Explanation:
Given:
[tex]A_1[/tex] = 4.5 cm[tex]^2[/tex]
[tex]v_1[/tex] = 40 cm/s
[tex]v_2[/tex] = 90 cm/s
[tex]A_2[/tex] = ?
a) The continuity equation is given by
[tex]A_1v_1 = A_2v_2[/tex]
Solving for [tex]A_2[/tex],
[tex]A_2 = \dfrac{v_1}{v_2}A1 = \left(\dfrac{40\:\text{cm/s}}{90\:\text{cm/s}}\right)(4.5\:\text{cm}^2)[/tex]
[tex]= 2\:\text{cm}^2[/tex]
b) If the cross-sectional area is reduced by 50%, its new area [tex]A_2'[/tex] now is only 1 cm^2, which gives us a radius of
[tex]r = \sqrt{\dfrac{A_2'}{\pi}} = 0.564\:\text{cm}[/tex]
