Ask Question
24 July, 03:06

What hanging mass will stretch a 2.1-m-long, 0.43 mm - diameter steel wire by 1.5 mm? the young'? s modulus of steel is 20*1010n/m2. express your answer using two significant figures?

+5
Answers (1)
  1. 24 July, 05:56
    0
    The force (weight) due to the hanging mass is

    F = mg

    where

    m = mass, kg

    g = 9.8 m/s²

    The cross sectional area of the wire is

    A = (π/4) * (0.43*10⁻³ m) ² = 1.4522*10⁻⁷ m²

    The stress in the wire is

    σ = F/A = (9.8m N) / (1.4522*10⁻⁷ m²) = 6.7484x10⁷m Pa

    The length of the wire is 2.1 m and the extension is 1.5 mm. Therefore the induced strain is

    ε = (1.5x10⁻³ m) / (2.1 m) = 7.1429x10⁻⁴

    By definition, Young's modulus is E = σ/ε. Therefore

    (6.7484x10⁷m Pa) / (7.1429x10⁻⁴) = 20x10¹⁰ Pa

    9.4477x10¹⁰m = 20x10¹⁰

    m = 2.117 kg

    Answer: 2.12 kg (to two sig. figures)
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “What hanging mass will stretch a 2.1-m-long, 0.43 mm - diameter steel wire by 1.5 mm? the young'? s modulus of steel is 20*1010n/m2. ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers