Answer:
713.51 N/m
Explanation:
Hook's Law: This law states that provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the applied force.
From hook's law,
F = ke ...........................Equation 1
Where F = Force exerted on the bowstring, e = Extension/compression of the bowstring, k = Spring constant of the bow.
Make k the subject of the equation,
k = F/e ............................ Equation 2
Given: F = 264 N, e = 0.37 m.
Substitute into equation 2
k = 264/0.37
k = 713.51 N/m
Hence the spring constant of the bow = 713.51 N/m
Answer:
hard and brittle
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7.5 is the answer. You have to move the decimal 2 places to the right.
2.5 moles of silver is 269.6705 grams