1. Lecture 1
MEASUREMENTS
and
PHYSICAL QUANTITIES

2. MEASUREMENT
 We obtain an understanding of
our environment by observing
the physical world through
measurements.
 Measurement is a comparison
of physical quantity with a
standard unit.

3. MEASUREMENT
 It is expressed with numbers and units.
 Physical characteristics can be
expressed in terms of fundamental
quantities.
 Our comprehension(level of
undersatnding) of the physical world is
based on fundamental quantities.

4. FUNDAMENTAL QUANTITIES
Four quantities have been identified and
specified as
fundamental.
These are:
 Length
 Mass
 Time
 Electric charge
They form the basis for other quantities
necessary to describe and understand the
physical sciences.

5. Length
 We use length to measure location or size.
 Length is defined as the measurement of
space in any direction.
Note: Space has three dimensions, each is
measured by length.
 Consider a box:
Length (l)
Width (w)
Height (h)

6. Time
 Time is the continuous forward
flow of events.
 Events allow us to precise time,
without events, we have no innate
awareness of time.

7. Mass
 Mass quantifies matter.
 Mass refers to the amount of matter
an object contains.

8. Electric Charge
 Electric Charge is the property
associated with some particles, that
gives rise to electric forces and
electrical phenomena.

9. STANDARD UNITS
 A standard unit is a fixed and
reproducible value for the purpose
of taking accurate measurements.
 A set of standard units is referred
to as a system of units.

10. International System of Units
There are seven base units of the
International
System of Units.
 Meter, m, measures length.
 Kilogram, kg, measures mass.
 Second, s, measures time.
 Ampere, A, measures electric current.
 Kelvin, K, measures temperature.
 Mole, mol, measures amount of
substance.
 Candela, cd, measures luminous
intensity.

11. Measurement Errors
Systematic Errors:
1. Systematic errors are associated
with instruments or the technique
used.
They are as a result of:
 Improperly calibrated instrument
 Error incurred from observer’s
reading

12. Measurement Errors
Random Errors:
2. Random Errors result from
unknown and unpredictable
variations in experimental situations.
 Such as fluctuation in electrical
voltage.
 Changes in temperature, pressure,
etc.

13. Accuracy / Precision
 Accuracy indicates how close
measurement comes to the true
value.
 Precision refers to the agreement
among repeated measurement, that
is, how close they are together.

14. 14

15. 15
 Physical quantity is defined as a ……………………………….
 It can be categorized into 2 types
 Basic (base) quantity
 Derived quantity
 Basic quantity is defined as …………………………………………….
 ………………………………………………………………………………..
 Table 1.1 shows all the basic (base) quantities.
1.1 Physical Quantities and Units
Quantity Symbol SI Unit Symbol
Length l metre m
Mass m ………………. kg
Time t second s
Temperature T/ kelvin K
Electric current I ampere …………..
Amount of substance ………. mole mol
Table 1.1

16. 16
 Derived quantity is defined as a quantity which can be expressed
in term of base quantity.
 Table 1.2 shows some examples of derived quantity.
Derived quantity Symbol Formulae Unit
Velocity v s/t m s-1
Volume …….. l  w  t m 3
Acceleration a v/t m s-2
Density  m/V …………….
Momentum p ………… kg m s-1
Force ……… m  a kg m s-2 @ N
Work W F  s ……….. @ J
Pressure P F/A N m-2 @ ……
Frequency f 1/T s-1 @ ……..
Table 1.2

17. 17
 It is used for presenting larger and smaller values.
 Table 1.3 shows all the unit prefixes.
 Examples:
 5740000 m = 5740 km = 5.74 Mm
 0.00000233 s = 2.33  106 s = 2.33 s
Prefix Multiple Symbol
tera  1012 T
giga  ……. G
mega  106 M
kilo  103 ………..
deci  101 d
centi  102 c
milli  103 m
micro  106 ………
nano  ,,,,,,, n
pico  1012 p
1.1.1 Unit Prefixes
Table 1.3

18. 18
Solve the following problems of unit conversion.
a. 15 mm2 = ? m2 b. 65 km h1 = ? m s1
c. 450 g cm3 = ? kg m3
Solution :
a. 15 mm2 = ? m2
b. 65 km h-1 = ? m s-1
1st method :







