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    Physics obj
    1-10 CAABADACC
    11-20 BAADDCBADA
    21-30 CCBACBCBAB
    31-40 ACADDBBBAC
    41-50 BCDABCABCA

     

    SECTION A ANS 5QUESTION

    ================================
    1)
    decrease in height (h)
    increase in time (t)
    this shows that distance changes with time
    ==================================

    3)
    i)Brownian motion
    ii)Atomic size
    iii)Expansion

    ================================

    2)
    given that U=15m3^-1
    R=U* sqr2h/g
    h=20m
    g=10
    R=15*sqr2*20/10
    R=15*sqr40/10
    R=15*sqr4
    R=15*2=30m
    R=30m

    =============================

    4)
    L1=14cm F1=4N,F2=6N
    L2=4cm, Lo=?
    Using hooke’s law
    4/14-Lo*6/4-Lo
    4(4-Lo)=6(14-Lo)
    16-4Lo=64-6Lo
    -4Lo+6Lo=64-16
    2Lo=48
    Lo=48/2
    Lo=24Cm

    =============================

    5a)
    i)force
    ii)length
    5b)
    by increasing the density mosquitoes
    lavrva than the density of water

    =================================

    6)
    A)Energy efficient- so far the best light for
    interior lighting
    B)Low production cost (of tubes, not of the
    ballasts)
    C)Long life of tubes

    ================================

    7)
    i)It requires low voltage sources for it to
    function
    ii)It cannot be easily broken
    iii)It is cheaper and quite easy to
    manufacture

    ==================================

    8b)
    If their distance and masses are doubled
    their corresponding force would also be
    doubled and the work done respectively
    I.e.Workdone=Force * distance
    Force= Mass * Acceleration
    W= Mg * d
    |
    Increase

    8c)
    -Acceleration due to gravity
    – At maximum height, final velocity is zero
    8d)
    i)pushing someone on a swing.
    ii)Tuning a radio to a particular frequency.

    ============

    10a)
    A wave front is an imaginary surface
    passing through the points of a medium
    oscillatory in phase

    ============

    11aii) GIVEN: A = 9.6 * 10^-3 M^2
    d = 2.25 * 10^4
    Eo = 8.85 * 10^-12 F/m
    Er = 900
    C = /
    using; C = EoErA/d
    capacity, C = 8.85*10^-12 * 900 *
    9.6*10^-3/2.25 * 106-4
    C = 33984 * 10^-11
    C = 3.3984 8 10^7 F
    C =0.344F

    11bi) A voltmeter has a higher resistance
    because it is designed to limit the flow of
    current to practically zero to avoid voltage
    drop or less across it.
    11bii) combined resistance Rt
    Rt = (400 * 400/400 + 400) + (800 * 800/800+
    800)
    Rt = 200 + 400
    Rt = 600ohms
    current supplied by battery = r/Rt
    = 6/600
    = 0.01A
    voltage drop across the resistance = 200 *
    0.01
    =2V
    therefore the voltage reading is 6 -2
    = 4V
    11c) Equivalent resistance is br /> Rt = 2+2+(2*2/2+2)
    Rt = 4+1
    Rt = 5ohms
    ====================

    SECTION B ANS 3 QUESION

    ================================

    12ai)
    Nuclear fission is a nuclear reaction in
    which a heavy nucleus splits spontaneously
    or on impact with another particle, with
    the release of energy.

    12aii)
    A graphite reactor is a nuclear reactor that
    uses carbon as a neutron moderator, which
    allows un-enriched uranium to be used as
    nuclear fuel.

    12aiii)
    Boron rods are used in
    nuclear reactors to control the fission rate
    of uranium and plutonium. They are
    composed of chemical elements such as
    boron, silver, indium and cadmium that are
    capable of absorbing many neutrons
    without themselves fissioning.

    ==============

    COMPLETED

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    Master Solution April 10, 2017 Categories: WAEC 1589


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    6 Comments

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      April 10, 2017 at 2:23 pm

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