CHAPTER 1.1 and 1.2

CHAPTER 1 PHYSICS

PHYSICAL
QUANTITIES

AND
MEASUREMENTS

CHAPTER 1 PHYSICS

1.1 Physical Quantities and Units

Learning Outcome:

At the end of this chapter, students should be able to:
➢ State basic quantities and their respective SI units: length (m), time (s), mass (kg), electrical

current (A), temperature (K), amount of substance (mol).

➢ State derived quantities and their respective units and symbols: velocity (m s-1), acceleration
(m s-2), work (J), force (N), pressure (Pa), energy (J), power (W) and frequency (Hz).

➢ State and convert units with common SI prefixes.

CHAPTER 1 PHYSICS

1.1 Physical Quantities and Units

1) Physical quantity is defined as a quantities that can be measured.

2) It can be categorized into 2 types
a) Basic (base) quantity
b) Derived quantity

3) Basic quantity is defined as a physical quantity that cannot be defined
in terms of other physical quantities.

CHAPTER 1 PHYSICS

1.1 Physical Quantities and Units

Quantity Symbol SI Unit Symbol
Length metre
Mass l ……Ki…log…ra…m …. m
Time m second kg
Temperature t kelvin s
Electric current ampere K
Amount of substance T/ mole ……A……..
mol
I
…n…….

Table 1.1 shows all the basic (base) quantities.

4) Derived quantity is defined as a
quantity which can be expressed in term of base quantity.

For example

Derived quantity Derive from base quntity of Derived unit

Area length x length m2

Volume length x length x length m3

Density mass kg m-3
volume

For example

Derived quantity Derive from base quntity of Derived unit

Velocity l m s-1
Acceleration t m s-2

velocity
time

Frequency 1 s-1/hz
T

For example

Derived quantity Derive from base quntity of Derived unit

Momentum Mass x velocity Kg ms-1

Force Mass x acceleration Kg ms-2

Pressure force N m-2
Area

Energy 1 mv2 Kg m2 s-2
2

CHAPTER 1 PHYSICS

1.1.1 Unit Prefixes

It is used for presenting larger and smaller values.

Prefix Multiple Symbol
tera
giga  1012 T
mega  … … . G
kilo  106 M
deci  103 …… …..
centi  10−1 d
milli  10−2 c
micro  10−3 m
nano  10−6 …… …
pico  … … − n
 10−12 p

Table 1.3 shows all the unit prefixes.

Example :

Write 2 x 10-7 in a suitable prefix.

2 10−7
= 2 10−6 10−1

= 2 10−1

= 0.2 Prefix tera giga mega Kilo deci centi mili micro nano pico

Factor 1012 109 106 103 10-1 10-2 10-3 10-6 10-9 10-12
Symbol T G M K d c m µ n P

EXEMPLE :

The derived unit change
7854 kg m-3 change into g cm-3.

Solution: 7854 103 g 7854 103 gcm −3
100cm 100cm 100cm 106
7854kg
1m 3

7.854 g cm-3

1.2 Dimensions and Physical Quantities

At the end of this chapter, students should be able to:

(a) use dimensional analysis to determine the dimensions of
derived quantities;

(b) check the homogeneity of equations using dimensional analysis;
(c) construct empirical equations using dimensional analysis;

The dimension of a physical quantity is a product of the
basic physical dimensions each raised to a rational power.

1) Each derive quantity in physic can be represent by basic quantity.
The dimension of a physical quantities is the relation between the
physical quantity and the base quantities

2) The Bracket ‘[ ]’ meant the dimension of a physical quantity

1.2 Dimensions and Physical Quantities

BASE QUANTITY DIMENSION
Mass M
Length L
Time T
I
Electric current
Temperature N
Amount of substance

Dimension of base quantity

Example BASE DIMENSION
QUANTITY
a) Dimension of area = [Area] M
= [Length x Breadth] Mass L
= LxL T
= L2 Length I

Time N

Electric current

Temperature

Amount of
substance

Unit of area = m2

b) Dimension of velocity = [Velocity] Unit of velocity = ms−1

= Displacement
Time

= L
T

Example BASE DIMENSION
QUANTITY
c) Dimension of acceleration = [Acceleration] M
Mass L
T
Length I

Time N

Electric current

Temperature

Amount of
substance

= Change of velocity
Time
Unit of acceleration = ms−2
= LT−1
T

= LT−2

d) Dimension of force = [Force] Unit of force = kgms−2 Or N

= Mass x Acceleration
= M LT−2

Example BASE DIMENSION
QUANTITY
e) Dimension of energy = [Energy] M
= [Force x Displacement] Mass L
T
Length I

Time N

Electric current

Temperature

Amount of
substance

= M LT−2L Unit of energy = km2s−2
= ML2T−2 Or J

f) Dimension of electric charge = [charge] Unit of electric charge = As
= [Current x Time] Or C
=AT

Use of dimensions

•To check the homogeneity of physical equations

Concept of homogeneous

➢ A physical equation is true in all systems of unit.

➢ Two physical quantities can only be equated, added or subtracted if they
have the same dimension.

➢ For correct physical equation, the dimensions of all terms in the
equation are the same.

➢ The dimension on the left side of the equation and the dimension on the
right side of the equation are the same.

Use of dimensions

Concept of homogeneous
➢ The equation is said to be dimensionally consistent or homogeneous.
➢ For example :

The equation of motion with uniform acceleration a, the initial velocity u,
final velocity v and displacement s are related by :

2 = 2+ 2

Determine the homogeneous of the equation v2 =u2 +2as.

Solution:

Left hand side :
[v2] = [v]2 = (LT-1)2 = L2 T-2

Right hand side :
[u2 + 2as] = L2 T-2 + L T-2 . L = L2 T-2

Conclusion ; the RHS dimension as same as the LHS dimension,
meaning that the above equation is homogeneous.





EXERCISES PHYSICS

1) Under uniform acceleration, motion of an object with velocity, v,
is represented by 2 = + where a and b are constants and
x is a variable for displacement. If both a and b have
dimensions, find the dimension of
a) a
b) bx
c) b

EXERCISES PHYSICS

2) The power P required to overcome external resistances when a

vehicle is travelling at a speed v is given by the expression
= + 2 where a and b are constants. Derive the

dimensions for the constants a and b. Then deduce the units for

a and b in terms of the base SI units.

EXERCISES PHYSICS

SOLUTION :

PHYSICS

Quiz

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