Given data
*The given horizontal piper of diameter is D = 3.0 cm = 0.03 m
*The given depth is d = 5.0 m
*The given water desnsity is
![p=1.0*10^3kg/m^3](https://img.qammunity.org/2023/formulas/physics/high-school/2yox6fpgyfbeeu9sbsxdx7ltewvq4ybyvi.png)
(a)
The magnitude of the frictional force between the plug and pipe wall is given as
![\begin{gathered} F=\text{pgd}* A \\ =\text{pgd}*\pi r^2 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/high-school/wirz5zohf89a3rm4xe692izdy75qhmomcy.png)
*Here r = d/2 is the radius
Substitute the values in the above expression as
![\begin{gathered} F=(1.0*10^3)(9.8)(5.0)*3.14*((0.03)/(2))^2 \\ =34.61\text{ N} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/high-school/2dj9evkzkrd2kfkdzuyw1iyl6b2zhxeo04.png)
(b)
The volume flow rate is calculated as
![\begin{gathered} V=Av \\ =\pi r^2\sqrt[]{2gd} \\ =3.14*((0.03)/(2))^2*\sqrt[]{2*9.8*5.0} \\ =6.99*10^(-3) \end{gathered}](https://img.qammunity.org/2023/formulas/physics/high-school/ie5vbx0t1gfmy38hyefku0e3oucyacb98m.png)
The total volume is calculated as
![\begin{gathered} V_{T_{}}=V* t \\ =6.99*10^(-3)*1.0*60 \\ =0.4194m^3 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/high-school/95kuc51cgb2fe08do1x4ad7g9x5pvcdwrh.png)