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andriy [413]
3 years ago
9

Please help as fast you can

Physics
1 answer:
Ilya [14]3 years ago
7 0

Answer:

d a

Explanation:

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Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point
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Answer:

The fluids speed at a) 0.105\,m^{2}  and b) 0.047\,m^{2} are 2.33\,\frac{m}{s^{2}}  and 5.21\,\frac{m}{s^{2}} respectively

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Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

\Delta Q=0(1)

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A_{2}v_{2}-A_{1}v_{1}=0 (2)

With A the cross sectional areas and v the velocities of the fluid.

a) Here, we use that point 2 has a cross-sectional area equal to A_{2}=0.105\,m^{2}, so now we can solve (2) for v_{2}:

v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}

b) Here we use point 2 as A_{2}=0.047\,m^{2}:

v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}

c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is Q=\frac{V}{t}, so we can write:

A_{1}v_{1}=\frac{V}{t}, solving for V:

V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}

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= 3.99*10^-3N/C

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