Homework 1 Solutions
2.
Picture the Problem: This is simply a units conversion problem.
Strategy: Multiply the given number by conversion factors to obtain the desired units.
Solution: (a) Convert the units:
70  m 
1.0 106 m
 7.0 105 m
m
(b) Convert the units again:
70  m 
1.0 106 m 1 km

 7.0 108 km
m
1000 m
Insight: The inside back cover of the textbook has a helpful chart of the metric prefixes.
14. Picture the Problem: This is a significant figures question.
Strategy: Follow the given rules regarding the calculation and display of significant figures.
Solution:
Round to the 3rd digit:
2.9979 108 m/s  3.00 108 m/s
Insight: It is important not to round numbers off too early when solving a problem because
excessive rounding can cause your answer to significantly differ from the true answer.
15. Picture the Problem: The parking lot is a rectangle.
144.3 m
Strategy: The perimeter of the parking lot is the sum of
the lengths of its four sides. Apply the rule for addition
47.66 m
of numbers: the number of decimal places after addition
equals the smallest number of decimal places in any of the
47.66 m
144.3 m
individual terms.
Solution: 1. Add the numbers:
144.3 + 47.66 + 144.3 + 47.66 m = 383.92 m
2. Round to the smallest number of decimal
383.92 m
 383.9 m
places in any of the individual terms:
Insight: Even if you changed the problem to
 2 144.3 m   2  47.66 m
you’d still have to
report 383.9 m as the answer; the 2 is considered an exact number so it’s the 144.3 m that limits
the number of significant digits.
16. Picture the Problem: The weights of the fish are added.
Strategy: Apply the rule for addition of numbers, which states that the number of decimal places
after addition equals the smallest number of decimal places in any of the individual terms.
Solution: 1. Add the numbers:
2.35 + 12.1 + 12.13 lb = 26.58 lb
2. Round to the smallest number of decimal
26.58 lb
 26.6 lb
places in any of the individual terms:
Insight: The 12.1 lb rock cod is the limiting figure in this case; it is only measured to within an
accuracy of 0.1 lb.
17. Picture the Problem: This is a significant figures question.
Strategy: Follow the given rules regarding the calculation and display of significant figures.
Solution: 1. (a) The leading zeros are not significant: 0.0000 5 4 has 2 significant figures
2. (b) The middle zeros are significant:
5
3.0 0 1×10 has 4 significant figures
Insight: Zeros are the hardest part of determining significant figures. Scientific notation can
remove the ambiguity of whether a zero is significant because any trailing zero to the right of the
decimal point is significant.
20. Picture the Problem: This is a units conversion problem.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
Solution: 1. (a) Find the length in feet:
 631 m  
3.28 ft 
  2069 ft
 1m 
2. Find the width in feet:
 707 yd  
3. Find the volume in cubic feet:
V  LWH   2069 ft  2121 ft 110 ft   4.83 108 ft 3
4. (b) Convert to cubic meters:
1m 
7
3
 4.83 108 ft 3   3.28
  1.37  10 m
ft 
 3 ft 
  2121 ft
 1 yd 
3
Insight: Conversion factors are conceptually equal to one, even though numerically they often
equal something other than one. They are often helpful in displaying a number in a convenient,
useful, or easy-to-comprehend fashion.
21. Picture the Problem: This is a units conversion problem.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
Solution: 1. Find the length in feet:
 2.5 cubit  
17.7 in  1 ft 

  3.68 ft
 1 cubit  12 in 
2. Find the width and height in feet:
1.5 cubit  
3. Find the volume in cubic feet:
V  LWH   3.68 ft  2.21 ft  2.21 ft   18 ft 3
17.7 in  1 ft 

  2.21 ft
 1 cubit  12 in 
Insight: Conversion factors are conceptually equal to one, even though numerically they often
equal something other than one. They are often helpful in displaying a number in a convenient,
useful, or easy-to-comprehend fashion.
22. Picture the Problem: This is a units conversion problem.
Strategy: Convert the frequency of cesium-133 given on page 4 to units of microseconds per
megacycle, then multiply by the number of megacycles to find the elapsed time.
Solution: Convert to micro
seconds per megacycle and
multiply by 1.5 megacycles:

  1 106 cycles   1  s 
1s
s



  108.7827757
6
Mcycle
 9,192, 631, 770 cycles   Mcycle   1 10 s 
s
108.7827757
 1.5 Mcycle  160  s  1.6 10  4 s
Mcycle
Insight: Only two significant figures remain in the answer because of the 1.5 Mcycle figure given
in the problem statement. The metric prefix conversions are considered exact and have an
unlimited number of significant figures, but most other conversion factors have a limited number
of significant figures.
23. Picture the Problem: This is a units conversion problem.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
Solution: Convert feet to kilometers:
 3212 ft  
1 mi  1.609 km 

