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WAEC 2018 Physics Practical Questions and Answers/Expo/Dubs/Runs
WAEC March 27, 2018 • 3 years ago • One Comment

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# 👑 2018 MAY/JUNE WAEC PHYSICS PRACTICAL ANSWERS👑

(1a)
TABULATE:
S/N: 1, 2 , 3, 4, 5
L(cm): 90.00, 80.00, 70.00, 60.00, 50.00
T(s): 38.00, 36.00, 34.00, 31.00, 28.00
T = t/20(sqt): 1.90, 1.80, 1.70, 1.55, 1.40
√L(cm^1/2): 9.49, 8.94, 8.37, 7.73, 7.07
(1aix)
Slope(s) = Δvertical/Δhorizontal
=1.90 – 1.55/9.49 – 7.75
= 0.35/1.74
S = 0.201
(1ax)
g = 4π²/S²
= 4 ×(3.142)²/(0.201)²
= 977.4cm/s²
(1axi)
(i) I ensured the angle of oscillation is relatively small.
(ii) I ensured that no external force is added to the system of oscillation.
(iii) I ensured the oscillation is perfect
1bi) by the law of conversation of energy total energy remains constant ie ½ Mv2 = mgh or U2 =2gh (M is constant) similarly, kc2=P.E converstion also takes place in simple pendulum bulb
1bii) given that
U= x
F=20
V=5
Using
I/u + I/v = I/F
1/x = 1/v = 1/f
1/x = 1/5 = 1/20
Multiply by the L.C.M
5/x = 5/20
x/5 = 4/1 = 20
Endpoint acceleration Is maximum and velocity is zero
acceleration is zero velocity is maximum
ibi) foeal length of an object is 20 and means that the distance between the optical center or middle of the lens and its principal focus is 20cm
==============================
(3a)
TABULATE
S/N:1,2,3,4,5
d(cm):25.00,35.00,50.00,65.00,80.00
I(A):0.80,0.75,0.65,0.60, 0.55
I^-1(A^-1):1.2500,1.3333,1.4286,1.5385,1.6667
(3aviii)
Slope=Change in logI^-1/Change in d(cm)
(Y2-Y1)/(X2-X1)=(0.23-0.06)/(80.0-13.0)
=0.17/67=0.0025
S=0.0025
(3aix)
-I ensure tight connections
-I avoided zero error of the ammeter
(3bii)
-Length of the wire
-temperature
-cross sectional area
-Nature of the material

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