Note Chapter1 0809

8/3/2019 Note Chapter1 0809

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PHYSICS CHAPTER 1

1

CHAPTER 1:Physical quantities and

measurements

( 3 Hours)

- CIK SHARIZA BT SHAHARI

- BACHELOR OF ELECTRICAL ENGINEERING (University of Malaya)


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PHYSICS CHAPTER 1

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

a) State basic quantities and their respective SI units: length(m), time (s), mass (kg), electrical current (A), temperature(K), amount of substance (mol) and luminosity (cd).

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

c) State and convert unit with common SI prefixes

Learning Outcome:

1.1 Physical Quantities and Units (1 hours)


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Physical quantity is defined as a quantity which can be measured.

It can be categorised into 2 types

Basic (base) quantity

Derived quantity

Basic quantity is defined as a quantity which cannot be derivedfrom any physical quantities.

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 kilogram kg

Time t second s

Temperature T/ kelvin K

Electric current I ampere A

Amount of substance N mole mol

Luminous Intensity candela cdTable 1.1


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Derived quantity is defined as a quantity which can be expressedin 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 V l w t m3

Acceleration a v/t m s-2

Density m/V kg m-3

Momentum

p m vkg m s-1

Force F m a kg m s-2 @ N

Work W F s kg m2 s-2 @ JTable 1.2


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Unit is defined as a standard size of measurement of

physical quantities. Examples :

1 second is defined as the time required for9,192,631,770 vibrations of radiation emitted by acaesium-133 atom.

1 kilogram is defined as the mass of a platinum-iridiumcylinder kept at International Bureau of Weights andMeasures Paris.

1 meter is defined as the length of the path travelled bylight in vacuum during a time interval of

s458,792,299

1


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oo

o

57.296180rad1

180rad

The unit of basic quantity is called base unit

addition unit for base unit: unit of plane angle - radian (rd)

unit of solid angle- steradian (sr)

The common system of units used today are S.I unit (SystemInternational/metric system) and cgs unit - UK.

The unit of derived quantity

called derived unit


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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 = 2.33 106 s = 2.33 s

Prefix Multiple Symbol

tera 1012 T

giga 109 G

mega 106

Mkilo 103 k

deci 101 d

centi 102 c

milli 103 m

micro 106

nano 109 n

pico 1012 p

1.1.1 Unit Prefixes

Table 1.3


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Solve the following problems of unit conversion.

a. 15 mm2 = ? m2 b. 65 km h1 = ? m s1

c. 450 g cm3 = ? kg m3 d. 29 cm = ? in

e. 12 mi h1 = ? m s1

Solution :

a. 15 mm2 = ? m2

b. 65 km h-1 = ? m s-1

1st method :

h1m1065hkm65

31

Example 1.1 :

s3600

m1065hkm65

31

11 sm81hkm65

232 m10mm1 262 m10mm1

25262m101.5orm1015mm15


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2nd method :

c. 450 g cm-3 = ? kg m-3

h1

km65hkm65 1

11sm18hkm65

332

33

3

3

m10

cm1

g1

kg10

cm1

g450cmg450

353mkg10.54cmg450

s3600

h1

km1

m1000

h1

km65hkm65 1


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d. 29 cm = ? in

e. 12 mi h-1 = ? m s-1

cm1

in

cm29cm292.54

1

in.411cm29

s3600

h1km1

m1000mi1

km1.609h1mi12hmi12 1

11sm365hmi12 .


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

a) Define scalar and vector quantities, unit vectors in Cartesiancoordinate.

b) Perform vector addition and subtraction operationsgraphically.

c) Resolve vector into two perpendicular components (2-D)and three perpendicular components (3-D):

Components in the x, y and z axes.

Components in the unit vectors.

Learning Outcome:

1.2 Scalars and Vectors (3 hours)

kji ,,


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

d) Define and use dot (scalar) product;

e) Define and use cross (vector) product

Direction of cross product is determined by corkscrewmethod or right hand rule.

Learning Outcome:

1.2 Scalars and Vectors (2 hours)

ABBABA coscos

ABBABA sinsin


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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 quantityis 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


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Table 1.6 shows written form (notation) of vectors.

Notation of magnitude of vectors.

1.2.1 Vectors

Vector A Length of an arrowmagnitude of vector A

displacement velocity acceleration

s

v

a

s av

vv

aa

s (bold) v (bold) a (bold)

Direction of arrow direction of vector A

Table 1.6


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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

ifk = +ve, the vector is in the same direction as vector A.

ifk = -ve, the vector is in the opposite direction of vector A.

P

Q

QP

Figure 1.1

Ak

Ak

A

A


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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 abovehorizontal.

1.2.2 Direction of Vectors

50v

x

y

0


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c) Cartesian coordinates

2-Dimension (2-D)

m)5m,1(),( yxs

s

y/m

x/m

5

10


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3-Dimension (3-D)

s

2

3

4

m2)3,4,(),,( zyxs

y/m

x/m

z/m

0


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d) Polar coordinates

e) Denotes with + or signs.

