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Sagot :
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
[tex]p=1.0\times10^3kg/m^3[/tex](a)
The magnitude of the frictional force between the plug and pipe wall is given as
[tex]\begin{gathered} F=\text{pgd}\times A \\ =\text{pgd}\times\pi r^2 \end{gathered}[/tex]*Here r = d/2 is the radius
Substitute the values in the above expression as
[tex]\begin{gathered} F=(1.0\times10^3)(9.8)(5.0)\times3.14\times(\frac{0.03}{2})^2 \\ =34.61\text{ N} \end{gathered}[/tex](b)
The volume flow rate is calculated as
[tex]\begin{gathered} V=Av \\ =\pi r^2\sqrt[]{2gd} \\ =3.14\times(\frac{0.03}{2})^2\times\sqrt[]{2\times9.8\times5.0} \\ =6.99\times10^{-3} \end{gathered}[/tex]The total volume is calculated as
[tex]\begin{gathered} V_{T_{}}=V\times t \\ =6.99\times10^{-3}\times1.0\times60 \\ =0.4194m^3 \end{gathered}[/tex]
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