 


h
1
m
10
65
h
km
65
3
1
Example 1.1 :







 


s
..........
m
10
65
h
km
65
3
1
1
1
s
m
........
h
km
65 


   2
2
......m
mm
1 
2
6
2
m
10
mm
1 


19. 19
2nd method :
c. 450 g cm-3 = ? kg m-3








h
1
km
65
h
km
65 1
1
1
s
m
18
h
km
65 



3
cm
g
450
3
5
3
m
kg
10
.5
4
cm
g
450 























s
3600
h
....
......
1
.......m
h
1
km
65
h
km
65 1

20. 20
Follow Up Exercise
1. A hall bulletin board has an area of 250 cm2. What is this area in
square meters ( m2 ) ?
2. The density of metal mercury is 13.6 g/cm3. What is this density
as expressed in kg/m3
3. A sheet of paper has length 27.95 cm, width 8.5 cm and
thickness of 0.10 mm. What is the volume of a sheet of paper in
m3 ?
4. Convert the following into its SI unit:
(a) 80 km h–1 = ? m s–1
(b) 450 g cm–3 = ? kg m–3
(c) 15 dm3 = ? m3
(d) 450 K = ? ° C

21. Any valid physical equation must be dimensionally
consistent – each side must have the same dimensions.
From the Table:
Distance = velocity × time
Velocity = acceleration ×
time
Energy = mass × (velocity)2
Dimensional Analysis (1)

22. Dimensional Analysis
 Fundamental Quantities
 Length - [L]
 Time - [T]
 Mass - [M]
 Derived Quantities
 Velocity - [L]/[T]
 Density - [M]/[L]3
 Energy - [M][L]2/[T]2

23. The period P [T] of a swinging
pendulum depends only on the length
of the pendulum d [L] and the
acceleration of gravity g [L/T2].
Which of the following formulas for P
could be correct ?
P
d
g
 2
P
d
g
 2
(a) (b) (c)
P  2 (dg)2
Example:

24. L
L
T
L
T
T






  
2
2 4
4
Dimensional Analysis (3)
L
L
T
T T
2
2
 
Remember that P is in units of time (T), d is
length (L) and g is acceleration (L/T2).
The both sides must have the same units
 
P dg
 2
2

(a) (b) (c)
P
d
g
 2
Try equation (a). Try equation (b). Try equation (c).
T
T
T
L
L 2
2


P
d
g
 2

25. 25
 Scalar quantity is defined as a quantity with magnitude only.
 e.g. mass, time, temperature, pressure, electric current,
work, energy and etc.
 Mathematics operational : ordinary algebra
 Vector quantity is defined as a quantity with both magnitude
& direction.
 e.g. displacement, velocity, acceleration, force, momentum,
electric field, magnetic field and etc.
 Mathematics operational : vector algebra
1.2 Scalars and Vectors

26. 26
 Table 1.4 shows written form (notation) of vectors.
 Notation of magnitude of vectors.
1.2.1 Vectors
Vector A
Length of an arrow– magnitude of vector A
displacement velocity acceleration
s

v

a

s a
v
v
v 

a
a 

s (bold) v (bold) a (bold)
Direction of arrow – direction of vector A
Table 1.4

27. 27
 Two vectors equal if both magnitude and direction are the same.
(shown in figure 1.1)
 If vector A is multiplied by a scalar quantity k
 Then, vector A is
 if k = +ve, the vector is in the same direction as vector A.
 if k = -ve, the vector is in the opposite direction of vector A.
P

Q

Q
P



Figure 1.1
A
k

A
k

A

A



28. 28
 Can be represented by using:
a) Direction of compass, i.e east, west, north, south, north-east,
north-west, south-east and south-west
b) Angle with a reference line
e.g. A boy throws a stone at a velocity of 20 m s-1, 50 above
horizontal.
1.2.2 Direction of Vectors
50
v

x
y
0

29. 29
c) Cartesian coordinates
 2-Dimension (2-D)
m)
5
m,
1
(
)
,
( 
 y
x
s

s

y/m
x/m
5
1
0

30. 30
 3-Dimension (3-D)
s

2
3
4
m
2)
3,
4,
(
)
,
,
( 
 z
y
x
s

y/m
x/m
z/m
0
...i +...j + ..k
s 

31. 31
Unit vectors
A unit vector is a vector that has a magnitude of 1 with no units.
Are use to specify a given direction in space.
i , j & k is used to represent unit vectors
pointing in the positive x, y & z directions.
| | = | | = | | = 1
iˆ ĵ k̂

32. 32
d) Polar coordinates
e) Denotes with + or – signs.
 


