Ch01 ssm

Chapter 1
Measurement and Vectors
Conceptual Problems
1 • Which of the following is not one of the base quantities in the SI
system? (a) mass, (b) length, (c) energy, (d) time, (e) All of the above are base
quantities.
Determine the Concept The base quantities in the SI system include mass, length,
and time. Force is not a base quantity. )(c is correct.
5 • Show that there are 30.48 cm per foot. How many centimeters are
there in one mile?
Picture the Problem We can use the facts that there are 2.540 centimeters in
1 inch and 12 inches in 1 foot to show that there are 30.48 cm per ft. We can then
use the fact that there are 5280 feet in 1 mile to find the number of centimeters in
one mile.
Multiply 2.540 cm/in by 12 in/ft to
find the number of cm per ft:
cm/ft48.30
ft
in
12
in
cm
540.2 =⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
Multiply 30.48 cm/ft by 5280 ft/mi to find the number of centimeters in one
mile:
cm/mi10609.1
mi
ft
5280
ft
cm
48.30 5
×=⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
Remarks: Because there are exactly 2.54 cm in 1 in and exactly 12 inches in 1 ft, we
are justified in reporting four significant figures in these results.
11 • A vector A
r
points in the +x direction. Show graphically at least three
choices for a vector B
r
such that AB
rr
+ points in the +y direction.
Determine the Concept The following figure shows a vector A
r
pointing in the
positive x direction and three unlabeled possibilities for vector .B
r
Note that the
choices for B
r
start at the end of vector A
r
rather than at its initial point. Note
further that this configuration could be in any quadrant of the reference system
shown.
1

Chapter 12
x
y
A
r
choicesSeveralB
r
13 • Is it possible for three equal magnitude vectors to add to zero? If so,
sketch a graphical answer. If not, explain why not.
Determine the Concept In order for
the three equal magnitude vectors to
add to zero, the sum of the three vectors
must form a triangle. The equilateral
triangle shown to the right satisfies this
condition for the vectors A
r
, B
r
, and
for which it is true that A = B = C,
whereas
C
r
.0=++ CBA
rrr
A
r
B
r
C
r
Estimation and Approximation
15 • Some good estimates about the human body can be made if it is
assumed that we are made mostly of water. The mass of a water molecule is
29.9 ×10−27
kg. If the mass of a person is 60 kg, estimate the number of water
molecules in that person.
Picture the Problem We can estimate the number of water molecules in a person
whose mass is 60 kg by dividing this mass by the mass of a single water
molecule.
Letting N represent the number of
water molecules in a person of mass
mhuman body, express N in terms of
mhuman body and the mass of a water
molecule mwater molecule:
moleculewater
bodyhuman
m
m
N =

Measurement and Vectors 3
Substitute numerical values and
evaluate N:
molecules100.2
molecule
kg
109.29
kg60
27
27
×=
×
=
−
N
19 •• A megabyte (MB) is a unit of computer memory storage. A CD has a
storage capacity of 700 MB and can store approximately 70 min of high-quality
music. (a) If a typical song is 5 min long, how many megabytes are required for
each song? (b) If a page of printed text takes approximately 5 kilobytes, estimate
the number of novels that could be saved on a CD.
Picture the Problem We can set up a proportion to relate the storage capacity of
a CD to its playing time, the length of a typical song, and the storage capacity
required for each song. In (b) we can relate the number of novels that can be
stored on a CD to the number of megabytes required per novel and the storage
capacity of the CD.
(a) Set up a proportion relating the
ratio of the number of megabytes on
a CD to its playing time to the ratio
of the number of megabytes N
required for each song:
min5min70
MB700 N
=
Solve this proportion for N to
obtain:
( ) MB50min5
min70
MB700
=⎟
⎠
⎞
⎜
⎝
⎛
=N
(b) Letting n represent the number of
megabytes per novel, express the
number of novels that can be
stored on a CD in terms of the
storage capacity of the CD:
novelsN
n
N
MB700
novels =
Assuming that a typical page in a
novel requires 5 kB of memory,
express n in terms of the number of
pages p in a typical novel:
pn ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
page
kB
5
Substitute for n in the expression
for to obtain:novelsN
p
N
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
page
kB
5
MB700
novels

