Water is flowing in a 90ο pipe with uniform cross section area of A=0.0083 m2. The velocity along the horizontal direction is v1=6.0 m/s. The pressure at the section 1 is p1=500kPa. Elevation difference and energy losses can be neglected. The density of water is ρ=1000kg/m3. An external force is needed to hold the pipe in equilibrium. (2) Select the correct expression of the external force component in the vertical direction Ry_________ A. B. C. D. Water is flowing in a 90° pipe with uniform cross section area of A=0.0083 m². Thevelocity along the horizontal direction is v1-6.0 m/s. The pressure at the section 1 isP1=500kPa. Elevation difference and energy losses can be neglected. The density of wateris p=1000kg/m³. An external force is needed to hold the pipe in equilibrium.(2) Select the correct expression of the external force component in the vertical directionRyP₁A₁•1+XP₂ A₂OAR =P₂A₂+POV 21OB.R=-PQ¹ 2OCR = P₂A₂-POV 21OD.R =PQ¹ 2Rx

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Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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Water is flowing in a 90^ο pipe with uniform cross section area of A=0.0083 m². The velocity along the horizontal direction is v_{1=6.0 m/s}. The pressure at the section 1 is p_1=500kPa. Elevation difference and energy losses can be neglected. The density of water is ρ=1000kg/m³. An external force is needed to hold the pipe in equilibrium.

(2) Select the correct expression of the external force component in the vertical direction R_y_________

A.

B.

C.

D.

Water is flowing in a 90° pipe with uniform cross section area of A=0.0083 m². The
velocity along the horizontal direction is v1-6.0 m/s. The pressure at the section 1 is
P1=500kPa. Elevation difference and energy losses can be neglected. The density of water
is p=1000kg/m³. An external force is needed to hold the pipe in equilibrium.
(2) Select the correct expression of the external force component in the vertical direction
Ry
P₁A₁
•1
+X
P₂ A₂
OAR =P₂A₂+POV 21
OB.R=-PQ¹ 2
OCR = P₂A₂-POV 21
OD.R =PQ¹ 2
Rx

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