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01 instructors solutions manual
Physics 2 (PHY 2054)
University of Florida
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- IDENTIFY: Convert units from mi to km and from km to ft.
SET UP: 1 in 2 54 cm,.=. 1 km 1000 m,= 12 in 1 ft,.= 1 mi 5280 ft.=
EXECUTE: (a) 23
5280 ft 12 in 2 54 cm 1 m 1 km 100 mi (100 mi) 161 km 1 mi 1 ft 1 in 10 cm 10 m
⎛⎞⎛⎞..⎛ ⎞⎛⎞⎛⎞
.=.⎜⎟⎜⎟⎜ ⎟⎜⎟⎜⎟=.
⎝⎠⎝⎠⎝ ⎠. ⎝⎠⎝⎠
(b)
32 10 m 10 cm 1 in 1 ft 3 100 km (100 km) 328 10 ft 1 km 1 m 2 54 cm 12 in
⎛⎞⎛ ⎞⎛⎞. ⎛⎞
.=.⎜⎟⎜ ⎟⎜⎟⎜⎟=.×
⎜⎟⎜ ⎟⎝⎠..⎝⎠
⎝⎠⎝ ⎠
EVALUATE: A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km.
- IDENTIFY: Convert volume units from L to
3 in ..
SET UP:
3 1 L 1000 cm .= 1 in 2 54 cm.=.
EXECUTE:
3 3 1000 cm 1 in 3 0473 L 289 in 1 L 2 54 cm
⎛⎞⎛⎞.
.×⎜⎟×⎜⎟=...
⎜⎟⎝⎠.
⎝⎠
EVALUATE:
3 1 in. is greater than
3
1 cm , so the volume in
3
in. is a smaller number than the volume in
3
cm , which is
3
473 cm.
1. IDENTIFY: We know the speed of light in m/s. tdv= / .Convert 1 ft to m and t from s to ns.
SET UP: The speed of light is
8
v= 10 m/s.× 1 ft 0 3048 m.=.
9
1 s 10 ns.=
EXECUTE:
9 8
03048 m
102 10 s 102 ns
300 10 m/s
t
. −
==.×=.
.×
EVALUATE: In 1 s light travels
85 5
300 10 m 300 10 km 186 10 mi..× =.× =.×
1. IDENTIFY: Convert the units from g to kg and from
3
cm to
3
m.
SET UP: 1 kg 1000 g.= 1 m 100 cm.=
EXECUTE:
3 4 33
g1kg100cm g k 19 3 1 93 10 cm 1000 g 1 m m
⎛⎞⎛ ⎞
.×⎜⎟×⎜ ⎟=.×
⎝⎠⎝ ⎠
EVALUATE: The ratio that converts cm to m is cubed, because we need to convert
3 cm to
3 m.
- IDENTIFY: Convert volume units from
3 in. to L.
SET UP:
3 1 L 1000 cm .= 1 in 2 54 cm..=.
EXECUTE:
333 (327 in ) (2 54 cm/in ) (1L/1000 cm ) 5 36 L.×. .× =.
EVALUATE: The volume is
3 5360 cm.
3 1 cm is less than
3
1 in ,. so the volume in
3 cm is a larger number
than the volume in
3
in ..
UNITS, PHYSICAL QUANTITIES, AND VECTORS
1
1-2 Chapter 1
- IDENTIFY: Convert
2 ft to
2 mand then to hectares.
SET UP:
42
100 hectare 100 10 m ..=.× 1 ft 0 3048 m.=.
EXECUTE: The area is
2 2
42
43 600 ft 0 3048 m 1 00 hectare (12 0 acres) 4 86 hectares. 1 acre 1 00 ft 100 10 m
⎛⎞, ⎛⎞..⎛ ⎞
.=⎜⎟⎜⎟⎜ ⎟.
⎜⎟⎝⎠. ⎝ ⎠.×
⎝⎠
EVALUATE: Since 1 ft 0 3048 m,=.
222 1 ft (0 3048) m .=.
- IDENTIFY: Convert seconds to years. 1 gigasecond is a billion seconds.
SET UP: 9
1 gigasecond 1 10 s.=× 1 day 24 h.= 1 h 3600 s.=
EXECUTE:
9 1 h 1 day 1 y
100 gigasecond (100 10 s) 317 y.
3600 s 24 h 365 days
⎛⎞⎛⎞⎛⎞
.=.×⎜⎟⎜⎟⎜⎟=.
⎝⎠⎝⎠⎝⎠
EVALUATE: The conversion
7
1 y 3 156 10 s=. × assumes 1 y 365 24 d,= is the average for one
extra day every four years, in leap years. The problem says instead to assume a 365-day year.
1. IDENTIFY: Apply the given conversion factors.
SET UP: 1 furlong 0 1250 mi and 1 fortnight 14 days=. =. 1 day 24 h=.
EXECUTE:
0125 mi 1 fortnight 1 day
(180 000 furlongs fortnight ) 67 mi/h
1 furlong 14 days 24 h
,/
⎛⎞. ⎛ ⎞⎛⎞
=⎜⎟⎜ ⎟⎜⎟
⎝⎠⎝ ⎠⎝⎠
EVALUATE: A furlong is less than a mile and a fortnight is many hours, so the speed limit in mph is a
much smaller number.
