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Our 2019 Waec physics practical answers expo.
(1aviii)
In a tabular form
Under y(cm)
20.0, 30.0, 40.0, 50.0, 60.0
Under t(s)
36.61, 33.66, 31.32, 28.62, 25.93
Under T = t/20(s)
1.8305, 1.6830, 1.5660, 1.4310, 1.2965
Under T²(s²)
3.3507, 2.8325, 2.4524, 2.0478, 1.6809
(1ax)
Slope s = ΔT²/Δy
S = 4 – 2.5/0 – 39
S = 1.5/-39
S = -0.0385
Intercept, C = 4.0
(1axi)
SR = c
R = c/s = 4/-0.0385 = -103.9
(1axii)
(i) I ensured that the retort stand was well clamped.
(ii) I avoided error due to parallax when reading the metre rule.
(1 GRAPH)
SUBSCRIBE FOR THE IMAGE
(1bi)
Draw the diagram
(α) K.E is maximum
(β) acceleration is maximum.
NUMBER ONE (B)
(1bii)
Period, T = 2π√m/k
Given : weight, W = 120N
Mass = w/g = 120/10 = 12
T = 45
π = 3.142
4 = 2(3.142)√12/k
0.6365 = √12/k
Square both sides
0.405 = 12
K = 12/0.405 = 29.6
=========================
(3ai)
E.m.f, E = 3.00v
(3avii)
In a tabular form
Under R(ohms)
2.0, 5.0, 10.0, 12.0, 15.0, 20.0
Under V(v)
0.86, 0.62, 0.43, 0.38, 0.32, 0.25
Under V-¹(v-¹)
1.1628, 1.6129, 2.3256, 2.6316, 3.1250, 4.0000
(3aix)
Slope s = ΔR/Δv-¹
=17.5 – 4.25/3.5 – 1.5
=13.25/2
=6.625
Intercept, C = -5.75
(3ax)
S = Roα
6.625 = 0.85(α)
α = 6.625/0.85 = 7.79
C = -(Ro + β)
= -5.75 = -(0.85 + β)
β = 5.75 – 0.85
β = 4.9
(3axi)
(i) I ensured tight connections.
(ii) I avoided error due to parallax when reading the voltmeter.
(3bi)
Vab = 12 – 8 = 4v
Rab = 4×5/4+5 = 20/9ohms
Iab = Vab/Rab = 4/20/9 = 4 ×9/20 = 9/5
=1.8A
Power = I²ab Rab
=1.8² * 20/9
= 7.2watts
(3bii)
Current in line = power/voltage = 3600/240
=15A
Circuit breaker remain closed because current in line is less than 20A
(3)
SUBSCRIBE FOR THE IMAGE
(3x)
(i) I avoided parallax error when taking readings from the voltmeter
(ii) I ensured that the key was opened when reading is not been taking
2020 WAEC Physics Practical Expo (Runs) Questions And Answers
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