9. Santa brings a special baby a bouncy swing. The child bounces in a harness, with a spring constant k, and is suspended off of the ground. a. If the spring stretches x1 = 0.27 m from equilibrium while supporting an 7.15-kg child, what is its spring constant, in newtons per meter?

259.52N

Explanation:

Force initiated on a spring that causes an extension is related by the expression below;

F= K×e

Where F is the Force

K is the spring constant

e is the extension caused by the spring.

By change of subject formula for K;

K = F/e

Now F is the same as weight and is given by mass× acceleration. In this case acceleration is g=9.8m/s2; this is because the child's mass is going to be under the influence of gravity as it swings up the harness}

Hence W =7.15×9.8=70.07N

Hence K = 70.07/0.27 =259.5185N/m

=259.52N/m to 2 decimal place.

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explanation:

real, smaller and upside-down.   so the answer should be b.