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Topic: 2017/2018 waec physics practical questions and answers posted here 2017
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(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) I ensured supports of pendula were rigid
(ii) I avoided parallax error on the meter rule

(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


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



(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) I ensured neat and tight terminals
(ii) I opened the key when reading was taken

(3bi)
(i)The brightness of the bulb increases
(ii) The voltage and current through the bulb increases

(3bii)
(i) diode
(ii) transistor
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