Tài liệu Physics exercises_solution: Chapter 14 pdf

14.1:
ρVgmgw 

 
 
   
N8.41sm80.9m1043.1m858.0mkg108.7
2
2
233


π
or 42 N to two places. A cart is not necessary.
14.2:
 
 
.mkg1033.3
m1074.1
kg1035.7
33
3
6
3
4
22
3
3
4





π
πr
m
V
m
ρ
14.3:
 
 
.mkg1002.7
3
3
mm0.300.150.5
kg0158.0
3


V
m
ρ
You were cheated.
14.4: The length L of a side of the cube is
cm.3.12
mkg104.21
kg0.40
3
1
3

1
3
1
33





















m
VL
14.5:
ρπrρVm
3

3
4

Same mass means
 
leadlaluminum,a
1
3
1a
3
a
 ρrρr

6.1
107.2
103.11
31
3
3
31
a
1
1
a






















ρ
ρ
r
r
14.6: a)
 
327
30
3
8
3
4
30
sun
sun

m10412.1
kg1099.1
m1096.6
kg1099.1






π
V
M
D

33
mkg10409.1 
b)
 
317
313
30
3
4
3
4
30
mkg10594.0
m10351.3
kg1099.1

m1000.2
kg1099.1







π
D

316
mkg1094.5 
14.7:
ρghpp 
0

m91.9
)sm80.9()mkg1030(
Pa1000.1
23
5
0





ρg

pp
h
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(
mN5980Pa5980
23
2

gh
h
14.9: a)
  
 
Pa.706m12.0sm80.9mkg600
23
ρgh
b)
  
 
Pa.1016.3m250.0sm80.9mkg1000Pa706
323

14.10: a) The pressure used to find the area is the gauge pressure, and so the total area
is




2
3
3
cm805
)Pa10205(
)N105.16(
b) With the extra weight, repeating the above calculation gives
2
cm1250
.
14.11: a)
Pa.1052.2)m250)(sm80.9)(mkg1003.1(
6
2
33
ρgh
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()(Pa1052.2(
526
 π
14.12:
atm.61.9Pa1027.6)m640)(sm80.9)(mkg1000.1(
6233
 ρghp
14.13: a)


)m1000.7)(sm80.9)(mkg106.13(Pa10980
22332

2a
ρgyp
Pa.1007.1
5

b) Repeating the calcultion with
00.4
12
 yyy
cm instead of
Pa.101.03gives
5
2
y
c) The absolute pressure is that found in part (b), 1.03
Pa.10
5

d)
Pa1033.5)(
3
12
 ρgyy
(this is not the same as the difference between the results of
parts (a) and (b) due to roundoff error).
14.14:
Pa.100.6)m1.6)(sm80.9)(mkg1000.1(
4233
ρgh
14.15: With just the mercury, the gauge pressure at the bottom of the cylinder

is


mm0
ghppp
With the water to a depth
w
h
, the gauge pressure at the bottom of the
cylinder is
.
wwmm0
ghpghρpp 
If this is to be double the first value, then
m.mww
ghρghρ 

m680.0)1000.1106.13)(m0500.0()(
33
wmmw
 ρρhh
The volume of water is
33424
cm816m108.16)m10m)(12.0(0.680A 

hV
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
23
2
h
To keep the bottom gauge pressure at 2500 Pa after the 1500 Pa increase at the surface,
the pressure due to the water’s weight must be reduced to 1000 Pa:
m102.0
)sm80.9)(mkg1000(
mN1000
23
2
h
Thus the water must be lowered by
m0.153m102.0m255.0 
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()(
5223
 wAρghF
14.18:
 
)m(2.00Pa1093m)2.14)(sm71.3)(mkg1000.1(Pa10130
232333


N.1079.1
5

14.19: The depth of the kerosene is the difference in pressure, divided by the product
,
V
mg
ρg 
m.14.4
)m250.0()smkg)(9.80205(
Pa1001.2)m(0.0700N)104.16(
32
523



h
14.20:
atm.64.1Pa1066.1
m)15.0(
)smkg)(9.801200(
)2(
5
2
2
2

πdπ
mg

A
F
p
14.21: The buoyant force must be equal to the total weight;
so,gg
icewater
mgVρVρ 
,m563.0
mkg920mkg1000
kg0.45
3
33
water





ice
ρρ
m
V
or
3
m56.0
to two figures.
14.22: The buoyant force is
and N,30.6N20.11N17.50 B
.m1043.6
)sm80.9)(mkg1000.1(

N)30.6(
34
233
water




gρ
B
V
The density is
.mkg1078.2
30.6
50.17
)mkg1000.1(
3333
water
water








B
w
ρ

gρB
gw
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
volume V, and the total volume be

fluid0
so,Then,.
fluido0
ρ
ρ
V
V
VρVV

The fraction
above the fluid is then
0,If .1
fluid
 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),
%.3232.0
mkg1030
)m103.04.0(5.0kg)042.0(
11
3
3-6
fluid





