$(1)$ \[A_1V_1=A_2V_2=Q\] \begin{align} V_1=\frac{Q}{A_1}=\frac{4Q}{\pi d_1^2}&=\frac{4\times0.2}{\pi\times0.3^2}\\ &=2.8294\\ &=2.83\,\rm{m/s} \end{align} \begin{align} V_2=\frac{4Q}{\pi d_2^2}&=\frac{4\times0.2}{\pi\times0.15^2}\\ &=11.318\\ &=11.32\,\rm{m/s}\end{align} $(2)$ \[ \frac{p_1}{\rho}+\frac{V_1^2}{2}=\frac{p_2}{\rho}+\frac{v_2^2}{2} \] \begin{align} p_2&=\frac{\rho}{2}(V_1^2-V_2^2)+p_1\\ &=\frac{1000}{2}(2.829^2-11.318^2)+300\times10^3\\ &=239953\\ &=240.0\,\rm{kpa} \end{align} $(3)$ \begin{align}F_x&=\rho QV_1-\rho QV_2\cos60^\circ +p_1\frac{\pi}{4}d_1^2-p_2\frac{\pi}{4}d_2^2\cos60^\circ \\ &=1000\times0.2\times(2.8294-11.318\cos60^\circ )\\ &+(300\times0.3^2-239.95\times0.15^2\times\cos60^\circ )\times\frac{\pi}{4}\times1000\\ &=18520\\ &=18.52\,\rm{kN} \end{align} \begin{align}F_y&=0-(-\rho QV_2\sin60^\circ )+P^2\frac{\pi}{4}d_2^2\sin60^\circ \\ &=1000\times0.2\times11.318\sin60^\circ \\ &+239.95\times1000\times\frac{\pi}{4}\times0.15^2\sin60^\circ \\ &=5632.5\\ &=5.63\,\rm{kN} \end{align} $(4)$ \begin{align}F=\sqrt{F_x^2+F_y^2}&=\sqrt{18.52^2+5.633^2}\\ &=19.36\rm{kN} \end{align} \begin{align}\beta=\tan^{-1}\frac{F_y}{F_x}&=\tan^{-1}\frac{5.633}{18.52}\\ &=16.92^\circ \end{align}