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5. REASONING In order to calculate d, the units of a and b must be, respectively, cubed and
squared along with their numerical values, then combined algebraically with each other and
the units of c. Ignoring the values and working first with the units alone, we have
a3
( m ) = m 3 2 = m2
→
cb2
( m/s )( s )2 m / s ⋅ s 2 1 s
3 d= ( ) Therefore, the units of d are m2/s. SOLUTION With the units known, the numerical value may be calculated:
d= 6. ( 9.7 )3
( 69 )( 4.2 )2 m2 /s = 0.75 m2 /s 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. Chapter 1 Problems L The dimensions of the right hand side are T 2 L= L
T 5 2
2 , while the dimensions of the left [ L] [ L] . The right side will equal the left side only when n = 2 .
hand side are = [T ] [T ]n n n ____________________________________________________________________________________________ 7. SSM 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: (1.0 jeroboam ) 0.792 gallons 1.0 jeroboam 3.785 liters 1.0 gallon 1.0 magnum = 2.0 magnums 1.5 liters ____________________________________________________________________________________________ 8. REASONING In the expression for the volume flow rate, the dimensions on the left side of
the equals sign are [L]3/[T]. If the expression is to be valid, the dimensions on the right side
of the equals sign must also be [L]3/[T]. Thus, the dimensions for the various symbols on
the right must combine algebraically to yield [L]3/[T]. We will substitute the dimensions for
each symbol in the expression and treat the dimensions of [M], [L], and [T] as algebraic
variables, solving the resulting equation for the value of the exponent n.
SOLUTION We begin by noting that the symbol π and the number 8 have no dimensions.
It follows, then, that Q= π R n ( P2 − P )
1
8η L [ L]3 = [ L]n
[T ] [ L] [T ] or [ L]
[T] or = [ L]n
[ L] =
[ L]
3 [M]
[ L][ T ]2 = [ L]n [ T ] = [ L]n
[M]
[ L][T ]2 [ L][T ]
[ L]
[ L] [T] [ L]n 3 or [ L]3 [ L] = [ L]4 = [ L]n 6 INTRODUCTION AND MATHEMATICAL CONCEPTS Thus, we find that n = 4 . 9. 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 .
____________________________________________________________________________________________ 10. REASONING To convert from gallons to cubic meters, use the equivalence
1 U.S. gal = 3.785×10−3 m3. To find the thickness of the painted layer, we use the fact that
the paint’s volume is the same, whether in the can or painted on the wall. The layer of paint
on the wall can be thought of as a very thin “box” with a volume given by the product of the
surface area (the “box top”) and the thickness of the layer. Therefore, its thickness is the
ratio of the volume to the painted surface area: Thickness = Volume/Area. That is, the larger
the area it’s spread over, the thinner the layer of paint.
SOLUTION
a. The conversion is ( 0.67 3.785 × 10−3 m3 U.S. gallons = 2.5 × 10−3 m3 U.S. gallons ) b. The thickness is the volume found in (a) divided by the area, Thickness = Volume 2.5 × 10−3 m3
=
= 1.9 × 10−4 m
2
Area
13 m Chapter 1 Problems 11. 7 SSM 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 = 4 π 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 ____________________________________________________________________________________________ 12. REASONING The shortest distance between the tree and the termite mound is equal to the
magnitude of the chimpanzee's displacement r. SOLUTION
a. From the Pythagorean theorem, we have
r= 51 m θ (51 m) 2 + (39 m) 2 = 64 m b. The angle θ is given by θ = tan −1 39
F mI =...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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