Homework 10 Fluids Solutions

duong (jld3967) Homework 10: Fluids gust (57605) 1

This print-out should have 18 questions.Multiple-choice questions may continue onthe next column or page find all choicesbefore answering.

001 (part 1 of 3) 10.0 pointsA column of water has a diameter of 1 m anda depth of 9.4 m.How much pressure is at the bottom of

the column? The acceleration of gravity is9.8 N/kg .

Correct answer: 92120 Pa.

Explanation:

Let : = 1000 kg/m3 ,

g = 9.8 N/kg , and

d = 9.4 m .

The pressure is

P = d g

= (1000 kg/m3) (9.4 m) (9.8 N/kg)

= 92120 Pa .

002 (part 2 of 3) 10.0 pointsWhat is the weight of this column of water?

Correct answer: 72350.9 N.

Explanation:

Let : r = 0.5 m .

The weight of this column of water is

W = P A = P (pi r2)

= (92120 Pa) pi (0.5 m)2

= 72350.9 N .

003 (part 3 of 3) 10.0 pointsWhat would be the pressure if the column hada radius of 6 m and the same depth?

Correct answer: 92120 Pa.

Explanation:

The pressure doesnt depend on the area ofthe column, just on the depth.

P = 92120 Pa .

004 10.0 pointsThe Mariana Trench, in the Pacific Ocean, isabout 11 km deep.If atmospheric pressure at sea level is

1.01 105 Pa and the density of sea wateris 1025 kg/m3, how much pressure would asubmarine need to be able to withstand toreach this depth? The acceleration of gravityis 9.81 m/s2 .

Correct answer: 1.10709 108 Pa.

Explanation:

Let : h = 11 km = 11000 m ,

P0 = 1.01 105 Pa ,

sea water = 1025 kg/m3 , and

g = 9.81 m/s2 .

P = P0 + g h

= 1.01 105 Pa

+ (1025 kg/m3) (9.81 m/s2)

(11000 m)

= 1.10709 108 Pa .

005 10.0 pointsAn Olympic skier moving at 30 m/s down a19 slope encounters a region of wet snow ofcoefficient of kinetic friction = 0.8.The acceleration of gravity is 9.8 m/s2 .How far down the slope does she travel

before coming to a halt?

Correct answer: 106.577 m.

Explanation:Align the x axis along the incline, and let

down the incline be positive. The normal


force is N = mg cos . Hence the frictionalforce is

f = N = mg cos

Applying Newtons second law to the skieralong the incline yields

F = mg sin mg cos = ma ,

or solving for a,

a = g [sin cos ]

= (9.8 m/s2)

[sin 19 (0.8) cos 19]

= 4.2223 m/s2 .

To find how far the slope the skier travelsbefore stopping, use the kinematic equation

v2 = v20 + 2 a d .

Setting v = 0 and solving for d,

d = v20

2 a

= (30 m/s)2

2 (4.2223 m/s2)= 106.577 m .

006 10.0 pointsOne method of measuring the density of a liq-uid is illustrated in the figure. One side of theU-tube is in the liquid being tested, and theother side is in water of density 1000 kg/m3.The air is partially removed at the upper

part of the tube and the valve is closed. Theheight of the water above its pool surface is0.57 m . The height of the liquid above itspool surface is 0.36 m . The difference in theheights of the pool surfaces is 0.06 m .

Valve

test liquid water

0.36 m0.57 m

0.06 m

Find the density of the liquid on the left.

Correct answer: 1583.33 kg/m3.

Explanation:

Let : w = 1000 kg/m3 ,

hw = 0.57 m ,

h = 0.36 m , and

h = 0.06 m .

The pressure at the upper surface of eachliquid is given by

P = Patm w g hw = Patm g h.

Therefore,

=hwh

w =(0.57 m)(1000 kg/m3)

0.36 m

= 1583.33 kg/m3 .

