Tài liệu Physics exercises_solution: Chapter 14 pdf

b The pressure difference is the gauge pressure, and the net force due to the water and the air is N... The pressure difference between the surface of the water and the bottom is due to

3 6 3

4

22 3

m V

m ρ

14.3:    7.02 103 kg m3

mm 0 30 0 15 0 5

kg 0158 0



V m  

14.4: The length L of a side of the cube is

cm

3.12m

kg104.21

kg0

3 3

L

14.5: m ρV πr3ρ

3 4





Same mass means 3 1a aluminum,l lead

1 a

103

3 3 3

8 3

4

30 sun

sun

m10412.1

kg1099.1m1096.6

kg1099.1

M D

1.409103 kg m3

30 3

4 3

4

30

mkg10594.0m10351.3

kg1099.1m1000.2

kg1099

5.941016 kg m3

14.7: p p0  ρgh

)sm80.9(mkg1030(

Pa1000.1

2 3

14.8: The pressure difference between the top and bottom of the tube must be at least

5980 Pa in order to force fluid into the vein:

Pa5980



ρgh

m581.0)sm80.9(mkg1050(

mN5980Pa

5980

2 3

)N105.16(b) With the extra weight, repeating the above calculation gives 1250cm2

14.11: a)ρgh(1.03103kg m3)(9.80m s2)(250m)2.52106Pa b) The pressure difference is the gauge pressure, and the net force due to the water and the air is

N

1078.1))m15.0()(

14.15: With just the mercury, the gauge pressure at the bottom of the cylinder

isp p0  pmghm With the water to a depth hw, the gauge pressure at the bottom of the cylinder isp p0 ρmghm pwghw If this is to be double the first value, then

h

The volume of water is

3 3

4 2

4m ) 8.16 10 m 816cm10

m)(12.0(0.680

V

14.16: a) Gauge pressure is the excess pressure above atmospheric pressure The

pressure difference between the surface of the water and the bottom is due to the weight

of the water and is still 2500 Pa after the pressure increase above the surface But the surface pressure increase is also transmitted to the fluid, making the total difference from atmospheric 2500 Pa+1500 Pa = 4000 Pa

b) The pressure due to the water alone is 2500 Paρgh Thus

m255.0)sm80.9(mkg1000(

mN2500

2 3

mkg1000(

mN1000

2 3

2





h

Thus the water must be lowered by 0.255m0.102m0.153m

14.17: The force is the difference between the upward force of the water and the

downward forces of the air and the weight The difference between the pressure inside and out is the gauge pressure, so

N.1027.2N300)m75.0(m)30(sm80.9(1003.1()

14.19: The depth of the kerosene is the difference in pressure, divided by the product

m250.0()smkg)(9.80205

(

Pa1001.2)m(0.0700N)

104.16(

3 2

5 2

)smkg)(9.801200

()2(

5 2

π

mg A

F

p

14.21: The buoyant force must be equal to the total weight; ρwaterVgρiceVgmg,so

,m563.0mkg920m

kg1000

kg0

3 3

m V

or 0.56m3 to two figures

14.22: The buoyant force is B17.50N11.20N6.30N, and

.m1043.6)sm80.9)(

mkg1000.1(

N)30.6

2 3

3 water

B V

The density is

.mkg1078.230.6

50.17)mkg1000.1

water water

B

g w

V

m

ρ

14.23: a) The displaced fluid must weigh more than the object, so  fluid b) If the ship does not leak, much of the water will be displaced by air or cargo, and the average density of the floating ship is less than that of water c) Let the portion submerged have

fluid 0

so,Then,

above the fluid is then 1 If 0,

P P p the entire object floats, and if  fluid, none

of the object is above the surface d) Using the result of part (c),

%

3232.0m

kg1030

)m103.04.0(5.0kg)042.0(1

3 -6 fluid

N6370

N5470and

water water

w V

V g ρ

w V

14.25: a) ρoilghoil 116Pa

b)  790kg m3 0.100m1000kg m3 0.0150m 9.80m s2921Pa

sm80.9

m100.0Pa805

2

2 top

g

w m

The density of the block is  3 822m 3

kg m

10 0 kg 822



p Note that is the same as the average density of the fluid displaced, 0.85 790kg m30.15(1000kg m3)

