SECTION A ANS 5QUESTION
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1)
decrease in height (h)
increase in time (t)
this shows that distance changes with time
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3)
i)Brownian motion
ii)Atomic size
iii)Expansion
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2)
given that U=15m3^-1
R=U* sqr2h/g
h=20m
g=10
R=15*sqr2*20/10
R=15*sqr40/10
R=15*sqr4
R=15*2=30m
R=30m
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4)
L1=14cm F1=4N,F2=6N
L2=4cm, Lo=?
Using hooke's law
4/14-Lo*6/4-Lo
4(4-Lo)=6(14-Lo)
16-4Lo=64-6Lo
-4Lo+6Lo=64-16
2Lo=48
Lo=48/2
Lo=24Cm
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5a)
i)force
ii)length
5b)
by increasing the density mosquitoes lavrva than the density of water
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6)
A)Energy efficient- so far the best light for interior lighting
B)Low production cost (of tubes, not of the ballasts)
C)Long life of tubes
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7)
i)It requires low voltage sources for it to function
ii)It cannot be easily broken
iii)It is cheaper and quite easy to manufacture
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SECTION B ANS 3 QUESION
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12ai)
Nuclear fission is a nuclear reaction in which a heavy nucleus splits spontaneously or on impact with another particle, with the release of energy.
12aii)
A graphite reactor is a nuclear reactor that uses carbon as a neutron moderator, which allows un-enriched uranium to be used as nuclear fuel.
12aiii)
Boron rods are used in
nuclear reactors to control the fission rate of uranium and plutonium. They are composed of chemical elements such as boron, silver, indium and cadmium that are capable of absorbing many neutrons without themselves fissioning
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8b)
If their distance and masses are doubled their corresponding force would also be doubled and the work done respectively
I.e.Workdone=Force * distance
Force= Mass * Acceleration
W= Mg * d
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Increase
8c)
-Acceleration due to gravity
- At maximum height, final velocity is zero
8d)
i)pushing someone on a swing.
ii)Tuning a radio to a particular frequency.
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10a)
A wave front is an imaginary surface passing through the points of a medium oscillatory in phase
10b)
i)Priam periscope used in harmod car
ii)prismatic bihocular to see round corner
10bii)
i)refractive indege of the glass prism
ii)the angle of incident of th
11aii) GIVEN: A = 9.6 * 10^-3 M^2
d = 2.25 * 10^4
Eo = 8.85 * 10^-12 F/m
Er = 900
C = /
using; C = EoErA/d
capacity, C = 8.85*10^-12 * 900 * 9.6*10^-3/2.25 * 106-4
C = 33984 * 10^-11
C = 3.3984 8 10^7 F
C =0.344F
11bi) A voltmeter has a higher resistance because it is designed to limit the flow of current to practically zero to avoid voltage drop or less across it.
11bii) combined resistance Rt
Rt = (400 * 400/400 + 400) + (800 * 800/800+ 800)
Rt = 200 + 400
Rt = 600ohms
current supplied by battery = r/Rt
= 6/600
= 0.01A
voltage drop across the resistance = 200 * 0.01
=2V
therefore the voltage reading is 6 -2
= 4V
11c) Equivalent resistance is:
Rt = 2+2+(2*2/2+2)
Rt = 4+1
Rt = 5ohms
