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## COMPLETE WAEC GCE PRACTICAL PHYSICS 2017 EXPO ANSWERS

COMPLETE WAEC GCE PRACTICAL PHYSICS 2017 EXPO ANSWERS | Get verified, approved, real and simplified today’s physic practical past questions and answers on physics 2017/2018 expo runz….. This page contains answers, expo and runz like: waec physics question 2016.
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waec physics practical questions and answers?

(1a)
Tabulate.
Firstly Under s/n: 1,2,3,4,5
Under M(N): 140,120,110,84,66
Under tita(o): 24,32,38,51,62
Lastly Under Sin tita: 0.4067, 0.5299,0.6156, 0.7771,0.8829.
Note that 1cm =20N.

(1aviii)
– I would ensure that the meter rules balanced before the readings.
– I would avoid error due to parallax.

(1bi)
i. Total forces in one direction are equal to the forces in opposite direction.
ii. The algebraic sum of the moment of all forces about any point should be zero.

(1bii.)
The moment of force at equilibrium point o is equal to,
sum of clockwise moment = sum of anti- clockwise moment
OC*M=CD*W.

====================================
(3a)
(i)d1=1.40cm, d2=1.55cm, d3=1.75cm, d4=2.1cm, d5=2.3cm
Real values of di
d1=1.4*0.5=0.7mm
d2=1.55*0.5=0.77mm
d3=1.75*0.5=0.875mm
d4=2.1*0.5=1.05mm
d5=2.3*0.5=1.15mm.

(ii)Ia1=2.4A, Ia2=3.8A, Ia3=5.0A, Ia4=7.4A, Ia5=11.6A.

Ib1=2.4A, Ib2=3.6A, Ib3=5.2A, Ib4=7.5A, Ib5=11.6A.

(iii)I=(Ia+Ib)/2
I1=(2.4+2.4)/2=2.4A
I2=(3.8+3.6)/2=3.7A
I3=(5.0+5.2)/2=5.1A
I4=(7.4+7.5)/2=7.45A
I5=(11.6+11.6)/2=11.6A
(3aiv)logd1=log0.7=-0.15mm
logd2=log0.775=0.11mm
logd3=log0.875=0.06mm
logd4=log1.05=0.02mm
logd5=log1.15=0.06mm
logI1=log2.4=0.38
logI2=log3.7=0.57
logI3=log5.1=0.71
logI4=log7.45=0.87
logI5=log11.6=1.06.

3av)
TABULATE
S/N; 1,2,,3,4,5
di(cm);1.40,1.55,1.75,2.10,2.30
di(mm);0.70,0.78,0.88,1.05,1.15
Ia(A);2.4,3.8,5.0,7.4,11.6
Ib(A);2.4,3.6,5.2,7.5,11.6
I(A);2.4,3.7,5.1,7.5,11.6
logd!(mm);-0.15,-0.11,-0.06,0.02,0.06
logI1;0.38,0.57,0.71,0.87,1.06.

(3avi)
|Slope(s)=change in logI/change in logd
=(0.87-0.57)/(0.02-(-0.11))
=0.3/0.13=2.3A.

(3avii)
-I will ensure the circuit is open when no readings are not taken
-i will ensure tight connection.

(3bi)
diameter(d)=1.09mm
(ii)
R=eL/A
R=(5.0*10^-2*0.01)/(pie/4)*(0.001)^2
R=(5*10^-4)/(0.7855*1*10^-6)
=6.37*10^2ohm
P=I^2R
=10^2*6.37*10^2
=6.37*10^4W.

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