NABTEB 2019 Physics Practical Question and Answer Expo
NABTEB 2019 Physics Solution to Question Answer – May/JuneExpo
NABTEB PHYSICS PRACTICAL SOLUTIONS???
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(1a)
Initial position of the pointer on the metre rule
Xo = 37.20cm
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Table of values
In a tabular form
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S/N: 1.0, 2.0, 3.0, 4.0, 5.0
M(g): 70.00, 90.00, 110.00, 130.00, 150.00
X1(cm): 39.80, 40.00, 42.00, 43.10, 44.20
e = X1 – X0(cm): 2.60, 3.70, 4.80, 5.90, 7.00
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(1aviii)
(i) I ensured that I avoided parallax error when taking readings on the metre rule.
(ii) I ensured firm suspension of the scale pan before and after leading the masses.
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(1bi)
Hooke’s law states that the extension experienced by an elastic material is directly proportional to the force applied, provided the elastic limit is not exceeded
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(1bii)
Pls deduce from the graph.
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(1biii)
1/2mv² = Fe
M = 5g = 0.005kg
V = ?
F = 70N
e = 7cm = 0.07m
1/2 × 0.005 × v² = 70 ×0.07
V² = 70×0.07×2/0.005 = 1960
V = √1960 = 44.272 = 44.27m
Instantaneous velocity of the stone when released = 44.27m
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(2ai)
The focal length of the concave mirror provided is approximately equal to 15.0cm
Table of value
In a tabular form
S/N: 1.0, 2.0, 3.0, 4.0, 5.0
y(cm): 10.00, 20.00, 30.00, 40.00, 50.00
x(cm): 75.00, 60.00, 30.00, 24.00, 21.50
1/x(cm-¹): 0.0133, 0.0167, 0.0333, 0.0417, 0.0465
1/y(cm-¹): 0.100, 0.050, 0.033, 0.025, 0.020
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(2avi)
Plot a graph
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(2aviii)
(i) I ensured that the image of the cross wires were sharp on the screen before taking my readings.
(ii) I ensured that I avoided parallax error when taking readings on the metre rule by placing my eye vertically above the scale.
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(2bi)
K is the focal length of the mirror
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(2bii)
(i) They are used as shaving mirrors.
(ii) They are used as car driving mirrors.
(iii) They are used as dentist mirrors.
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(2biii)
(i) The image is magnified (ie larger than the object)
(ii) The image is real.
(iii) The image is inverted.
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(3i)
EMF=2.0V
Tabulate
K(Ω) | I(A) | I^-1(A^-1)
1.0 | 0.44 | 2.573
2.0 | 0.34 | 2.941
3.0 | 0.28 | 3.571
4.0 | 0.24 | 4.167
5.0 | 0.22 | 4.545
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(3avii)
Slope (s) =ΔI^-1(A^-1)/ΔR(Ω) =4.85-2.60/4.90-1.50
=2.25/3.40 =0.662A^-1Ω^-1
Intercept (c) =2.65A^-1
3aviii) K=E*C=2*2.65=5.3A^-1V
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(3aix)
– I ensured that I tightened the connection
– I avoided parallax error in reading ammeter/voltmeter
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(3bi)
The value of K gives the overall resistance R(Ω)
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(3bii)
Terminal p.d is the potential difference a cell in a circuit unit. When no current is flowing the terminal p.d is equal to the e.m.f of the cell. If a current is flowing then the terminal p.d of the cell will be lower than it’s EMF as some loss of energy takes place due to its internal resistance
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NABTEB Geography Obj And Theory/Essay Expo Answer – May/June 2017
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