Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 9, Problem 83P

(a)

To determine

The speed with which the water leave the showerhead.

(a)

Expert Solution
Check Mark

Answer to Problem 83P

The speed with which the water leave the showerhead is 26m/s.

Explanation of Solution

Find the equation for the speed with which the water leave the showerhead.

P1+ρgy1+12ρv12=P2+ρgy2+12ρv2212ρ(v12v22)=P2P1+ρg(y2y1)v12A12A22v12=2Pgaugeρ2ghv1=2Pgaugeρ2gh1[Nr12r22]2 (I)

Here, P2P1=Pgauge is the gauge pressure, y2y1=h is the height of the showehead, ρ is the density of water, g is the acceleration due to gravity, r2 is the radius of the output pipe, r1 is the radius of the showehead, A2 is the area of the output pipe, A1 is the area of the showehead and N is the number of holes.

Conclusion:

Substitute 6.3mm for r2, 0.33mm for r1, 36 for N, 410kPa for Pgauge, 6.7m for h, 1.00×103kg/m3 for ρ and 9.80m/s2 for g in equation (I) to find v1.

v1=2((410kPa)(103Pa1kPa))1.00×103kg/m32(9.80m/s2)(6.7m)1362[0.33mm6.3mm]4=2(410×103Pa)1.00×103kg/m32(9.80m/s2)(6.7m)1362[0.33mm6.3mm]4=26m/s

Thus, the speed with which the water leave the showerhead is 26m/s.

(b)

To determine

The speed with which the water moves through the output pipe of the pump.

(b)

Expert Solution
Check Mark

Answer to Problem 83P

The speed with which the water moves through the output pipe of the pump is 2.6m/s.

Explanation of Solution

Find the equation for the speed with which the water moves through the output pipe of the pump.

v2=A1A2v1=Nr12r22v1 (II)

Here, v2 is the speed with which the water moves through the output pipe of the pump.

Conclusion:

Substitute 6.3mm for r2, 0.33mm for r1, 26m/s for v1 and 36 for N in equation (II) to find v2.

v2=36(0.33mm6.3mm)2(26m/s)=2.6m/s

Thus, the speed with which the water moves through the output pipe of the pump is 2.6m/s.

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Chapter 9 Solutions

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