1. IDENTIFY: Convert miles/gallon to km/L.
SET UP: 1 mi 1 609 km.=. 1 gallon 3 788 L=..
EXECUTE: (a)
1609 km 1 gallon 55 0 miles/gallon (55 0 miles/gallon) 23 4 km/L. 1 mi 3 788 L
⎛⎞. ⎛⎞
.=.⎜⎟⎜⎟=.
⎝⎠⎝⎠.
(b) The volume of gas required is
1500 km 64 1 L. 23 4 km/L
=.
.
64 1 L
14 tanks. 45 L/tank
.
=.
EVALUATE: 1 mi/gal 0 425 km/L.=. A km is very roughly half a mile and there are roughly 4 liters in a
gallon, so 2
4
1 mi/gal km/L,∼ which is roughly our result.
1. IDENTIFY: Convert units.
SET UP: Use the unit conversions given in the problem. Also, 100 cm 1 m= and 1000 g 1 kg.=
EXECUTE: (a)
mi 1 h 5280 ft ft
60 88
h3600s 1mi s
⎛⎞⎛⎞⎛ ⎞
⎜⎟⎜⎟⎜ ⎟=
⎝⎠⎝⎠⎝ ⎠
(b)
22
ft 30 48 cm 1 m m
32 9 8
ss1ft 100 cm
⎛⎞ ⎛ ⎞⎛⎞.
⎜⎟ ⎜ ⎟ =⎜⎟.
⎝⎠ ⎝ ⎠⎝⎠
(c)
3 3 33
g100 cm 1 kg g k 10 10 cm 1 m 1000 g m
⎛⎞⎛⎞⎛⎞
⎜⎟.=⎜⎟⎜⎟
⎝⎠⎝⎠⎝⎠
EVALUATE: The relations 60 mi/h 88 ft/s= and
33 3
1 g/cm 10 kg/m= are exact. The relation
22
32 ft/s 9 8 m/s=. is accurate to only two significant figures.
1. IDENTIFY: We know the density and mass; thus we can find the volume using the relation
density mass/volume / .==mV The radius is then found from the volume equation for a sphere and the
result for the volume.
SET UP:
3
Density 19 5 g/cm=. and mcritical=. .60 0 kg For a sphere
4 3 3
Vr= π.
EXECUTE:
3 critical 3
60 0 kg 1000 g /density 3080 cm. 19 5 g/cm 10 kg
Vm
⎛⎞. ⎛⎞
==⎜⎟⎜⎟=
⎜⎟.
⎝⎠. ⎝⎠
1-4 Chapter 1
EXECUTE:
24 h 3600 s 7 (365 24 days/1 yr) 3 15567 10 s; 1 day 1 h
⎛⎞⎛⎞
.=⎜⎟⎜⎟....×
⎝⎠⎝⎠
77
π×10 s 3 14159 10 s=. ...×
The approximate expression is accurate to two significant figures. The percent error is 0%. EVALUATE: The close agreement is a numerical accident.
- IDENTIFY: To asses the accuracy of the approximations, we must convert them to decimals.
SET UP: Use a calculator to calculate the decimal equivalent of each fraction and then round the numeral to the specified number of significant figures. Compare to π rounded to the same number of significant figures. EXECUTE: (a) 22/7 = 3 (b) 355/113 = 3 (c) The exact value of π rounded to six significant figures is 3. EVALUATE: We see that 355/113 is a much better approximation to π than is 22/7.
- IDENTIFY: Express 200 kg in pounds. Express each of 200 m, 200 cm and 200 mm in inches. Express
200 months in years. SET UP: A mass of 1 kg is equivalent to a weight of about 2 lbs in 2 54 cm..=. 1 y 12 months.=
EXECUTE: (a) 200 kg is a weight of 440 lb. This is much larger than the typical weight of a man.
(b)
43 1 in 200 m (2 00 10 cm) 7 9 10 inches. 254 cm
⎛⎞.
=.× ⎜⎟=.×
⎝⎠.
This is much greater than the height of a person.
(c) 200 cm 200 m 79 inches 66 ft.=. = =. Some people are this tall, but not an ordinary man. (d) 200 mm 0200 m 79 inches.=. =. This is much too short.
(e)
1 y 200 months (200 mon) 17 y. 12 mon
⎛⎞
==⎜⎟
⎝⎠
This is the age of a teenager; a middle-aged man is much
older than this. EVALUATE: None are plausible. When specifying the value of a measured quantity it is essential to give the units in which it is being expressed.
- IDENTIFY: Estimate the number of people and then use the estimates given in the problem to calculate the
number of gallons.
SET UP: Estimate 8 310 × people, so 8 210 × cars. EXECUTE: (Number of cars miles/car day)/(mi/gal) gallons/day×=
88 (2 10 cars 10000 mi/yr/car 1 yr/365 days)/(20 mi/gal) 3 10×× × =× gal/day
EVALUATE: The number of gallons of gas used each day approximately equals the population of the U.
