Question

1. The initial pressure reading on an ocean-faring submarine is 105 kPa at a depth of 0.35 m. The submarine descends and the new pressure reading is 125 kPA. At what depth is the submarine now located? The density of seawater is 1023 kg/m3 .

1B. The area of a hose on the left end is 3.5 x10-3 m3 while the area on the right end of the hose is 3.0 x 10-3 m3 . A liquid will travel a constant 10 cm in 400 ms on the left end of the hose. What will be the velocity of the liquid as it exits the right end of the hose? Assume a constant density.

I know we are only to post one at a time, but I really need help with both of these, I will rate!! Please explain answers fully, so that I may rework and understand. Thank you!

Answer 1

1) Gauge pressure at a depth h inside a liquid due yo the liquid column =(rho)gh

DATE PAGE No. Soph. 0 hange pressure due to quid o column AB =) Let Po be atmospheric pressure Pressure on submarine at a depth h P = lot .3gh liven that Ps 105 kPa P pot Bgh - Tosx103 lot 10 23x9.810.35=105X103 Po = 101491.11 Pa At depth ĥ, P = 125 kPa S-4023 TS Po tigh s 125x103 101491.11 t 1023x9.8 h = 125000 = 2.345m

DATE 1 PAGE NO 11 _ II- _Abably equ of Continuity T Rate of flow = constant 163,5X20-3) v = (3x1063) Vo =0 400 ms 10 X 105 0.400 s 0.25 m/s (35*703)0.25 = 38103 V = 0.292 mlo = 29.2 cm/8