Monday 3rd April, 2017

    Physics 3 (Practical) (Alternative A) 09.30am – 12.15pm (1st Set)
    Physics 3 (Practical) (Alternative A) 12.40pm – 3.25pm (2nd Set)
    ============================

    2)

    2b)
    bi)The boiling point of a liquid is the temperature
    at which the vapor pressure of the
    liquid equals the pressure surrounding the liquid
    and the
    liquid changes into a vapor.
    bii) impurities increases the temperature range a
    substance can exist as a liquid


    (1)
    TABULATE:
    i:1,2,3,4,5,
    L(cm):80.00,70.00,60.00,50.00,40.00
    t1(s):17.87,16.22,16.09,14.08,13.28
    t2(2):17.56,16.46,15.09,14.77,13.36
    t(s)mean:17.715,16.340,15.590,14.425,13.320
    T(t/10)s:1.7715,1.6340,1.5590,1.4425,1.3320
    LogT(s):0.2483,0.2133,0.1928,0.1591,0.1212
    Log L(cm):1.9031,1.8451,1.7782,1.6990,1.6021
    Log T*10^-2(s):24.83,21.33,19.28,15.91,12.12
    Log L*10^-1(cm):19.031,18.451,17.782,16.990,16.021
    SLOPE(s)=(LogT2*10^-2 – LogT1*10^-2)(/Log L2 *10^-1 – LogL1*10^-1)
    =(21.33-12.12(s))*10^-2/(18.50-16.02(cm)*10^-1)
    =3.714*10^-1
    =0.3714cm^-1


    1axi)
    i)ensured supports of pendula were rigged
    ii)avoided parallax error on meter rule/stop watch

    1bi)
    simple harmonic motion is the motion of a body whose acceleration is always direct towards a fixed point and is proportional to the displacement from the fixed points

    1bii)
    T=1.2secs
    Log 1.2=0.079
    0079 shown on graph with corresponding Log L read L correctly determined

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

    2 xii) two precaution

    1. Avoided splas hing of water.
    2Avoided parallax error on thermometer/

    (2b i) This is the temperature of a liquid at which the saturated vapour pres sure of the liquid equals the external atmospheric
    pressure.

    2bii) Impurities increase the boiling point of a liquid.
    =================================

    (3a)
    TABULATE:
    x(cm):10,20,30,40,50,60
    V(v):0.65,0.75,1.00,1.20,1.45,1.55
    I(A):0.20,0.30,0.35,0.40,0.45,0.55
    LogV(v):-0.187,-0.125,0.000,0.079,0.161,0.190
    LogI(ohm):-0.699,-0.523,-0.456,-0.396,-0.391,-0.360
    SLOPE(s):=( LogI2-LOgI1)/(LogV2-LogV1)
    =-0.2-(-0.7)/0.3-(-0.2)
    =0.5/0.5
    =1AV^-1

    3axi)
    i)ensured clean/tight terminals
    ii)open key when reading was taken

    3bi)
    i)brightness of the bulb increases
    ii)voltage/current through the bulb increases

    3bii)
    i)diode
    ii)transistor

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