 ## WAEC GCE Physics Obj And Essay/Theory Questions and Answer – NOV/DEC 2017 Expo Runz.

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PHYSICS OBJ:
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Completed

PHYSICS THEORY
SECTION B - ANSWER 3 QUESTIONS
11a)
i)hammering method
ii)electrical method

11b)
electrical line of force- this is the line drawn such that the tangent to it at a given point is in the direction of the force acting on a small positive change at the point

11c)
diagrams
at beaker point
potential different accross Q=p d are p
IiQ=I2------eq1
p d across = p d across R
IiS=I2R------eq2
divide eq 1by2
IiQ/IiS=I2P/I2R
R=SP/Q

11d)
resistivity- this is the resistance of a wire unit length and unit cross-sectional area
i;e P=RA/L

11e)
Given: F = 50Hz
L = 10 mH (10×10^-³H)
Inductive reactance, XL = 2πFL
= 2×3.14×50×10×10^-³
= 3..14 ohms

11f)
If voltmeter reads 12V, over the 500 ohms, then using I = V/R
current in the circuit is I = 12/500
I = 0.024A
Hence over the 200 ohms, V = IR
V = 0.024×200
V = 4.8 VOLTS
Voltmeter will read 4.8 volts

9a)
i)Poor conductors of heat application
ii)Coat (Trapped air in between the fabric

9b)
Renewable: Water, corn stalk
Non-renewable: coal, natural gasgas

9c)
A closed is a physical system so far removed from other systems that it does not interact with them.

9d)
heat required to melt = mc∆ + ml
Not sure of the temperature given, if it is -3°C or -5°C, However, *Using -3°C*
Heat = (0.1×2200×3)+(0.1×2.26×10^6)
= 226660 J

9ei)
Given: 75 heart beats in 1 min
Therefore in 1 hour, we have (75×60) heart beats = 4500 heart beats
Also given: 2J of energy is expended in 1 heart beat
Therefore for 4500 heart beats, energy expended = (4500×2) J = 9000J
9eii)
Energy = mc∆
9000 = (0.25×4200×∆)
9000 = 1050∆
∆ = 9000/1050
Temperature rise = 8.57°C

8a)
force is a push or pull which changes a body state of rest or of uniform motion in a straight time
8aii)
i)constant force
ii)field force

8bi)
bodies are streamline to make sure that the surface are in head on contact with the fluid during motion
8bii)
i)aircrafts
ii)ship

8ci)
diagrams
8cii)
mass of bucket =1.3kg
speed of loop(v)=7.0ms^-1
(x)acceleration (a)
a=rw^2
v=rw
w=v/r
a=r(v/r)^2
a=rv^2/r^2
a=v^2/r
a=(2.0)^2/1.2=49/1.2
a=40.8ms^-2
b)net force (F)
F=F1-Fg
F=Ma-Mg
F=(1.3*40.8)-1.3*19i
F=53-04-13
F=40N

SECTION A - ANSWER 5 QUESTIONS
2)
i)used in eye surgery
ii)used for the precise cutting of flat material
iii)used for measuring distances

3)
i)iron
ii)steel
iii)nickel

4a)
Time (t)=4secs
V=0ms^-1
pie call,V^2=U^2+2gh
O^2=20^2+2(-10)h
0=400-20h
20h/20=400/20
h=20m

4b)
the magnitude of the initial speed (u) is equal to the magnitude of the height attained

5)
i)availability of constant sunlight
ii)nearness to power grid
iii)must be closer to both skilled and unskilled labour

6)
P-type semi conductors are created by dopung an intrisic semi conductor with acceptor impurities.

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