007 10.0 pointsA simple U-tube that is open at both ends ispartially filled with a heavy liquid of density1000 kg/m3 . A liquid of density 800 kg/m3 isthen poured into one arm of the tube, forminga column 7 cm in height, as shown.

h

7 cm light liquid800 kg/m3

heavy liquid1000 kg/m3

What is the difference in the heights of thetwo liquid surfaces?


Correct answer: 1.4 cm.

Explanation:

Let : ` = 7 cm ,

` = 800 kg/m3 , and

h = 1000 kg/m3 .

Because the liquid in the U-tube is static,the pressure exerted by the heavy liquid col-umn of height ` h in the left branch of thetube must balance the pressure exerted bythe liquid of height h poured into the rightbranch, so

P0 + (` h) h g = P0 + ` ` g .

h = `

(1

`h

)

= (7 cm)

(1

800 kg/m3

1000 kg/m3

)

= 1.4 cm .

008 10.0 points

The small piston of a hydraulic lift has across-sectional area of 1.8 cm2 and the largepiston has an area of 47 cm2, as in the figurebelow.

F

1.8 cm2area

47 cm2

What force F must be applied to the smallpiston to maintain the load of 63 kN at aconstant elevation?

Correct answer: 2412.77 N.

Explanation:

Let : A1 = 1.8 cm2 ,

A2 = 47 cm2 , and

W = 63 kN .

According to Pascals law, the pressure ex-erted on A1 must be equal to the one exerted

on A2 . The pressure P1 =F

A1must be equal

to the pressure P2 =W

A2due to the load.

F

A1=

W

A2,

F =A1A2

W =(1.8 cm2)

(47 cm2)(63000 N)

= 2412.77 N .

009 10.0 pointsA fountain sends a stream of water 23.1 m upinto the air.If the base of the stream is 5.5 cm in diame-

ter, what power is required to send the waterto this height? The acceleration of gravity is9.8 m/s2 .

Correct answer: 11444.2 W.

Explanation:

Let : h1 = 23.1 m ,

d = 5.5 cm = 0.055 m ,

r = 2.75 cm = 0.0275 m , and

g = 9.8 m/s2 .

Using conservation of energy,

1

2mv2 = mg h

v =2 g h

=2 (9.8 m/s2) (23.1 m)

= 21.2782 m/s

is the velocity of the stream at its base. Theflow rate at the base is

pi r2 v = pi (0.0275 m)2(21.2782 m/s)

= 0.0505533 m3/s .


Using the density of water as 1000 kg/m3, wesee that 50.5533 kg of water emerges from thefountain each second.

m

t= pi r2 v

= (1000 kg/m3) pi (0.0275 m)2

(21.2782 m/s)

= 50.5533 kg .

Thus, each second, sufficient work is done toraise 50.5533 kg to a height of 23.1 m or

P =W

t= mg

h

t

= pi r2 v gh

t= (1000 kg/m3) pi (0.0275 m)2

(21.2782 m/s) (9.8 m/s2)23.1 m

1 s

= 11444.2 W .

010 10.0 pointsA non-uniform disk of mass M and radius Rhas its mass distributed according to

(r) = b

[( rR

)2+ 1

],

where b is a constant to be determined and ris measured from the center of the disc.What is the rotational inertia of the disk

about an axis through the center of mass,perpendicular to the plane of the disc? A con-venient form for the area differential consistsof a ring of radius r and thickness dr, so thatdA = 2 pi r dr .

1.7

8M R2

2.1

2M R2

3.1

3M R2

4.2

5M R2

5.5

9M R2 correct

6.2

3M R2

7. 2M R2

8. M R2

Explanation:The mass per unit area of the disc is

dM

dA= (r) = b

[( rR

)2+ 1

], so

M =

dm =

dM

dAdA .

dA = 2 pi r dr , so

dM

dAdA = b

(r2

R2+ 1

)(2 pi r dr)

= 2 pi b

(r3

R2+ r

)dr and

M = 2 pi b

R0

(r3

R2+ r

)dr

= 2 pi b

(r4

4R2+r2

2

)R

0

= 2 pi bR2(1

4+

2

4

)=

3 pi bR2

2

b =2M

3 piR2.

