WAEC PHYSICS PRACTICAL ANSWER
=====================================
Monday, 3rd April, 2017
PHYSICS PRACTICAL - 9.30 am. – 12.15 pm
PHYSICS PRACTICAL - 9.30 am. – 12.15 pm
=====================================
NOTE: ^ means raise to power.i: meansthat under i, u will get 1,2,3,4,5..
=====================================
=====================================
(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
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
=====================================
(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
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
No3) Graph visit: http://k003.kiwi6.com/hotlink/zcsplikpj7/IMG-waec-phy-pra3-gidifans.jpg===========================
FOR MORE ANSWER DROP YOUR COMMENT IN THE COMMENT BOX
SCORE ABOVE 270+ IN UR UPCOMING 2017 JAMB CBT EXAM & Also Make A's, B's & C's In NECO/NABTEB Exam VISIT www.willrunz.blogspot.com to Subscribe
Tell UrFriends About Us
SCORE ABOVE 270+ IN UR UPCOMING 2017 JAMB CBT EXAM & Also Make A's, B's & C's In NECO/NABTEB Exam VISIT www.willrunz.blogspot.com to Subscribe
Tell UrFriends About Us
No comments: