Answer
The pressure at the narrow end of the segment is $1.12\times 10^5~Pa$
Work Step by Step
We can use the continuity equation to find the speed $v_2$ that the water moves through the narrow end of the segment:
$A_2~v_2 = A_1~v_1$
$v_2 = \frac{A_1~v_1}{A_2}$
$v_2 = \frac{(50.0\times 10^{-4}~m^2)~(0.040~m/s)}{0.50\times 10^{-4}~m^2}$
$v_2 = 4.0~m/s$
The speed of the water through the narrow end of the segment is $4.0~m/s$
We can use Bernoulli's equation to find the pressure $P_2$ at the narrow end of the segment:
$P_2 + \frac{1}{2}\rho~v_2^2 = P_1+\frac{1}{2}\rho~v_1^2$
$P_2 = P_1+\frac{1}{2}\rho~(v_1^2-v_2^2)$
$P_2 = (1.20\times 10^5~Pa)+(\frac{1}{2})(10^3~kg/m^3)[(0.040~m/s)^2-(4.0~m/s)^2]$
$P_2 = 1.12\times 10^5~Pa$
The pressure at the narrow end of the segment is $1.12\times 10^5~Pa$.