- IDENTIFY: Estimate the number of blinks per minute. Convert minutes to years. Estimate the typical
lifetime in years. SET UP: Estimate that we blink 10 times per minute y 365 days.= 1 day 24 h,= 1 h 60 min.= Use 80
years for the lifetime.
EXECUTE: The number of blinks is
60 min 24 h 365 days 8 (10 per min ) (80 y/lifetime) 4 10 1 h 1 day 1 y
⎛⎞⎛⎞⎛ ⎞
⎜⎟⎜⎟⎜ ⎟ =×
⎝⎠⎝⎠⎝ ⎠
EVALUATE: Our estimate of the number of blinks per minute can be off by a factor of two but our
calculation is surely accurate to a power of 10.
- IDENTIFY: Approximate the number of breaths per minute. Convert minutes to years and 3 cm to 3 mto
find the volume in 3 m breathed in a year.
SET UP: Assume 10 breaths/min.
24 h 60 min 5 1 y (365 d) 5 3 10 min. 1 d 1 h
⎛⎞⎛ ⎞
==⎜⎟⎜ ⎟.×
⎝⎠⎝ ⎠
2 10 cm 1 m= so
63 3 10 cm 1 m .= The volume of a sphere is 4133 36
Vr d==ππ,where r is the radius and d is the diameter.
Don’t forget to account for four astronauts.
EXECUTE: (a) The volume is
5 63 53 10 min 43 (4)(10 breaths/min)(500 10 m ) 1 10 m /yr. 1 y
− ⎛⎞.×
×⎜⎟=×
⎜⎟
⎝⎠
Units, Physical Quantities, and Vectors 1-
(b)
1/3 43 1/ 66 [110 m] 27 m
V
d
ππ
⎛⎞⎛⎞×
==⎜⎟⎜⎟=
⎝⎠⎜⎟
⎝⎠
EVALUATE: Our estimate assumes that each
3 cm of air is breathed in only once, where in reality not all the oxygen is absorbed from the air in each breath. Therefore, a somewhat smaller volume would actually be required.
- IDENTIFY: Estimation problem.
SET UP: Estimate that the pile is 18 in 18 in 5 ft 8 in.× .× .. Use the density of gold to calculate the mass of gold in the pile and from this calculate the dollar value.
EXECUTE: The volume of gold in the pile is 3
V= in 18 in 68 in 22,000 in×.×.=. Convert to.
3 cm :
333 5 3 V=,000 in (1000 cm /61 02 in ) 3 6 10 cm..=.×.
The density of gold is 3 19 3 g/cm ,. so the mass of this volume of gold is
3536 m=.(19 3 g/cm )(3 6 10 cm ) 7 10 g.× =×.
The monetary value of one gram is $10, so the gold has a value of
67 ($10/gram)(7 10 grams) $7 10 ,×=×
or about 6 $100 10× (one hundred million dollars).
EVALUATE: This is quite a large pile of gold, so such a large monetary value is reasonable.
- IDENTIFY: Estimate the number of beats per minute and the duration of a lifetime. The volume of blood
pumped during this interval is then the volume per beat multiplied by the total beats. SET UP: An average middle-aged (40 year-old) adult at rest has a heart rate of roughly 75 beats per minute. To calculate the number of beats in a lifetime, use the current average lifespan of 80 years.
EXECUTE:
9 beats
60 min 24 h 365 days 80 yr (75 beats/min) 3 10 beats/lifespan 1 h 1 day yr lifespan
N
⎛⎞⎛⎞⎛ ⎞⎛ ⎞
==⎜⎟⎜⎟⎜ ⎟⎜ ⎟×
⎝⎠⎝⎠⎝ ⎠⎝ ⎠
9 37 blood 3
1 L 1 gal 3 10 beats (50 cm /beat) 4 10 gal/lifespan 1000 cm 3788 L lifespan
V
⎛⎞⎛⎞⎛⎞×
==⎜⎟⎜⎟⎜⎟×
. ⎜⎟
⎝⎠⎝⎠⎝⎠
EVALUATE: This is a very large volume.
- IDENTIFY: Estimate the diameter of a drop and from that calculate the volume of a drop, in 3 m. Convert 3 mto L. SET UP: Estimate the diameter of a drop to be d=2 mm. The volume of a spherical drop is 413333 36
Vr d==ππ. 10 cm 1 L.=
EXECUTE:
1 333 6
V π(0 2 cm) 4 10 cm.
− =. =× The number of drops in 1 L is
3 5 33
1000 cm 210 410 cm −
=×
×
EVALUATE: Since 3 Vd∼ , if our estimate of the diameter of a drop is off by a factor of 2 then our
estimate of the number of drops is off by a factor of 8.
- IDENTIFY: Draw the vector addition diagram to scale.
SET UP: The two vectors A
G
and B
G
are specified in the figure that accompanies the problem.
EXECUTE: (a) The diagram for R=+AB
G G G
is given in Figure 1. Measuring the length and angle of
R
G
gives R= 0 mand an angle of 34 .θ=°
(b) The diagram for E=AB−
GGG
is given in Figure 1. Measuring the length and angle of E
G
gives
D=22 mand an angle of 250 .θ=°
(c) −−AB AB=−(+),
G G
so −−AB
G G
has a magnitude of 9 m (the same asAB+
G G
) and an angle with the
+x axis of 214 °(opposite to the direction of AB+ ).