Thus the moment of inertia is

ICM =

R0

r2 dm =

R0

r2dM

dAdA

= 2 pi b

R0

(r5

R2+ r3

)dr

= 2 pi

(2M

3 piR2

) (r6

6R2+r4

4

)R

0

=4M

3R2R4

(1

6+

1

4

)

=4M R2

3

(2

12+

3

12

)=

5

9M R2

011 10.0 points


A water hose 2.26 cm in diameter is used tofill an 18 L bucket.If it takes 2.28 min to fill the bucket, what is

the speed at which the water leaves the hose?One liter is equal to 1000 cm3.

Correct answer: 32.8005 cm/s.

Explanation:

Let : r = 1.13 cm ,

V = 18 L , and

t = 2.28 min .

The cross-sectional area of the hose is

A = pi r2 = pi (1.13 cm)2

= 4.0115 cm2 .

According to the data given, the flow rate isequal to

F =V

t= v A

v =V

A t

=18 L

(4.0115 cm2) (2.28 min)

1000 cm3

1 L

1 min

60 s

= 32.8005 cm/s .

012 10.0 pointsA balloon filled with helium plus air at atmo-spheric pressure is designed to support a massof 106 kg (payload + empty balloon).Air density is 1.29 kg/m3 , helium plus air

density in the balloon is 0.956 kg/m3 , andhelium density is 0.179 kg/m3 .Ignoring the volume of the payload, find

the volume of the balloon.

Correct answer: 317.365 m3.

Explanation:

Let : a = 1.29 kg/m3 ,

b = 0.956 kg/m3 ,

He = 0.179 kg/m3 , and

M = 106 kg .

The buoyant force must equal the totalweight (payload, balloon, and helium ):

a g V =M g + b g V

V =M

a b

=106 kg

1.29 kg/m3 0.956 kg/m3

= 317.365 m3 .

013 (part 1 of 2) 10.0 pointsAn incompressible, non-viscous liquid of den-sity flows with speed vx into a pipe of diam-eter dx. The diameter of the pipe decreasesto du at its exit end. The elevation of the en-trance is h above the elevation of the exit endof the pipe. The pressure at the exit of thepipe is Pu.

Px

vxdx

Pu

vudu

h

What is the velocity vu of the liquid flowingout of the exit end of the pipe?

1. vu =

(dxdu

)vx

2. vu =

(d2ud2x

)vx

3. vu =

(Axd2u

)vx

4. None of these

5. vu =

(Audx

)vx

6. vu =

(dudx

)vx

7. vu =

(Axdu

)vx


8. vu =

(d2xd2u

)vx correct

9. vu =

(Aud2x

)vx

Explanation:From the continuity equation, the volume

of liquid entering the pipe during a certaintime interval must equal the volume of liquidleaving the pipe during that time interval:

Ax vx = Au vu

vu =AxAu

vx =pi r2zpi r2x

vx =

(dxdu

)2v1 ,

since the ratio of the radii is the same as theratio of the diameters.

014 (part 2 of 2) 10.0 pointsApplying Bernoullis principle, what is thepressure Px at the entrance end of the pipe?Patm is atmospheric pressure.

