Answer:
The pressure of the water in the pipe is 129554 Pa.
Explanation:
There are wrongly written values on the proposal, the atmospheric pressure must be 101105 Pa, and the density of water 1001.03 kg/m3, those values are the ones that make sense with the known ones.
We start usign the continuity equation, and always considering point 1 a point inside the pipe and point 2 a point in the nozzle:
[tex]A_1v_1=A_2v_2[/tex]
We want [tex]v_2[/tex], and take into account that the areas are circular:
[tex]v_2=\frac{A_1v_1}{A_2}=\frac{\pi r_1^2 v_1}{\pi r_2^2}=\frac{r_1^2 v_1}{r_2^2}[/tex]
Substituting values we have (we don't need to convert the cm because they cancel out between them anyway):
[tex]v_2=\frac{r_1^2 v_1}{r_2^2}=\frac{(1.8cm)^2 (0.56m/s)}{(0.49cm)^2}=7.56m/s[/tex]
For determining the absolute pressure of the water in the pipe we use the Bernoulli equation:
[tex]P_1+\frac{\rho v_1^2}{2}+\rho gh_1=P_2+\frac{\rho v_2^2}{2}+\rho gh_2[/tex]
Since the tube is horizontal [tex]h_1=h_2[/tex] and those terms cancel out, so the pressure of the water in the pipe will be:
[tex]P_1=P_2+\frac{\rho v_2^2}{2}-\frac{\rho v_1^2}{2}=P_2+\frac{\rho (v_2^2-v_1^2)}{2}[/tex]
And substituting for the values we have, considering the pressure in the nozzle is the atmosphere pressure since it is exposed to it we obtain:
[tex]P_1=101105 Pa+\frac{1001.03Kg/m^3 ((7.56m/s)^2-(0.56m/2)^2)}{2}=129554 Pa[/tex]
