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Two springs of stiffness Ka, and Kb , are placed one inside the other so as to compress by the same amount under an applied axial load. The combined stiffness of the two springs will be E

AKa+Kb

B(KaKb) / (Ka+Kb)

C(Ka+Kb)/2

D1/Ka + 1/Kb

Answer:

A. Ka+Kb


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