  0.9788 km
5280
ft  1 mi 

Insight: Conversion factors are conceptually equal to one, even though numerically they often
equal something other than one. They are often helpful in displaying a number in a convenient,
useful, or easy-to-comprehend fashion.
26. Picture the Problem: This is a units conversion problem.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
Solution: Convert carats to pounds:
0.20 g   1 kg   2.21 lb 

  0.23 lb

 ct   1000 g   kg 
 530.2 ct  
Insight: Conversion factors are conceptually equal to one, even though numerically they often
equal something other than one. They are often helpful in displaying a number in a convenient,
useful, or easy-to-comprehend fashion.
27. Picture the Problem: This is a units conversion problem.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
Solution: 1. (a) The speed must be greater than 55 km/h because 1 mi/h = 1.609 km/h.
2. (b) Convert the miles to kilometers:
mi  1.609 km 
km

 55

  88
h
mi
h



Insight: Conversion factors are conceptually equal to one, even though numerically they often
equal something other than one. They are often helpful in displaying a number in a convenient,
useful, or easy-to-comprehend fashion.
28. Picture the Problem: This is a units conversion problem.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
Solution: Convert m/s to miles per hour:

8 m  1 mi  3600 s 
8 mi
 3.00 10


  6.7110
s  1609 m  1 h 
h

Insight: Conversion factors are conceptually equal to one, even though numerically they often
equal something other than one. They are often helpful in displaying a number in a convenient,
useful, or easy-to-comprehend fashion.
29. Picture the Problem: This is a units conversion problem.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
Solution: Convert to ft per second per second: 
m  3.28 ft 
ft
 98.1 2 
  322 2
s  1 m 
s

Insight: Conversion factors are conceptually equal to one, even though numerically they often
equal something other than one. They are often helpful in displaying a number in a convenient,
useful, or easy-to-comprehend fashion.
30. Picture the Problem: This is a units conversion problem.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units. In
this problem, one “jiffy” corresponds to the time in seconds that it takes light to travel one
centimeter.
Solution: 1. (a): Determine the magnitude
of a jiffy:
1s
s
jiffy

 1 m 
11
1


  3.3357  10
8
cm
cm
 2.9979 10 m  100 cm 
1 jiffy  3.3357 1011 s
2. (b) Convert minutes to jiffys:
1 minute  
60 s 
1 jiffy

12

  1.7987 10 jiffy
11
 1 min  3.3357 10 s 
Insight: Conversion factors are conceptually equal to one, even though numerically they often
equal something other than one. They are often helpful in displaying a number in a convenient,
useful, or easy-to-comprehend fashion.
32. Picture the Problem: The volume of the oil is spread out into a slick that is one molecule thick.
Strategy: The volume of the slick equals its area times its thickness. Use this fact to find the
area.
Solution: Calculate the area for
A
the known volume and thickness:
V
1.0 m3  1  m 
6
2


  2.0 10 m
h 0.50  m  1106 m 
Insight: Two million square meters is about 772 square miles!
33. Picture the Problem: This is a units conversion problem.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
Then use a ratio to find the factor change in part (b).
Solution: 1. (a) Convert square inches to square meters:
2. (b) Calculate a ratio to find the new area:
 1 m2 
A  8.5 in  11 in  
 0.060 m2
2 
1550
in


Anew LnewWnew  12 Lold  12 Wold  1



Aold
LoldWold
LoldWold
4
Anew 
1
4
Aold
Insight: If you learn to use ratios you can often make calculations like these very easily. Always
put the new quantity in the numerator and the old quantity in the denominator to make the new
quantity easier to calculate at the end.
34. Picture the Problem: This is a units conversion problem.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
Solution: 1. Convert m/s to ft/s:
m  3.28 ft 

 20.0 
  65.6 ft/s
s  m 

2. (b) Convert m/s to mi/h:
m  1 mi  3600 s 

 20.0 

  44.7 mi/h
s  1609 m  1 h 

Insight: Conversion factors are conceptually equal to one, even though numerically they often
equal something other than one. They are often helpful in displaying a number in a convenient,
useful, or easy-to-comprehend fashion.
35. Picture the Problem: This is a units conversion problem.
Strategy: Multiply the known quantity by appropriate conversion factors to change the units.
Solution: Convert meters to feet:
m  3.28 ft 

2
 9.81 2 
  32.2 ft/s
s  1 m 

Insight: Conversion factors are conceptually equal to one, even though numerically they often
equal something other than one. They are often helpful in displaying a number in a convenient,
useful, or easy-to-comprehend fashion.