Answer:
power = 5.7 kW
Explanation:
given data
plant flow Q = 3,780 m³/day
velocity gradient = [tex]10^4[/tex] Â /s
viscosity (m) is 0.001307 N-s/m²
solution
we get here first volume of mining vessel that is express as
volume = Q × ∅
volume = 3,780 × 1 sec  × [tex]\frac{1}{60}[/tex]  × [tex]\frac{1}{1440}[/tex] Â
volume = 0.044 m³ Â
and
now we get here power that is
power = velocity gradient² × viscosity × volume  ......................1
power =  0.001307 ×  ([tex]10^4[/tex])² × 0.044
power = 5.7 kW