1.1:
     
 
km61.1cm10km1.incm54.2ftin.12mift5280mi1
5

Although rounded to three figures, this conversion is exact because the given conversion
from inches to centimeters defines the inch.
1.2:
.in9.28
cm54.2
in1
L1
cm1000
L473.0
3
3
3














4
3
3



















1.5:
 
 
 
L.36.5cm1000L1incm54.2in327
3
3
3





















bottles2112bottles9.2111 
The daily consumption must then be
.
da
bottles
78.5
da24.365
yr1
yr
bottles

mile1
fortnight
furlongs
000,180




























1.10: a)
s
ft
88
mi1
ft5280
s3600
h1
hr
mi
60



























c)
3
3
3
3
m
kg
10
g1000
kg1
m1
cm100
cm
g
1.0








Vrr
1.12:
%58.0100)s10(3.16s)10πs1016.3(
777

1.13: a)
%.101.1
m10890
m10
3
3



b) Since the distance was given as 890 km, the total distance should be 890,000
meters.
To report the total distance as 890,010 meters, the distance should be given as
890.01 km.
1.14: a)
   
2
mm72mm98.5mm12 
(two significant figures).
b)
mm12
mm98.5
= 0.50 (also two significant figures).
c) 36 mm (to the nearest millimeter).
d) 6 mm.
e) 2.0.

 
3
2
cm8.2cm050.0
4
cm50.8
π
(two significant figures) and the uncertainty in the volume, found from the extreme
values of the diameter and thickness, is about
3
cm3.0
, and so the volume of a
cookie is
.cm3.08.2
3

(This method does not use the usual form for progation of errors,
which is not addressed in the text. The fractional uncertainty in the thickness is so much
greater than the fractional uncertainty in the diameter that the fractional uncertainty in the
volume is
%10
, reflected in the above answer.)
b)
.20170
05.
50.8

1.18: (Number of cars

miles/car

minutes per day, 365 days per
year, and a lifespan of fourscore (80) years, the total volume of air breathed in a lifetime
is about
35
m102
. This is the volume of a room
m20m100m100 
, which is kind of
tight for a major-league baseball game, but it’s the same order of magnitude as the
volume of the Astrodome.
1.23: This will vary from person to person, but should be of the order of
5
101
.
1.24: With a pulse rate of a bit more than one beat per second, a heart will beat 10
5
times per day. With 365 days in a year and the above lifespan of 80 years, the number of
beats in a lifetime is about
9
103
. With
20
1
L (50 cm
3
) per beat, and about
4
1
gallon per
liter, this comes to about

324318
cm104m104 
. Characterizing
the size of a “drop” is a personal matter, but 25
3
cmdrops
is reasonable, giving a total of
26
10
drops of water in the oceans.
1.27: This will of course depend on the size of the school and who is considered a
"student''. A school of thousand students, each of whom averages ten pizzas a year
(perhaps an underestimate) will total 10
4
pizzas, as will a school of 250 students
averaging 40 pizzas a year each.
1.28: The moon is about
mm104m104
118

away. Depending on age, dollar
bills can be stacked with about 2-3 per millimeter, so the number of bills in a stack
to the moon would be about 10
12
. The value of these bills would be $1 trillion (1
terabuck).
1.29:
   
.billsofnumber billArea USAofArea 
   

m.6.937.0cosm0.12m,2.737.0sinm0.12; 


yx
AAA
   
   
m.2.560.0sin m0.6m,0.360.0cosm0.6;
m.6.940.0sin m0.15m,5.1140.0cosm0.15;






yx
yx
CC
BB
C
B
1.36:
500.0
m2.00
m00.1
tan (a)



X

1
1
















θ
A
A
θ
θ
A
A
θc
θ
A
A
θ

21

xxx
FFF
and
N275
21

yyy
FFF
The resultant force is 1190 N in the direction

13.4
above the forward direction.
1.38: (The figure is given with the solution to Exercise 1.31).
The net northward displacement is (2.6 km) + (3.1 km) sin 45
o
= 4.8 km, and the
net eastward displacement is (4.0 km) + (3.1 km) cos 45
o
= 6.2 km. The
magnitude of the resultant displacement is
22
)km2.6()km8.4( 
= 7.8 km, and
the direction is arctan
 
