Answer:
The answer for (1) 2.07 m/s (2) F= 63.8kN
Step-by-step explanation:
Solution
By applying the energy equation from the approach section 1 to the contacted section 2 with negligible head losses and assuming a horizontal bottom
we have y₁ + V₁²/ 2g = y₂ + q₂²/2gy₂²
Where
q is =flow rate
y = the depth
g = the acceleration
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