Chapter 14
Assuming that a typical novel has
200 pages:
novels107
novel
pages
200
page
kB
5
MB
kB10
MB700
2
3
novels
×=
⎟
⎠
⎞
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
=N
Units
23 •• In the following equations, the distance x is in meters, the time t is in
seconds, and the velocity v is in meters per second. What are the SI units of the
constants C1 and C2? (a) x = C1 + C2t, (b) x = 1
2 C1t2
, (c) v2
= 2C1x,
(d) x = C1 cos C2t, (e) v2
= 2C1v – (C2x)2
.
Picture the Problem We can determine the SI units of each term on the right-
hand side of the equations from the units of the physical quantity on the left-hand
side.
(a) Because x is in meters, C1 and
C2t must be in meters:
m/sinism;inis 21 CC
(b) Because x is in meters, ½C1t2
must be in meters:
2
1 m/sinisC
(c) Because v2
is in m2
/s2
, 2C1x
must be in m2
/s2
:
2
1 m/sinisC
(d) The argument of a trigonometric
function must be dimensionless; i.e.
without units. Therefore, because x
is in meters:
1
21 sinism;inis −
CC
(e) All of the terms in the expression
must have the same units. Therefore,
because v is in m/s:
1
21 sinism/s;inis −
CC
Conversion of Units
33 •• You are a delivery person for the Fresh Aqua Spring Water Company.
Your truck carries 4 pallets. Each pallet carries 60 cases of water. Each case of
water has 24 one-liter bottles. You are to deliver 10 cases of water to each
convenience store along your route. The dolly you use to carry the water into the
stores has a weight limit of 250 lb. (a) If a milliliter of water has a mass of

1 g, and a kilogram has a weight of 2.2 lb, what is the weight, in pounds, of all the
water in your truck? (b) How many full cases of water can you carry on the cart?
Picture the Problem The weight of the water in the truck is the product of the
volume of the water and its weight density of 2.2 lb/L.
(a) Relate the weight w of the water
on the truck to its volume V and
weight density (weight per unit
volume) D:
DVw =
Find the volume V of the water:
L5760
)
case
L
24()
pallet
cases
(60pallets)4(
=
=V
Substitute numerical values for D
and L and evaluate w:
( )
lb103.1
lb10267.1L5760
L
lb
2.2
4
4
×=
×=⎟
⎠
⎞
⎜
⎝
⎛
=w
(b) Express the number of cases of
water in terms of the weight limit of
the cart and the weight of each case
of water:
waterofcaseeachofweight
carttheoflimitweight
=N
evaluate N:
cases7.4
case
L
24
L
lb
2.2
lb250
=
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=N
cases.4carrycanYou
35 •• In the following, x is in meters, t is in seconds, v is in meters per
second, and the acceleration a is in meters per second squared. Find the SI units
of each combination: (a) v2
/x, (b) ax , (c) 1
2 at2
.
Picture the Problem We can treat the SI units as though they are algebraic
quantities to simplify each of these combinations of physical quantities and
constants.
(a) Express and simplify the units of
v2
/x:
( )
22
22
s
m
sm
m
m
sm
=
⋅
=

Chapter 16
(b) Express and simplify the units of
ax :
ss
m/s
m 2
2
==
(c) Noting that the constant factor
2
1
has no units, express and simplify
the units of
2
2
1
at
:
( ) ms
s
m 2
2
=⎟
⎠
⎞
⎜
⎝
⎛
Dimensions of Physical Quantities
41 •• The momentum of an object is the product of its velocity and mass.
Show that momentum has the dimensions of force multiplied by time.
Picture the Problem The dimensions of mass and velocity are M and L/T,
respectively. We note from Table 1-2 that the dimensions of force are ML/T2
.
Express the dimensions of
momentum:
[ ]
T
ML
T
L
M =×=mv
From Table 1-2:
[ ] 2
T
ML
=F
Express the dimensions of force
multiplied by time:
[ ]
T
ML
T
T
ML
2
=×=Ft
Comparing these results, we see that momentum has the dimensions of force
multiplied by time.
43 •• When an object falls through air, there is a drag force that depends on
the product of the cross sectional area of the object and the square of its velocity,
that is, Fair = CAv2
, where C is a constant. Determine the dimensions of C.
Picture the Problem We can find the dimensions of C by solving the drag force
equation for C and substituting the dimensions of force, area, and velocity.
Solve the drag force equation for the
constant C: 2
air
Av
F
C =
Express this equation dimensionally:
[ ] [ ]
[ ][ ]2
air
vA
F
C =

