Given,
The initial inside diameter of the pipe, d₁=4.50 cm=0.045 m
The initial speed of the water, v₁=12.5 m/s
The diameter of the pipe at a later position, d₂=6.25 cm=0.065 m
From the continuity equation,
![\begin{gathered} A_1v_1=A_2v_2 \\ \pi(\frac{d_1}{2})^2v_1=\pi(\frac{d_2}{2})^2v_2 \\ \Rightarrow d^2_1v_1=d^2_2v_2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A_1v_1%3DA_2v_2%20%5C%5C%20%5Cpi%28%5Cfrac%7Bd_1%7D%7B2%7D%29%5E2v_1%3D%5Cpi%28%5Cfrac%7Bd_2%7D%7B2%7D%29%5E2v_2%20%5C%5C%20%5CRightarrow%20d%5E2_1v_1%3Dd%5E2_2v_2%20%5Cend%7Bgathered%7D)
Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.
On substituting the known values,
![\begin{gathered} 0.045^2\times12.5=0.065^2\times v_2 \\ \Rightarrow v_2=\frac{0.045^2\times12.5}{0.065^2} \\ =5.99\text{ m/s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%200.045%5E2%5Ctimes12.5%3D0.065%5E2%5Ctimes%20v_2%20%5C%5C%20%5CRightarrow%20v_2%3D%5Cfrac%7B0.045%5E2%5Ctimes12.5%7D%7B0.065%5E2%7D%20%5C%5C%20%3D5.99%5Ctext%7B%20m%2Fs%7D%20%5Cend%7Bgathered%7D)
Thus, the flow rate of the water at the later position is 5.99 m/s
Explanation:
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Answer:
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