14.26: a) Neglecting the density of the air,

mkg107.2sm80.9

N

3 3

g w ρ

m V

00.11N891

V gρ w B w

T

14.27: a) The pressure at the top of the block is p p0 gh,where his the depth of the top of the block below the surface h is greater for block  , so the pressure is greater

at the top of block 

b)BflVobjg The blocks have the same volume Vobjso experience the same buoyant force

c) T wB so T wB

w ρVg The object have the same V but ρis larger for brass than for aluminum so

w is larger for the brass block Bis the same for both, so Tis larger for the brass block,blockB

14.28: The rock displaces a volume of water whose weight is 39.2N-28.4N10.8N.The mass of this much water is thus 10.8N 9.80m s2 1.102kgand its volume, equal to the rock’s volume, is

3 3 3

mkg101.00

kg102

The weight of unknown liquid displaced is 39.2N18.6N20.6N,and its mass is

kg

102.2sm

(0.80)s)

14.30:    

2

3 2

2 2

1 1 2

sm245.0)m0700.0s)(

m50.3(

Α Α

Α

Α v v

a) (i) 0.1050m , 2.33m s (ii) 0.047m2, 2 5.21m s

2 2

v

14.31: a) 16.98

)m150.0(

)sm20.1(

dt dV v

b) r2 r1 v1 v2  (dV dt) πv2 0.317m

14.32: a) From the equation preceding Eq (14.10), dividing by the time interval dt

gives Eq (14.12) b) The volume flow rate decreases by 1.50% (to two figures)

14.33: The hole is given as being “small,”and this may be taken to mean that the

velocity of the seawater at the top of the tank is zero, and Eq (14.18) gives

))((

= 2((9.80m s2)(11.0m)(3.00)(1.013105 Pa) (1.03103 kg m3))

28.4m s

Note that y = 0 and p pa were used at the bottom of the tank, so that p was the given

gauge pressure at the top of the tank

14.34: a) From Eq (14.18), v 2gh  2(9.80m s2)(14.0m) 16.6m s

b)vΑ(16.57m s)(π(0.3010 2m)2)4.6910 4 m3 s Note that an extra figure was kept in the intermediate calculation

14.35: The assumption may be taken to mean that v10in Eq (14.17) At the

maximum height,v2 0, and using gauge pressure for p1and p2,p2 0(the water is open

to the atmosphere), 1.47 105Pa

2

p

()m(16.2

m 10 00 2

s m 10 30 1

1

y y ρg v v ρ p

2 3

2 2

m 10 32 1

s m 10 65 4

14.40: From Eq (14.17), with y1  y2,

1 1

2 1 2 1 1

2 2

2 1 1

34

2

12

1

ρv p

v v ρ p v v ρ p

v

v  has been used

14.41: Let point 1 be where r1 4.00cm and point 2 be wherer2 2.00cm The

volume flow rate has the value 7200cm3 s at all points in the pipe

sm43.1so,cm

2 1 1

1

v

sm73.5so,cm

2 2 2

2

v

2 2 2

2

2 1 1

12

1

ρv ρgy

p ρv ρgy

2

1so

Pa,1040.2

2

2 1 1

2

5 2

14.44: a) The weight of the water is

m

N1061.2

, to

h      where AH b) The torque on a strip of vertical thickness dh about the bottom is dτ dFH hgwhH hdh, and

integrating from h to0 hH gives τ  ρgwH3 6 ρgAH2 6 c) The force depends

on the width and the square of the depth, and the torque about the bottom depends on the width and the cube of the depth; the surface area of the lake does not affect either result (for a given width)

14.47: The acceleration due to gravity on the planet is

d

p ρd

p g

gR M

2





14.48: The cylindrical rod has mass M radius ,, R and length L with a density that is proportional to the square of the distance from one end,  Cx2.

a)M dV Cx2dV The volume element dV πR2dx Then the integral

0

R C dx x R C

for C,C3M R2L3

b) The density at the xL end is 2  32 3  2  3 2

L πR

M L

14.49: a) At r 0,the model predicts   A12,700kg m3 and at r R,the model predicts

.mkg103.15m)

1037.6)(

mkg1050.1(mkg700,

3 2

4

33

44

34]

[4

mkg1050.1(3mkg700,123

m)1037.6(

3 3

44

33

B

A πG g

3

24

33

234

)mkg700,12(kgmN10673.6(

4 3

2 3 2

2 11

14.50: a) Equation (14.4), with the radius rinstead of height y becomes ,

.)(

s r R dr ρg

ρg dr r R

Pa

1071.1m)

