 exam runs

Waec GCE 2019 Physics Solutions & Answer Available Here  Written by
Read Time2 Minutes, 29 Seconds

*Physics solutions*

*PHYSICS-OBJ*
1-10: BCCDBDCDDB
11-20: ABDBACABDA
31-40: DBCACACBCB
======================

EL

1 60 marks)

A
rdelt questions all.
Five questions from Part I and three questions from Pat

PART
1 15 marks

Answer any live questions from this part

All questions carry equal marks
Using an energy level diagram, explain how an acceptor impurity fits into the atomic structure of an intrinsic
semiconductor

An object is projected at 20 mp horizontally from the top of an 80hich cliff. Calculate the

(a) time the object hits the ground:
1 range
110 m2]

==============

(2a)
H=ut+½gt²
Where H=80m, t=?
80=0(t) + ½(10)(t²)
80=0 + 5t²
80=5t²
80/5=t²
16=t²
t=√16
t=4s

(2b)
Range = U²/g
20²/10
=400/10
=40m
========================
(3)
F = KAV/L
FL = KAV
K = FL/AV
Where F = ma
= MLT-²
r = M, V = LT-¹
And A = M²

K = MLT-² × M/m² × LT-¹
K = LT-² ×L-¹T-¹
K = T-¹
====================================
6a) Ferromagnetism is the basic mechanism by which certain materials (such as iron) form permanent magnets, or are attracted to magnets.

6b)
i) Iron
ii) nickel
iii) cobalt
====================================
8ai)Pressure (symbol: p or P) is the force applied perpendicular to the surface of an object per unit area over which that force is distributed. … Pressure may also be expressed in terms of standard atmospheric pressure; the atmosphere (atm) is equal to this pressure, and the torr is defined as 1⁄760 of this

8ii)
i) They are the depth of the fluid and its density.
ii) A fluid exerts more
pressure at greater depths
=====================
(11a)
An electric field is a region where an electric force is experienced by a charged body.

(11b)
In a tabular form
Under gravitational potential
(i) Scalar quantity
(ii) Unit is Joule per coulomb

Under gravitational field strength
(i) Vector quantity.
(ii) Unit is Newton per coulomb.

(11ci)
Equivalent capacitance = 5uf//(2uf + 2uf)
= 5uf // 4uf
= 5×4/5+4 uf
= 20/9 uf
= 2.22uf

(11cii)
Total energy stored = 1/2cv²
= 1/2×20/9×10^-6×12×12
=160×10^-6
= 0.00016Joules.

(11di)
Given: Im = 20×10^-3A; V = 25sin100πl
Capacitive reactance Xl = Vm/Im = 25/26×10^-3 = 1250ohms

(11dii)
1/2πfc = Xl
Where f = 50Hz
1/2×22/7×50c = 1250
314.286c × 1250 = 1
392857 c = 1
C = 1/392857 = 2.545×10^-6F
= 2.545uf

(11e)
A phasor diagram is a graphical way of representing the magnitude and directional relationship between two or more alternating quantities.

===============
(12a)
Thermoelectrons are electrons emitted at a high temperature such as one produced in a thermionic valve. They are negatively charged.

(12b)
(i) Low voltage battery
Used to heat the heater or heating element.
(ii) Target metal
Used to collect the emitted electrons.

(12Ci)
α – particles has a helium nuclei He while a β-particle has high energy electrons.

(12Cii)
α – particle has a charge of +2e while β-particle has a charge of -e.

(12di)
The radioactive decay law states that the probability per time that a nucleus will decay is constant, independent of time.

(12dii)
(i) Damage to skin.
(ii) Causes cancer.

(12ei)
Given 238 U –> m X + 2α
92 n
= 238 U –> m X + 2 4He
92 n 2
m+(2×4) = 238
m + 8 = 238
m = 238 – 8 = 230
Also
N + (2×2) = 92
N + 4 = 92
N = 92 – 4
n = 88

(12eii)
230
X
88

No of neutrons = 230 – 88 = 142
Neutron – Proton ratio = 142/88 = 1.61

(12eiii)
It is unstable.
======================== 0 %
Happy
0 % 