SHE1215 1 CHAPTER 1 PHYSICAL QUANTITIES AND VECTORS 1.1 THE NATURE OF PHYSICS The word ‘physics’ comes from the Greek word which means nature. Physics was conceived as a study of the natural phenomena around us. Each theory in physics involves: a. A few concept or physical quantities b. Assumptions in order to obtain a mathematical model c. Procedures to relate mathematical models to actual measurement from experiments d. Relationships between various physical concepts e. Experimental proofs to devise explanations to natural phenomena 1.2 BASIC QUANTITIES AND SI UNITS Physics is based on quantities known as physical quantities. Example: length, mass and time. A physical quantity is clearly defined with a numerical value and a unit. In this text, we emphasize the system of units known as SI units, which stands for the French phrase”Le Systeme International d’Unites”. 1.2.1 Base quantities and SI units In the International System of Units (SI), six physical quantities are selected as base quantities. Units for these base quantities are known as base units. Basic quantity Base unit Symbol Length Meter m Mass Kilogram kg Time Second s Electric current Ampere A Thermodynamic temperature Kelvin K Quantity of matter Mole mol Luminous intensity Candela cd Table 1: Base quantities and their SI base units
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SHE1215
1
CHAPTER 1
PHYSICAL QUANTITIES AND VECTORS
1.1 THE NATURE OF PHYSICS
The word ‘physics’ comes from the Greek word which means nature. Physics was conceived as
a study of the natural phenomena around us.
Each theory in physics involves:
a. A few concept or physical quantities
b. Assumptions in order to obtain a mathematical model
c. Procedures to relate mathematical models to actual measurement from experiments
d. Relationships between various physical concepts
e. Experimental proofs to devise explanations to natural phenomena
1.2 BASIC QUANTITIES AND SI UNITS
Physics is based on quantities known as physical quantities. Example: length, mass and time. A
physical quantity is clearly defined with a numerical value and a unit. In this text, we emphasize
the system of units known as SI units, which stands for the French phrase”Le Systeme
International d’Unites”.
1.2.1 Base quantities and SI units
In the International System of Units (SI), six physical quantities are selected as base quantities.
Units for these base quantities are known as base units.
Basic quantity Base unit Symbol
Length Meter m
Mass Kilogram kg
Time Second s
Electric current Ampere A
Thermodynamic temperature Kelvin K
Quantity of matter Mole mol
Luminous intensity Candela cd
Table 1: Base quantities and their SI base units
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1.2.2 Derived Quantities and Derived Units
Physical quantities other than the base quantities are known as derived quantities.
Derived quantity Unit and Symbol In Terms of Base Units
Force Newton, N N = kgms-2
Pressure Pascal, Pa Pa = Nm-2
= kgm-1
s-2
Energy Joule, J J = Nm = kgm2s
-2
Power Watt, W W = Js-1
= kgm2s
-2
Charge Coulomb, C C = As
Voltage Volt, V V = JC-1
= kgm2s
-3A
-1
Resistance Ohm, = VA-1
s = kgm2s
-3A
-2
Capacitance Farad, F F = CV-1
= kg-1
m-2
s4A
2
Inductance Henry, H H = VA-1
s = kgm2s
-2A
-2
Frequency Hertz, Hz Hz = s-1
Table 2: Derived quantities and their units
1.2.3 Other Systems of Units / The Conversion of Units
Sometimes, it is necessary to convert one system of units to another. This is done using the
conversion factors. Some conversion factors between the SI system of units and the C.G.S (cm
gram second) system are shown in table 3. In any conversion, if the units do not combine
algebraically to give the desired result, the conversion has not been carried out properly.
SI to C.G.S C.G.S to SI
1m = 100 cm 1 cm = 10-2
m
1kg = 1000g 1 g = 10-3
kg
1m2
= 104
cm2 1 cm
2 = 10
-4 m
2
1 m3 = 10
6 cm
3 1cm
3 = 10
-6 m
3
Table 3: Conversion factors
EXAMPLE 1.1
An acre is defined such that 640 acres=1 mi2. How many square meters are in 1 acre?
