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A molecule of DNA (deoxyribonucleic acid) is 2.33 µm long. The ends of the molecule become singly ionized: negative on one end, positive on the other. The helical molecule acts like a spring and compresses 1.01% upon becoming charged. Determine the effective spring constant of the molecule.I've seen a couple examples of this problem, including the one already solved on cramster. I tried that method but it didn't work. I got 8.114E-6, then 9.55 E-6, and webassign said that both answers were wrong. I need a fool proof method. One of my calculators said 9.544E-30 but I'm afraid to try it because I only get 10 chances to try an answer and I've tried about 5 times.

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princessesther2011

Answer:

The effective spring constant is (4.25×10^7F) Newton per meter

Explanation:

From Hookes law of elasticity,

Force required to compress a spring (F) = force constant (k) × compression (c)

k = F/c

Length of the spring = 2.33 micrometers = 2.33×10^-6 meters

Compression (c) = 1.01% × 2.33×10^-6 = 0.0101 × 2.33×10^-6 = 2.3533×10^-8

k = F(Newton)/2.3533×10^-8 meters = (4.25×10^7F) Newton per meter

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