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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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 the force constant of this strip of aortal material

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Delwyn 6 months 2021-07-23T05:59:55+00:00 1 Answers 14 views 0

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    2021-07-23T06:01:06+00:00

    Answer:

    53.41 N/m

    Explanation:

    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

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