1038.6(8

)smkg)(9.8010

97.5(3)0

2 6

2 24

c) While the same order of magnitude, this is not in very good agreement with the estimated value In more realistic density models (see Problem 14.49 or Problem 9.99), the concentration of mass at lower radii leads to a higher pressure

where R and h are the radius and height of the tank (the fact that 2Rhis more or less coincidental) Using the given numerical values givesF 5.07108 N

14.53: For the barge to be completely submerged, the mass of water displaced would

)mkg10

14.54: The difference between the densities must provide the “lift” of 5800 N (see

Problem 14.59) The average density of the gases in the balloon is then

.mkg96.0)m2200)(

sm80.9(

)N5800(m

kg23

3 2

)kg900(

3 3

3 water

water

V ρ

m V

g ρ w V

V

b) As the car is about to sink, the weight of the water displaced is equal to the weight

of the car plus the weight of the water inside the car If the volume of water inside the car

w V

V g p V w g Vρ

water water

3 d) The melted water takes up more volume than the salt water displaced, and so 0.46cm3 flows over A way of considering this situation (as a thoughtexperiment only) is that the less dense water “floats” on the salt water, and as there is insufficient volume to contain the melted ice, some spills over

14.57: The total mass of the lead and wood must be the mass of the water displaced, or

;)( Pb wood waterwood

wood Pb

solving for the volume VPb,

water Pb

wood water

3 3

3 3

2

mkg1000.1mkg103.11

mkg600mkg1000.1)m102.1(

14.58: The fraction f of the volume that floats above the fluid is 1 ,

f ρ ρ

and

2

1 1 2 2

f ρ ρ f f





 In this form, it’s clear that a larger f2

corresponds to a larger density; more of the stem is above the fluid Using

097.0

,242.0

) cm 2 13 (

) cm cm)(0.400 20 3 ( 2 )

mkg0899.0mkg(1.20

N000,

2 3

000,120

H air

He air

ρ ρ

This increase in lift is not worth the hazards associated with use of hydrogen

14.60: a) Archimedes’ principle states ,so

A

M L Mg gLA



b) The buoyant force is gA(Lx)MgF, and using the result of part (a) and

solving for x gives x ρgA F

c) The “spring constant,” that is, the proportionality between the displacement x and the applied force F, is k  ρgA, and the period of oscillation is

.2

2

ρgA

M π k

M π

kg0.70

2 3

m ρgA

mg ρgA

w x

b) Note that in part (c) of Problem 14.60,M is the mass of the buoy, not the mass of the man, and Ais the cross-section area of the buoy, not the amplitude The period is then

1.03 10 kg m 9.80m s  0.450m 2.42s.

kg950





π π

T

14.62: To save some intermediate calculation, let the density, mass and volume of the

life preserver be0,mandv, and the same quantities for the person be ρ1,M andV.Then, equating the buoyant force and the weight, and dividing out the common factor of

M v ρ

v

ρ

1 water

ρ

ρ v

M ρ

mkg101.03(0.80)1

m0.400

kg0.75mkg1003

1

732kg m3

14.63: To the given precision, the density of air is negligible compared to that of brass,

but not compared to that of the wood The fact that the density of brass may not be known the three-figure precision does not matter; the mass of the brass is given to three figures The weight of the brass is the difference between the weight of the wood and the buoyant force of the air on the wood, and canceling a common factor of

and)

air wood

wood brass

wood wood

ρ ρ

ρ M

V ρ M

kg

0958.0m

kg150

mkg20.11)kg0950.0(

1 3

14.64: The buoyant force on the mass A, divided

byg,must be7.50kg1.00kg1.80kg4.70kg(see Example 14.6), so the mass block

is 4.70kg3.50kg 8.20kg.a) The mass of the liquid displaced by the block is

3 3

the mass of the block, 8.20kg,as found above Scale E will read the sum of the masses of

the beaker and liquid, 2.80kg

14.65: Neglecting the buoyancy of the air, the weight in air is

N

0.45)(ρAuVAu ρA1VA1 

g

and the buoyant force when suspended in water is

N

6.0N39.0N45.0)

Au Au

Au

ρ ρ ρ

ρ gV

00.1(

)3.19



 33.5N

Note that in the numerical determination of wAu,specific gravities were used instead of densities