N,150
30

F
F

150
+
+
-
-

33. 33
 There are two methods involved in addition of vectors graphically i.e.
 Parallelogram
 Triangle
 For example :
1.2.3 Addition of Vectors
Parallelogram Triangle
B

A

B

A

B
A



O
B
A



B

A

B
A



O

34. 34
 Triangle of vectors method:
a) Use a suitable scale to draw vector A.
b) From the head of vector A draw a line to represent the vector B.
c) Complete the triangle. Draw a line from the tail of vector A to the
head of vector B to represent the vector A + B.
A
B
B
A






 Commutative Rule
B

A

A
B



O

35. 35
 If there are more than 2 vectors therefore
 Use vector polygon and associative rule. E.g. R
Q
P





R

Q

P

R

Q

P
  
Q
P



   
R
Q
P
R
Q
P










 Associative Rule
  R
Q
P






36. 36
 Distributive Rule :
a.
b.
 For example :
Proof of case a: let  = 2
  B
A
B
A






 


  A
A
A






 


number
real
are
,

   
B
A
B
A






 2

B

A

B
A



O
 
B
A



2

37. 37
A

2
O
B

2
B
A


2
2 
  B
A
B
A




2
2
2 

 
B
A
B
A




2
2 




38. 38
Proof of case b: let  = 2 and  = 1
A

    A
A
A



3
1
2 


 

A

3
A
A
A
A




1
2 

 

A

2 A


A

3

  A
A
A



1
2
1
2 

 

39. 39
 For example :
1.2.4 Subtraction of Vectors
Parallelogram Triangle
D

C

O
D
C



O
......
 
D
C
D
C








C

D


D
C



C

D


D
C




40. 40
 Vectors subtraction can be used
 to determine the velocity of one object relative to another object
i.e. to determine the relative velocity.
 to determine the change in velocity of a moving object.
1. Vector A has a magnitude of 8.00 units and 45 above the positive x
axis. Vector B also has a magnitude of 8.00 units and is directed along
the negative x axis. Using graphical methods and suitable scale to
determine
a) b)
c) d)
(Hint : use 1 cm = 2.00 units)
Exercise 1 :
B
A


 B
A



B
2
A


 B
A
2




41. 41
 1st method :
1.2.5 Resolving a Vector
R

y
R

x
R


0
x
y
θ
R
Rx
cos
 ..........
Rx 

θ
R
Ry
sin
 θ
Rsin
....

 2nd method :
R

y
R

x
R


0
x
y

sin

R
Rx

sin
R
Rx 


cos

R
Ry
..........
..........



42. 42
 The magnitude of vector R :
 Direction of vector R :
 Vector R in terms of unit vectors written as
......
..........
or 
R
R

x
y
R
R
θ 
tan or








 
x
y
R
R
θ 1
tan
.....
..........

R


43. 43
A car moves at a velocity of 50 m s-1 in a direction north 30 east.
Calculate the component of the velocity
a) due north. b) due east.
Solution :
Example 1.2 :
N
E
W
S
N
v

E
v

v

30
60
a)
b)
or

60
v
vN sin


44. 44
A particle S experienced a force of 100 N as shown in figure above.
Determine the x-component and the y-component of the force.
Solution :
Example 1.3 :
150
F

S
x
150
30
F

S
x
y
y
F

x
F

Vector x-component y-component
…………………………
…………………………

30
cos
F
Fx 

N
6
.
6
8


x
F
or
F


150
cos
F
Fx 
...
..........

x
F

150
cos
100

x
F
N
0
5

y
F
or

150
sin
F
Fy 
........

y
F

150
sin
100

y
F

45. 45
The figure above shows three forces F1, F2 and F3 acted on a particle
O. Calculate the magnitude and direction of the resultant force on
particle O.
Example 1.4 : y
30o
O
)
N
30
(
2
F

