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, Chapter 1: Overview
Chapter 1: Overview
Concept Checks
1.1. a 1.2. a) 4 b) 3 c) 5 d) 6 e) 2 1.3. a, c and e 1.4. b 1.5. e 1.6. a) 4th b) 2nd c) 3rd d) 1st
Multiple-Choice Questions
1.1. c 1.2. c 1.3. d 1.4. b 1.5. a 1.6. b 1.7. b 1.8. c 1.9. c 1.10. b 1.11. d 1.12. b 1.13. c 1.14. a 1.15. e 1.16. a
Conceptual Questions
1.17.
(a) In Europe, gas consumption is in L/ 100 km . In the US, fuel efficiency is in miles/gallon. Let’s
relate these two: 1 mile = 1.609 km, 1 gal = 3.785 L.
1 mile 1.609 km 1.609 1
(100) L = ( 0.00425) L/100 km = 235.24 L/100 km
km 1 1
= =
gal 3.785 L 3.785 100
Therefore, 1 mile/gal is the reciprocal of 235.2 L/100 km.
12.2 L 1L 1
(b) Gas consumption is . Using = from part (a),
100 km 100 km 235.24 miles/gal
12.2 L 1L 1 1
= 12.2 = 12.2 235.24 miles/gal = .
100 km 100 km 19.282 miles/gal
Therefore, a car that consumes 12.2 L/100 km of gasoline has a fuel efficiency of 19.3 miles / gal .
(c) If the fuel efficiency of the car is 27.4 miles per gallon, then
27.4 miles 27.4 1
= = .
gal 235.24 L/100 km 8.59 L/100 km
Therefore, 27.4 miles / gal is equivalent to 8.59 L/ 100 km .
(d)
1.18. A vector is described by a set of components in a given coordinate system, where the components are
the projections of the vector onto each coordinate axis. Therefore, on a two-dimensional sheet of
paper there are two coordinates and thus, the vector is described by two components. In the real
three-dimensional world, there are three coordinates and a vector is described by three components.
A four-dimensional world would be described by four coordinates, and a vector would be described
by four components.
1
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Bauer/Westfall: University Physics, 3E
1.19. A vector contains information about the distance between two points (the magnitude of the vector).
In contrast to a scalar, it also contains information direction. In many cases knowing a direction can
be as important as knowing a magnitude.
1.20. In order to add vectors in magnitude-direction form, each vector is expressed in terms of component
vectors which lie along the coordinate axes. The corresponding components of each vector are added
to obtain the components of the resultant vector. The resultant vector can then be expressed in
magnitude-direction form by computing its magnitude and direction.
1.21. The advantage to using scientific notation is two-fold: Scientific notation is more compact (thus saving
space and writing), and it also gives a more intuitive way of dealing with significant figures since you
can only write the necessary significant figures and extraneous zeroes are kept in the exponent of the
base.
1.22. The SI system of units is the preferred system of measurement due to its ease of use and clarity. The
SI system is a metric system generally based on multiples of 10, and consisting of a set of standard
measurement units to describe the physical world. In science, it is paramount to communicate results
in the clearest and most widely understood manner. Since the SI system is internationally recognized,
and its definitions are unambiguous, it is used by scientists around the world, including those in the
United States.
1.23. It is possible to add three equal-length vectors and obtain a vector sum of zero. The vector
components of the three vectors must all add to zero. Consider the following arrangement with
T1 = T2 = T3 :
The horizontal components of T1 and T2 cancel out, so the sum T1 + T2 is a vertical vector whose
magnitude is T cos + T cos = 2T cos . The vector sum T1 + T2 + T3 is zero if
2T cos − T = 0
1
cos =
2
= 60
Therefore it is possible for three equal-length vectors to sum to zero.
1.24. Mass is not a vector quantity. It is a scalar quantity since it does not make sense to associate a direction
with mass.
1.25. The volume of a sphere is given by V = ( ) r 3 . Doubling the volume gives
2V = 2( 4/ 3) r3 = ( 4/ 3) (23/3 )r3 = ( 4/ 3) (21/3 r)3 . Now, since the distance between the flies is the
diameter of the sphere, d = 2r , and doubling the volume increases the radius by a factor of 2 , the
1/ 3
distance between the flies is then increased to 2(21/3 r) = 21/3 (2r) = 21/3 d. Therefore, the distance is
increased by a factor of 21/ 3.
2
, Chapter 1: Overview
1.26. The volume of a cube of side r is Vc = r 3 , and the volume of a sphere of radius r is Vsp = ( ) r 3 .
The ratio of the volumes is:
Vc r3 3
= = .
Vsp 4 r 3 4
3
The ratio of the volumes is independent of the value of r.
1.27. The surface area of a sphere is given by 4 r 2 . A cube of side length s has a surface area of 6s2 . To
determine s set the two surface areas equal:
4 r 2 2
6s2 = 4 r 2 s = =r .
6 3
1.28. The mass of Sun is 2 1030 kg, the number of stars in the Milky Way is about 100 109 = 1011 , the number
of galaxies in the Universe is about 100 109 = 1011 , and the mass of an H-atom is 2 10−27 kg.
(a) The total mass of the Universe is roughly equal to the number of galaxies in the Universe
multiplied by the number of stars in a galaxy and the mass of the average star:
M universe = (1011 )(1011 )(2 1030 ) = 2 10(11+11+30) kg = 2 1052 kg.
M universe 2 1052 kg
(b) nhydrogen = = 1079 atoms
M hydrogen 2 10−27 kg
1.29. The volume of 1 teaspoon is about 4.93 10−3 L , and the volume of water in the oceans is about
1.35 1021 L.
1.35 1021 L
= 2.74 1023 tsp
4.93 10−3 L/tsp
There are about 2.74 1023 teaspoons of water in the Earth’s oceans.
1.30. The average arm-span of an adult human is d = 2 m. Therefore, with arms fully extended, a person
takes up a circular area of r 2 = ( d / 2) = (1 m)2 = m 2 . Since there are approximately 6.5 109
2
humans, the amount of land area required for all humans to stand without being able to touch each
other is 6.5 109 m 2 ( ) = 6.5 109 m 2 (3.14) = 2.0 1010 m 2 . The area of the United States is about
3.5 106 square miles or 9.11012 m 2 . In the United States there is almost five hundred times the
amount of land necessary for all of the population of Earth to stand without touching each other.
1.31. The diameter of a gold atom is about 2.6 10−10 m. The circumference of the neck of an adult is roughly
0.40 m. The number of gold atoms necessary to link to make a necklace is given by:
circumference of neck 4.0 10−1 m
n= = = 1.5 109 atoms.
diameter of atom 2.6 10−10 m/atom
The Earth has a circumference at the equator of about 4.008 107 m . The number of gold atoms
necessary to link to make a chain that encircles the Earth is given by:
circumference of Earth 4.008 107 m
N= = = 1.5 1017 atoms.
diameter of a gold atom 2.6 10−10 m
Since one mole of substance is equivalent to about 6.022 1023 atoms , the necklace of gold atoms has
( )( )
1.5 109 atoms / 6.022 1023 atoms/mol = 2.5 10−15 moles of gold. The gold chain has
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