Respuesta :
Answer:
[tex]3.3\times10^{4} Pa[/tex]
[tex]6.67\times10^{5} Pa[/tex]
Explanation:
Complete statement of the question is
A relaxed biceps muscle requires a force of 25.0 N for an elongation of 3.0 cm; the same muscle under maximum tension requires a force of 500 N for the same elongation. Find Young's modulus (Pa) for the muscle tissue under each of these conditions .The muscle is assumed to be a uniform cylinder with length 0.200 m and cross-sectional area 50.0 cm^2.
For relaxed muscle :
[tex]F[/tex] = force required = 25 N
[tex]L[/tex] = Normal length = 0.2 m = 20 cm
[tex]\Delta L[/tex] = elongation in length = 3 cm
[tex]A[/tex] = Area of cross-section = 50 cm² = 50 x 10⁻⁴ m²
[tex]Y[/tex] = Young's modulus for the muscle
Young's modulus is given as
[tex]Y = \frac{FL}{A \Delta L} \\Y = \frac{(25)(20)}{(50\times10^{-4}) (3)} \\Y = 3.3\times10^{4} Pa[/tex]
Under maximum tension :
[tex]F'[/tex] = force required = 500 N
[tex]L[/tex] = Normal length = 0.2 m = 20 cm
[tex]\Delta L[/tex] = elongation in length = 3 cm
[tex]A[/tex] = Area of cross-section = 50 cm² = 50 x 10⁻⁴ m²
[tex]Y[/tex] = Young's modulus for the muscle
Young's modulus is given as
[tex]Y = \frac{F'L}{A \Delta L} \\Y = \frac{(500)(20)}{(50\times10^{-4}) (3)} \\Y = 6.67\times10^{5} Pa[/tex]
