Answer:
46.67 N/m
Explanation:
mass, m = 0.1 kg
distance, y = 2.1 cm = 0.021 m
Let K be the spring constant.
F = mg = Ky
0.1 x 9.8 = K x 0.021
K = 46.67 N/m
Thus, the spring constant of the spring is 46.67 N/m.
Answer:
The mass is 
Explanation:
From the question we are told that
The extension of the rod is 
The area is 
The density increase as follows 
The equation 
at

So

=> 
So at
, 
So

=> 
Now

![m = 8 [{2.5 +\frac{ 1.27x^2}{2} } ]\left | 13} \atop {0}} \right.](https://tex.z-dn.net/?f=m%20%20%3D%20%208%20%20%20%5B%7B2.5%20%2B%5Cfrac%7B%201.27x%5E2%7D%7B2%7D%20%7D%20%5D%5Cleft%20%20%7C%2013%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![m = 8 [{2.5 +\frac{ 1.27(13)^2}{2} } ]](https://tex.z-dn.net/?f=m%20%20%3D%20%208%20%20%20%5B%7B2.5%20%2B%5Cfrac%7B%201.27%2813%29%5E2%7D%7B2%7D%20%7D%20%5D)


Answer:
g ≈ 2.82 m/s^2
Explanation:
By W = mg,
W = weight (in newtons)
m = mass (in kg)
g = gravitational acceleration (in m/s^2)


g ≈ 2.82 m/s^2
I took this test before, aren't there answer choices? :)