Answer:
The answer is "[tex]18.62 \ \frac{m}{s}[/tex]"
Explanation:
Given value:
[tex]P_A= 101.3 \frac{kN}{m^2}\\\\P_B= 200 \frac{kN}{m^2}\\\\Z_A=0.5 \ m\\\\Z_B= 7 \ m\\\\r= 0.9 \times 10^3= 900 \frac{kg}{m^3}\\\\r=8.825 \frac{KN}{m^3}\\\\[/tex]
[tex]\frac{P_A}{r} + \frac{V A^2}{2g} + Z_A = \frac{P_B}{r}+ \frac{V B^2}{2g} +Z_B\\\\\ where \ \ \ \ \ \ \ \frac{V B^2}{2g}=0\\\\\frac{101.3}{8.825} + \frac{V A^2}{2 \times 9.81} + 0.5 = \frac{200}{8.825}+ 7\\\\11.478 +\frac{V A^2}{19.62} +0.5 = 22.668+7\\\\\frac{V A^2}{19.62} = 29.668+11.978\\\\V A^2 = 17.684 \times 19.62\\\\V A^2 = 346.96\\\\V A = 18.62 \ \frac{m}{s}\\\\[/tex]