Answer:[tex]A_2=0.349\times 10^{-4} m^2[/tex]
Step-by-step explanation:
Given
velocity [tex]v_1=0.72 m/s[/tex]
Area of cross-section [tex]A_1=1\times 10^{-4} m^2[/tex]
velocity after Falling 0.19 m
[tex]v_2^2-v_1^2=2g\cdot h[/tex]
[tex]v_2^2=v_1^2+2g\cdot h[/tex]
[tex]v_2^2=(0.72)^2+2\times 9.8\times 0.19[/tex]
[tex]v_2^2=0.518+3.724[/tex]
[tex]v_2=\sqrt{4.24}[/tex]
[tex]v_2=2.05 m/s[/tex]
conserving Flow
[tex]A_1v_1=A_2v_2[/tex]
[tex]1\times 10^{-4}\times 0.72=A_2\times 2.05[/tex]
[tex]A_2=0.349\times 10^{-4} m^2[/tex]
