This is given by the equation Q = A * v, where A is the cross-sectional area of the hose and v is the velocity of the water. We are given that Q = 13.3 m^3/s and A = 3 m^2, so we can solve for v:
\[v = \frac{Q}{A} = \frac{13.3 \, \text{m}^3/\text{s}}{3 \,
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