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Water is flowing in a straight horizontal pipe of variable cross section. Where the cross-sectional area of the pipe is 3.70·10-2 m2, the pressure is 6.10·105 Pa and the velocity is 0.260 m/s. In a constricted region where the area is 9.50·10-4 m2, what is the velocity?

Respuesta :

Answer:

v = 10.1 m/s

Explanation:

As we know that by the law of conservation of volume the rate of volume flowing through the pipe will remain conserved

so here we have flow rate given as

[tex]Q = Area\times velocity[/tex]

now we have

[tex]A_1 v_1 = A_2 v_2[/tex]

now we have

[tex]A_1 = 3.70 \times 10^{-2} m^2[/tex]

[tex]v_1 = 0.260 m/s[/tex]

[tex]A_2 = 9.50 \times 10^{-4} m^2[/tex]

now from above equation we have

[tex]v_2 = \frac{A_1}{A_2} v_1[/tex]

[tex]v_2 = \frac{3.70\times 10^{-2}}{9.50\times 10^{-4}}(0.260)[/tex]

[tex]v_2 = 10.1 m/s[/tex]

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