Substitute the dimensions of force,
area, and velocity and simplify to
obtain:
[ ] 32
2
2
L
M
T
L
L
T
ML
=
⎟
⎠
⎞
⎜
⎝
⎛
=C
Scientific Notation and Significant Figures
45 • Express as a decimal number without using powers of 10 notation: (a)
3 × 104
, (b) 6.2 × 10–3
, (c) 4 × 10–6
, (d) 2.17 × 105
.
Picture the Problem We can use the rules governing scientific notation to
express each of these numbers as a decimal number.
(a) 000,30103 4
=× (c) 000004.0104 6
=× −
(b) 0062.0102.6 3
=× −
(d) 000,2171017.2 5
=×
47 • Calculate the following, round off to the correct number of significant
figures, and express your result in scientific notation: (a) (1.14)(9.99 × 104
),
(b) (2.78 × 10–8
) – (5.31 × 10–9
), (c) 12π /(4.56 × 10–3
), (d) 27.6 + (5.99 × 102
).
Picture the Problem Apply the general rules concerning the multiplication,
division, addition, and subtraction of measurements to evaluate each of the
given expressions.
(a) The number of significant figures
in each factor is three; therefore the
result has three significant figures:
( )( ) 54
1014.11099.914.1 ×=×
(b) Express both terms with the same
power of 10. Because the first
measurement has only two digits
after the decimal point, the result can
have only two digits after the
decimal point:
( ) ( )
( )
8
8
98
1025.2
10531.078.2
1031.51078.2
−
−
−−
×=
×−=
×−×
(c) We’ll assume that 12 is exact.
Hence, the answer will have three
significant figures:
3
3
1027.8
1056.4
12
×=
× −
π

Chapter 18
(d) Proceed as in (b): ( )
2
2
1027.6
627
5996.271099.56.27
×=
=
+=×+
49 • A cell membrane has a thickness of 7.0 nm. How many cell
membranes would it take to make a stack 1.0 in high?
Picture the Problem Let N represent the required number of membranes and
express N in terms of the thickness of each cell membrane.
Express N in terms of the thickness
of a single membrane: nm7.0
in1.0
=N
Convert the units into SI units and simplify to obtain:
66
9
106.31063.3
m10
nm1
cm100
m1
in
cm2.540
nm7.0
in1.0
×=×=×××= −
N
51 •• A square peg must be made to fit through a square hole. If you have a
square peg that has an edge length of 42.9 mm, and the square hole has an edge
length of 43.2 mm, (a) what is the area of the space available when the peg is in
the hole? (b) If the peg is made rectangular by removing 0.10 mm of material
from one side, what is the area available now?
Picture the Problem Let sh represent the side of the square hole and sp the side
of the square peg. We can find the area of the space available when the peg is in
the hole by subtracting the area of the peg from the area of the hole.
(a) Express the difference between
the two areas in terms of sh and sp:
2
p
2
hΔ ssA −=
evaluate ΔA:
( ) ( )
2
22
mm62
mm9.42mm2.43Δ
≈
−=A
(b) Express the difference between
the area of the square hole and the
rectangular peg in terms of sh ,sp and
the new length of the peg lp:
pp
2
hΔ lssA' −=

Substitute numerical values and evaluate ΔA′:
( ) ( )( ) 22
mm03mm10.0mm9.42mm9.42mm2.43Δ ≈−−=A'
Vectors and Their Properties
53 • Determine the x and y components of the following three vectors in the
xy plane. (a) A 10-m displacement vector that makes an angle of 30° clockwise
from the +y direction. (b) A 25-m/s velocity vector that makes an angle of −40°
counterclockwise from the −x direction. (c) A 40-lb force vector that makes an
angle of 120° counterclockwise from the −y direction.
Picture the Problem The x and y components of these vectors are their
projections onto the x and y axes. Note that the components are calculated using
the angle each vector makes with the +x axis.
(a) Sketch the displacement vector
(call it A
r
) and note that it makes
an angle of 60° with the +x axis:
A
r
x
y
°30
Ay
Ax
Find the x and y components of A
r
: ( ) m0.560cosm10 =°=xA
and
( ) m.7860sinm10 =°=yA
(b) Sketch the velocity vector (call
it v
r
) and note that it makes an
angle of 220° with the +x axis:
v
r
°40
y
xxv
yv

