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You need to siphon water from a clogged sink. The sink hasan area of 0.38 $\mathrm{m}^{2}$ and is filled to a height of 4.0 $\mathrm{cm} .$ Yoursiphon tube rises 45 $\mathrm{cm}$ above the bottom of the sink and then descends 85 $\mathrm{cm}$ to a pail as shown in Fig. $59 .$ The siphontube has a diameter of 2.0 $\mathrm{cm} .$ (a) Assuming that the waterlevel in the sink has almost zero velocity, estimate thewater velocity when it enters the pail. (b) Estimate howlong it will take to cmpty the sink.

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(a) 2.9 $\mathrm{m} / \mathrm{s}$(b) 16$s$

Physics 101 Mechanics

Chapter 13

Fluids

Fluid Mechanics

University of Michigan - Ann Arbor

University of Winnipeg

McMaster University

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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so we need to siphon water in order to clear a clock sink. So we need to apply Bernoulli's equation immediately. The pressure at the top well, plus 1/2 times the density times of velocity at the top squared plus 1/2 times the density times G. Why at the top, I will be able to pee at the bottom. Well, it's a bot for bottom, plus 1/2 times the density times the velocity of the bottom squared plus the density times g times Wyatt the bottom. So we can say that the velocity at the bottom ah would be equal to two g times. Why at the top minus y at the bottom, Uh, to the 1/2 power and we can solve. So the velocity at the bottom would be equal to two times 9.8 times 0.44 meters and this will be to the 1/2 power. This is giving us 2.94 meters per second. So this would be the answer for a party and then for part B. They want us to find out the time. So how long it takes to empty out this clogged sink so the volumetric flow rate of the lower section, Times T would be equal to the volume of the sink. Therefore, the time it takes would be equal to the volume of the sink divided by the cross sectional area of the lower section times of velocity of the lower section. And so we can say that this is gonna be equal to the volume of the sink. So 0.38 meters times 0.4 meters, divided by pi times 0.1 meters Quantity squared times 2.397 meters per second and we find that two Delta T is gonna be equal to 16.47 seconds. So that's how long it takes to empty out the sink. This will be our final answer for part B. That is the end of the solution. Thank you for watching

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