Finally, we find the mass-flow rate Rm using Eq. 10.31 to be:Rm = ρA1 v1 = ρRV = (103 kg/m3 )(6.735 × 10−5 m3 /s) = 0.067 kg/s70 g/sExample 10.16Water flowing from a faucet of cross-sectional area A = 2 cm2 is used to fill abucket of volume V = 30 liters = 30 × 103 cm3 , see Fig. 10.24. What is the[r]