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WAEC Physics Answers 2023 – Tuesday 23rd May 2023

WAEC Physics Answers 2023: Are you a 2023 Waec candidate searching for 100% verified questions and answers for both objective and essay that will guarantee you A or B in the WAEC Physics 2023 examination? look no further, we have all it takes to make you score A in this Waec examination.

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Waec Physics Answers 2023 for 23rd May 2023 (Objective and Essay): Access free live 2023 WAEC May/June Physics Essay (Theory) and Objectives (OBJ) Questions and Answers for school candidates, without any charges! Join the WAEC May/June Free Physics Questions and Answers EXPO on the 23rd of May, 2023. Join the WAEC May/June 2023 FREE Physics ANSWER ROOM for school candidates now!

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PHYSICS OBJ

01-10: CBAACCBCBB
11-20: DBDDCDACAA
21-30: BDCDADDCCB
31-40: CBBABDCBBC
41-50: CDDCACCCCD

COMPLETED

NUMBER 4
(4)
(PICK ANY THREE)
(i) Geostationary satellites orbit the Earth directly above the equator in a geosynchronous orbit. They remain fixed relative to a specific point on the Earth’s surface WHILE Polar satellites orbit the Earth in a polar orbit, passing over or near the North and South poles on each orbit.

(ii) Geostationary satellites have an inclination of 0 degrees, meaning their orbital plane aligns with the equatorial plane WHILE Polar satellites have a high inclination angle (usually around 90 degrees) to achieve polar orbits.

(iii) Geostationary satellites provide coverage over a large portion of the Earth’s surface (typically one-third to half) but are limited to a specific longitude range WHILE Polar satellites provide global coverage, as they pass over or near the poles on each orbit, covering the entire Earth’s surface over time.

(iv) Geostationary satellites observe the Earth from a fixed position, providing a continuous view of a specific region WHILE Polar satellites observe the Earth from a changing perspective as they orbit, resulting in different views of the Earth’s surface with each pass.

(v) Geostationary satellites primarily collect data for meteorological observations, weather forecasting, and telecommunications WHILE Polar satellites collect data for a wide range of applications, including weather monitoring, climate research, environmental monitoring, and scientific research.

(vi) Geostationary satellites are positioned at an altitude of approximately 35,786 kilometers (22,236 miles) above the Earth’s surface WHILE Polar satellites generally operate at lower altitudes, typically between 700 and 1,500 kilometers (435 to 932 miles.

(vii) Geostationary satellites provide a continuous stream of data and observations for a specific region, allowing for real-time monitoring WHILE Polar satellites have a lower data refresh rate since they pass over a given area less frequently. However, they provide a broader coverage over time

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

NUMBER 5

5b)

Wave: Diffraction of light through a narrow slit or around an obstacle is a phenomenon that demonstrates that light behaves as a wave. This is because the diffraction pattern of light shows interference fringes, which are characteristic of wave interference.

Particle: The photoelectric effect is a phenomenon that demonstrates that light behaves as a particle. This is because the photoelectric effect involves the emission of electrons from a metal surface when it is illuminated by light. The energy of the emitted electrons is proportional to the frequency of the light, which is consistent with the idea that light is made up of particles, or photons, with discrete energy levels.

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

NUMBER 7

7(a) Fibre optics is a technology that uses thin, flexible fibres made of glass or plastic to transmit data in the form of light signals over long distances.

7(b) :

– Higher bandwidth: Optical fibres can carry more data over longer distances than copper cables without degradation of the signal. This makes them ideal for high-speed internet connections, video streaming, and other data-intensive applications.
– Immunity to electromagnetic interference: Unlike copper cables, optical fibres are not affected by electromagnetic interference from other devices or sources. This makes them more reliable and less prone to signal loss or distortion.

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

NUMBER 9

No.9a

(i) A pivot is a point around which a lever or other rigid body rotates.
(ii) A load is the weight or force that is being moved or lifted by a machine.
(iii) Efficiency is the ratio of output work to input work, expressed as a percentage.

Waec Physics Answers 2023

Tuesday, 23rd May, 2023

  • Physics 2 (Essay) – 2:00pm – 3:30pm
  • Physics 1 (Objective) – 3:30pm – 4:45pm

Attention all WAEC candidates! Are you prepared for your upcoming Physics exam on May 23rd, 2023? Don’t leave anything to chance – make sure you have the advantage you need to ace this test with our WAEC Physics Answers and Questions for 2023!