G G
(d) B−AAB=−(−),
GGGG
so B−A
G G
has a magnitude of 22 m and an angle with the +x axis of 70° (opposite
to the direction ofAB−
G G
).
EVALUATE: The vector −A
G
is equal in magnitude and opposite in direction to the vector A.
G
Units, Physical Quantities, and Vectors 1-
- IDENTIFY: For each vector ,V
G
use that VVx= cosθ and VVy= sin ,θ when θ is the angle V
G
makes
with the +x axis, measured counterclockwise from the axis.
SET UP: For ,A
G
2 θ=.°700. For ,B
G
θ=.°60 0. For ,C
G
θ=.°205 0. For ,D
G
θ=.°143 0.
EXECUTE: Ax=0, Ay=−800 m.. Bx= m,By= 0 m=−10 9 m,. Cy=−507 m..
Dx=−799 m,. Dy= m.
EVALUATE: The signs of the components correspond to the quadrant in which the vector lies.
- IDENTIFY: tan ,
y
x
A
A
θ= for θ measured counterclockwise from the +x-axis.
SET UP: A sketch of ,Ax Ay and A
G
tells us the quadrant in which A
G
lies.
EXECUTE:
(a)
100 m tan 0 500. 200 m
y
x
A
A
θ
−.
== =−.
.
1 θ tan ( 0 500) 360 26 6 333. − =−.=°−.°=°
(b)
100 m tan 0 500. 200 m
y
x
A
A
θ
.
== =.
.
1 θ tan (0 500) 26 6. − =.=.°
(c)
100 m tan 0 500. 200 m
y
x
A
A
θ
.
== =−.
−.
1 θ tan ( 0 500) 180 26 6 153. − =−.=°−.°=°
(d)
100 m tan 0 500. 200 m
y
x
A
A
θ
−.
== =.
−.
1 θ tan (0 500) 180 26 6 207 − =.=°+.°=°
EVALUATE: The angles 26 6.°and 207°have the same tangent. Our sketch tells us which is the correct value of .θ
- IDENTIFY: Given the direction and one component of a vector, find the other component and the
magnitude. SET UP: Use the tangent of the given angle and the definition of vector magnitude.
EXECUTE: (a) tan32.
x
y
A
A
°=
Ax=°(9 m)tan32 6 m.=Ax=−6 m.
(b) 22 AAA=+=xy11 m.
EVALUATE: The magnitude is greater than either of the components.
- IDENTIFY: Given the direction and one component of a vector, find the other component and the
magnitude. SET UP: Use the tangent of the given angle and the definition of vector magnitude.
EXECUTE: (a) tan34.
x
y
A
A
°=
16 m 23 m tan34 tan34.
x y
A
A ===
°°
Ay=−23 m.
(b)
22 AAA=+=xy28 m.
EVALUATE: The magnitude is greater than either of the components.
- IDENTIFY: If CAB=+,
GGG
then CABx=+xxand CAByyy=+. Use Cxand Cyto find the magnitude and
direction of .C
G
1-8 Chapter 1
SET UP: From Figure E1 in the textbook,Ax=0, Ay=−800 m. and BBx=+sin30 0 7 50 m,. °=.
BBy=+cos30 0 13 0 m.. °=.
EXECUTE: (a) CAB=+
GGG
so CABxxx=+= 50 mand CAByyy=+=+.500 m.C= m.
500 m tan 750 m
y
x
C
C
θ
.
==
.
and 33 7 .θ=.°
(b) ,B+=+AAB
GGGG
so B+A
G G
has magnitude 9 m and direction specified by 33 7 ..°
(c) DAB=−
GGG
so DABxxx=−=−7 50 m. and DAByyy=−=−21 0 m.. D= 3 m.
21 0 m tan 750 m
y
x
D
D
φ
−.
==
−.
and 70 3 .φ=.°D
G
is in the rd 3 quadrant and the angle θcounterclockwise from the
+x axis is 180 70 3 250 3 .°+. °=. °
(d) ()B−AAB=− −,
GGGG
so B−A
G G
has magnitude 22 m and direction specified by 70 3 .θ=.°
EVALUATE: These results agree with those calculated from a scale drawing in Problem 1.
- IDENTIFY: Find the vector sum of the three given displacements.
SET UP: Use coordinates for which +x is east and +y is north. The driver’s vector displacements are:
ABC= km, 0 of north; 40 km, 0 of east; 31 km, 45 north o° =. ° =. ° f east.
KKK
EXECUTE: RABCxxxx=++=+. +.040 km(31 km)cos(45)62 km;°=. RABCyyyy=++=
26 km 0 (31 km)(sin45 ) 48 km;.++. °=. 22 RRR=+= km; 1 θ tan [(4 8 km)/(6 2 km)] 38 ; − =. .=°
R= km, 38 north of east°.
K
This result is confirmed by the sketch in Figure 1.
EVALUATE: Both Rx and Ry are positive and R
G
is in the first quadrant.
Figure 1.
