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2017 WAEC GCE Physics Practical Expo — Physics Practical Answers/Runs/Dubs
WAEC September 26, 2017 • 3 years ago • No Comment Yet

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? 2017 WAEC GCE PHYSICS PRACT. ANSWERS?

100% VERIFIED PHYSICS PRACT. THEORY ANSWERS
(1ai)
Real values of masses(msi)g
msi=50.0
ms2=52.5
ms3=54.5
ms4=58.0
ms5=60.0
Real values of di(cm)
d1=5.50
d2=5.80
d3=6.00
d4=6.30
d5=6.60
(1aii)
Real values of ti(secs)
t1=6.45
t2=6.80
t3=6.85
t4=7.40
t5=7.45
(1aiii)
Real values of mli(g)
ml1=>msi-mso=50-20=30
ml2=>ms2-mso=52.5-20=32.5
ml3=>ms3-mso=54.5-20=34.5
ml4=>ms4-mso=58.0-20=38.0
ml5=>ms5-mso=60.0-20=40.0
CLICK HERE FOR THE 1ai to 1aiii) IMAGE
(1aiv)
TABULATE
i:1,2,3,4,5
msi(g):50.0,52.5,54.5,58.0,60.0
di(cm):5.50,5.80,6.00,6.30,6.60
ti(s):6.45,6.80,6.85,7.40,7.45
mli(g):30.0,32.5,34.5,38.0,40.0
T=t/n(s):0.645,0.680,0.685,0.740,0.745 where n=10
li(g/cm):5.4545,5.6034,5.7500,6.0317,6.0606
T^2(S^2):0.4160,0.4624,0.4692,0.5476,0.5550
CLICK HERE FOR COMPLETE NUMBER 1 IMAGE
(1vi) CLICK HERE FOR THE GRAPH
(1aix)
(i) I avoided parallax error in reading clock/weight balance
(ii) I ensue repeated readings are taken to avoid random error
(1bi)
(i) weight
(ii) upthrust
(iii) viscous force/drag or liquid friction
(1bii)
The amplitude of oscillation of the loaded test tube decreases with time because of the viscous drag(force) which opposes the motion.
============================
ANSWER NO2
b(ii) given : f = 10cm
u = 20cm
v = ?
Using:
1/f = 1/u + 1/v
1/10 = 1/20 + 1/v
Multiplying through with 20v, we have:
2v = v + 20
2v – v = 20
v = 20cm
Magnification, m = v/u
m = 20/20
m = 1
===============================
3a)
Table W1
s/n: 1,2,3,4,5,6
xn(cm)/d1=0.009m: 3.06,3.50,3.70,4.30,4.50,4.80
xm(cm)converted/- 15.30,16.50,18.50,21.50,22.50,24.40
xm(cm)converted d2=0.005m- 84.70,83.50,81.50,78.50,77.50,75.60
Ri(n): 11.07,10.12,8.81,7.30,6.89,620
Tabulate W2
s/n: 1,2,3,4,5,6
xmi(cm): 1.40,1.65,1.80,2.20,2.55,2.94
converted/xmi(cm)7.00,8.25,9.00,11.00,12.75,14.70
xni(cm): 93.00,91.75,91.00,89.00,87.25,85.3
R2i(n): 26.57,22.24,20.22,16.18,13.69,11.61
slope
dy/dx=26.57-13.69/11.07-6.89
=12.88/4.18
=3.08
3ix)
k=d2/d1*squareroot5
=0.005/0.009*squareroot13.08
=0.56*1.75
=0.98
CLICK HERE FOR NUMBER 3 GRAPH
3x)
i)i would ensure that the terminal are well tightened to avoid partial contact
ii)i would obey the connect connection of positive to positive to negative to negative when connecting to circuit
3bi)
when temperature increases the molecules tends to vibrate more intense there by opposing the flow of current
3bii)
P=I^R
P=V^2/R
power=1000W
Voltage=200V
R=?
1000=200^2/R
R=40000/1000
R=40
CLICK HERE FOR NUMBER 3 COMPLETE IMAGE

=====COMPLETED=====

 

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