SOLUTION:
1 mile=1.609 km=1609 m
1 acre=640
1mi
2=
640
1(1609m)
2=4.05 x 10
3 m
2
EXAMPLE 1.2:
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A geologist finds that a rock sample has a volume of 2.40 in3. Express this volume in cubic
centimeters and in cubic meters.
Solution:
1 in = 2.54 cm, so V = 2.40 in3 = 2.40(2.54cm)
3
1 cm = 10-2
m, so V = 39.3 cm3 = (39.3) (10
-2m)
3=3.93 x 10
-5 m
3
EXAMPLE 1.3
The mass of the parasitic wasp Caraphractus cintus can be as small as 5x10-6
kg. What is this
mass in (a) grams (g), (b) milligrams (mg), and (c) micrograms (μg)
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
(1.0 103
g)/(1.0 kg)= 1. Therefore,
5x10-6
kg = kg
gx
0.1
100.1 3
x 5x10-6
kg=5x10-3
g
b. Since 1.0 103 milligrams = 1.0 gram,
5x10-3
g= 5x10-3
g x g
mgx
0.1
100.1 3
=5 mg
c. Since 1.0 106
micrograms = 1.0 gram,
5x10-3
g=5x10-3
g x g
gx
0.1
100.1 6=5x10
3g
EXAMPLE 1.4
An engineering student wants to buy 18 gal of gas, but the gas station has installed new pumps
that are measured in liters. How many liters of gas (rounded off to a whole number) should he
ask for?
SOLUTION 18 gal = (18 gal) 3.785 L
1 gal = 68 L .
EXAMPLE 1.5
An automobile speedometer is shown.
(a) What would be the equivalent scale readings (for each empty box) in kilometers per
hour?
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(b) What would be the 70-mi/h speed limit in kilometers per hour?
km/h
5060
70
80
90
1000 mi/h
10
20
30
40
km/h 0
Speedometer readings
SOLUTION:
(a) 10 mi/h = (10 mi/h) 1.609 km
1 mi = 16 km/h for each 10 mi/h .
(b) 70 mi/h = (70 mi/h) 1.609 km
1 mi = 113 km/h .
1.3 DIMENSIONS OF PHYSICAL QUANTITIES / DIMENSIONAL ANALYSIS
In Physics, the term dimension is used to refer to the physical nature of a quantity and the type of
unit used to specify it. The dimension of a physical quantity relates the physical quantity to the
base quantities such as:
Mass (M)
Length (L)
Time (T)
Electric current (I)
Temperature (θ)
Quantity of matter (N)
EXAMPLE 1.6:
a)
1
)(
)( LTT
L
time
ntdisplacemevelocity
b) [force] = [mass] x [acceleration]
=M x LT-2
=MLT-2
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USE OF DIMENSIONS
1. To check the homogeneity of physical equations
Homogeneous-the dimension on both sides of an equation must also be the same
EXAMPLE 1.7: Show that this equation s = ut + 2
1at
2 is dimensionally homogeneous.
Left side: [s] = L
Right side: [ut] = T
L.T = L
[at2]=
T
L 2.T
2 = L
Since all the terms in the equation have the same dimension, the equation is dimensionally
homogeneous.
2. To derive a physical equation
Using dimensions, an equation can be derived to relate a physical quantity to the variables that
the quantity is dependent on.
EXAMPLE 1.8:
The period T of a simple pendulum depends on its length l and the acceleration due to gravity g.
T lxg
y T=k l
xg
y
Since the dimensions on both sides of the equation must be the same;
[T] = [k lxg
y] T = L
x(LT
-2)y
Equating the indices of T; -2y = 1 y = -2
1
Equating the indices of L; x+y=0 x-2
1 =0 x=2
1
Hence; T=kl1/2
g-1/2
So, the value of the constant k can be determined experimentally
EXAMPLE 1.9
The following are dimensions of various physical parameters. Here [L], [T] and [M] denote,
respectively, dimensions of length, time and mass.
Parameter Dimension
Distance (x) [L]
Time (t) [T]
Mass (m) [M]
Speed (v) [L]/[T]
Acceleration (a) [L]/[T]2
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Force (F) M[L]/[T]2
Energy (E) M[L]2/[T]
2
Which of the following equations are dimensionally correct?