2018 NECO PHYSICS ANSWER

PHYSICS OBJECTIVE:
1BBBBAAEECD
11CCBCDDAADC
21EEACBAECAB
31EAAECEECBD
41AECCABEECE
51CEABBCABEB


7a)
i)used in car headlamps
ii)used as a shaving mirror
given
object distance U =13cm
image distance V = -4cm
focal length F = ?
Using the real is positive convention
1/f=1/u+1/v
1/f=1/13-1/4
1/f=4-13/52
1/f= -9/52
f= -52/9= -5.8cm

2)
the distance covered S by a body with initail speed U and final speed V in time t is
distance = average speed x time
S = (u+v/2)t
first equation
acceleration = change in velocity/time
a=v-u/t
v-u=at
v=u+at
putting the first equation into S = (u+v/2)t we have
S = (u+u+at/2)t
S = (2u+at/2)t
S = (u+at/2/2)t
S = ut+1/2at^2
this is the second equation

14a)
The law of reflection state that the angle which the incident ray makes with the normal is equal to the angle which the reflected ray makes to the same normal.

14b)
i)Effect of density
ii)Effect of temperature:- In a gas.

14c)
CLICK HERE FOR THE DIAGRAM

14d)
using the lens formula
1/f=1/u+1/v
1/5=1/6+1/v
1/5-1/6=1/v
6-5/30=1/v
1/30=1/v
V=30cm (image distance from lens)
Total distance between image of len And image of mirror is
=30+12+16
=48cm

10a)
i)Mercury dropped on glass from droplets because the cohesion of mercury is stronger than it adhesion

ii)Water rise in capillary tube because the force of adhesion at water molecules with glass is greater than their cohesion to each other.

10b)
Surface tension, the ability of water to behave an elastic skin make it possible for an insect to crawl on water

6a). The incident ray, reflected ray and normal at the point of incidence all lie on the same plane

6b i) myopia can be corrected by placing a suitable concave or diverging lens in front of the eye

ii) the action of the lens is to diverge the light rays entering the eye from a distant object and make them appear coming from the persons far Point

8)b Any two correct advantages potentiometer over the voltmetre e.g
– High accuracy
– Null deflection (does not draw current at balance point)

– Range can be increased as required
– No friction in moving parts.

9a) (i)Plate spacing
ii)plate area

12a) i) trajectory or flight path is the path that a massive
object in motion follows through space as a function of time.

ii) -a baseball that has been pitched, batted, or thrown
-a bullet the instant it exits the barrel of a gun or rifle
– a moving airplane in the air with its engines and wings disabled.

(13a)
Specific latent heat of fusion is the quantity of heat energy need to convert a mass 1kg of a body from ice to liquid without a change in temperature

(13bi) Conventional current can be defines as the flow of electrons through the circuit and into the negative terminal of the source

(13bii)
Practical application:
-For correcting change in electron flow
-For creating electrons for positive terminals

(13c)
Mass of water=500g=0.5kg
MC(change in tita),ML
0.5*4.2*10^3*100=[(0.5+x)*3.2*10^5]
210000=1.6*10^5+3.2*10^5x
50000=3.2*10^5x
x=50000/3.2*10^5
x=0.16kg
x=1600g

(13a)
Specific latent heat of fusion is the quantity of heat energy need to convert a mass 1kg of a body from ice to liquid without a change in temperature

(13bi) Conventional current can be defines as the flow of electrons through the circuit and into the negative terminal of the source

(13bii)
Practical application:
-For correcting change in electron flow
-For creating electrons for positive terminals

(13c)
Mass of water=500g=0.5kg
MC(change in tita),ML
0.5*4.2*10^3*100=[(0.5+x)*3.2*10^5]
210000=1.6*10^5+3.2*10^5x
50000=3.2*10^5x
x=50000/3.2*10^5
x=0.16kg
x=1600g

3a.
Basically forces are of two types:
Contact forces

Non contact forces
Types of contact forces
some types of contact forces are given in the list below:
Applied force
Normal force
Frictional force
Tension force
Air resistance force
Spring force

Non contact force examples
Examples of non contact or action at a distance forces,which are given in the list given below:
Gravitational force
Electromagnetic force
Weak nuclear force
Strong nuclear force

1a)Electricity is produced from a dam by first restricting the flow of water from one end of the dam to the other, concentrating the water and making it so that the water flowing through the dam has a higher velocity. The water flows through a turbine, or series of turbines. The lateral velocity of the water turns the turbines. The water escapes out the other end of the dam.
The turbine is connected to a shaft and a generator, essentially a large magnet with a coil of wires around it. Moving the magnet or the coils around generates the electricity. The basic premise is to get the magnet or coils inside the generator to move, and in a hydroelectric power plant, the energy in the natural movement of water in streams and rivers is harnessed to do so.

