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## 5. Tests performed on a 16.0 cm strip of the donated aorta reveal that it stretches 3.37 cm when a 1.80 N pull is exerted on it. (a) What is

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## Answers ( )

Answer:53.41 N/mExplanation:From Hooke’s law,Applying,

F = ke…………. Equation 1

Where F = Force, e = extension, k = force constant of the aortal material

Make k the subject of the equation

k = F/e…………. Equation 2

From the question,

Given: F = 1.8 N, e = 3.37 cm = 0.0337 m

Substitute these values into equation 2

k = 1.8/(0.0337)

k = 53.41 N/m

Hence the force constant of the aortal material is 53.41 N/m