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#Post#: 230--------------------------------------------------
Free 100% Correct Waec Gce Physics (Alternative To Practical) A
nswers
By: Ebenezer Date: September 4, 2013, 3:34 am
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USE THIS ARRANGED WORK:
(3)
(3i.) lo = 0, l1= 0.5A, l2= 1.0A, l3= 1.5A, l4=2.0A, l5=2.5A,
l6=3.0A
(3ii.) Vo= 1.3v, V1= 1.28v, V2=1.25v, V3=1.23v,
V4=1.2v,V5=1.17v, V6=1.15v
(3iii.)
Tabulate
SN: 0,1,2,3,4,5,6
V(v): 1.30,1.28,1.25,1.23,1.20,1.17,1.15
l(A): 0.00,0.50,1.00,1.50,2.00,2.50,3.00
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NOTE THAT comma means NEXT LINE, Draw a table, under I, put
0.50, 1.00, 1.50 etc..
======================================
(3viii)
(a) i will ensure that the key is opened when no reading is
taken to avoid batterywastage
(b) i will ensure that the circuit element are tightly connected
to ensure accurate result
(3bi)
i. Temperature
ii. Nature of material
iii. cross-Sectional Area
iv. Length of conductor
(3bii) Electromotive force is the potential difference
between the terminals of a cell when no current is
flowing in the circuit.
(1i.)Experimental Result
h1=5.1cm, h2=4.25cm, h3=3.65cm, h4= 2.95cm, h5=1.7cm
h1=0.051m, h2= 0.0425m, h3=0.0365m, h4=0.0295,h5=0.017m
Real values of h
h1=0.2295m, h2=0.19125m, h3=0.1643m, h4=0.13275m, h5=0.0765m
(1ii.)
first Set
L1=2.7cm, L2=2.9cm, L3=3.1cm, L4=3.4cm,L5=3.55cm
Second Set
L1=2.8cm, L2=2.9cm, L3=3.0cm, L4= 3.15cm, L5=3.6cm
(1iii) Mean Value
L1 = 2.7+2.8/2 = 2.75cm
L2= 2.9+ 2.9/ 2 = 2.9cm
L3 = 3.1 +3.0/2 = 3.05cm
L4 = 3.4+3.15/2 = 3.275cm
L5= 3.5+3.6/2= 3.575cm
(1iv)
e= Lo - Lmean
e1= 6 - 2.75= 3.25cm
e2= 6 -2.9 = 3.1cm
e3= 6 - 3.05= 2.95cm
e4= 6 - 3.275 = 2.725cm
e5= 6 - 3.575 = 2.425cm
e1^2= 10.56cm^2, e2^2= 9.61cm^2, e3^2 = 8.7025cm^2, e4^2 =
7.426cm^2, e5^2= 5.881cm^2
NOTE THAT ^ MEANS "RAISE TO POWER SIGN"
(1v) Tabulate
SN: 1, 2,3,4,5
h(m): 0.2295,0.19125,0.1643,0.1327,0.0765
h(cm): 22.95,11.13,16.43,13.28,7.65
L(cm): 2.70,2.90,3.10,3.40,3.55
L(cm): 2.80,2.90,3.00,3.15,3.60
e(cm): 3.25,3.10,2.95,2.73,2.43
e^2(cm): 10.56,9.61,8.70,7.43,5.88
NOTE THAT ^ MEANS "RAISE TO POWER SIGN"
(1ix)
a.) i will ensure the spring is not slanted to avoid error of
reading length L
b.) i will avoid error of parallax due to reading the
instrument.
(1bi)
Given, Force constant(K)= 300N/m
Extension(e)= 3.0cm = 0.03m
Energy stored (E)= 1/2Ke^2
E= 1/2 *300*0.03*0.03
E = 150*0.03*0.03
E =0.135J
(1bii)
F1/L1-Lo = F2/L2-Lo = F3/L3-Lo
5/12-8 = 10 / 30-12 = F/30-12
5/4 =F/18
F = 5*18/4 = 22.5N
#Post#: 233--------------------------------------------------
Re: Free 100% Correct Waec Gce Physics (Alternative To Practica
l) Answers
By: Ebenezer Date: September 5, 2013, 9:50 am
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(2a)
sn:1,2,3,4,5,6
M/g:20,30,40,50,60,70
tita(degree celcius):39.2,41.6,43.2,44.2,45.4,45.8
xi(degree):14.2,16.6,18.2,19.2,20.4,20.8
yi(degree):10.8, 8.4.6.8,5.8,4.6,4.2
Ti:1.31,1.98,2.68,3.31,4.43,4.95
(2v)
slope= deltay/delta x
=2.68-1.31/40-20
=0.69
(2vi)
-i would ensure that parallax errors were avoided
-i would ensure that i continuously stir the iron fillings
(2b)
-it means that iron requires 470 joules of heat energy to raise
1kg of iron by 1 degreekelvin of temperature
#Post#: 234--------------------------------------------------
Re: Free 100% Correct Waec Gce Physics (Alternative To Practica
l) Answers
By: feargod Date: September 5, 2013, 10:51 am
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All is okey but sm waiting till 2014 jamb i cnt wait to access
it
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