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Alika [10]
3 years ago
10

A 13 cm long tendon was found to stretch 3.7 mm by a force of 12.1 N . The tendon was approximately round with an average diamet

er of 8.8 mm . Calculate the Young's modulus of this tendon.
Physics
1 answer:
liq [111]3 years ago
8 0

Answer:

young's modulus=\frac{stress}{strain}=\frac{0.1990\times 10^{6}}{28.46\times 10^{-3}}=6.99\times 10^6N/m^2

Explanation:

We have given length of tendon L= 13 cm =0.13 m

Change in length \Delta L=3.7mm=3.7\times 10^{-3}m

Force = 12.1 N

Average diameter d =8.8 mm

So r=\frac{d}{2}=\frac{8.8}{2}=4.4mm=4.4\times 10^{-3}m

Area A=\pi\times  (4.4\times 10^{-3})^2=60.79\times 10^{-6}m^2

Now stress =\frac{force}{area}=\frac{12.1}{60.79\times 10^{-6}}=0.1990\times 10^6N/m^2

Strain =\frac{change\ in\ lenght}{length}=\frac{3.7\times 10^{-3}}{0.13}=28.46\times 10^{-3}

Now young's modulus=\frac{stress}{strain}=\frac{0.1990\times 10^{6}}{28.46\times 10^{-3}}=6.99\times 10^6N/m^2

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