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9.Water flows'along a horizontal pipe of cross sectional area 48cm2 which has a constriction of
cross sectional area 12cm2 at one place. If the speed of the water at the constriction is 36m/s,
calculate the speed in the wider section,
(3mk)
​

Respuesta :

hamzaahmeds

Answer:

v₁ = 9 m/s

Explanation:

Here, we can use the continuity equation, as follows:

[tex]A_1v_1 = A_2v_2[/tex]

where,

A₁ = Cross-sectional area of the wider section = 48 cm²

A₂ = Cross-sectional area of the constriction = 12 cm²

v₁ = speed of flow in wider section = ?

v₂ = speed flow in constriction = 36 m/s

Therefore,

[tex](48\ cm^2)v_1 = (12\ cm^2)(36\ m/s)\\\\v_1 = \frac{(12\ cm^2)(36\ m/s)}{(48\ cm^2)} \\\\[/tex]

v₁ = 9 m/s

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