The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.
Speed is defined as the rate of change of the distance or the height attained.
The given data in the problem is;
The initial diameter is,[tex]\rm d_1 = 2.2 \ cm[/tex]
initial radius,
[tex]r_1 = \frac{d_1}{2} \\\\ r_1 = \frac{2.2}{2} \\\\ r_1 = 1.1\ cm[/tex]
The initial crossection area;
[tex]\rm A_1 = \pi r_1^2 \\\\ \rm A_1 = 3.14 \times (1.1\times 10^{-2})^2 \\\\ \rm A_1 =3.8 \times 10^{-4} \ m^2[/tex]
The final crossection area;
[tex]\rm A_2 = \pi r_2^2 \\\\ \rm A_2 = 3.14 \times ( 2 \times 10^{-2})^2 \\\\ \rm A_2 = 12.56 \ m^2[/tex]
The initial flow rate is;
R = density ×velocity ×area
[tex]\rm R = \rho A V \\\\ 1.5 = 1000 \times V_1 \times 3.8 \times 10^{-4} \\\\ V_1 = 3.947 \ m/sec[/tex]
The speed of the water in the wider part will be;
From the continuity equation;
[tex]\rm A_1 V_1 = A_2V_2 \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec[/tex]
Hence, the speed of the water in the wider part will be 1.194 m/sec.
To learn more about the speed, refer to the link;
https://brainly.com/question/7359669
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