chapter 1 physical quantities(2)

7/31/2019 CHAPTER 1 Physical Quantities(2)

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1

KOLEJ MATRIKULASI PERAK


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1.1 Physical Quantities

and Units

1.2 Scalars and Vectors


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3

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).

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).

State and convert units with common SI prefixes.

1.1 Physical Quantities and Units (1 hours)


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How fast does light travel ? How much do you weigh ?What is the radius of the Earth?What temperature does ice melt at?

We can find the answers to all of thesequestions by measurement.Speed, mass, length and temperature are

all examples of physical quantities.

Measurement of physical quantities is an essential part of Physics.


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1.1 Physical Quantities and Units

Physical Quantities- Quantities that are measurable with instruments in laboratory or can be

derived from these measured quantities.

- consists of a precise numerical value & a unit.


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- categorized into 2:

1. Base Quantities

2. Derived Quantities

- are standards for measurement of physical quantities that need cleardefinitions to be useful.

Physical Unit

- ex: metre (m) unit for lengthsecond (s) unit for timeKelvin (K) unit for temperature

SI Unit- International System of Units

- has been agreed internationally.


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Base Quantities & units

- fundamental quantity that can not be derived in terms of other physicsquantities.

candela

mole

kelvin

ampere

second

kilogram

meter

Name of SI unit

KTemperature, T

mLength , l

kgMass, m

sTime, t

cdLuminous intensity

molAmount of substance, n

AElectric current, I

Unit symbolBase

Quantity


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Derived Quantities & units- Are the physical quantities other than the base quantities.

- Are derived from base quantities according to a defining equation.

Q = It

W = Fs

P = F / A

F = ma

f = 1 / T

= m / Vv = s / t

Defining equation

A s

kg m2 s2

kg m1 s2

kg m s2

s1

kg m3

m s1

SI unit

Pa (Pascal)Pressure

--Velocity

--Density

Hz (Hertz)Frequency

C (Coulomb)Charge

J (Joule)Work

N (Newton)Force

Special

name

Physical Quantity


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Unit is defined as a standard size ofmeasurement of physical quantities.

Examples : 1 second is defined as the time required for

9,192,631,770 vibrations of radiation emitted bya caesium-133 atom.

1 kilogram is defined as the mass of a platinum-iridium cylinder kept at International Bureau ofWeights and Measures Paris.

1 meter is defined as the length of the pathtravelled by light in vacuum during a timeinterval of

s,, 458792299

1


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Prefixes- can be added to SI base & derived units to

make larger or smaller units

- Some physical quantities have no units.- Example:refractive index, strain

MultiplePrefix ( &

symbol)

1012 tera- (T)

109 giga- (G)

106 mega- (M)

103 kilo- (k)


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102 hecto- (h)

101 deci- (d)

102 centi- (c)

103 milli- (m)

106 micro- ()

109 nano- (n)

1012 pico- (p)

1015 femto- (f)


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Unit Conversions

- SI unit system is predominant throughout the world.

- another metric system that has been used:(a) cgs system ( centimeter gram second )(b) British Engineering system ( foot slug second )

- Units in different systems or differentunits in the same system can

express the same quantity.

- Is necessary to change from oneset of units to another.


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Solve the following problems of unit conversion.

a. 30 mm2 = ? m2 b. 865 km h-1 = ? m s-1

c. 300 g cm-3 = ? kg m-3 d. 17 cm = ? in

e. 24 mi h-1 = ? km s-1

Solution :

a. 30 mm2 = ? m2

b. 865 km h-1 = ? m s-1

1st method :

h1m10865hkm8653

1

Example 1 :

s3600

m10865hkm865

31

11 sm240hkm865

232 m10mm1 262 m10mm1

25262m103.0orm1030mm30


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2nd method :

c. 300 g cm-3 = ? kg m-3

s3600

h1

km1

m1000

h1

km865hkm865 1

s3600

h1

km1

m1000

h1

km865hkm865 1

11 sm240hkm865

33

2-

33-

3

3-

m10

cm1

g1

kg10

cm1

g300cmg300

-353 mkg103.0cmg300


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d. 17 cm = ? in

e. 24 mi h-1 = ? km s-1

cm1

incm17cm17 2.54

1

in6.69cm17

s3600

h1

mi1

km1.609

h1

mi24hmi24 1-

-1-21 skm101.07hmi24


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1.2 Scalars and Vectors (2 hours)


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1.2 Scalars and Vectors (2 hours)

ABBABA coscos

ABBABA sinsin


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Scalar Quantity

- Quantity which has only magnitude.

