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A Type I muscle fibers have an average diameter of 98 µm, and can exert a compressive load of 532 µN[1]. If they are modeled as a perfect cylinder, what is the stress in an individual fiber?

Respuesta :

Answer:

stress is 70529.30 N/m²

Explanation:

given data

diameter = 98 µm = 98 × [tex]10^{-6}[/tex] m

compressive load = 532 µN = 532 × [tex]10^{-6}[/tex] N

to find out

stress in an individual fiber

solution

we know that stress formula that is

stress = [tex]\frac{load}{area}[/tex]     ................1

put here value we get  stress

stress = [tex]\frac{load}{area}[/tex]

stress = [tex]\frac{532*10^{-6}}{\frac{\pi }{4} (98*10^{-6})^2}[/tex]

stress = 70529.30 N/m²

so stress is 70529.30 N/m²

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