Answer:
1.31498 kg
0.72050 s
0.72050 s
Explanation:
m = Mass of block
g = Acceleration due to gravity = 9.81 m/s²
k = Spring constant = 100 N/M
x = Displacement = 0.129 m
The force balance is

The mass of the block is 1.31498 kg
Time period is given by

The period of oscillations is 0.72050 s
The time period does not depend on the acceleration due to gravity. It varies with the mass and the spring constant.
Hence, the time period would be the same
Answer:
Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.
Explanation:
Considering the complete question attached in figure below.
Time period for balance wheel is:


m = mass of balance wheel
R = radius of balance wheel.
Angular frequency is related to Time period as:

As dimensions of new balance wheel are one-third of their original values


<span>Notice for the Carbon question they were the same element and the shared the same number of protons. so i think d. is the answer</span>