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.
==============================
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
==============================
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
==============================
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
=======================================
Neco Physics (Objective and Essay) Runz, 2018/2019 Physics (Objective and Essay) Questions and Answers Runz – June/July Expo, 2018 Neco Physics (Objective and Essay) Questions, 2018 Neco Physics (Objective and Essay) Answers, 2018/2019 NECO SSCE Physics (Objective and Essay) Questions And Answers
Exam, Expo, Runz..
