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    A steel cable with a cross-sectional area of
    2.01 cm^2 has a mass of 3.3 kg/m . Its Young’s
    modulus is 2.36 × 10^11 N/m^2 .
    The acceleration of gravity is 9.8 m/s^2 .
    If 539 m of the cable is hung over a vertical
    cliff, how much does the cable stretch under
    its own weight? Answer in units of cm.

    My work.
    Converted square cm to square m, to conform with units on Young's modulus. .000201 sq cm. Multiply 3.3 kg/m by 539 m and 9.8 m/s*2 to get force.

    Now into E=\frac{F}{Ao}\times\frac{Lo}{\D  elta L}

    2.36\times10^{11}=\frac{17431.26  }{.000201}\times \frac{539}{\Delta L}

    I end up getting .1981 m. Convert to cm and input, but it's wrong. What am I missing?
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    I can't spot anything wrong with your working. Is your answer actually wrong or are just out by a factor of 10?
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    I have no way to tell. The homework site we use just says "wrong," not "wrong by X amount" :>.<:
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    Mind if I have a peek at the URL? And the only thing I can suggest is keep playing around with powers of 10. Since your arithmetic is otherwise correct.
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    (Original post by Dagrath)
    Mind if I have a peek at the URL? And the only thing I can suggest is keep playing around with powers of 10. Since your arithmetic is otherwise correct.
    Viewing anything other than the start page requires a username and password. The stuff in the OP before my work was copied and pasted directly from there. My only alteration was to add some missing ^ as in "2.01 cm^2."
 
 
 
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