- IDENTIFY: Vector addition problem. We are given the magnitude and direction of three vectors and are
asked to find their sum. SET UP:
A=3 km
B=2 km
C=1 km
Figure 1
1-10 Chapter 1
1. IDENTIFY: Vector addition problem. ABA B−=+()−.
G G G G
SET UP: Find the x- and y-components of A
G
and B.
G
Then the x- and y-components of the vector sum are
calculated from the x- and y-components of A
G
and B.
G
EXECUTE:
AAx=(60 0 )°
Ax=.(2 80 cm)cos(60 0 ) 1 40 cm.°=+.
AAy=(60 0 )°
Ay=.(2 80 cm)sin(60 0 ) 2 425 cm.°=+.
BBx=cos( 60 0 )−.°
Bx=.(1 90 cm) cos( 60 0 ) 0 95 cm−.°=+.
BBy=sin( 60 0 )−.°
By=.(1 90 cm) sin ( 60 0 ) 1 645 cm−.°=−.
Note that the signs of the components correspond to the directions of the component vectors.
Figure 1
(a) Now let RAB=+.
GGG
RABxxx=+=+.140 cm 095 cm 235 cm+. =+..
RAByyy=+=+.2425 cm 1645 cm 078 cm−. =+..
22 2 2
RRR=+=(2 35 cm) (0 78 cm)+.
R= cm
078 cm tan 0 3319 235 cm
y
x
R
R
θ
+.
== =+.
+.
θ=.°18 4
Figure 1
EVALUATE: The vector addition diagram for R=+AB
G G G
is
R
G
is in the 1st quadrant, with | | | | ,Ryx<R
in agreement with our calculation.
Figure 1
Units, Physical Quantities, and Vectors 1-
(b) EXECUTE: Now let R=AB−.
GGG
RABxxx=−=+.140 cm 095 cm 045 cm−. =+..
RAByyy=−=+.2425 cm 1645 cm 4070 cm+. =+..
22 2 2
RRR=+=(0 45 cm) (4 070 cm)+.
R= cm
4070 cm tan 9 044 045 cm
y
x
R
R
θ
.
== =+.
.
θ=.°83 7
Figure 1
EVALUATE: The vector addition diagram for R=+AB()−
G G G
is
R
G
is in the 1st quadrant, with | | | |,Rx<Ry
in agreement with our calculation.
Figure 1
(c) EXECUTE:
B−AAB=− −()
G G G G
B−A
G G
and AB−
G G
are equal in magnitude and opposite in direction. R= cm and 83 7 180 264θ=.°+ °= °
Figure 1
Units, Physical Quantities, and Vectors 1-
22
R=+RRx y
22 R=(1) (9) 9.− + =
- tan 9.
y
x
R
R
θ== =− −
θ=−83 180 96 .°+ °= °
Figure 1.
EVALUATE: Rx< 0 and Ry>0, so R
G
is in the 2nd quadrant.
1. IDENTIFY: Use trigonometry to find the components of each vector. Use RABxxx=++" and
RAByyy=++" to find the components of the vector sum. The equation Ai=+AAxˆˆyj
G
expresses a
vector in terms of its components.
SET UP: Use the coordinates in the figure that accompanies the problem.
EXECUTE: (a) Ai=.(3 60 m)cos70 0 (3 60 m)sin 70 0.°+.ˆˆ.°= (1 23 m) (3 38 m)iˆ+. ˆj
G
B=−(2 40 m)cos30 0 (2 40 m)sin30 0.. °iˆˆ−.. ° =j ( 2 08 m) ( 1 20 m)−. iˆ+−. ˆj
G
() (300) (400) (300)(123 m) (300)(338 m) (400)(208 m) (40b CAB i=. −. =.. ˆˆ ˆ ˆ+.. j−. −. i−.0)(120 m)−. j
GGG
Ci=+(12 m) (14 m)ˆˆj
G
(c) From =+
22
A AAxy and tanθ=
y
x
A
A
,
22 14 94 m (12 01 m) (14 94 m) 19 17 m, arctan 51 2 12 01 m
C
⎛⎞.
=. +. =. ⎜⎟=.°
⎝⎠.
EVALUATE: Cxand Cyare both positive, so θis in the first quadrant.
- IDENTIFY: We use the vector components and trigonometry to find the angles.
SET UP: Use the fact thattanθ= /
yx
A A.
EXECUTE: (a) θ =
−
=
6.
3.
tan /
yx
AA. θ = 117° with the +x-axis.
(b) θ= =
2.
tan /
7.
yx
BB. θ = 15°.
(c) First find the components of
G
C. Cx = Ax + Bx = ---3 + 7 = 4,
Cy = Ay + By = 6 + 2 = 8.
θ===
8.
tan / 2.
4.
CCyx. θ = 63°
EVALUATE: Sketching each of the three v ectors to scale will show that the answers are reasonable.
1. IDENTIFY: A
G
and B
G
are given in unit vector form. Find A, B and the vector difference AB−.
G G
SET UP: Ai=−200 300 400 ,. +. j+. k
GGGG
B=. +.300 100 300i j−.k
G G GG
Use 222 AAAA=++x yz to find the magnitudes of the vectors.
EXECUTE: (a) 222 2 2 2 AAAA=++=xyz(200) (300) (400) 538−.+.+.=.