N,15030F

F

150

+

+-

-


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There are two methods involved in addition of vectors graphically i.e.

Parallelogram Triangle

For example :

1.2.3 Addition of Vectors


B

A

B

A

BA

O

BA

B

A

BA

O


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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 thehead of vector B to represent the vector A + B.

ABBA

Commutative Rule

B

A

AB

O


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If there are more than 2 vectors therefore

Use vector polygon and associative rule. E.g. RQP

RQ

P

R

Q

P

QP

RQPRQP

Associative Rule

RQP


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Distributive Rule :

a.

b.

For example :

Proof of case a: let = 2

BABA

AAA

numberrealare,

BABA

2

B

A

BA

O BA

2


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PHYSICS CHAPTER 1

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A

2O

B

2

BA

22

BABA

222

BABA

22


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PHYSICS CHAPTER 1

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Proof of case b: let = 2 and = 1

A

AAA

312

A

3

AAAA

12

A

2 A

A3

AAA

1212


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For example :

1.2.4 Subtraction of Vectors


DC

O

DC

O

D

DCDC

C

D

DC

C

D

DC


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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 alongthe negative x axis. Using graphical methods and suitable scale todetermine

a) b)

c) d)(Hint : use 1 cm = 2.00 units)

Exercise 1.2 :

BA

BA

B2A

BA2


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1st method :

1.2.5 Resolving a Vector

R

yR

xR

0

x

y

R

Rx cos RRx cos

R

Rysin RRy sin

2nd method :

R

yR

xR

0x

y

sinR

Rx sinRRx

cosR

RycosRRy


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The magnitude of vectorR :

Direction of vectorR :

VectorR in terms of unit vectors written as

22or yx RRRR

x

y

R

Rtan or

x

y

R

R

1tan

jRiRR yx


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A car moves at a velocity of 50 m s-1 in a direction north 30 east.

Calculate the component of the velocitya) due north. b) due east.

Solution :

Example 1.6 :

N

EW

S

Nv

Ev

v30

60

a)

b)

30vvN cos

1sm43.3Nv

3050vN cos

or

60vvN sin

6050vN sin

30vvE

sin

1sm25Ev

3050vE sinor

60vvE cos

6050vE cos


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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 1.7 :

150F

Sx

15030

F

Sx

y

yF

xF

Vector x-component y-component

30cosFFx

N6.68xF

30cos100xF

orF

150cosFFx

N6.68x

F

150cos100xF

30sinFFy

N05yF

30sin100yF

or150sinFFy

N05yF

150sin100yF


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The figure above shows three forces F1, F2and F3 acted on a particleO. Calculate the magnitude and direction of the resultant force onparticle O.

Example 1.8 : y

30o

O

)N30(2F

)N10(1F

30o x

)N40(3F


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30o

Solution :

O

y

x

3F

30o

y3F

321 FFFFFr

yxr FFF

xxxx FFFF 321

yyyy FFFF 321

xF2

1F2F

60o

yF2

x3F


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Solution :

Vector x-component y-component

1F

3F

2F

N01 xF 11FFy

N011 yF

60cos302

xF N152 xF

60sin302

y

F

N622 yF30cos403 xF

N34.63 xF

30sin403 yFN203 yF

Vectorsum

34.6510 xFN49.6 xF

20.02601 yFN16 yF

S CS C


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PHYSICS CHAPTER 1

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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.

22 yxr FFF

N.125

rF

22 1649.6 rF

x

y

F

F

1tan

1849.6

16tan 1

yF

xF

162

rF

18

PHYSICS CHAPTER


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1. Vector has componentsAx = 1.30 cm,Ay = 2.25 cm; vector

has componentsBx = 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 vectorresolution 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 1.3 :

BA

A

BA

AB

AB

B

A

B

BA

AB

BA

AB

Figure 1.2

y

x0

37.0

-1sm18.0B

-1sm12.0A

PHYSICS CHAPTER 1


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3. Vector points in the negativex direction. Vector points at an

angle of 30 above the positivex axis. Vector has a magnitude of

15 m and points in a direction 40 below the positivex axis. Giventhat , determine the magnitudes of and .(Walker,pg.78,no. 65)

ANS. : 28 m; 19 m

4. Given three vectors P,Q andR as shown in Figure 1.3.

Calculate the resultant vector of P,Q andR.

ANS. : 49.4 m s2; 70.1 above + x-axis

Exercise 1.3 :

CA

B

0 CBA

A

B

Figure 1.3

y

x0

50 2sm10 R

2sm35 P

2sm24 Q

PHYSICS CHAPTER 1


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notations

E.g. unit vector a a vector with a magnitude of 1 unit in the direction

of vectorA.

Unit vectors are dimensionless.

