1e10
Area = 4.6 mm2 = 0.46 cm2
Force = 5 kN = 5000 N
L0 = 2.1 m = 2100 cm

I converted all my units to cm because that's what the answer should be in.

Then I plugged that info into my Young's Modulus equation:

9.1e10 = [(5000)/(0.46)] / [(\DeltaL)/2100)]
\DeltaL = (5000)(2100) / (.46)(9.1e10)
\DeltaL = 2.5e-4

I thought I set this up right, but maybe I'm missing something.
 
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I'd keep everything in meters.

Then convert for the answer.
 

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