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A fluid enters the control volume of the Figure at a mass flow rate of 2.50 slugs/s with a speed of 6.6 ft/s at an angle of 4

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Given IW in = 2.5 slugs/s V = 6.6 ft/s H L as swefan, V is perpendicular to surface area (A) WL know, m = FAV 2.5 = fx (WL) x. E cs dadz -66 2 ਦੇ ( i ) 2. 4 A ਕਿ 6-1 ( Fy: 11-6-6) (dx da), (2) Fy: Il s() OL) - ) ( ) ( - ) $ : 0:325 ) (-) A : 6. ੨੫8C-for any doubts, please comment..

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