Question: 17% Part (e) Young's modulus for bone is about Y = 1.6 x 1010 N/mz. The tibia (shin bone) of a man is 0.2 m

 17% Part (e) Young's modulus for bone is about Y =

17% Part (e) Young's modulus for bone is about Y = 1.6 x 1010 N/mz. The tibia (shin bone) of a man is 0.2 m long and has an average cross sectional area of 0.02 ml. What is the effective spring constant of the tibia in N/m? k = 1.6 * 10( 9) k = 1.6E+9 u/ Correct! Q 17% Part (f) Given part (e), if a man weighs 750 N, how much is the tibia compressed if it supports half his weight? Give your answer in meters. d = l cos() m

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