Question

Part C please!

Open 5. Piping A tube with flowing water has a vertical tube with an opening in order to determine the pressure in the tube given the height of the water. As the water flows, the pipe moves up 4.0 m through a narrower tube before exiting at a speed of 4.0 m/s as shown. muullis 5.0 cm 4.0 m/s 4,0m amis 10 cm u-amis A. Where the water enters from the left, what is the speed of the water? A. = 10 cm- -> 01 Aivi AzV2 Az= 5.0cm 0.05. V,=? Vi= A2V2 -(0.05 naty u Om/s) V₂ = 4.0m Continuity B. The diagonal tube has the same cross-sectional area as the outgoing horizontal opening. What is the speed of the water at a height of 2.0 m? Avi = A/V2 it has the same speed 80 4ms P2 - atmpress C. What is the absolute pressure at the point P at the bottom of the vertical tube? Pit gogh, tal v2= P2 + 8gh the pure Pipe = 2 pv, 2-pgh. = 28 12 22 (2) 2 - 3 g h = 28 (4) 2 P apluz -1,2) - 2x1000x (10-4)

Answer 1

1. Here As = 1o em? ... A2=5 emr =10 110-2² m 2 .. =5x(10-3 m2 . = loxio "in2 = 5810-4 m2 2 = 4 m/s, y = 5x109 x 4 10x10-4 w = 2mis Let the point at height 4 m' is a. Applying Bernoulli's the open at Þ and g - na hizo, hq = 4 m Pq + Pg hit 2 eg²=P2 + P g hg the per Pj-P2 = 1/30 ( 620 ) + egi(he ha) y = 214222] +.pgx4 =e[£*12 +4x9-8] 1. g=9.8 m/s2 = 10² [6 +39.2] P = 10² kg/m3 != 45.2X30 3 Pascal (N/m2) P2 = 1 atm = 1-0 13X 105 Pa .". Po = 10.452 +1-013)X165 pa = 1.465 X 105 Pa - Please give a rating. Comment any concern about solution. CSscanned with CamScanner