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.

the skeletons of most mammals reach a certain size and then stop growing. however, many animals, including some mammals, keep growing throughout their lives. kangaroos, for example, just keep growing and growing until they die. also, the longer an animal lives the greater the chance it might catch some deadly disease.