Well, this seems like a sticky situation, doesn't it? It's like the water is going through its own little adventure park – with a constriction thrown in for added excitement.
To find the speed in the wider section, we can use the principle of continuity. Since the volume flow rate remains constant (like the height of an inept tightrope walker), we can use the equation A1v1 = A2v2, where A1 and A2 are the cross-sectional areas, and v1 and v2 are the speeds.
Given that A1 = 48 cm^2 and A2 = 12 cm^2 (or 0.048 m^2 and 0.012 m^2 respectively), and v2 = 4 m/s, we can solve for v1.
48 cm^2 * v1 = 12 cm^2 * 4 m/s
v1 = (12 cm^2 * 4 m/s) / 48 cm^2
= 1 m/s
So, in the wider section, the speed of the water is 1 m/s. It's like a leisurely stroll in comparison to the constriction.
Now, let's move on to the pressure at the constriction. Since the speed of the water increases, according to Bernoulli's principle (the master of fluid dynamics), the pressure decreases. So, the pressure at the constriction will be lower than in the wider section.
Given that the pressure in the wider section is 1.0 x 10^5 Pa, we can say that the pressure at the constriction is less than that. Perhaps we can call it the "constricted pressure"?
But, if you want an actual numerical value for the pressure at the constriction, we'll need to know the height or elevation of the pipe. Without that information, I'm afraid I'm as clueless as a balloon without air.
So, we can conclude that the speed in the wider section is 1 m/s, but the pressure at the constriction will remain a mystery until we have more information. Good luck with your hydrodynamic adventures!