Explanation:
The given data is as follows.
radius (r) = 3.25 cm, 
Now, we will calculate the tangential acceleration as follows.
Putting the given values into the above formula as follows.
= 
= 37.7 
Thus, we can conclude that the tangential acceleration of a point on the rim of the flywheel during this spin-up process is 37.7 .
THE INTERSTELLAR MEDIUM COMPOSED GAS AND DUST
Answer:
<h2>1567.09 N/m</h2>
Explanation:
Step one:
given data
mass m=5kg
compression x= 3.13cm to m= 0.0313m
<em>According to Hooke's law, provided the elastic limit of an elastic material is not exceeded the extension e is directly proportional to the applied force</em>
F=ke
where
k= spring constant in N/m
e= extension/compression in
Step two:
assume g= 9.81m/s^2
F=mg
F=5*9.81
F=49.05N
substitute in the expression F=ke
49.05=k*0.0313
k=49.05/0.0313
k=1567.09 N/m
<u>The force constant (in N/m) of the spring is 1567.09 N/m</u>
