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Finger [1]
3 years ago
14

A 19

Physics
1 answer:
Hoochie [10]3 years ago
4 0

Answer:

Young's modulus of this tendon is 9.03\times 10^6\ N/m^2.

Explanation:

Given that,

Length of the tendon, l = 19 cm

It is stretched by 4.5 mm, \Delta l=4.5\ mm

Force, F = 11.3 N

Average diameter, d = 8.2 mm

Radius, r = 4.1 mm

The formula of Young's modulus of this tendon is given by :

Y=\dfrac{Fl}{\Delta l A}\\\\Y=\dfrac{11.3\times 0.19}{4.5\times 10^{-3}\times \pi (4.1\times 10^{-3})^2}\\\\Y=9.03\times 10^6\ N/m^2

So, the Young's modulus of this tendon is 9.03\times 10^6\ N/m^2. Hence, this is the required solution.                

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