)
N
10
(
1
F

30o
x
)
N
40
(
3
F


46. 46
30o
Solution :
O
y
x
3
F

30o
y
3
F

x
F2

1
F

2
F

60o
y
F2

x
3
F


47. 47
Solution :
Vector x-component y-component
1
F

3
F

2
F

N
0
1 
x
F 1
1 F
F y 
N
0
1
1 
y
F

60
sin
30
2 
y
F
N
6
2
2 
y
F

30
cos
40
3 

x
F
N
34.6
3 

x
F
Vector
sum
.......
..........

 x
F ........
..........

 y
F

48. 48
y
x
O
Solution :
The magnitude of the resultant force is
and
Its direction is 162 from positive x-axis OR 18 above negative x-axis.
   2
2

 
 y
x
r F
F
F
52.1 ....
r
F 
................
r
F 












x
y
F
F
θ 1
tan

18
49.6
16
tan 1









 
θ
 y
F

 x
F


162
r
F

18

49. 49
1. Vector has components Ax = 1.30 cm, Ay = 2.25 cm; vector
has components Bx = 4.10 cm, By = -3.75 cm. Determine
a) the components of the vector sum ,
b) the magnitude and direction of ,
c) the components of the vector ,
d) the magnitude and direction of . (Young & freedman,pg.35,no.1.42)
ANS. : 5.40 cm, -1.50 cm; 5.60 cm, 345; 2.80 cm, -6.00 cm;
6.62 cm, 295
2. For the vectors and in Figure 1.2, use the method of vector
resolution to determine the magnitude and direction of
a) the vector sum ,
b) the vector sum ,
c) the vector difference ,
d) the vector difference .
(Young & freedman,pg.35,no.1.39)
ANS. : 11.1 m s-1, 77.6; U think;
28.5 m s-1, 202; 28.5 m s-1, 22.2
Exercise 2 :
B
A



A

B
A



A
B



A
B



B

A

B

B
A



A
B



B
A



A
B



Figure 1.2
y
x
0
37.0
 
-1
s
m
18.0
B

 
-1
s
m
12.0
A


50. 50
3. Vector points in the negative x direction. Vector points at an
angle of 30 above the positive x axis. Vector has a magnitude of
15 m and points in a direction 40 below the positive x axis. Given
that , determine the magnitudes of and .
(Walker,pg.78,no. 65)
ANS. : 28 m; 19 m
4. Given three vectors P, Q and R as shown in Figure 1.3.
Calculate the resultant vector of P, Q and R.
ANS. : 49.4 m s2; 70.1 above + x-axis
Exercise 2 :
C

A

B

0


 C
B
A



A

B

Figure 1.3
y
x
0
50
 
2
s
m
10 
R

 
2
s
m
35 
P

 
2
s
m
24 
Q


51. 51
 notations –
 E.g. unit vector a – a vector with a magnitude of 1 unit in the direction
of vector A.
 Unit vectors are dimensionless.
 Unit vector for 3 dimension axes :
1.2.6 Unit Vectors
A

â
c
b
a ˆ
,
ˆ
,
ˆ
1
ˆ 

A
A
a 

  1
ˆ 
a
)
(
@
ˆ
⇒
- bold
j
j
axis
y 1
ˆ
ˆ
ˆ 

 k
j
i
)
(
@
ˆ
⇒
- bold
i
i
axis
x
)
(
@
ˆ
⇒
- bold
k
k
axis
z

52. 52
 Vector can be written in term of unit vectors as :
 Magnitude of vector,
x
z
y
k̂
ĵ
iˆ
k
r
j
r
i
r
r z
y
x
ˆ
ˆ
ˆ 



     2
z
2
y
2
x r
r
r
r 



53. 53
 E.g. :  m
ˆ
2
ˆ
3
ˆ
4 k
j
i
s 



      m
5.39
2
3
4
2
2
2




s
ĵ
3
x/m
y/m
z/m
0
s

i
ˆ
4
k̂
2

54. 54
Two vectors are given as:
Calculate
a) the vector and its magnitude,
b) the vector and its magnitude,
c) the vector and its magnitude.
Solution :
a)
The magnitude,
Example 1.5 :
a
b



