Answer
$V^{}= 289.9$ m/s
$m^{.}= 1310$ kg/s
Work Step by Step
The velocity at the nozzle exit is the sonic speed, which is determined to be $$
V=c=\sqrt{k R T}=\sqrt{(1.33)(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})\left(\frac{1000 \mathrm{~m}^2 / \mathrm{s}^2}{1 \mathrm{~kJ} / \mathrm{kg}}\right)(220 \mathrm{~K})}=289.8 \mathrm{~m} / \mathrm{s}
$$ Noting that thrust $F$ is related to velocity by $F=\dot{m} V$, the mass flow rate of combustion gases is determined to be
$$ \dot{m}=\frac{F}{V}=\frac{380,000 \mathrm{~N}}{289.8 \mathrm{~m} / \mathrm{s}}\left(\frac{1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^2}{1 \mathrm{~N}}\right)=1311 \mathrm{~kg} / \mathrm{s} \cong 1310 \mathrm{~kg} / \mathrm{s}
$$
