4/12/2016

WAEC SSCE 2016 EXPO: PHYSICS Objectives and Essay Theory Questions And Answers

Today is Tuesday 12th April 2016
PHYSICS OBJECTIVES & ESSAY QUESTIONS AND ANSWERS

WAEC SSCE 2016 EXPO: PHYSICS OBJECTIVES & ESSAY Theory Questions And Answers
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PHYSICS OBJ:
1-10: CCBBCCDBBD
11-20: CBCCADACBB
21-30: BBABDBBDBD
31-40: DBBDDABABC
41-50: ACBDDABADB
(1)
(a)
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
(b)
Acceleration(a)=v/t=(d/t)/t
a=d/t^2= L/T^2= LT^-2
(c)
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)
-for exploration of space
-it is used for war fare
-it is used for launching of satellite
================================
(6)
(a)doping is the process of producing an impure semiconductor by adding foreign elements to the pure semiconductor
(b)Doping a semiconductor improves the number of charge carriers
=============================
(8a)
Net force is the resultant force acting on the body
(8b)The principle of conservation of linear momentum states that is an isolated system,the total momentum before colision is equal to the total momentum after collision
example:collision of a bullet with a target
(8c)
Mass(m)200g=200/1000=0.2kg
height(h)=2.0m
New height(h)=1.8m
Impulse=F*t=m(v-u)
I=mgh*t=m(v-u)
I=0.2*9.8*(2.0 1.8)
I=7.45N
(8d)
Given; mass(m)=20g=20/1000
m=0.02kg
Frequency(f)=5Hz
Distance(x)=10cm=10/1000=0.1m
Velocity(v)=200cms^-1=2ms^-1
(i)
V=wsqroot(A^2-x^2)
V^2=w^2(A^2-x^2)
w=2*pie*f=2*3.14*5
w=31.4rad/s
2^2=3.14^2(A^2-0.1^2)
A^2-0.1^2=4/985.96
A^2-0.1^2=0.0041
A^2=0.00414 0.01=0.0141
A=sqroot(0.0141)
=0.12m
(ii)
Maximum velocity(v)
V=wA
V=3.14*0.12
V=3.768m/s
=3.8m/s
(iii)
Maximum potential energy
P.Emax==1/2MV^2max
P>Emax=1/2*0.02*3.8^2
=0.14J
----------------------------
(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
(10c)
Given D=46.2degrees
Refractive index(n)=?
n=sin1/2(A D)/sin1/2A
Refracting angle(A)=60degrees
n=sin1/2(60 46.2)/sin1/2*60
n=sin1/2(106.2)/sin30
n=sin53.1/sin30
n=0.7997/0.5
=1.5994
n=1.6
(10d)
-The object is placed at the principal focus
-The object is placed beyond the centre of curvature
(10e)
Object(u)=75cm
focal length(f)=30cm
(10ei)
Image distance(v)=?
1/u 1/v=1/f
1/75 1/v=1/30
1/v=1/30-1/75
1/v=(5-2)/150
1/v-3/150
v=150/3
v=50cm
(10eii)
Object distance(u)=75 25=100cm
1/u 1/v=1/f
1/100 1/50=1/f
1/f=(1 2)/100
1/f=3/100
f=100/3
f=33.3cm
================================
(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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