(a) 5.66 m/s
The flow rate of the water in the pipe is given by
[tex]Q=Av[/tex]
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
[tex]Q=1.20 m^3/s[/tex]
the radius of the pipe is
r = 0.260 m
So the cross-sectional area is
[tex]A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2[/tex]
So we can re-arrange the equation to find the speed of the water:
[tex]v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s[/tex]
(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:
[tex]Q_1 = Q_2\\A_1 v_1 = A_2 v_2[/tex]
where we have
[tex]A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s[/tex]
and where [tex]A_2[/tex] is the cross-sectional area of the pipe at the second point.
Solving for A2,
[tex]A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2[/tex]
And finally we can find the radius of the pipe at that point:
[tex]A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m[/tex]
The speed of the water at point one is equal to 5.66 meter per seconds.
Given the following data:
Radius = 0.260 m.
Flow rate = 1.20 [tex]m^3/s[/tex].
First of all, we would determine the area of the pipe by using this formula:
[tex]A=\pi r^2\\\\A= 3.142 \times 0.260^2\\\\A=0.212\;m^2[/tex]
Mathematically, the flow rate of a liquid is given by this formula:
Q = AV
Where:
Substituting the parameters into the formula, we have;
1.20 = 0.212 × V
V = 1.20/0.212
V = 5.66 m/s.
In order to determine the radius of the pipe at the second point, we would determine its area as follows:
[tex]Q_1=Q_2\\\\A_1V_1=A_2V_2\\\\A_2=\frac{A_1V_1}{V_2} \\\\A_2=\frac{0.212 \times 5.66}{3.60} \\\\A_2 = 0.333 \;m^2[/tex]
For the radius, we have:
[tex]A=\pi r^2\\\\0.333= 3.142 \times r^2\\\\r=\sqrt{\frac{0.333}{3.142} }[/tex]
r = 0.326 meter.
Read more on area here: brainly.com/question/14478195
