... c+ddeVD− ρ ∆ M: 1 = c+dL : -1 = a + b - 3c - d + e T : -2 = - a - dsuy ra: e = e ; d = d; c = 1 – d; b = -d - e; a = 2 - dVậy τmax=KV2-d.D-d-e. ρ1-d. µd. ∆e0maxrJ2τ = γMặt khácde2max2VDK ... F=1. 144 (2)(1 ,4) 2 222 21 1 11 12 2 2 2 2 21 12. 144 2 2 2 2V VL Q LQH Q QgA K gA gA K gAξ ξ = + + = + + 31212 2 20.027 /2. 144 2 2VHQ m sLgA K gAξ= = + + 111Q 144 .2FQQQ =+= (4) Một ... zC= 2,4m; Q3=50lít/s; zB=3,04m. Tìm Q1; Q2; zA.Cho: L1=1250m; d1=0,4m; n1=0,016. ⇒A1=0,1256 m2L2= 140 0m; d2=0,32m; n2=0,016. ⇒A2=0,08 04 m2L3=800m; d3=0,24m; n3=0,02. ⇒A3=0, 045 2 m2Giải:Theo...