A shower head has 25 circular openings, each | vedclass

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(C) The radius of each circular opening is $r_1 = 1 \,mm = 10^{-3} \,m$.The cross-sectional area of each opening is $A_1 = \pi r_1^2 = \pi 	imes (10^

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(C) The radius of each circular opening is $r_1 = 1 \,mm = 10^{-3} \,m$.The cross-sectional area of each opening is $A_1 = \pi r_1^2 = \pi 	imes (10^{-3})^2 = \pi 	imes 10^{-6} \,m^2$.The radius of the pipe is $r_2 = 2 \,cm = 0.02 \,m$.The cross-sectional area of the pipe is $A_2 = \pi r_2^2 = \pi 	imes (0.02)^2 = 4\pi 	imes 10^{-4} \,m^2$.The speed of water in the pipe is $v_2 = 25 \,cm/s = 0.25 \,m/s$.Let $v_1$ be the speed of water as it exits the $n = 25$ openings.According to the equation of continuity, the total flow rate is constant:$n A_1 v_1 = A_2 v_2$$25 	imes (\pi 	imes 10^{-6}) 	imes v_1 = (4\pi 	imes 10^{-4}) 	imes 0.25$$v_1 = \frac{4\pi 	imes 10^{-4} 	imes 0.25}{25 	imes \pi 	imes 10^{-6}}$$v_1 = \frac{10^{-4}}{25 	imes 10^{-6}} = \frac{100}{25} = 4 \,m/s$.

$A$ shower head has $25$ circular openings, each with radius $1 \,mm$. The shower head is connected to a pipe with radius $2 \,cm$. If the speed of the water in the pipe is $25 \,cm/s$, what is its speed as it exits the shower head openings (in $\,m/s$)?

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