The flow rate $Q$ is related to the velocity $v$ and the cross-sectional area $A$ by: \[ Q = A \times v \] The area of the pipe is: \[ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.1}{2}\right)^2 = 7.85 \times 10^{-3} \, \text{m}^2 \] Thus, \[ v = \frac{Q}{A} = \frac{0.5}{7.85 \times 10^{-3}} \approx 5 \, \text{m/s} \]