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WAEC 2016/2017 Physics Objectives & Theory Answers Posted Here [12th April, 2016]

CURRENT EXAMINATION:

WAEC Physics Objectives & Theory Answers ;
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PHYSIC OBJ
1CCBBCBDBCD
11DACADCAADB
21BACBBBDDBD
31DDDBDABABA
41BCAADABCDC


PHYSICS THEORY


1a)
impulse(I)=F*t
F=ma= m(v/t)= M((d/t)/(t))
f=M(l/T)/T)= MLT^-2
I=MLT^-2 * T
I=MLT^-1

1b)
Acceleration(a)=v/t=(d/t)/t
a=d/t^2= L/T^2= LT^-2

1c)
WOrk(w)=f*d
W=MLT^-2 * L
W=ML^2T^-2

 
2)
Velocity(u)=20ms^-1
Angle(tita)=40 degree
At maximum height, h
V^2=u^2 + 2gh
U^2 sin^2 tita+u
u= 20^2* (sin 40 degree)^2
u=20^2*(sin 4o degree)
u=400* 0.3455
u=138.2ms^-1

3)
-smoke
-aerosol
-dust particle


4) Given flux density(B)=0.217
Force(F)=9.6*10^-12N
Speed(v)=?
F=Bqv
9.6*10^-12= 0.12* 1.6* V
V=(9.6*10^-12)/0.192*10^-19
=50*10^-12+9
v=50*10^7
=5*10^8ms^-1

5)
i) For launching astronauts into the space
ii) For launching satellite
iii) For Scientific research

6a)
Doping is the addition of small quantity of an element to a semi conductor to change its characteristics

6b)
Doping introduce impurities into an extremely pure instrintic semi conductor for the purpose of modulating its elctrical properties
================================
7)
X = Electron gun
Y = Accelerating mode
Z = Deflection plate

9a)
i) thermal equilibrium is the point at which the temperature of an object remain constant or steady

ii)fundamental interval is the distance between upper and lower fixed point of a thermometre


8a)
Net force: The net force is the vector sum of all the forces that act upon an object. It is also the sum of all the forces, taking into account the fact that a force is a vector and two forces of equal magnitude and opposite direction will cancel each other out.

8b)
The law of conservation of linear momentum states that the total
momentum of a system of particles remains constant, so long as no
external forces act on the system.
e.g Firing a bullet from a gun or a radioactive nucleus spontaneously emitting an alpha particle.

8c)
M = 200g = 0.2kg, h= 2cm, h2= 1.8m, u = 0, v=?
P.E = mgh
P.E = K.E = 1/2MV^2
4 = 1/2 x 0.2 x V^2
V = √40 = 6.32m/s^-1

Impulse = M(V-U)
Impulse = 0.2 ( 6.32 - (-o))
= 1.264Kgms^-1

8d)
M= 0.02kg , f= 5Hz, x= 10cm = 0.1 , V = 200cm = 2m/s

8dii)
r = ? , w = 2πf = 2x3.142x5
W= 31.4
From V= w √r^2 - 0.1^2
2 = 31.4 √ r^2 - 0.1^2
= r^2 - 0.1^2 = 0.6369^2
r^2 = 0.41564161
r = √ 0.41564161
r = 0.64m

Maximum vel
V = wr (x = 0 at maximum)

8diii)
maximum potential energy
=1/2 MVm^2
=1/2*20^10 *(372.3)^2
=10*138596.4
=1385964J

9b)
i) it can be used to raise heavy load
ii) it is used in printing press

9c)
magnet

9di)
by using calometry law
heat last = heat gain
MmCmDθm=CaDθ + Mw Cw Dθ
300 * Cm * ( 85 - 40 ) = 90 + 500 * 4200
13500Cm = 90 + 29400000
13500CM = 29400090/13500
Cm=2177.78Jg^-K^-1

9dii) §
i) by convection
ii) by conduction

9diii) §
i) power generation
ii) chemical processing
================================

10a)
Critical angle is the angle of incidence in the optically densed medium when the angle of refraction is 90degrees

10b)
Anti-note are created by region of maximum displacement in a wave

11a)
Dielectric strength is the property of an insulating material to retain material to retain electrical properties

11bi)
-Infra-red wave
-Geiger muller counter
-Scintillation counter

11bii)
Irradiation of solar energy

11c)
Charge(q)=1.0*10^-19C
Electric field intensity(E)=1200NC^-1
Weight of the oil drop=?
E=F/q
1200=F/1.0*10^-19
F=1200*1.0*10^-19
F=1.2*10^3*10^-19
F=1.2*10^-16N

11d)
R=100ohm, L=0.05H, C=0.25uF
V=220v, f=50Hz

11di)
Impedance(z)=sqroot(R^2+(Xl-Xc)^2)
Xl=2*pie*fl
Xl=2*(22/7)*50*0.05
Xl=15.71ohm
Xl=(1/2*pie*fc)
Xc=1/78.55*10^-6
Xc=10^6/78.55
=12730.7ohm
Z=sqroot(100^2+(15.71^2-12730.7^2)
Z=sqroot(10000+2.63*10^16
Z=sqroot(2.63*10^16)
Z=1.62*10^16ohm

11dii)
Vrms=IZ
I=220/1.62*10^16
=1.36*10^-14
================================
 
 

  


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