- Example: mass, distance, speed, work.

Vector Quantity

- Quantity which has both magnitude and direction.

- Example: displacement, velocity, force, momentum


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- Magnitude of the vector is written as |A|A

- Symbols for vectors : A or A

- A vector can be represented by an arrow. The length of the arrowindicatesits magnitude & arrow head shows the direction.

A

Representing vectors

Head of vector

Tail ofvector

A

magnitude

direction


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Equality of two vectors

- 2 vectors & are equal if they have the same magnitude and point inthe same direction.

A

B


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Negative Vector- Negative vector is a vector with the same magnitude as

but points in opposite direction.A

A


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Vector Addition & SubtractionAddition

- The addition of 2 vector, and result in a third vector called

resultant vector.

A

B

R

- Resultant vector is a single vector that will have the same effect as 2 ormore vectors.

- 2 methods of vector addition:

(1) Drawing / Graphical method - tail to head & Parallelogram

(2) Mathematic Calculation unit vector & trigonometry


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(1) vectors in the same directions

(2) vectors in the opposite directions

The direction ofresultant vector R is inthe direction of thebigger vector

Adding Parallel Vectors

NA 3

NB 7

NR 10Resultant,

NA 3

NB 7

NR 4Resultant,


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Resultant = 9 N East

Resultant = 40 N East


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(a) Tail to head method

Two equivalent ways to add vectors graphically: the tail-to-head method and

the parallelogram method.

Placing the tail of Bso that it meets the head of A

The Resultant, R=( A + B), is the vector from the tail of A to the head of B

A

B


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Placing the tail of each successive arrow at the head of theprevious one. The resultant vector is the arrow drawn from the tailof the first vector to the head of the last vector.

A

B C

D

+ + += ?

How to add vector A, B, C and D ?


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DCBAR

A

B

C

D

R

Tail of firstvector

Head of last vector


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(b) Parallelogram method

Resultant vector, : diagonal of a parallelogram formed with & as

two of its 4 sides.

R

A

B

A

B


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Resolving vector into 2 perpendicular compoments (2D)

A vector may be expressed in terms of its components.

A

Ax

Ay

x

y


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with the aid of trigonometry:

22|| yx AAA

A

Ax

cos cosAAx

A

Aysin sinAAy

Magnitude of vector A :

Direction of vector A :

x

y

A

Atan

* is alwaysmeasured from +x

axis.


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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.

, & is used to represent unit vectorspointing in the positive x, y & z directions.

i j k

| | = | | = | | = 1i j k
http://d/Physics_SF017/Chap1/unitvector.swfhttp://d/Physics_SF017/Chap1/unitvector.swfhttp://d/Physics_SF017/Chap1/unitvector.swf


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jAiAA yx

The vector can also written in unit vector form:A


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Example

A force of 800 N is exerted on a bolt A as shown in Fig. below. Determine thehorizontal and vertical components of the force.

x

y

yF

xF


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Solution

with the aid of trigonometry:

cosFFx

35cos800

NFx

655

sinFFy

35sin800

NFy 459

We may write in the unit vector formF

jNiNF )459()655(


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Example

The magnitudes of the 3 displacement vectors shown in drawing. Determine theresultant value when these vectors are added together.

