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A suspension cable is made of steel with Young's modulus 210 times ten to the ninth newtons per square meter, has an unsupported length of 3020 m, a diameter of 5.6 cm, and carries a tension of 3.0 times ten to the sixth newtons. What percentage of its unsupported length does the cable stretch by?
AAbout 0.6 percent
BAbout 6 percent
CAbout 60 percent
DAbout 0.06 percent
Answer & Solution
Correct answer: A. About 0.6 percent
1. The cross-sectional area is pi times the radius squared, with radius 2.8 cm, giving about 2.46 times ten to the minus three square meters.
2. The change in length is delta L equal to (1/Y)(F/A)(L0), using the given force, area, and length.
3. Substituting the numbers gives a stretch of about 17.5 m, which the source rounds to 18 m.
4. Dividing that stretch by the original length of 3020 m and converting to a percentage gives about 0.6 percent.
5. The source confirms this exact figure, ruling out the options that are off by a factor of ten in either direction.
_Source: OpenStax College Physics (CC BY 4.0), Ch 5 "Further Applications of Newton's Laws: Friction, Drag, and Elasticity", section 5.3 Elasticity: Stress and Strain_
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