Answer:[tex]P_2=3.30\times 10^5 Pa[/tex]
Explanation:
Given
For First Floor
[tex]P_1=3.7\times 10^5 Pa[/tex]
[tex]v_1=2.3 m/s[/tex]
[tex]z_1=0 m[/tex]
For second Floor
[tex]z_2=3.6 m[/tex]
[tex]v_2=3.6 m/s[/tex]
Applying Bernoulli's equation
[tex]P_1+\frac{1}{2}\rho v_1^2+\rho gz_1=P_2+\frac{1}{2}\rho v_2^2+\rho gz_2[/tex]
[tex]P_2=P_1+\frac{1}{2}(\rho v_1^2-\rho v_2^2)+\rho g(z_1-z_2)[/tex]
[tex]P_2=3.30\times 10^5 Pa[/tex]