Explanation:
Given that,
Mass = 0.254 kg
Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]
Force = 0.5 N
y = 0.628
We need to calculate the A and d
Using formula of A and d
.....(I)
....(II)
Put the value of
in equation (I) and (II)


From equation (II)


Put the value of
in equation (I) and (II)


From equation (II)


Put the value of
in equation (I) and (II)


From equation (II)


Put the value of
in equation (I) and (II)


From equation (II)


Hence, This is the required solution.
(a) 5.66 m/s
The flow rate of the water in the pipe is given by

where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have

the radius of the pipe is
r = 0.260 m
So the cross-sectional area is

So we can re-arrange the equation to find the speed of the water:

(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:

where we have

and where
is the cross-sectional area of the pipe at the second point.
Solving for A2,

And finally we can find the radius of the pipe at that point:

A= v²/R
a = 12²/30 =4.8 m/s²