Respuesta :
Answer:
4.55×10^-5m
Explanation:
Young modulus of the material is equal to the ratio of the tensile stress to tensile strain of the elastic material.
Young modulus = Tensile stress/Tensile strain
Tensile stress = Force/cross sectional area
Give mass = 73kg
Force = mg = 73×10 = 730N
Cross sectional area = 5.2×10^-4m²
Tensile stress = 730/5.2×10^-4
Tensile stress = 1.4×10^6N/m
Strain = extension/original length
Given original length = 52cm = 0.52m
Tensile strain = extension(e)/0.52
Substituting the values given into the young modulus formula we have;
1.6×10^10 = 1.4×10^6/{e/0.52}
1.6×10^10 = 1.4×10^6×0.52/e
e = 1.4×10^6×0.52/1.6×10^10
e = 7.28×10^5/1.6×10^10
e = 4.55×10^-5m
This shows that the femur compresses by 4.55×10^-5m
The ground's normal on the femur while running can reach several larger
than when standing.
- Amount of compression of the femur is approximately 1.343 × 10⁻⁴ m.
Reasons:
The length of the bone, L = 52 cm = 0.52 m
Mass of the runner = 73 kg
Cross sectional area of the femur = 5.2 × 10⁻⁴ m²
Young's modulus = 1.6 × 10¹⁰ N/m²
Normal force = 3 × Body weight
Solution:
[tex]Young's \ modulus \ of \ elasticity = \dfrac{Stress}{Strain}[/tex]
[tex]Tensile \ stress \ on \ femur, \ \sigma = \dfrac{73 \times 9.81 \times 3}{5.2 \times 10^{-4} } \approx 4.132 \times 10^6[/tex]
[tex]Strain = \dfrac{Stress}{Young's \ modulus} \approx \dfrac{4.132 \times 10^6}{1.6 \times 10^{10}} \approx 2.582 \times 10^{-4}[/tex]
[tex]Strain = \dfrac{\Delta L}{L}[/tex]
Compression, ΔL = L × Strain
Therefore;
ΔL = 0.052 m × 2.582 × 10⁻⁴ ≈ 1.343 × 10⁻⁴ m
The compression of the femur when running, ΔL ≈ 1.343 × 10⁻⁴ m.
Learn more here:
https://brainly.com/question/13257353
Question parameter obtained from a similar online question:
The normal force of the ground on the leg can be up to thrice the runners weight.