By= 5 sin 30

Bx = 5 cos 30Ax=10 sin 45

Ay = 10 cos 45


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1.572.74Resultant

R

80C

5 sin 30 = 2.505 cos 30 =4.33B

10 cos 45 = 7.0710 sin 45 = 7.07A

Component yComponent xVector

22yx RRR

m16.3

)57.1()74.2(22

Magnitude of resultant vector

74.2x

R

57.1yR

R


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x

y

RRtan 573.0

74.257.1

xabove81.29

Direction of resultant vector

Or can write in unit vector form

jmimR )57.1()74.2(

Resultant vector , R = 3.16 m at 29.81 above +x


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Example

Let :

jib

jia

35

52

Find : (a)

(b)

(c)

ba

ba

32 |2| a

Solution

)

3

5()

5

2( jijiba

(a)

ji 27


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To find the magnitude of , 1st we have to calculate

(b) )35(3)52(232 jijiba

jiji 915104

ji

19

11

(c) |2| a

a

2

)52(22 jia

)104 ji

22 104|2| a

77.10


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Multiplying a vector by a vector

Dot Product ( )BA

cos|||| BABA

where |A| : magnitude of vector|B|: magnitude of vector

: angle between &

A

B

A

B

0 180


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is the magnitude of multiplied bythe component of parallel to .

BA

A

B

A

B cos

cos|||| BABA


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= zero when = 90 because cos 90 = 0BA

= maximum value when = 0because cos 0 = 1BA

Commutative law applied to dot product :

ABBA

Example of physical quantity : sFW

Dot product Calculation

Given 2 vector :

kBjBiBB

kAjAiAA

zyx

zyx

How to perform ? BA


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zzyyxx BABABABA

Remember :

090cos)1)(1(

10cos)1)(1(

kjkiji

kkjjii

)()( kBjBiBkAjAiABA zyxzyx
http://d/Physics_SF017/Chap1/unitvector.swf


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Example

Given 2 vectors :

)285(

)423(

kjiB

kjiA

Calculate(a) the value of(b) the angle between 2 vectors

BA

Solution

)285()423( kjikjiBA

)2)(4()8)(2()5)(3(

9BA

(a)

produces a scalar


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(b)

||||cos

BA

BA

)64.9)(39.5(

9

cos:from BABA

03.80

222

)()()(|| zyx AAAA

222 )()()(|| zyx BBBB

39.5)4()2()3( 222

64.9)2()8()5( 222


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Example

Find the scalar product of the two vectors in figure. Themagnitude of the vectors are A = 4.0 N and B = 5.0 m

BA

A

B

130

53

Answer : 4.50


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Cross Product ( )BA

sin|||||| BABA

- produce a third vector, which is perpendicular to both of theoriginal vectors.

- The magnitude of the cross product is given by:

0 180

- Also called vector product.

A

B

BA


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is equals the magnitude of multiplied by thecomponent of perpendicular to .

|| BA

A

B

A

AB

sinB


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-- if is parallel @ anti parallel ( =0 @ 180 ) BA

& 0|| BA

-- if is 90 max|| BA

BA

&

Example of physical quantity :

BvqFm

Force acting on a charge moving in magnetic field

00sin|||||| BABA

ABBABA

90sin||||||

1

0


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BA

- the direction of new vector ( ) is normal to the plane that contain vector

& given by Right Hand RuleBA

A

B


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A

B

BA

Directed

upwards

A

B

AB

Directeddownwards

)( ABBA

C d t C l l ti


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Cross product Calculation

zyx

zyx

BBBAAA

kji

BA

jABBAiABBA zxzxzyzy ][][ kABBA yxyx ][

Keep In mind (RHR) :

0 kkjjii

kBB

AA

jBB

AA

iBB

AA

yx

yx

zx

zx

zy

zy

jkijik

kijkji

ijkikj
http://d/Physics_SF017/Chap1/unitvector.swf


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Fig. (a) Fig. (b)


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Example

Given 2 vector :

Calculate : (a)(b)

)085(

)423(

kjiB

kjiA

Solution

)085()423( kjikjiBA

(a)

BA

|| BA


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222 342032|| BA

085

423

kji

BA

i)]4(8)0(2[

kjiBA 342032

79.50

j)]4)(5()0(3[

k

)]2)(5()8(3[

(b)

produce a vector

chapter 1 physical quantities(2)

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