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Four lawn sprinkler heads are fed by a 1.9 -cm-diameterpipe. The water comes out of the heads at an angle of $35^{\circ}$to the horizontal and covers a radius of 7.0 $\mathrm{m} .$ (a) What isthe velocity of the water coming out of each sprinklerhead? (Assume zero air resistance.) (b) If the output diam-eter of each head is 3.0 $\mathrm{mm}$ , how many liters of water dothe four heads deliver per second? (c) How fast is thewater flowing inside the 1.9 -cm-diameter pipe?

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(a) 8.5 $\mathrm{m} / \mathrm{s}$(b) 0.24$L / s$(c) 0.85 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 13

Fluids

Fluid Mechanics

Rutgers, The State University of New Jersey

University of Washington

Simon Fraser University

University of Sheffield

Lectures

03:45

In physics, a fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases, plasmas and, to some extent, plastic solids.

09:49

A fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases and plasmas. Fluids display properties such as flow, pressure, and tension, which can be described with a fluid model. For example, liquids form a surface which exerts a force on other objects in contact with it, and is the basis for the forces of capillarity and cohesion. Fluids are a continuum (or "continuous" in some sense) which means that they cannot be strictly separated into separate pieces. However, there are theoretical limits to the divisibility of fluids. Fluids are in contrast to solids, which are able to sustain a shear stress with no tendency to continue deforming.

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Four lawn sprinkler heads …

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A garden hose with an inte…

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Garden Hose A garden hose …

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A garden hose with a diame…

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Water flows through a fire…

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Water is pumped through a …

09:31

Water at $20^{\circ} \math…

04:28

A garden hose having an in…

or per day. We can assume that the water was launched at ground level and lands at ground level so we can use the range formula from example 3 10 and we can say that then the ranges equaling the initial square times sign of two theta divided by G. And so the initial velocity could be given as the square root of our G, the range times acceleration due to gravity divided by sign of tooth data. And so we can say that the initial initial velocity rather would be equaling the square root of the range 7.0 meters multiplied by 9.80 meters per second squared. This would be divided bye sign of 70 degrees and this is equaling eight 0.54 meters per second. This would be our final answer for part A. We know that then, for part B, we know that the volume we confined the volumetric flow rate where the volume rate of flow is the area of the flow times the speed of the flow. We're going to multiply this by four for the four heads and so you can say que the volumetric flow rate would be equaling the cross sectional area times the speed of the flow and this would be equaling 24 times pi r squared times v and so this would be equal to four pi multiplied by 1.5 times 10 to the negative third meters quantity squared multiplied by eight 0.5 for four meters per second. And so the volumetric floor eight q would be equaling 22.416 times 10 to the negative fourth meters cubed per second. For every one leader, there are 10 to the negative third cubic meters and so we can then say that this is gonna be equaling approximately 0.24 two leaders her second. This would be the volumetric flow rate for part B and then for part C, we can use the equation of continuity at a constant density. And this is giving us that the area times the velocity in the supply pipe would be equal in the area times the velocity times in the heads. And so we can then say that the velocity through the supply pipe would be equaling the volumetric flow rate through the heads, which we have already calculated divided by the area of the supply pipe and so this would be equaling 2.416 times 10 to the negative forth meters cubed per second, divided by pi times 0.95 times 10 to the negative second meters quantity squared. This is giving us 0.85 meters per second. The speed through the supply pipe. This would be our final answer for part C. That is the end of a solution. Thank you for

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