Free Nabteb Gce 2016 Physics Practical Answers – Nov/Dec Expo
Free Nabteb Gce 2016 Physics Practical Answers – Nov/Dec Expo
1a
Tablet
S/N: 123456
L: 31.00, 27.50, 24.00, 20.75,17.50, 14.00
I: 8.00, 12.00, 14.00, 17.00, 18.00, 20.00
T: 0.80, 1.20, 1.40, 1.70, 1.80, 2.00
Log T: 1.90, 0.08, 0.15, 0.23, 0.26, 0.30
Log L: 1.49, 1.14, 1.38, 1.32, 1.24, 1.15
1v)
1vii)
– I Will Ensure That No External Force Is Added To The System
-I Will Ensure That No Pallarax Error In Taking The Readings Of The Stop Watch
– I Will Ensure That No External Force Is Added To The System
-I Will Ensure That No Pallarax Error In Taking The Readings Of The Stop Watch
1bi)
Period Of Oscillation Of A Simple Pendulum Is The The Taking The Pendulum To Complete One Oscillation
Period Of Oscillation Of A Simple Pendulum Is The The Taking The Pendulum To Complete One Oscillation
1bii)
1bii)
NaA=15
NB=20
TA=1.5sec
NaA=15
NB=20
TA=1.5sec
But TA= TA/NA
TA =Ta*Na
=1.5*15
TA=22.5sec
TA =Ta*Na
=1.5*15
TA=22.5sec
The Tb= TB/Nb
But TA=TB=22.5sec
Tb=22.5/20
=1.125sec
================
But TA=TB=22.5sec
Tb=22.5/20
=1.125sec
================
3a)
TABULATE
S/N:1,2,3,4,6
E1:22.00,34.00,48.00,66.00,80.00,92.00
I1:8*10^-5,6*10^-4,1*10^-3,1.42*10^-3,1.84*10^-3,2.4*10^-3
I^-1:12500,1667,1000,704,543,417
E^-1:0.045,0.029,0.021,0.015,0.013,0.011
TABULATE
S/N:1,2,3,4,6
E1:22.00,34.00,48.00,66.00,80.00,92.00
I1:8*10^-5,6*10^-4,1*10^-3,1.42*10^-3,1.84*10^-3,2.4*10^-3
I^-1:12500,1667,1000,704,543,417
E^-1:0.045,0.029,0.021,0.015,0.013,0.011
3avi)
S=AB/BC= 0.0225-0.01/1200-400
=0.0125/800
=1.5626*10^-5
S=AB/BC= 0.0225-0.01/1200-400
=0.0125/800
=1.5626*10^-5
3avii)
– I Will Ensure That There Are No Partial Connections
– I Will Ensure That The Key Is Removed After Each Reading.
– I Will Ensure That There Are No Partial Connections
– I Will Ensure That The Key Is Removed After Each Reading.
3bi)
1) Red Light Has Higher Wavelenght Than Violet Light
1) Red Light Has Higher Wavelenght Than Violet Light
2) Red Light Has Lower Energy Than Violet Light
3bii)
E=H C/N
E= 6.6*10^-34 * 3*10^8/700*10^-9
= 2.83*10^-19 J
E=H C/N
E= 6.6*10^-34 * 3*10^8/700*10^-9
= 2.83*10^-19 J
3biii)
– Cross Sectional Area
– Length
– Temperature.
– Cross Sectional Area
– Length
– Temperature.
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