1-14 Chapter 1
222 2 2 2 BBBB=++=.+.+xyz(3 00) (1 00) ( 3 00) 4 36−.=.
(b) AB i−=(200 300 400) (300 100 300)−. +. +.ˆˆˆ ˆˆˆj ki−. +.j−.k
G G
AB−=(200 300) (300 100) (400 (300)) 500 200 700−.−. +.iˆˆ ˆ−. +.j −−.ki=−. +. +. .ˆˆj kˆ
G G
(c) Let CAB=−,
GGG
so Cx=−5 00,. Cy=+ .200,Cz=+. 700
222 2 2 2 CCCC=++=xyz(500) (200) (700) 883−.+.+.=.
B−AAB=− −(),
GGGG
so AB−
G G
and B−A
G G
have the same magnitude but opposite directions.
EVALUATE: A, B, and C are each larger than any of their components.
- IDENTIFY: Target variables are AB⋅
G G
and the angle φ between the two vectors.
SET UP: We are given A
G
and B
G
in unit vector form and can take the scalar product using
AB⋅=AB AB ABx xyyzz+ +
G G
. The angle φ can then be found from AB⋅=ABcosφ
G G
.
EXECUTE: (a) Ai=+4 7 ,ˆˆj
G
B=5 2 ;iˆˆ−j
G
A=8,B=5.
AB i j i j⋅=(4 7 ) (5 2 ) (4)(5) (7)( 2)ˆˆ ˆˆ+ ⋅ − = + − =
G G
20 14 6.−=+
(b)
6.
cos 0; AB (8)(5)
φ
⋅
== =
AB
G G
φ=°82..
EVALUATE: The component of B
G
along A
G
is in the same direction as ,A
G
so the scalar product is
positive and the angle φ is less than 90 .°
1. IDENTIFY: AB⋅=ABcosφ
G G
SET UP: For A
G
and ,B
G
φ=.°150 0. For B
G
and ,C
G
φ=.°145 0. For A
G
and ,C
G
φ=.°65 0.
EXECUTE: (a) 2 AB⋅=.(8 00 m)(15 0 m)cos150 0. .°=−104 m
G G
(b) 2 BC⋅= .(15 0 m)(12 0 m) cos145 0. .°=−148 m
G G
(c) 2 AC⋅=.(8 00 m)(12 0 m)cos65 0 40 6 m. .°=.
GG
EVALUATE: When 90φ<° the scalar product is positive and when 90φ>° the scalar product is negative.
- IDENTIFY: Target variable is the vector AB×
G G
expressed in terms of unit vectors.
SET UP: We are given A
G
and B
G
in unit vector form and can take the vector product using
iiˆˆ ˆˆ×=jj×= 0 , ˆˆ ˆi×j=k, and ˆˆ ˆj×ik=−.
EXECUTE: Ai=+4 7 ,ˆˆj
G
B=5 2 .iˆˆ−j
G
AB i×=(4 7 ) (5 2 ) 20ˆˆ ˆˆ ˆ+ j× i− j= ii i×−ˆˆ8׈ˆjj+35 ×−iˆˆ14 .jj׈
G G
But iiˆˆ ˆˆ×=jj×= 0
and iˆˆ ˆ×j=k,ˆˆ ˆj×ik=−, so AB k k k×=−8 35( ) 43 .ˆˆˆ+ −=−
G G
The magnitude of AB×
G G
is 43.
EVALUATE: Sketch the vectors A
G
and B
G
in a coordinate system where the xy-plane is in the plane of the
paper and the z-axis is directed out toward you. By the right-hand rule AB×
G G
is directed into the plane of the paper, in the -direction.−z This agrees with the above calculation that used unit vectors.
Figure 1.
1-16 Chapter 1
EXECUTE: (a) The earth has volume
44363213 33
Vr==.ππ(6 37 10 m) 1 0827 10 m× =.×. Its density is
24 3 3 33 3 21 3 2
597 10 kg 10 g 1 m density (5 51 10 kg/m ) 5 51 g/cm 10827 10 m 1 kg 10 cm
m
V
.× ⎛⎞⎛⎞
== =.× ⎜⎟⎜⎟=.
.× ⎜⎟⎝⎠
⎝⎠
(b)
44363213 33
Vr==.ππ(7 5 10 m) 1 77 10 m× =.×
30 3 93 63 21 3 3
199 10 kg 1 g/cm density (1 1 10 kg/m ) 1 1 10 g/cm 1 77 10 m 1000 kg/m
m
V
.× ⎛⎞
== =.× ⎜⎟=.×
.× ⎜⎟
⎝⎠
(c)
44343123 33
Vr==.ππ(1 0 10 m) 4 19 10 m× =.×
30 3 17 3 14 3 12 3 3
199 10 kg 1 g/cm density (4 7 10 kg/m ) 4 7 10 g/cm 419 10 m 1000 kg/m
m
V
.× ⎛⎞
== =.× ⎜⎟=.×
.× ⎜⎟
⎝⎠
EVALUATE: For a fixed mass, the density scales as
3
1/r. Thus, the answer to (c) can also be obtained
from (b) as 3 6 63 143 4
750 10 m (1 1 10 g/cm ) 4 7 10 g/cm 10 10 m
⎛⎞.×
.× ⎜⎟= .×.