1b)
Given:
M=10
U=?
V=mu
F=+5cm
V=10u(virtual image)
using the real is positive convention
1/f=1/u+1/v
1/5=1/u-1/10u
1/5=10-1/10u
1/5=9/10u
10u=45
u=45/10
u=4.5cm
the object must be placed 4.5cm from the microscope.


17a)
i)Stay on top of large machinery operator training

ii)Keep large machinery clean, and maintain a clean environment

17bi)
-Compound Microsoft-
for viewing samples at high magnification (40 - 1000x), which is achieved by the combined effect of two sets of lenses: the ocular lens (in the eyepiece) and the objective lenses

17bii)
-Film projector-
an optical device that projects an image (or moving images) onto a surface, commonly a projection screen.

17c)
Artificial satellite orbits the earth and observes it. A rocket is designed to go to different planets, either taking people with it, or taking a space prob to take pictures of that planet.

17d)
data given
Rg=5n,Ig=10mA Vg=40V Rx=?
Ig=10mA =10*10^-3A
Ig=10^-2A
For Conversion Of galvanometer into voltmeter.
Vg=Ig(Rg+Rx)
Vg=Ig Rg + Ig Rx
Vg-Ig Rg =Ig Rx
Rx=Vg-IgRg/Ig
Rx=Vg/Ig-Rg
=40/10^-2 - 5/1
Rx=40/0.01 - 5/1
Rx=40-0.05/0.01
Rx=39.95/0.01
Rx=3995
the resistance of resistor is 3995

15ai)
Force field is a region around a body in which it experiences an effect (force) due to the presence of another body

15aii)
i)gravitational field
ii)magnetic field

15b)
Show that the escape velocity(Ve) of an object from the earth surface is √2Gme/r
To prove the escape velocity (Ve) of an object from the earth surface we can say let the kinetic energy of the object of mass M is
Ek=1/2MVe^2
When an object is at any given point P which is at a distance X from the centre of the earth the force of unity between the object and earth is
F=Gmm/X^2
Work done is taken the body against gravitational attraction DW=Fdx=Gmm/X^2 DX
Where W=work done in taking the body against gravitational attraction
W={R DW= {R Gmm/X^2 DM
=Gmm {R X^2 DM = Gmm [X^2/Y {R
= -Gmm [1/X]R = - Gmm [1/~ - 1/R]
Or, W=Gmm/R
For the object to escape from the earth surface
Ek=workdone,W
Ek=W
1/2MVe=Gmm/R
Ve=√2Gm/R(proved)

15ci)
i)Power supply smoothing applications
ii)RF coupling or filter capacitor applications

15cii)
Data given
d =4mm=4*10^-3 , 0.004m
A= 0.2m^2, V=300V
[Eo=8.85*10^-12fm^-1]

i)capacitance of the capacitor
using the approprate formula
C=EoAv/d
C=8.85*10^-12*0.2*200/0.004
C=8.85*10^-12*2*10^-1*2*10^2/4*10^3
C= 35.4*10^-11/4*10^-3
C= 8.85*10^-10^-11+3
C=8.85*10^-8F
the capacitance of the capacitor is 8.85*10^-8F

ii)Electric Field intensity between the plates
E=F/q =K.q.d/d^2/q
E=kQ/d^2
E=8.85*10^-12*Q/d^2
but Q=CV
Q=8.85*10^-8*2*10^2
Q=17.7*10^-6 C,1.77*10^-7C
E=8.85*10^-12*1.77*10^-7/(4*10^-3)^2
E=15.7*10^-19/16*10^-6
E=1.57*10^-20/1.6*10^-5
E=0.98*10^-15
E=9.8*10^-1*10^-15
E=9.8*10^-16N|C or E=9.8*10^-16NC
the electric field intensity have the plates is 9.8*10^-16 NC
=======================================
Pls Keep Refreshing Page for More..

Comments

Popular posts from this blog

2018 NECO ENGLISH LANGUAGE ANSWER

2019 waec literature in english answer