Answer:
805.48N/m
Explanation:
According to Hookes law
F = Ke
F is the force = mg
F = 2.4×9.8 = 23.52N
e is the extension = 2.92cm = 0.0292m
Force constant K = F/e
K = 23.52/0.0292
K = 805.48N/m
Hence the force constant of the spring is 805.48N/m
The answer is d.heuristic
Answer:
C.
would adapt if the dress were held completely still
Explanation:
By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.
<h3>What are the limits?</h3>
First, we need to find the limits.
We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.
So if 58 in is the 100%, the 26% and 43% of that are:
- 26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
- 43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.
But we know that the CG is found to be 45.5% MAC, then it measures:
(45.5%/100%)*58in = 0.455*58in = 26.39 in
We need to compare it with the largest limit, so we get:
26.39 in - 24.94 in = 1.45 in
This means that the CG is 1.45 inches out of limits.
If you want to learn more about percentages, you can read:
brainly.com/question/14345924