14.24: a)
   
N.6370m650.0sm80.9mkg1000.1
3233
water
 gVρB
b)
kg.558
2
sm9.80
N900-N6370



g
TB
g
w
m
c) (See Exercise 14.23.) If the submerged volume is
,V

%.9.858590.
N6370
N5470
and
waterwater




gVρ
w
V
V
g
ρ
w
V
14.25: a)
Pa.116
oiloil

ghρ
b)
 
 
 
 
  
Pa.921sm80.9m0150.0mkg1000m100.0mkg790
233

c)
 
  
 
kg.822.0
sm80.9
m100.0Pa805
2
2
topbottom



g
App
g
w
m
The density of the block is
 

.822
33
m
kg
m10.0
kg822.0
p
Note that is the same as the average
density of the fluid displaced,
 
 
 
.)mkg1000(15.0mkg79085.0
33

14.26: a) Neglecting the density of the air,
 
  
,m1036.3
mkg107.2sm80.9
N89
33
332




gρ
w
ρ

gw
ρ
m
V
33
m104.3or


to two figures.
b)
 
N.0.56
7.2
00.1
1N891
aluminum
water
water


















ρ
ρ
ω
VgρwBwT
14.27: a) The pressure at the top of the block is
,
0
ghpp


where
h
is 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)
.
objfl
gVB



The blocks have the same volume
obj
V
so experience the same
buoyant force.
c)
. so 0 BwTBwT 

.ρVgw 
The object have the same
V
but
ρ
is larger for brass than for aluminum so
w
is larger for the brass block.
B
is the same for both, so
T
is larger for the brass block,
block
B.
14.28: The rock displaces a volume of water whose weight is
N.10.8N28.4-N2.39 
The mass of this much water is thus
kg102.1sm9.80N8.10
2


and its volume, equal to
the rock’s volume, is
33
33
m10102.1
mkg101.00
kg102.1



The weight of unknown liquid displaced is
N,20.6N18.6N2.39 
and its mass is
kg.102.2sm9.80N6.20
2

The liquid’s density is thus
33
m101.102kg102.2


,mkg1091.1
33

or roughly twice the density of water.
14.29:
)(,
21122211
ΑΑvvΑvΑv 

2
2
2
1
cm)10.0(20,cm)80.0( πΑπΑ 

sm6.9
)10.0(20
(0.80)
s)m0.3(
2
2
2

π
π
v
14.30:

2
3
2
2
2
1
12
sm245.0)m0700.0s)(m50.3(
ΑΑΑ
Α
vv

a) (i)
s.m21.5,m047.0(ii) s.m33.2,m1050.0
2
2
22
2
2
 vΑvΑ
b)
.m882s)3600()sm(0.245
33
2211
 tΑυtΑv
14.31: a)
.98.16
)m150.0(
)sm20.1(
2
3

πA
dtdV
v
b)
.m317.0)(
22112
 πvdtdVvvrr
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 ρpgyv 
=
))mkg10(1.03Pa)1013(3.00)(1.0m)0.11)(sm80.9((2
3
352


s.m4.28
Note that y = 0 and
a
pp 
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),
s.m6.16m)0.14)(sm80.9(22
2
 ghv
b)
s.m1069.4)m)10(0.30s)(m57.16(
3422 
 πvΑ
Note that an extra figure
was kept in the intermediate calculation.
14.35: The assumption may be taken to mean that
0
1
v
in Eq. (14.17). At the
maximum height,

,0
2
v
and using gauge pressure for
0,and
221
ppp
(the water is open
to the atmosphere),
Pa.1047.1
5
21
 ρgyp
14.36: Using
1
4
1
2
vv 
in Eq. (14.17),
















)(
32
15
)()(
2
1
21
2
1121
2
2
2
112
yygυρpyyρgvvρpp








m)0.11)(sm80.9()sm00.3(
32
15

)mkg1000.1(Pa1000.5
22334

Pa.1062.1
5

14.37: Neglecting the thickness of the wing (so that
21
yy 
in Eq. (14.17)), the pressure
difference is
Pa.078)(2)1(
2
1
2
2
 vvρp
The net upward force is then
N.496)smkg)(9.801340()m(16.2Pa)780(
22

14.38: a)
  
s.kg30.1
s0.60
kg355.0220

b) The density of the liquid is
,mkg1000
3

m100.355
kg355.0
33



and so the volume flow rate is
s.L1.30sm1030.1
33
mkg1000
skg30.1
3


This result may also be obtained
from
  
s.L30.1
s0.60
L355.0220

c)
24
33
m1000.2
sm1030.1
1





v
s.m63.14s,m50.6
12
 vv
d)
 
 
12
2
1
2
221
2
1
yyρgvvρpp 

 
 
   
 
