1.
REASONING AND SOLUTION
a. The SI unit for x is m. The SI units for the quantity vt are
FG IJ
H K
m
§¨ m ·¸
(s) = m
(s) = m
©s¹
s
Therefore, the units on the left hand side of the equation are consistent with the units on the
right hand side.
b. As described in part a, the SI units for the quantities x and vt are both m. The SI units for
1
the quantity at2 are
2
FG m IJ (s
Hs K
2
2
)=m
Therefore, the units on the left hand side of the equation are consistent with the units on the
right hand side.
c. The SI unit for v is m/s. The SI unit for the quantity at is
FG m IJ (s) = m
Hs K s
2
Therefore, the units on the left hand side of the equation are consistent with the units on the
right hand side.
d. As described in part c, the SI units of the quantities v and at are both m/s. The SI unit of
1
the quantity at3 is
2
FG m IJ (s ) = m ⋅ s
Hs K
3
2
Thus, the units on the left hand side are not consistent with the units on the right hand side.
In fact, the right hand side is not a valid operation because it is not possible to add physical
quantities that have different units.
e. The SI unit for the quantity v3 is m3/s3. The SI unit for the quantity 2ax3 is
FG m IJ (m
Hs K
2
2
)=
m3
s2
Therefore, the units on the left hand side of the equation are not consistent with the units on
the right hand side.
f. The SI unit for the quantity t is s. The SI unit for the quantity
m
=
(m / s 2 )
FG s IJ =
H mK
2
m
2x
is
a
s2 = s
Therefore, the units on the left hand side of the equation are consistent with the units on the
right hand side.
3.
REASONING AND SOLUTION It is not always possible to add two numbers that have
the same dimensions. In order to add any two physical quantities they must be expressed in
the same units. Consider the two lengths: 1.00 m and 1.00 cm. Both quantities are lengths
and, therefore, have the dimension [L]. Since the units are different, however, these two
numbers cannot be added.
5.
REASONING AND SOLUTION For the equation to be valid, the dimensions of the left
hand side of the equation must be the same as the dimensions on the right hand side. Since
the quantity c has no dimensions, it does not contribute to the dimensions of the right hand
side, regardless of the value of n. Therefore, the value of n cannot be determined from
dimensional analysis.
1.
SSM REASONING When converting between units, we write down the units explicitly
in the calculations and treat them like any algebraic quantity. We construct the appropriate
conversion factor (equal to unity) so that the final result has the desired units.
SOLUTION
a. Since 1.0 × 103 grams = 1.0 kilogram, it follows that the appropriate conversion factor is
3
(1.0 × 10 g)/(1.0 kg) = 1. Therefore,
c5 × 10
−6
F 1.0 × 10 g IJ =
kg h G
H 1.0 kg K
3
5 × 10 −3 g
b. Since 1.0 × 103 milligrams = 1.0 gram,
c5 × 10 g h FGH 1.0 ×1.010gmg IJK =
3
−3
5 mg
c. Since 1.0 × 106 micrograms = 1.0 gram,
c5 × 10
−3
F 1.0 × 10 µg IJ =
gh G
H 1.0 g K
6
5 × 10 3 µg
6.
REASONING This problem involves using unit conversions to determine the number of
magnums in one jeroboam. The necessary relationships are
1.0 magnum = 1.5 liters
1.0 jeroboam = 0.792 U. S. gallons
1.00 U. S. gallon = 3.785 × 10 –3 m 3 = 3.785 liters
These relationships may be used to construct the appropriate conversion factors.
SOLUTION By multiplying one jeroboam by the appropriate conversion factors we can
determine the number of magnums in a jeroboam as shown below:
gallons I F 3.785 liters I F 1.0 magnum I
b1.0 jeroboam g FGH 0.792
J=
JG
JG
1.0 jeroboam K H 1.0 gallon K H 1.5 liters K
2.0 magnums
7. REASONING AND SOLUTION
a. F = [M][L]/[T]2; ma = [M][L]/[T]2 = [M][L]/[T]2 so F = ma is dimensionally correct .
b. x = [L]; at3 = ([L]/[T]2)[T]3 = [L][T] so x = (1/2)at3 is not dimensionally correct .
c. E = [M][L]2/[T]2; mv = [M][L]/[T] so E = (1/2)mv is not dimensionally correct .
d. E = [M][L]2/[T]2; max = [M]([L]/[T]2)[L] = [M][L]2/[T]2 so E = max
is dimensionally correct .
e. v = [L]/[T]; (Fx/m)1/2 = {([M][L]/[T]2)([L]/[M])}1/2 = {[L]2/[T]2}1/2 = [L]/[T] so
v = (Fx/m)1/2 is dimensionally correct .
8.
REASONING AND SOLUTION
x has the dimensions of [L], v has the dimensions of
[L]/[T], and a has the dimensions of [L]/[T]2. The equation under consideration is vn = 2ax.
2
L
L
L =
, while the dimensions of the left
The dimensions of the right hand side are
2
2
T
T
FLI
hand side are G
H T JK
n
=
L
T
n
n
. The right side will equal the left side only when
n=2 .
10. REASONING The dimension of the spring constant k can be determined by first solving
the equation T = 2π m / k for k in terms of the time T and the mass m. Then, the
dimensions of T and m can be substituted into this expression to yield the dimension of k.
SOLUTION Algebraically solving the expression above for k gives k = 4π 2 m / T 2 . The
term 4π 2 is a numerical factor that does not have a dimension, so it can be ignored in this
analysis. Since the dimension for mass is [M] and that for time is [T], the dimension of k is
Dimension of k =
[M]
[T ]2