1. Px = g h+1

2 (vu

2 vx2)

2. Px = Pu Patm + g h+1

2 (vu

2 vx2)

3. Px = Pu + g h+1

2 (vx

2 vu2)

4. Px = Pu g h+1

2 (vx

2 vu2)

5. Px = Pu g h+1

2 (vu

2 vx2) cor-

rect

6. Px = Pu + g h

7. Px = Pu Patm g h+1

2 (vx

2 vu2)

8. Px = Pu + g h+1

2 (vu

2 vx2)

9. Px = g h+1

2 (vx

2 vu2)

10. Px = Pu g h

Explanation:

y = yu yx = h, since the entranceheight yx is greater than the exit height yu .Applying Bernoullis principle to the fluid flowat the entrance and exit of the pipe,

Px + g yx +1

2 v2x = Pu + g yu +

1

2 v2u

Px = Pu + g (yu yx) +1

2 (vu

2 vx2)

= Pu g h+1

2 (vu

2 vx2) .

015 10.0 pointsA bridge of length 48 m and mass 40000 kgis supported at each end. A truck of mass41400 kg is located 26 m from the left end.

A B

26 m48 m

What is the force on the bridge at the leftpoint of support? The acceleration of gravityis 9.8 m/s2 .

Correct answer: 3.81955 105 N.

Explanation:

Let : ` = 48 m ,

M = 40000 kg ,

m = 41400 kg ,

x = 26 m , and

g = 9.8 m/s2 .

For equilibrium,

~F = 0 and

~ = 0 .

Let NA and NB be the normal forces at thepoints of support. Choosing the origin atpoint A,

= mg x

1

2M g `+NB ` = 0 ,


NB =M g

2+mg

x

`and

Fy = NA +NB M g mg = 0 .

NA =M g +mg NB

NA =M g +mg M g

2mg

x

`

=M g

2+mg

(1

x

`

)

=(40000 kg) (9.8 m/s2)

2+ (41400 kg) (9.8 m/s2)

(1

26 m

48 m

)

= 3.81955 105 N .

016 10.0 pointsCalculate the depth in the ocean at which thepressure is three times atmospheric pressure.The acceleration of gravity is 9.81 m/s2 andthe density of sea water is 1025 kg/m3 .

Correct answer: 20.089 m.

Explanation:

Let : P = 3P0 ,

P0 = 1.01 105 Pa ,

= 1025 kg/m3 , and

g = 9.81 m/s2 .

P = P0 + h g

h =P P0g

=3P0 P0

g =

2P0g

=2 (1.01 105 Pa)

(9.81 m/s2) (1025 kg/m3)

= 20.089 m .

017 10.0 pointsA beaker of water is sitting on a sensitivespring scale.

If you dip your finger into the water, butdo not touch the side of the beaker, and thebeaker does not overflow, what happens tothe reading of the scale? (Part of your fingeris completely underwater.)

1. The reading of the scale remains preciselythe same, of course.

2. The reading of the scale decreases.

3. It depends on just how much of yourfinger is submerged. Unless we are told, wecant answer.

4. The reading of the scale increases becauseof the extra pressure. correct

5. It depends on how deep the water is. Insome cases, the scale reading might increase,in other cases it might decrease.

Explanation:The reading of the scale increases because

your finger displaces water. The water muststill support itself, but must also exert anadditional buoyant force on your finger. ByNewtons Third Law, this extra upward forceexerted on your finger by the water must bepaired with a downward force exerted on thewater by your finger. This extra downwardforce is passed onto the supporting springscale, which must balance it, so the reading ofthe scale increases.Another way to understand this is that the

level of the water in the beaker increases, sothe pressure it exerts on the bottom of thebeaker increases.

018 10.0 pointsAn object of volume V is floating at the in-terface between water and a denser liquid.2

3of the objects volume is displacing water,

while1

3of the objects volume is displacing

the heavier liquid.


w

If the density of the heavier liquid is andthe density of water is w, what is the totalbuoyant force on the object?

1.2 w V g

3

2.(+ 2 w)V g

3correct

3. ( w)V g

4. (+ w)V g

5.w V g

3

6. V g

3

7.(2 + w)V g

3

Explanation:The buoyant force is the density of the fluid

displaced times the volume displaced, so

FB =2

3w V g +

1

3 V g =

(+ 2 w) V g

3.

Homework 10 Fluids Solutions

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