⎜⎟.×
⎝⎠
1. IDENTIFY: Area is length times width. Do unit conversions.
SET UP: 1 mi 5280 ft.=
3 1 ft 7 477 gal.=.
EXECUTE: (a) The area of one acre is
11 1 2 88 0 6 40
mi× mi= mi ,so there are 640 acres to a square mile.
(b)
2 2 1 mi 5280 ft 2 (1 acre) 43, 560 ft 640 acre 1 mi
⎛⎞⎛⎞
××⎜⎟⎜⎟=
⎜⎟⎝⎠
⎝⎠
(all of the above conversions are exact).
(c) (1 acre-foot)
35 3
7477 gal (43,560 ft ) 3 26 10 gal, 1 ft
⎛⎞.
=×⎜⎟=.×
⎝⎠
which is rounded to three significant figures.
EVALUATE: An acre is much larger than a square foot but less than a square mile. A volume of 1 acre-
foot is much larger than a gallon.
1. IDENTIFY: The density relates mass and volume. Use the given mass and density to find the volume and
from this the radius.
SET UP: The earth has mass
24
mE= 10 kg× and radius
6
rE= 10 m.× The volume of a sphere is
4 3 3
Vr= π.
33 ρ = g/cm 1760 km/m .=
EXECUTE: (a) The planet has mass
25
mm=. 55 E= 10 kg.×
25 22 3 3
328 10 kg 186 10 m. 1760 kg/m
m
V
ρ
.×
== =.×
1/3 1/ 22 3 33 [18610 m] 74 164 10 m 164 10 km 44
V
r
ππ
⎛⎞⎛⎞.×
==⎜⎟⎜⎟=.×=.×
⎜⎟
⎝⎠⎝⎠
(b) rr=. 257 E
EVALUATE: Volume V is proportional to mass and radius r is proportional to
1/
V , so r is proportional to
1/
m. If the planet and earth had the same density its radius would be
1/
(5 5).=rrEE1 8 .. The radius of the
planet is greater than this, so its density must be less than that of the earth.
1. IDENTIFY and SET UP: Unit conversion.
EXECUTE: (a)
9 f= 10 cycles/s,× so 10 9
1
s70410 s 1420 10
− =.× .×
for one cycle.
(b)
12 10
3600 s/h 511 10 cycles/h 704 10 s/cycle −
=.×
.×
Units, Physical Quantities, and Vectors 1-
(c) Calculate the number of seconds in 4600 million
9 years 4 6 10 y=.× and divide by the time for 1 cycle: 97 26 10
(4 6 10 y)(3 156 10 s/y) 21 10 cycles 704 10 s/cycle −
.×. ×
=.×
.×
(d) The clock is off by 1 s in
5 100,000 y 1 10 y,=× so in
9 460 10 y.× it is off by
9 4 5
460 10
(1 s) 4 6 10 s 110
⎛⎞.×
⎜⎟=.×
⎜⎟×
⎝⎠
(about 13 h).
EVALUATE: In each case the units in the calculation combine algebraically to give the correct units for the
answer.
1. IDENTIFY: Using the density of the oxygen and volume of a breath, we want the mass of oxygen (the
target variable in part (a)) breathed in per day and the dimensions of the tank in which it is stored.
SET UP: The mass is the density times the volume. Estimate 12 breaths per minute. We know 1 day = 24 h,
1 h = 60 min and 1000 L = 1 m 3
. The volume of a cube having faces of length l is
3
Vl=.
EXECUTE: (a) ()
60 min 24 h 12 breaths/min 17,280 breaths/day. 1 h 1 day
⎛⎞⎛⎞
⎜⎟⎜⎟=
⎝⎠⎝⎠
The volume of air breathed in
one day is 1 3 ( L/breath)(17,280 breaths/day) 8640 L 8 m. 2 == The mass of air breathed in one day is the
density of air times the volume of air breathed: 33
m==(1 kg/m )(8 m ) 11 kg. As 20% of this
quantity is oxygen, the mass of oxygen breathed in 1 day is (0)(11 kg) 2 kg 2200 g.==
(b) V =
3 8 m and
3
Vl= , so
1/ lV==2 m.
EVALUATE: A person could not survive one day in a closed tank of this size because the exhaled air is
breathed back into the tank and thus reduces the percent of oxygen in the air in the tank. That is, a person cannot extract all of the oxygen from the air in an enclosed space.
1. IDENTIFY: Use the extreme values in the piece’s length and width to find the uncertainty in the area.
SET UP: The length could be as large as 7 cm and the width could be as large as 1 cm.
EXECUTE: (a) The area is 14 ± 0 cm
2 .
(b) The fractional uncertainty in the area is
2
2
0 cm 0%, 14 cm
= and the fractional uncertainties in the
length and width are
0 cm 0% 7 cm
= and
0 cm 0%. 1 cm
= The sum of these fractional uncertainties is
0% 0% 0%,+= in agreement with the fractional uncertainty in the area.
EVALUATE: The fractional uncertainty in a product of numbers is greater than the fractional uncertainty in
any of the individual numbers.
