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You're holding a hose at waist height and spraying water horizontally with it. The hose nozzle has a diameter of 1.80 $\mathrm{cm},$ and the water splashes on the ground a distance of 0.950 m horizontally from the nozzle. Suppose you now constrict the nozzle to a diameter of 0.750 $\mathrm{cm}$ ; how far horizontally from the nozzle will the water travel before hitting the ground? (Ignore air resistance.)

$d_{2}=5.472 m$

Physics 101 Mechanics

Chapter 13

Fluid Mechanics

Temperature and Heat

University of Washington

Hope College

University of Sheffield

University of Winnipeg

Lectures

04:16

In mathematics, a proof is…

04:48

In mathematics, algebra is…

14:15

You hold a hose at waist h…

02:25

A garden hose is attached …

03:43

(a) What is the pressure d…

07:45

What is the pressure drop …

01:57

. The nozzle of a fountain…

0:00

A jet of water squirts out…

03:22

One arm of a U-shaped tube…

07:25

A drinking fountain shoots…

A fire hose exerts a force…

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01:39

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. A water hose is used to …

01:49

Water flows through a fire…

03:20

06:22

A nozzle at $A$ discharges…

okay. And this problem, we have a person shooting water from a hose horizontally att the level of their waste. We're given the initial diameter of the hose, which we want to convert to a radius expressed in meters. That's given by this expression. And we know how far the water travels at this radius of hose. It goes 0.95 meters, then were asked what happens if we constrict the radius to a new value given by this expression that I'm circling? Um, we're asked how far the water is going to go now. So the principle that we want to use to tackle this problem is continuity Cross sectional area of the hose times. Its velocity is equal to any new cross sectional area times the velocity of that area. So because it's a hose, the cross section is in the shape of a circle, and so the area will be given by pi r squared. So we recast in this form and then kinnah magically, we know that the velocity is equal to the horizontal distance that the water travels divided by the time that it spends in the air and because the water is being shot horizontally from the same height before and after the hoses constricted. The time it's going to spend in the air is the same. And so we can replace 31 with X one over tea and be two x two over T, where the two teas are the same. So we can cancel these from the equation. And what we have then is that it fto is equal toe are one squared times x one divided by our two squared where our wine is thiss value. Our two is this value on X one. Is this values now? It's just a matter of plugging in the numbers.

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