(1) The difference in pressure between the ends of the pipe is 611 kPa.
(2) The Reynold's number for this torpedo is 1.2 x 10⁷.
(3) The flow of water around the torpedo is turbulent.
The difference in pressure is calculated by applying Poiseuille formula.
[tex]V = \frac{\pi \Delta P r^4 t}{8\eta L} \\\\\frac{V}{t} = \frac{\pi \Delta P r^4 }{8\eta L}\\\\Q = \frac{\pi \Delta P r^4 }{8\eta L}[/tex]
where;
[tex]\Delta P = \frac{8Q\eta L}{\pi r^4}[/tex]
[tex]\Delta P = \frac{8(0.1)(1.2)(200)}{\pi (0.1)^4} \\\\\Delta P = 611,155 \ Pa\\\\\Delta P \approx 611 \ kPa[/tex]
[tex]Re = \frac{Vl \rho }{\mu} \\\\Re = \frac{(10)(1.2)(1000)}{1 \times 10^{-3}} \\\\Re = 1.2 \times 10^7[/tex]
The calculated Reynold's number (1.2 x 10⁷) is greater than 3500. Thus, the flow is turbulent.
Learn more about turbulent flow here: https://brainly.com/question/12081428
#SPJ1