Answer
a) See solution
b) $v_1=0.165\frac{m}{s}$
Work Step by Step
a) $P_2+\frac{1}{2}\rho v_2^2=P_1+\frac{1}{2}\rho v_1^2$
$(v_1^2-v_2^2)=\frac{2(P_2-P_1)}{\rho}$
$v_1^2-A_1^2\frac{v_1^2}{A_2^2}=v_1^2(1-\frac{A_1^2}{A_2^2})=v_1^2(\frac{A_2^2-A_1^2}{A_2^2})$
$v_1^2=\frac{2(P_2-P_1)A_2^2}{\rho (A_2^2-A_1^2)}$
$v_1=A_2\sqrt{\frac{2(P_1-P_2)}{\rho(A_1^2-A_2^2)}}$
b) $r_1=0.035m$
$r_2=0.01m$
$A_1=3.85\times10^{-3}m^2$
$A_2=3.14\times10^{-4}m^2$
$P_1-P_2=(18mmHg)\Bigg(\frac{113\frac{N}{m^2}}{mmHg}\Bigg)=2034\frac{N}{m^2}$
$A_1^2-A_2^2=1.482\times10^{-5}m^2-9.860\times10^{-8}m^2=1.472\times10^{-5}m^2$
$v_1=A_2\sqrt{\frac{2(P_1-P_2)}{\rho(A_1^2-A_2^2)}}=0.165\frac{m}{s}$
