Respuesta :
Answer:
The answer is "[tex]16 \ \frac{m}{s}[/tex]"
Explanation:
[tex]Pressure \ (P) = 3.00 \times 10^{5}\ Pa\\\\Velocity\ V_1 = 1.00 \ \frac{m}{s}[/tex]
Using the formula for equating the continuity equation:
[tex]\to V_1 \times A_1 = V_2 \times A_2\\\\\to A_2 = \frac{A_1}{16} \\\\\therefore \\\\\to V_2 = V_1 \times \frac{A_1}{A_2} \\\\\to V_2 = 1 \times 16 = 16 \ \frac{m}{s}[/tex]
The speed of the flow in the narrow section is [tex]16m/s[/tex].
To find the speed of the flow in the narrow section, We use equation of continuity.
[tex]V_{1}*A_{1}=V_{2}*A_{2}[/tex]
Given that , initial speed [tex]V_{1}=1m/s[/tex]
Let us consider that, initial cross-section area is[tex]A_{1}=A[/tex].
When the pipe narrows to one-fourth its original diameter.
Then, Area [tex]A_{2}=\frac{A}{4^{2} }=\frac{A}{16}[/tex]
Substituting values in above equation.
[tex]1*A=V_{2}*\frac{A}{16}\\\\V_{2}=16m/s[/tex]
Hence, the speed of the flow in the narrow section is [tex]16m/s[/tex].
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