Waec 2019 Physics Practical Answer Now Available.

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(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.

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(1bi)
Draw the diagram
(α) K.E is maximum
(β) acceleration is maximum

(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

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2)

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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

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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.

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(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

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