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Waec 2016 Physics practical Question And Answers - May/ June Expo
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(4)
Ohm’s law state that the current flow through a metallic conductor is directly proportional to the potential
difference provided that temperature and other physical conditions are constant.
(4b)
Factors that determine resistance of
conduction.
PL Loading. . .
A
i– Nature Of the materials .
ii – Length of the conducting wire.
iii – Cross setional area .
iv – Temperature .
How Temperature affects Conduction
Answers
How Temperature affects Conduction Is that The more the temperature will be the resistance (If neccessary).
(4c)
Since the lost volt have developed across the cell when the supplied volt across the
1i)
Centre of gravity =50.5
TABULATE
S/N-A(cm)-B(cm)
1-10.0-51.90
2-15.0-53.50
3-20.0-53.80
4-25.0-54.90
5-30.0-55.80
1xii)
-I avoided the effect of draught during the experiment
-I avoided error due to parallax on the metre rule
-I balanced the metre rule horizontally on the knife edge before taking my readings
1bi)
Moment of a force about a point is the product of the force and its perpendicular distance from the point to the line of action of the force. i.e moment=force*time and its unit is Nm
1bii)
-Sum of forces acting in one direction must equal to the forces acting in another direction.
-The algebric sum of moment acting about a point must be equal to zero
2bi)Heat Capacity: is the amount of heat required to change its temperature by one degree, and
has units of energy per
degree.
1b i Moment of a force about a point
The moment of a force about a point O is defined as the vector product (cross product)
Mo= r × F
where r = xi + yj + zk is the radius vector (position vector) drawn from O to the point of
application A of the force F , and F = Fx + Fy + Fz is the force vector acting on A.
(3a)
TABULATE
S/N|E(v)|V(v)|I(A)|V^-1(v^-1)|I^-1(A^-1)
1 |1.50|0.80|0.40| 1.250 |2.500
2 |3.00|1.40|0.70| 0.714 |1.429
3 |4.50|2.00|1.00| 0.500 |1.000
4 |6.00|2.50|1.20| 0.400 |0.833
5 |7.50|3.20|1.50| 0.313 |0.667
Precautions:
-I tightened all the wires tightly together
-I ensured clean terminals
-I avoided zero error on the ammeter used
(3bi)
-By using electric cells
-By using electrostatic method
(3bii)
For parallel connection=1/R=1/R+1/R
2Rt=R
Rt=R/2ohm
Total=(R/2)+2
4=(R/2)+2
4=(R+4)/2
8=R+4
8-4=R
R=4ohm -
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