  
 
m35.1sm80.9mkg1000
sm50.6sm63.1mkg100021kPa152
23
22
3



kPa119
14.39: The water is discharged at a rate of
s.m352.0
23
34
m1032.1
sm1065.4
1





v
The pipe is
given as horizonatal, so the speed at the constriction is
s,m95.82
2
12
 ρpvv
keeping an extra figure, so the cross-section are at the constriction is
,m1019.5
25
sm95.8
sm1065.4
34





and the radius is
cm.41.0

Ar
14.40: From Eq. (14.17), with
,
21
yy 

 
2
11
2
1
2
11
2
2
2
112
8
3
42
1
2
1
ρvp
v
v

ρpvvρpp 










 
 
Pa,1003.2sm50.2mkg1000.1
8
3
Pa1080.1
4
2
334

where the continutity relation
2
1
2
v
v

has been used.
14.41: Let point 1 be where

cm00.4
1
r
and point 2 be where
cm.00.2
2
r
The
volume flow rate has the value
scm7200
3
at all points in the pipe.
sm43.1so,cm7200
1
32
1111
 vπrvAv
sm73.5so,cm7200
2
32
2222
 vπrvAv
2
222
2
111
2
1
2
1

ρvρgypρvρgyp 
 
Pa1025.2
2
1
soPa,1040.2and
52
2
2
112
5
221
 vvρpppyy
14.42: a) The cross-sectional area presented by a sphere is
,
4
2
D
π
therefore
 
.
4
0
2
D
πppF 
b) The force on each hemisphere due to the atmosphere is
 
2

2
m1000.5

π
 
 
.776975.0Pa10013.1
5

14.43: a)
   
Pa.1010.1m1092.10sm80.9mkg1003.1
83233
ρgh
b) The fractional change in volume is the negative of the fractional change in density.
The density at that depth is then
 
     
111833
0
Pa108.45Pa1016.11mkg1003.11

 pkρρ

,mkg1008.1
33

A fractional increase of
%.0.5
Note that to three figures, the gauge pressure and absolute

pressure are the same.
14.44: a) The weight of the water is
  
    
N,1088.5m0.3m0.4m00.5sm80.9mkg1000.1
5233
ρgV
or
N109.5
5

to two figures. b) Integration gives the expected result the force is what it
would be if the pressure were uniform and equal to the pressure at the midpoint;
2
d
gAF



  
    
N,1076.1m50.1m0.3m0.4sm80.9mkg1000.1
5233

or
N108.1
5

to two figures.
14.45: Let the width be w and the depth at the bottom of the gate be

.H
The force on a
strip of vertical thickness
dh
at a depth
h
is then
 
wdhρghdF 
and the torque about
the hinge is
 
;2 dhHhρgwhdτ 
integrating from
Hhh  to0
gives
m.N1061.212
43
 Hg

14.46: a) See problem 14.45; the net force is
dF
from
,22, to0
2
gAHHgFHhh


where
.HA



b) The torque on a strip of
vertical thickness
dh
about the bottom is
   
,dhhHgwhhHdFdτ 
and
integrating from
Hhh  to0
gives
.66
23
ρgAHρgwHτ 
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
V
m





and so the planet’s mass is
mGd
pVR
G
gR
M
22


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,
2
Cx

.
a)
.
2
dVCxdVM 

The volume element
.
2

dxπRdV 
Then the integral
becomes
.
22
0
dxRCxM
L


Integrating gives
.
3
3
22
0
2
L
RCdxxRCM
L


Solving
for
.3,
32
LRMCC


b) The density at the

Lx 
end is
 
 
 
.
232
3
2
3
2
LπR
M
L
πR
M
LCxρ 
The denominator is
just the total volume
,V
so
,3 VM

or three times the average density,
.VM
So the
average density is one-third the density at the
Lx 
end of the rod.
14.49: a) At

,0r
the model predicts
3
mkg700,12 A

and at
,Rr 
the model
predicts
.mkg103.15m)1037.6)(mkg1050.1(mkg700,12
336433


BRA

b), c)
 
























R
BR
A
πRBRAR
πdrrBrAπdmM
0
343
2
4
3
3
4
43
4][4





















4
m)1037.6)(mkg1050.1(3
mkg700,12
3
m)1037.6(4
643
3
36
π

kg,1099.5
24

which is within 0.36% of the earth’s mass. d) If
m
)(r

is used to denote the mass
contained in a sphere of radius
,r
then
.)(
2
rrGmg 
Using the same integration as
that in part (b), with an upper limit of
r
instead of
R
gives the result.
e)
)kgmN10673.6()(,at and,0at 0
22112


RRGmgRrgrg
.sm85.9m)10(6.37kg)1099.5(
22624

f)
;
2
3
3
4
4
3

3
4
2





























Br
A
πGBr
Ar
dr
d
πG
dr
dg
setting ths equal to zero gives
m1064.532
6
 BAr
, and at this radius

































B
A
BA
B
A
πG
g
3
2
4
3
3
2

3
4

B
πGA
9
4
2


.sm02.10
)mkg1050.1(9
)mkg700,12()kgmN10673.6(4
2
43
232211






π