Answer:
2.68 m/s
Explanation:
volume] = [mass] / [density] so the volume flow rate of the gasoline is just [mass flow rate] / density
= [5.94 * 10^-2 kg/s] / [735 kg/m^3]
= 8.08 * 10^-5 m^3 /s
The cross sectional area (A) of the fuel line of radius r is given by
A = πr² = π * (3.10* 10^-3)²
A = 3.02* 10^-5 m²
Linear flow rate = [volume flow rate] / [cross sectional area]
= [8.08 * 10^-5 m^3/s] / [3.02 * 10^-5 m^2]
= 2.68 m/s