2.6
8.4
= 38

, and an angle of
.6.77
2.4
10.8








b) The magnitude and direction of A + B are the same as B + A.
c) The x- and y-components of the vector difference are – 26.4 m and
m,8.10
for a magnitude of
m28.5
and a direction arctan
 
.202
4.26
8.10




Note that

180
must be added to

arctanofangleandat m5.28m8.10m26.4Magnitude
22









1.40: Using Equations (1.8) and (1.9), the magnitude and direction of each of the given
vectors is:
a)
22
)cm20.5()cm6.8( 
= 10.0 cm, arctan
 
60.8
20.5

= 148.8
o
(which is
180
o
– 31.2
o
).
b)

).
1.41:
The total northward displacement is
km,75.1km50.1km3.25 
, and the total
westward displacement is
km4.75
. The magnitude of the net displacement is
   
km.06.5km75.4km1.75
22

The south and west displacements are the same, so
The direction of the net displacement is

69.80
West of North.
1.42: a) The x- and y-components of the sum are 1.30 cm + 4.10 cm = 5.40 cm,
2.25 cm + (–3.75 cm) = –1.50 cm.
b) Using Equations (1-8) and (1-9),
22
)cm50.1()cm04.5( 
= 5.60 cm, arctan
 
40.5
50.1


= 344.5
o














and the angle is
   
   



18
0.60coscm90.160.0coscm2.80
0.60sincm90.160.0sin cm2.80
arctan










and the angle is
   
   



84
0.60coscm90.160.0coscm2.80
0.60sincm90.160.0sin cm2.80
arctan











c)
 
.26418084isangle theandcm4.10ismagnitude the;







..
1.45:
       
jijiA
ˆ
m6.9
ˆ
m2.7
ˆ
37.0cosm0.12
ˆ
37.0sinm0.12 


       
       
jijiC
jijiB
ˆ
m2.5
ˆ
m0.3
ˆ
60.0sin m0.6
ˆ
60.0cosm0.6
ˆ
m6.9

ˆ
m08.2
ˆ
30.0sin m40.2
ˆ
30.0cosm40.2 


b)
   
BAC


00.400.3 

           
   
ji
jiji
ˆ
94.14
ˆ
m01.12
ˆ
m20.100.4
ˆ
m08.200.4
ˆ
m38.300.3
ˆ

00.5
ˆ
00.1
ˆ
00.200.5
ˆ
00.300.4 


c)
   

3.101
1.00-
5.00
arctan ,10.500.51.00
22








d)
1.48: a)
13111
ˆˆˆ
222

aa
A

20.0
0.5
1
a
1.49: a) Let
,
ˆˆ
jiA
yx
AA 

.
ˆˆ
jiB
yx
BB 

 
 
 
 
jiAB
jiBA
ˆˆ
ˆˆ
yyxx
yyxx



b)
 
 
 
kjiBA
ˆˆˆ
xyyxzxxzyzzy
BABABABABABA 


 
 
 
ˆˆˆ
kjiAB
xyyxzxxzyzzy
ABABABABABAB 


Comparison of each component in each vector product shows that one is the
negative of the other.
1.50: Method 1:
 
θcosmagnitudesofoduct Pr 
 
 
 
2

.00.1400.200.300.500.4  BA


b)
   
 
 
.7.58.5195arccos39.500.500.14arccosso,cosAB

 θθBA
1.52: For all of these pairs of vectors, the angle is found from combining Equations
(1.18) and (1.21), to give the angle

as
.arccosarccos























.
b)
,136,34,60  BABA



28
13634
60
arccos









.
c)
.90,0 


so
 
,
ˆ
00.23 kBA 


and the magnitude of the vector product is 23.00.