Unit vector for 3 dimension axes :

1.2.6 Unit Vectors

A

a

cba ,,

1 A

Aa

1 a

)(@- boldjjaxisy 1 kji)(@- boldiiaxisx

)(@- boldkkaxisz

PHYSICS CHAPTER 1


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Vector can be written in term of unit vectors as :

Magnitude of vector,

x

z

y

k

j

i

krjrirr zyx

2z2

y

2

x rrrr

PHYSICS CHAPTER 1


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E.g. : m234 kjis

m5.39234222

s

j

3

x/m

y/m

z/m

0

s

i4k2

PHYSICS CHAPTER 1


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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 1.9 :

ab

m62 kjia

ba

m34 kjib

ibaba xxx541

jbaba yyy532

m755 kjiba

kbaba zzz 716

m9.95755

222 ba

ba

2

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

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b)

The magnitude,

c)

The magnitude,

iabab xxx314

jabab yyy

23

m53 kjiab

kabab zzz 561

m5.92513 222 ab

ibaba xxx 641222

jbaba yyy 732222

m13762 kjiba

kbaba zzz 1316222

m15.913762

222 ba

PHYSICS CHAPTER 1


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Scalar (dot) product

The physical meaning of the scalar productcan 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

Bcos

Figure 1.4b

A

B

AcosFigure 1.4c

ABABA

toparallelofcomponent

BABBA

toparallelofcomponent

PHYSICS CHAPTER 1


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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.

BABA

cos

vectorsobetween twangle:

ABAB cos

900 scalar product is positive 18009 scalar product is negative

90 scalar product is zero

ABBA

PHYSICS CHAPTER 1


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Example of scalar product is work done by a constant force where theexpression is given by

The scalar product of the unit vectors are shown below :

Fs

sFsFW coscos

x

z

y

kj

i

111cos 2 o2 0iii

1 kkjjii

111cos 2 o2 0jjj

111cos 2

o2 0kkk

09cos o011ji0 kikjji

09cos

o

011ki

09cos o011kj

PHYSICS CHAPTER 1


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Calculate the and the angle between vectors and for the

following problems.a) b)

Solution :

a)

The magnitude of the vectors:

The angle ,

Example 1.10 :

A

BA

B

kkjjiiBA 312141

3111 222 A

kjiA

kjiA 34

kjiB 324

kjB 32

324 BA

3BA

29324222

BABBA cos

293

3coscos 11

AB

BA

2.71

ANS.:3; 99.4

PHYSICS CHAPTER 1


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Referring to the vectors in Figure 1.5,a) determine the scalar product between them.

b) express the resultant vector of CandD in unit vector.

Solution :

a) The angle between vectorsC

andD

is

Therefore

Example 1.11 :

2m991.DC

1741925180

Figure 1.5

y

x0

m1C

m2D19

25

CDDC cos

174cos21

PHYSICS CHAPTER 1


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b) Vectors CandD in unit vector are

and

Hence

jCiCCyx

ji 25sin125cos1 m42.0910 ji.C

jiDC 65.042.089.191.0

m23.098.0 ji

jiD 19sin219cos2

m65.0891 ji.D

PHYSICS CHAPTER 1


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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

krjqipB

kzjyixA

CBA

ABBACBA sinsin

vectorsobetween twangle:

RIGHT-HAND RULE

C

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

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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

CBA

CAB

ABBA

but ABBA

B)(C

A

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

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The vector product of the unit vectors are shown below :

Example of vector product is a magnetic force on the straightconductor carrying current places in magnetic field where theexpression is given by

x

z

y

k

j

i

ijkkj

kijji

jkiik

0 kkjjii

0in o2 0siii

0in o2 0sjjj

0in

o2

0skkk

BlIF

IlBF sin

PHYSICS CHAPTER 1


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The vector product can also be expressed in determinant form as

1st method :

2nd method :

Note :

The angle between two vectorscan only be determined byusing thescalar (dot) product.

rqp

zyx

kji

BA

kypxqjzpxrizqyrBA

kypxqjxrzpizqyrBA

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

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Given two vectors :

Determine

a) and its magnitude b)

c) the angle between vectors and .

Solution :

a)

The magnitude,

Example 1.12 :

501

123

kji

BA

BA

BA

kjiBA 120311530152

kjiBA 26110

kjiBA 20115010

19BA

kjiA 23

kiB

5

A

B

222 21610 BA

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

54

b)

c) The magnitude of vectors,

Using the scalar (dot) product formula,

kjikjiBA 5023

2BA

kkjjiiBA

51

02

13

503 BA

ABBA cos

2614

2coscos 11

AB

BA

84

14123 222 A

26501 222 B

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

55

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 ,

determinea) the direction of

b) the angle between and .

ANS. : U think, 92.8

Exercise 1.4 :

46;26;2k

jia += 53

jib += 42

ba

ba

bba

kjickjibkjia 22and24;233

cba

cba

cba

kji 9115;9;21

kjiP 23

kjiQ 342

QP

P

Q

PHYSICS CHAPTER 1


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PHYSICS CHAPTER 1

THE END

Next ChapterCHAPTER 2 :

Kinematics of Linear Motion

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Note Chapter1 0809