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    PHY 130 : FUNDAMENTAL PHYSICS

    COURSE WORKTest-20 %

    Quiz & Assignment-10 %Experiment-20 %

    FINALFinal exam - 50 %

    TOTAL = 100 %

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    SYSTEM OF UNITSCHAPTER 1

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    SYSTEM OF UNITS

    Basic Quantities

    Derived Quantities

    Units

    Prefixes

    Dimensional Analysis

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    PhysicalQuantities

    * Basis of physical

    quantities * Combination of

    one or more basic

    quantity quantities

    BASIC

    QUANTITIES

    DERIVED

    QUANTITIES

    Basis of physical quantitiesExample :

    Length (m)Mass (kg)

    Time (s)Temperature (K)Electric current (A)

    Combination of one or more

    basic quantities.

    Example :Area (m2)

    Volume (m3)

    Velocity (ms-1)

    Acceleration (ms-2)

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    BASIC

    QUANTITIES

    COMBINATION OF

    QUANTITIES

    DERIVED

    QUANTITIES

    Length (Length)2

    Area(m2

    )

    Length (Length)3 Volume(m3)

    Length, time Length/time Speed(ms-1)

    Length, time Length/(time)2 Acceleration(ms-2)

    Length, mass Mass/(length)3 Density(kgm-3)

    Mass, time (Mass x length)/(time)2 Force(kgms-2)

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    SCIENTIFIC NOTATION

    A way of writing numbers that accommodates values too largeor small to be conveniently written in standard decimal notation

    In scientific notation, numbers are written in the form:

    Example:

    An electron'smass is about

    0.000 000 000 000 000 000 000 000 000 000 910 938 22 kg.

    In scientific notation, this is written 9.10938221031kg.

    http://en.wikipedia.org/wiki/Electronhttp://en.wikipedia.org/wiki/Electron
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    Used to simplify big numbers.

    Replace powers of ten.

    POWER PREFIX ABBREV.

    x 10-12 pico p

    x 10-9 nano n

    x 10-6 micro

    x 10-3 milli m

    x 10-2 centi c

    x 103 kilo k

    x 106 Mega M

    x 109 Giga G

    x 1012 Tera T

    PREFIXES

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    3 km = ? m

    1 km = 1000 m

    3 km = 3 x 1000 m=3000 m

    or 3 km = 3 km x 1000 m

    1 km= 3000 m

    CONVERSION OF UNITS

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    45 cm = ? km

    km4.5x10cm45

    km45x10cm45

    m1000

    1km

    cm100

    1mxcm45cm45

    4

    5

    CONVERSION OF UNITS

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    35 km.hr-1= ? m.s-1

    11ms9.72km.hr35

    s

    m

    60x60

    35x1000

    1hr

    km35

    s60

    1min

    min60

    1hr

    km1

    m1000

    1hr

    km35

    hr1

    km35

    CONVERSION OF UNITS

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    20 kg.m-3= ? g.cm-3

    323

    33

    3

    3

    33

    3

    33

    cm.g10x2m.kg20

    cm

    g

    100x100x100

    1000x20

    m1

    kg20

    cm100

    m1

    kg1

    g1000

    m1

    kg20

    m1

    kg20

    cm100

    m1

    kg1

    g1000

    m1

    kg20

    m1

    kg20

    CONVERSION OF UNITS

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    Three basic ways to describe a physicalquantity the space it takes, the matter itcontains, and how long it persists.

    All measurements reduce tolength, time, and

    mass Three primary dimensions: Length(L),

    Mass(M), and Time(T)

    Additional dimensions: electric current(A) andTemperature().

    DIMENSIONS

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    Dimensions of a quantity are the base units that

    make it up; they are generally written using square

    brackets.

    Example: Speed = distance / time

    Dimensions of speed: [L/T]

    Quantities that are being added or subtracted

    must have the same dimensions.

    In addition, a quantity calculated as the solution

    to a problem should have the correct dimensions.

    DIMENSIONS DIMENSIONAL

    ANALYSIS

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    Check validity of equations/expressions Determine exponent of equations

    Example 1 :

    Given that an equation mass = density x area Is that

    equation correct?

    DIMENSIONAL ANALYSIS

    A

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    Answer :

    If [left hand side] = [right hand side], then equation

    is valid.

    Therefore: [mass] = [density] x [area]

    Density = mass / volumeVolume = length x width x height

    Dimension density : [] = [M/ L3] = [ML-3]

    Dimension area : [A] = [L2

    ] = [L2

    ]Dimension mass : [m] = [M]

    [left hand side] = [right hand side]

    [mass] = [density] x [area]

    [M] = [ML-3] x [L2][M] = [ML-1]

    Therefore, [left hand side] [right hand side].

    The equation is not correct.

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    ts as nm tkas

    nmm

    nmm

    n

    m

    nm

    TLL

    TTLL

    T

    T

    LL

    tkas

    2

    2

    2

    ][][

    Example 2 :

    Given that and and

    Where k is constant. Find the values of m and n by using

    the dimensional analysis.

    Answer :

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    If the equation is true, then the exponent of each

    dimension on each side of the equation must be equal.

    Hence, Looking at L;

    Looking at T:

    Solving for m and n ; m = 1 and n = 2

    Therefore the equation s = kat2

    The value of k is actually

    This is an equation for an object accelerating from rest

    i.e

    nm2m01

    TLTL

    m1

    nm20

    2at2

    1s

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