Chapter 110
Find the x and y components of v
r
: ( ) m/s19220cosm/s52 −=°=xv
and
( ) m/s16220sinm/s52 −=°=yv
(c) Sketch the force vector (call it
F
r
) and note that it makes an angle
of 30° with the +x axis:
x
°120
F
r
y
Find the x and y components of
F
r
:
( ) lb3530coslb04 =°=xF
and
( ) lb0230sinlb04 =°=yF
59 •• Calculate the unit vector (in terms of and ) in the direction opposite
to the direction of each of the vectors in Problem 57.
iˆ jˆ
Picture the Problem The unit vector in the direction opposite to the direction of a
vector is found by taking the negative of the unit vector in the direction of the
vector. The unit vector in the direction of a given vector is found by dividing the
given vector by its magnitude.
The unit vector in the direction of
A
r
is given by: 22
ˆ
yx AA +
==
A
A
A
A
r
r
r
Substitute for ,A
r
Ax, and Ay and
evaluate :ˆA ( ) ( )
ji
ji
A ˆ81.0ˆ59.0
7.44.3
ˆ7.4ˆ4.3ˆ
22
+=
+
+
=
The unit vector in the direction
opposite to that of A
r
is given by:
jiA ˆ81.0ˆ59.0ˆ −−=−
B
r
is given by: 22
ˆ
yx BB +
==
B
B
B
B
r
r
r

Substitute for B,B
r
B
x, and ByB and
evaluate :ˆB ( ) ( )
ji
ji
B
ˆ38.0ˆ92.0
2.37.7
ˆ2.3ˆ7.7ˆ
22
+−=
+−
+−
=
opposite to that of B
r
is given by:
jiB ˆ38.0ˆ92.0ˆ −=−
is given by:C
r
22
ˆ
yx CC +
==
C
C
C
C
r
r
r
Substitute for Cx, and Cy and
evaluate
,C
r
:ˆC ( ) ( )
ji
ji
C ˆ86.0ˆ51.0
1.94.5
ˆ1.9ˆ4.5ˆ
22
−=
−+
−
=
opposite that of is given by:C
r jiC ˆ86.0ˆ51.0ˆ +−=−
General Problems
61 • The Apollo trips to the moon in the 1960's and 1970's typically took 3
days to travel the Earth-moon distance once they left Earth orbit. Estimate the
spacecraft's average speed in kilometers per hour, miles per hour, and meters per
second.
Picture the Problem Average speed is defined to be the distance traveled divided
by the elapsed time. The Earth-moon distance and the distance and time
conversion factors can be found on the inside-front cover of the text. We’ll
assume that 3 days means exactly three days.
Express the average speed of Apollo
as it travels to the moon: timeelapsed
traveleddistance
av =v
Substitute numerical values to
obtain: d3
mi102.39 5
av
×
=v
Use the fact that there are 24 h in
1 d to convert 3 d into hours:
mi/h1032.3
mi/h10319.3
d
h24
d3
mi102.39
3
3
5
av
×=
×=
×
×
=v

