Correct answer is (a) 106.1 cm/s
To explain: The cross-sectional area of the hose will be given by:
A1 = π r^2 = π (2 cm/2)^2 = π cm^2
To find the velocity, v1, we use
Flow rate = A1 v1 = 20.0 L/min = \(\frac{20.0 × 10^3 cm^3}{60.0 s}\)
v1 = \(\frac{20.0 × \frac{10^3 cm^3}{60.0s}}{\pi cm^2}\)
= 106.1 cm/s