14.66: The ball’s volume is

3 3

3 (12.0cm) 7238cm3

43

cmg00.1)(

cm7238)(

16.0

N6

14.67: a) The weight of the crown of its volume V is w  ρcrowngV , and when

suspended the apparent weight is the difference between the weight and the buoyant force,

.)( crown water

1

or water

crown crown

crown water

f ρ

ρ fρ

ρ ρ

apparent weight is the same as the weight, which means that the buoyant force is

negligble compared to the weight, and the specific gravity of the crown is very large, as

reflected in the above expression b) Solving the above equations for f in terms of the

specific gravity, 1 ,

crown

water

ρ ρ

f   and so the weight of the crown would be

1 119.312.9N12.2N c) Approximating the average density by that of lead for a

“thin” gold plate, the apparent weight would be 1111.3 12.9N11.8N

14.68: a) See problem 14.67 Replacing f with, respectively, wwater wand wfluid w gives

,

- fluidfluid

steel

w w

w ρ

- waterfluid

steel

w w

w ρ

and dividing the second of these by the first gives

.-

water

-fluid water

fluid

w w

w w ρ

b) When wfluidis greater than wwater, the term on the right in the above expression is less than one, indicating that the fluids is less dense than water, and this is consistent with the buoyant force when suspended in liquid being less than that when suspended in water If the density of the fluid is the same as that of water wfluid  wwater, as expected Similarly,

if wfluid is less than wwater, the term on the right in the above expression is greater than one, indicating the the fluid is denser than water c) Writing the result of part (a) as

.1

1water

fluid water

fluid

f

f ρ

ρ





and solving for ffluid,

14.69: a) Let the total volume be V; neglecting the density of the air, the buoyant force

in terms of the weight is

,)(

0 m water

water    

ρ

g w g ρ gV ρ B

B V

w water

0b) 2.52 10 4m3

Cu water







 ρ w g g

ρ B Since the total volume of the casting is ,

14.70: a) Let d be the depth of the oil layer, h the depth that the cube is submerged in

the water, and L be the length of a side of the cube Then, setting the buoyant force equal

to the weight, canceling the common factors of g and the cross-section area and

sm80.9)(

mkg



pressure, found from the depths and densities of the fluids, is

Pa

539)sm80.9))(

mkgm)(1000025

.0()mkgm)(750

040

0

14.71: The ship will rise; the total mass of water displaced by the barge-anchor

combination must be the same, and when the anchor is dropped overboard, it displaces some water and so the barge itself displaces less water, and so rises

To find the amount the barge rises, let the original depth of the barge in the water be

0 m m A, where m andm

h    are the masses of the barge and the anchor, and

A is the area of the bottom of the barge When the anchor is dropped, the buoyant force

on the barge is less than what it was by an amount equal to the buoyant force on the anchor; symbolically,

 a steel water ,water

kg0

2 3

steel

a 0



or about 0.56 mm

14.72: a) The average density of a filled barrel is

,mkg875m

kg

m 0.120 kg 0 15 3

so the barrel floats

b) The fraction that floats (see Problem 14.23) is

%

0.15150.0mkg1030

mkg8751

3 water

0.120 kg 0 32 m

910   which means the barrel sinks

In order to lift it, a tension

N173)80.9)(

m120.0)(

1030()80.9)(

m120.0)(

1177

m 3

m

kg s

m 3

gL wgy

ω L B L

ρ ρ L ρ ρ

y  c)

m

046.0m)10

w A

g ρ w A

V y

water water 

.60)((

sm80.9)(

mkg(1.00x10

N)1050.2(

2 3

14.75: a) Consider the fluid in the horizontal part of the tube This fluid, with mass

,

Al

 is subject to a net force due to the pressure difference between the ends of the tube, which is the difference between the gauge pressures at the bottoms of the ends of the tubes This difference is ρg(yL yR), and the net force on the horizontal part of the fluid is

, )

g

a y

so the difference in heights between the columns is(2l 2)(l g)2l2 2g

Anticipating Problem, 14.77, an equivalent way to do part (b) is to break the fluid in

the horizontal part of the tube into elements of thickness dr; the pressure difference

between the sides of this piece is dp(2r)dr (see Problem 14.78), and integrating from r 0 torlgivesp2l2 2, giving the same result

c) At any point, Newton’s second law gives dpA pAdla from which the area A

cancels out Therefore the cross-sectional area does not affect the result, even if it varies

Integrating the above result from 0 to l gives p pal between the ends This is related to the height of the columns through p pgy from which p cancels out.