Chapter 112
Use the fact that 1 mi is equal to 1.609 km to convert the spacecraft’s average
speed to km/h:
km/h1034.5
h
km
10340.5
mi
km
609.1
h
mi
10319.3 333
av ×=×=××=v
Use the facts that there are 3600 s in 1 h and 1000 m in 1 km to convert the
spacecraft’s average speed to m/s:
m/s1049.1m/s10485.1
s3600
h1
km
m10
h
km
10340.5 33
3
6
av ×=×=×××=v
Remarks: An alternative to multiplying by 103
m/km in the last step is to
replace the metric prefix ″k″ in ″km″ by 103
.
71 ••• You are an astronaut doing physics experiments on the moon. You are
interested in the experimental relationship between distance fallen, y, and time
elapsed, t, of falling objects dropped from rest. You have taken some data for a
falling penny, which is represented in the table below. You expect that a general
relationship between distance y and time t is y = Bt
C
, where B and C are
constants to be determined experimentally. To accomplish this, create a log-log
plot of the data: (a) graph log(y) vs. log(t), with log(y) the ordinate variable and
log(t) the abscissa variable. (b) Show that if you take the log of each side of your
equation, you get log(y) = log(B) + Clog(t). (c) By comparing this linear
relationship to the graph of the data, estimate the values of B and C. (d) If you
drop a penny, how long should it take to fall 1.0 m? (e) In the next chapter, we
will show that the expected relationship between y and t is ,2
2
1
aty = where a is
the acceleration of the object. What is the acceleration of objects dropped on the
moon?
y (m) 10 20 30 40 50
t (s) 3.5 5.2 6.0 7.3 7.9
Picture the Problem We can plot log y versus log t and find the slope of the best-
fit line to determine the exponent C. The value of B can be determined from the
intercept of this graph. Once we know C and B, we can solve for t as a
function of y and use this result to determine the time required for an object to fall
a given distance on the surface of the moon.
C
Bty =

(a) The following graph of log y versus log t was created using a spreadsheet
program. The equation shown on the graph was obtained using Excel’s ″Add
Trendline″ function. (Excel’s ″Add Trendline″ function uses regression analysis
to generate the trendline.)
log y = 1.9637log t - 0.0762
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
0.50 0.60 0.70 0.80 0.90 1.00
log t
logy
(b) Taking the logarithm of both
sides of the equation yields:C
Bty =
( )
tCB
tBBty CC
loglog
loglogloglog
+=
+==
(c) Note that this result is of the
form:
mXbY +=
where
Y = log y, b = log B, m = C, and
X = log t
From the regression analysis
(trendline) we have:
log B = −0.076
Solving for B yields: 2076.0
m/s84.010 == −
B
where we have inferred the units from
those given for .C
Bty =
Also, from the regression analysis
we have:
0.296.1 ≈=C
(d) Solve for t to obtain:C
Bty = C
B
y
t
1
⎟
⎠
⎞
⎜
⎝
⎛
=

Chapter 114
evaluate t to determine how long it
would take a penny to fall 1.0 m: s1.1
s
m
84.0
m0.1
2
1
2
≈
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
=t
(e) Substituting for B and C in
yields:C
Bty =
2
2
s
m
84.0 ty ⎟
⎠
⎞
⎜
⎝
⎛
=
Compare this equation to 2
2
1
aty = to
obtain:
22
1
s
m
84.0=a
and
22
s
m
7.1
s
m
84.02 =⎟
⎠
⎞
⎜
⎝
⎛
=a
Remarks: One could use a graphing calculator to obtain the results in Parts
(a) and (c).
73 ••• The Super-Kamiokande neutrino detector in Japan is a large
transparent cylinder filled with ultra pure water. The height of the cylinder is
41.4 m and the diameter is 39.3 m. Calculate the mass of the water in the cylinder.
Does this match the claim posted on the official Super-K Web site that the
detector uses 50000 tons of water?
Picture the Problem We can use the definition of density to relate the mass of
the water in the cylinder to its volume and the formula for the volume of a
cylinder to express the volume of water used in the detector’s cylinder. To convert
our answer in kg to lb, we can use the fact that 1 kg weighs about 2.205 lb.
Relate the mass of water contained in
the cylinder to its density and
volume:
Vm ρ=
Express the volume of a cylinder in
terms of its diameter d and height h: hdhAV 2
base
4
π
==
Substitute in the expression for m to
obtain: hdm 2
4
π
ρ=
evaluate m: ( ) ( ) (
kg10022.5
m4.41m3.39
4
kg/m10
7
233
×=
⎟
⎠
⎞
⎜
⎝
⎛
=
π
m )

Convert 5.02 × 107
kg to tons:
ton104.55
lb2000
ton1
kg
lb2.205
kg10022.5
3
7
×=
×××=m
The 50,000-ton claim is conservative. The actual weight is closer to 55,000 tons.

Ch01 ssm

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