 m
ˆ
6
ˆ
2
ˆ k
j
i
a 



b
a



 m
ˆ
ˆ
3
ˆ
4 k
j
i
b 



  ........................
x
a b
 
  ........................
y
a b
 
.........................
a b
 
  k
b
a
b
a z
z
z
ˆ
7
1
6 







..................... 9.95 m
a b
  
b
a



2

55. 55
b)
The magnitude,
c)
The magnitude,
  ............
x x
x
b a b a
   
  ................
y y
y
b a b a
   
.............. m
b a
 
  ..................
z z
z
b a b a
   
.....................
b a
 
 
2 .....................
x
a b
 
 
2 .........................
y
a b
 
2 ....................... m
a b
 
    k
b
a
b
a z
z
z
ˆ
13
1
6
2
2
2 







      m
15.9
13
7
6
2
2
2
2





 b
a

56. 56
Scalar (dot) product
 The physical meaning of the scalar product can be explained by
considering two vectors and as shown in Figure 1.4a.
 Figure 1.4b shows the projection of vector onto the direction of
vector .
 Figure 1.4c shows the projection of vector onto the direction of
vector .
1.2.7 Multiplication of Vectors
A

B


A

B

A
 B

Figure 1.4a

A

B

A

B

θ
Bcos
Figure 1.4b

A

B

θ
Acos
Figure 1.4c
 
A
B
A
B
A




to
parallel
of
component


 
B
A
B
B
A




to
parallel
of
component



57. 57
 From the Figure 1.4b, the scalar product can be defined as
meanwhile from the Figure 1.4c,
where
 The scalar product is a scalar quantity.
 The angle  ranges from 0 to 180 .
 When
 The scalar product obeys the commutative law of multiplication i.e.
 
θ
B
A
B
A cos




vectors
o
between tw
angle
:
θ
 
θ
A
B
A
B cos






90
θ
0 
 scalar product is positive


180
θ
0
9 
 scalar product is negative

90
θ  scalar product is zero
A
B
B
A








58. 58
 Example of scalar product is work done by a constant force where the
expression is given by
 The scalar product of the unit vectors are shown below :
   
θ
F
s
θ
s
F
s
F
W cos
cos 





x
z
y
k̂
ĵ
iˆ
    1
1
1
cos
ˆ
ˆ 2



 o
2
0
i
i
i
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ 




 k
k
j
j
i
i
    1
1
1
cos
ˆ
ˆ 2



 o
2
0
j
j
j
    1
1
1
cos
ˆ
ˆ 2



 o
2
0
k
k
k
   0
9
cos
ˆ
ˆ 

 o
0
1
1
j
i
0
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ 




 k
i
k
j
j
i
   0
9
cos
ˆ
ˆ 

 o
0
1
1
k
i
   0
9
cos
ˆ
ˆ 

 o
0
1
1
k
j

59. 59
Calculate the and the angle  between vectors and for the
following problems.
a) b)
Solution :
a)
The magnitude of the vectors:
The angle  ,
Example 1.6 :
A

B
A


 B

ˆ ˆ
ˆ ˆ ˆ ˆ
......... ........ .......
A B i i j j k k
      
      3
1
1
1
2
2
2





A
k
j
i
A ˆ
ˆ
ˆ 



k
j
i
A ˆ
ˆ
3
ˆ
4 



k
j
i
B ˆ
3
ˆ
2
ˆ
4 



k
j
B ˆ
3
ˆ
2 


.............
A B
 
3

 B
A


      29
3
2
4
2
2
2






B
θ
AB
B
A cos




















 
 

29
3
3
cos
cos 1
1
AB
B
A
θ



2
.
71

θ
ANS.:3; 99.4

60. 60
Referring to the vectors in Figure 1.5,
a) determine the scalar product between them.
b) express the resultant vector of C and D in unit vector.
Solution :
a) The angle between vectors C and D is
Therefore
Example 1.7 :
1 99 .......
C D .
  
  
174
19
25
180 



θ
Figure 1.5
y
x
0
 
m
1
C

 
m
2
D

19
25
θ
CD
D
C cos




...................