1. IDENTIFY: Calculate the average volume and diameter and the uncertainty in these quantities.
SET UP: Using the extreme values of the input data gives us the largest and smallest values of the target
variables and from these we get the uncertainty.
EXECUTE: (a) The volume of a disk of diameter d and thickness t is
2
Vdt=.π(/2)
The average volume is
23
V=.π(8 50 cm/2) (0 050 cm) 2 837 cm. =.. But t is given to only two significant
figures so the answer should be expressed to two significant figures:
3
V= cm.
We can find the uncertainty in the volume as follows. The volume could be as large as 23
V=.π(8 52 cm/2) (0 055 cm) 3 1 cm ,. =. which is
3 03 cm. larger than the average value. The volume
could be as small as 23 V=.π(8 48 cm/2) (0 045 cm) 2 5 cm ,. =. which is 3 03 cm. smaller than the average
value. The uncertainty is 3
±.03 cm , and we express the volume as
3
V=.±.28 03 cm.
Units, Physical Quantities, and Vectors 1-
1. IDENTIFY: We know the magnitude and direction of the sum of the two vector pulls and the direction of
one pull. We also know that one pull has twice the magnitude of the other. There are two unknowns, the
magnitude of the smaller pull and its direction+=xxand Ay+=BCyy give two equations for these
two unknowns.
SET UP: Let the smaller pull be A
G
and the larger pull be B.
G
B=2=A B+
G G G
has magnitude 460 N
and is northward. Let +x be east and +y be north=−sin 21°and BBy=°cos21 .Cx=0,
Cy=460 N. A
G
must have an eastward component to cancel the westward component of B.
G
There are
then two possibilities, as sketched in Figures 1 a and b. A
G
can have a northward component or A
G
can have a southward component.
EXECUTE: In either Figure 1 a or b, ABCx+=xx and BA= 2 gives (2 )sin21°= sinφ and
φ=°45. In Figure 1, Ay+=BCyy gives 2cos21 cos45 460 NAA°+ °= , so 179 N= In
Figure 1, 2 cos21°− cos45 460 N°= and A=393 N. One solution is for the smaller pull to
be 45°east of north. In this case, the smaller pull is 179 N and the larger pull is 358 N. The other solution is for the smaller pull to be 45° south of east. In this case the smaller pull is 393 N and the larger pull is 786 N.
EVALUATE: For the first solution, with A
G
east of north, each worker has to exert less force to produce the
given resultant force and this is the sensible direction for the worker to pull.
Figure 1.
1. IDENTIFY: Let D
G
be the fourth force. Find D
G
such that ABCD+++ =0,
G G G G
so DABC=−( + + ).
GGGG
SET UP: Use components and solve for the components Dxand Dyof .D
G
EXECUTE: AAxy=+cos30 0. °=+ .86 6N, AA=+sin30 0. °=+ .50 00N.
BBxy=−sin30 0. °=−40 00N,. BB=+cos30 0. °=+ .69 28N.
CCxy=− cos53 0. °=−24 07 N,. CC=− sin53 0. °=−31 90N..
Then Dx=−22 53 N,. Dy=−87 34N. and
22
DDD=+= 2 N | / | 87 34/22 53.α==DDyx ..
α=.°75 54. 180 256,φ=°+= °α counterclockwise from the -axis+.x
EVALUATE: As shown in Figure 1, since Dxand Dyare both negative, D
G
must lie in the third quadrant.
Figure 1.
1-20 Chapter 1
1. IDENTIFY: Vector addition. Target variable is the 4th displacement.
SET UP: Use a coordinate system where east is in the -direction+x and north is in the +.y-direction
Let ,A
G
B,
G
and C
G
be the three displacements that are given and let D
G
be the fourth unmeasured
displacement. Then the resultant displacement is R=+++.ABCD
G G GGG
And since she ends up back where
she started, R=. 0
G
0,=+++ABCD
GGGG
so DABC=−()+ +
GGGG
DABCx=−()xxx+ + and Dyyyy=−(ABC+ + )
EXECUTE:
Ax=−180 m, 0 Ay=
BBx=°cos315 (210 m)cos315= °=+148 5 m.
BBy=°sin315 (210 m)sin315= °=−148 5 m.
CCx=°cos60 (280 m)cos60= °=+140 m
CCy=°sin60 (280 m)sin60= °=+242 5 m.
Figure 1
DABCxxxx=−()+ + =− −(180 m1485 m140 m)+. + =−1085 m.
DABCyyyy=−()+ + =− −(01485 m2425 m). +. =−940 m.
22
DDD=+x y
22
D=(1085 m) (940 m) 144 m−. +−. =
94 0 m tan 0 8664 108 5 m
y
x
D
D
θ
−.
== =.
−.
θ=°+.°= .°180 40 9 220 9
(D
G
is in the third quadrant since both
Dx and Dy are negative.)
Figure 1
The direction of D
G
can also be specified in terms of φ=θ−180 40 9 ;°= .°D
G
is 41° south of west.
EVALUATE: The vector addition diagram, approximately to scale, is
Vector D
G
in this diagram agrees qualitatively with our calculation using components.
Figure 1
01 instructors solutions manual
Course: Physics 2 (PHY 2054)
University: University of Florida
