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Nabteb GCE Physics Questions

NABTEB GCE 2019 Physics Practical Expo Question & Answer Now Available

 

NABTEB GCE 2019 Physics Practical Expo Answer – Nov/Dec Exam Runz

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Nabteb GCE Physics Practical Questions

Nabteb GCE Physics Practical Questions

Advanced NABTEB GCE 2019 Physics Practical Expo Answers

Table of Contents

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(1aiv)
IN A TABULAR FORM

S/N | H(cm) | L(m) | t(s) |
1 2.70 2.55 31.20
2 2.70 2.31 30.10
3 2.70 2.10 28.70
4 2.70 1.92 27.30
5 2.70 1.71 25.90

T(s) | T²(S²) | X(m)
1.56 2.43 0.15
1.51 2.28 0.39
1.44 2.07 0.60
1.37 1.88 0.78
1.30 1.69 0.99

(1av)
PLOT THE GRAPH: CLICK HERE FOR THE IMAGE

(1avi)
slope = ΔT²/ Δx = T₂² – T₁² / x₂ – x₁ = 2.65 – 1.00/0 – 1.65 = 1.65/-1.65 = -1.0S²/m
Intercept c = -2.65m

(1avii)
K = c/5 = 2.65/ -1.00 = -2.65m

(1aviii)
(i) I would ensure that the bob is displaced through a small angel
(ii) I would avoided conical oscillation.

(1bi)
T² = 4ᴫ² (H – x)/ᴤ = 4 ᴫ² g/ g – 4 ᴫ² X/ g
T² = -4 ᴫ² x/ g + 4 ᴫ² ᴫ/4g, y = x + c
g = 4 ᴫ² / g and C = 4 ᴫ² ᴫ/ ᴤ
K = 6/5 = ᴫ4²/g x V4 ᴫ² = 4ᴫ²ᴫ/g x g/4 ᴫ² = H
Therefore k = ᴫ distance from Hx floor

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(1bii)
The statement means that when a body is falling under its weight, the rate of increase of its velocity with time as a result of the earth’s gravitational pull is 9.8m/s²

(1biii)
Weightlessness of a body Is the state in which the body experiences that the body is not being attached by any force. i.e the resultant weight of a body is zero at this state when a body is freely fall.

(2av)
IN A TABULAR FORM
s/n|i(°)|e(°)|D(°)
1. | 50 | 45 | 48
2. | 53 | 45 | 53
3. | 60 | 55 | 70
4. | 40 | 43 | 38
5. | 48 | 45 | 45

(2av)
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(2avi)
Minimum deviation (dm) =38°
Angle of mediation (im) = 40°

(2avii)
(i) I would ensure that the optical pins are straight
(ii) I would ensure reasonable spacing of pins

(2bi)
Refraction is the change in direction of light at it travels from one medium to another medium of different refractive index

(2bii)
(i) Change in speed
(ii) Angle of incident ray

(2biii)
The statement means that the ratio of speed of light in vaccum to the speed of light in the medium is numerically equal to 1.33
i.e Refractive index = speed of light in a vaccum/speed of light in a vaccum = 1.33

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(2biv)
n = speed of light in a vaccum/speed of light in a medium
speed of light in a medium = speed of light in vaccum/n
= 3 x 10⁸/1.85 = 1.62 x 10⁸m/s

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NABTEB GCE Physics PRACTICAL SOLUTION

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