By what amount does the 52-cmcm-long femur of an 85 kgkg runner compress at this moment? The cross-section area of the bone of the femur can be taken as 5.2×10−4m25.2×10−4m2 and its Young's modulus is 1.6×1010N/m2.1.6×1010N/m2.

0.156 mm

Explanation:

Here is the complete question

The normal force of the ground on the floor can reach three times a runner's body weight when the foot strikes the pavement. By what amount does the 52-cm-long femur of an 85 kg runner compress at this moment? The cross-section area of the bone of the femur can be taken as 5.2 × 10⁻⁴  m²  and its Young's modulus is 1.6 × 10¹⁰ N/m²

The Young's modulus of the bone Y = stress/strain = σ/ε = F/A ÷ ΔL/L = FL/AΔL where F = force on bone = 3mg(since it is 3 times his weight) where m = mass of runner = 85 kg and g = acceleration due to gravity = 9.8 m/s². L = length of femur = 52 cm = 0.52 m, A = cross-sectional area of femur = 5.2 × 10⁻⁴  m² and ΔL = compression of femur.

Making ΔL subject of the formula,

ΔL = FL/AY

ΔL = 3mgL/AY

Substituting the values of the variables into the equation, we have

ΔL = 3mgL/AY

ΔL = 3 × 85 kg × 9.8 m/s² × 0.52 m/(5.2 × 10⁻⁴  m² × 1.6 × 10¹⁰ N/m²)

ΔL = 1299.48 kgm²/s² ÷ 8.32 × 10⁻⁶ N

ΔL = 156.1875 × 10⁻⁶ m

ΔL = 0.1561875 × 10⁻³ m

ΔL = 0.1561875 mm

ΔL ≅ 0.156 mm

true

explanation:

the answer is b

explanation: