<p>A shower head has 20 circular openings, each with radius&#160;<span>0.80</spa

Question

<p>A shower head has 20 circular openings, each with radius&#160;<span>0.80</span>&#160;<img src="http://session.masteringphysics.com/render?units=mm" alt="mm" align="middle" />. The shower head is connected to a pipe with radius&#160;<span>0.97</span>&#160;<img src="http://session.masteringphysics.com/render?units=cm" alt="cm" align="middle" />.</p>
<p>If the speed of water in the pipe is&#160;<span>3.4</span>&#160;<img src="http://session.masteringphysics.com/render?units=m%2Fs" alt="m/s" align="middle" />, what is its speed as it exits the shower-head openings?</p>

Explanation / Answer

The radius of the circular openings r2 = 0.8mm

the no.of openings n = 20

the radius of the pipe r1 = 0.97cm

the speed of water in pipe v1 = 3.4m/s

From equation of continuity

the volume rate of flow is constant

so A1v1 = A2v2

     r1^2 v1 = n(r2^2) v2

therefore the speed

     v2 = r1^2 v1 / nr2^2

          = [(0.97*10^-2m) ^2 / 20(0.8*10^-3)^2 ] (3.4m/s)

           = (7.35)(3.4m/s)

           = 24.99 or 25m/s

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