Answer:
[tex]P_2 = 2000 Pa[/tex]
Explanation:
we know from continuity equation
A1 V1 = A2 V2
[tex]2 \times 2 = 1 \times V2[/tex]
V2 = 4 m/s
[tex]P_1 +\frac{1}{2} \rho v_1^2 + mg y_1 = P_2 +\frac{1}{2} \rho v_2^2 + mg y_2[/tex]
we know[tex] y_1 = y_2[/tex], so we have
[tex]P_1 +\frac{1}{2} \rho v_1^2 = P_2 +\frac{1}{2} \rho v_2^2[/tex]
[tex]P_2 = P_1 + \frac{1}{2} \rho (v_1^2 -v_2^2)[/tex]
[tex]P_2 = 8000 + \frac{1}{2} \times 1000 (2^2 - 4^2)[/tex]
after solving we get
[tex]P_2 = 2000 Pa[/tex]