61. 61
b) Vectors C and D in unit vector are
and
Hence
j
C
i
C
C y
x
ˆ
ˆ 


ˆ ˆ
......... ..........
i j
 
 m
ˆ
42
.
0
ˆ
91
0 j
i
.
C 



   j
i
D
C ˆ
65
.
0
42
.
0
ˆ
89
.
1
91
.
0 







 m
ˆ
23
.
0
ˆ
98
.
0 j
i 

   j
i
D ˆ
19
sin
2
ˆ
19
cos
2 





.....................m
D 

62. 62
Vector (cross) product
 Consider two vectors :
 In general, the vector product is defined as
and its magnitude is given by
where
 The angle  ranges from 0 to 180  so the vector product always
positive value.
 Vector product is a vector quantity.
 The direction of vector is determined by
k
r
j
q
i
p
B ˆ
ˆ
ˆ 



k
z
j
y
i
x
A ˆ
ˆ
ˆ 



C
B
A





θ
AB
θ
B
A
C
B
A sin
sin 








vectors
o
between tw
angle
:
θ
RIGHT-HAND RULE
C


63. 63
 For example:
 How to use right hand rule :
 Point the 4 fingers to the direction of the 1st vector.
 Swept the 4 fingers from the 1st vector towards the 2nd vector.
 The thumb shows the direction of the vector product.
 Direction of the vector product always perpendicular
to the plane containing the vectors and .
A

C

B
 A

B

C

C
B
A





C
A
B





A
B
B
A






 but  
A
B
B
A








B

)
(C

A


64. 64
 The vector product of the unit vectors are shown below :
 Example of vector product is a magnetic force on the straight
conductor carrying current places in magnetic field where the
expression is given by
x
z
y
k̂
ĵ
iˆ
i
j
k
k
j ˆ
ˆ
ˆ
ˆ
ˆ 




k
i
j
j
i ˆ
ˆ
ˆ
ˆ
ˆ 




j
k
i
i
k ˆ
ˆ
ˆ
ˆ
ˆ 




0
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ 




 k
k
j
j
i
i
0
in
ˆ
ˆ 

 o
2
0
s
i
i
i
0
in
ˆ
ˆ 

 o
2
0
s
j
j
j
0
in
ˆ
ˆ 

 o
2
0
s
k
k
k
 
B
l
I
F





θ
IlB
F sin


65. 65
 The vector product can also be expressed in
determinant form as
 Note :
 The angle between two vectors can only be
determined by using the scalar (dot) product.
r
q
p
z
y
x
k
j
i
B
A
ˆ
ˆ
ˆ




     k
yp
xq
j
zp
xr
i
zq
yr
B
A ˆ
ˆ
ˆ 









66. 66
Given two vectors :
Determine
a) and its magnitude b)
c) the angle between vectors and .
Solution :
a)
The magnitude,
Example 1.8 :
ˆ
ˆ ˆ
i j k
A B
 
B
A


 B
A



A B
 
k
j
i
B
A ˆ
2
ˆ
6
1
ˆ
10 





19

 B
A


k
j
i
A ˆ
ˆ
2
ˆ
3 




k
i
B ˆ
5
ˆ 


A

B

A B
 

67. 67
b)
c) The magnitude of vectors,
…………………………………………..
………………………………………….
Using the scalar (dot) product formula,
   
k
j
i
k
j
i
B
A ˆ
5
ˆ
0
ˆ
ˆ
ˆ
2
ˆ
3 









2

 B
A


θ
AB
B
A cos





84

θ
14
A 
26
B 

68. 68
1. If vector and vector , determine
a) , b) , c) .
ANS. :
2. Three vectors are given as follow :
Calculate
a) , b) , c) .
ANS. :
3. If vector and vector ,
determine
a) the direction of
b) the angle between and .
ANS. : U think, 92.8
Exercise 3 :
46
;
26
;
ˆ
2k
j
i
a ˆ
+
ˆ
= 5
3

j
i
b ˆ
+
ˆ
= 4
2

b
a


 b
a


   b
b
a





k
j
i
c
k
j
i
b
k
j
i
a ˆ
ˆ
2
ˆ
2
and
ˆ
2
ˆ
4
ˆ
;
ˆ
2
ˆ
3
ˆ
3 












 
c
b
a




  
c
b
a




  
c
b
a





k
j
i ˆ
9
ˆ
11
ˆ
5
;
9
;
21 



k
j
i
P ˆ
ˆ
2
ˆ
3 



k
j
i
Q ˆ
3
ˆ
4
ˆ
2 




Q
P



P

Q


69. PHYSICS CHAPTER 1
69
THE END…
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CHAPTER 2 :
Kinematics of Linear Motion