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WAEC Areas of Concentration for Physics 2023/2024

Listed below are the areas of concentration and topics that will be covered in the WAEC 2023 Physics exam:

  1. Concepts of matter
  2. Fundamental and derived quantities and units
  3. Position, distance, and displacement
  4. Mass and weight
  5. Time
  6. Fluid at rest
  7. Motion
  8. Speed and velocity
  9. Rectilinear acceleration
  10. Scalars and vectors
  11. Equilibrium of forces
  12. Simple harmonic motion
  13. Newton’s laws of motion
  14. Energy
  15. Work, Energy and Power
  16. Heat Energy
  17. Production and propagation of waves
  18. Types of waves
  19. Properties of waves
  20. Light waves
  21. Electromagnetic waves
  22. Sound Waves
  23. Description property of fields
  24. Gravitational field
  25. Electric Field
  26. Magnetic field
  27. Electromagnetic field
  28. Simple a.c. circuits
  29. Structure of the atom
  30. Structure of the nucleus
  31. Wave-particle paradox

Tips for Candidates to Excel in the WAEC Physics Exam

  1. Start Early: Do not wait until the last minute to start preparing for the exam. Start early, make a study plan, and stick to it. This will give you enough time to cover all the topics on the list and revise your notes.
  2. Seek Help When Necessary: If you encounter difficulties with any of the topics, do not hesitate to seek help from your teachers or tutors. They are there to help you understand the subject matter and will be more than willing to assist you.
  3. Practice Past Questions: Practicing past questions is an excellent way to prepare for the exam. It gives you an idea of the type of questions that will be asked and helps you to identify areas where you need to improve.
  4. Stay Focused and Disciplined: Avoid distractions and stay focused on your studies. It is essential to stay disciplined and manage your time effectively. This will ensure that you cover all the topics on the list and are adequately prepared for the exam.
  5. Stay Positive: Finally, stay positive and believe in yourself. Remember that you have prepared adequately, and you are capable of performing well in the exam. Stay calm and confident during the exam, and you will do just fine.

Past Physics Answers

PAPER 1
(OBJECTIVES)

  1. To rectify a current simply means to
    A. increase the magnitude of the current.
    B. measure the current using an ammeter.
    C. convert the current from a.c to d.c.
    D. connect more resistors in series in a circuit.
  2. Which of the following statements is correct about laser light?
    A. It converges at all times.
    B. It diverges at all times.
    C. It converges and diverges at different intervals.
    D. It neither diverges nor converges.
  3. X Y Z
    0 0 0
    1 0 0
    0 1 0
    1 1 1

The truth table above represents a two input X, Y and a Z output device(s). The device(s) is
A. AND gate.
B. NAND gate.
C. OR gate.
D. AND gate + inverter.

  1. The dimensions of pressure is
    A. ML-1T-2
    B. ML T-2
    C. ML3T-2
    D. ML-1T
  1. The recoil of a gun after it has been shot is a demonstration of
    A. the principle of conservation of energy.
    B. Newton’s second law of motion.
    C. the principle of conservation of matter.
    D. Newton’s third law of motion.
  2. Two standard resistors of magnitudes 10 and 5 are connected in series to a series
    arrangement of secondary cells. If the current through the 10 resistor is 5A, the
    current through the 5 resistor is
    A. 2.5 A
    B. 5.0 A
    C. 7.5 A
    D. 10.0 A.

SECTION A
(Short Structures)

  1. (a) How does an active satellite differ from a passive satellite?
    (b) State one use of an active satellite. [3 marks ]
  2. (a) Draw a two input Nand gate.
    (b) State its truth table. [3 marks ]

SECTION B
(Essay)

  1. (a) Define a photoelectron. [ 2 marks ]

(b) The equation below represents a radioactive decay

(i) What are the values of x and p? [2 marks
(ii) State the conservation principles used in deriving the values of x and p. [2 marks ]
(c) (i) What is meant by the half life of a radioactive substance?
(ii) A radioactive substance has a half-life of 3.8 days.
Calculate its decay constant. [5 marks]
(d) State one function of each of the following materials utilized by an
operating nuclear reactor:
(i) Boron rods,
(ii) U235,
(iii) Graphite moderators [3 marks]

(1bii)
Weight in Air = 2.9N
Weight in water= 1.3N
Relative Density= ?
Relative Density = Weight in Air/ Change in weight of water
R.D = 2.9/(2.9 – 1.3)
= 2.9/1.6 = 1.8125
Relative Density = 1.813
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
(3vi)
[TABULATE]

R(Ω): 1.0 || 2.0 || 3.0 || 4.0 || 5.0

l (cm): 60.00 || 50.0 || 45.0 || 42.9 || 41.2

(100 – l)cm: 40.0 || 50.0 || 54.5 || 57.1 || 58.8

Z=(100 – l)/l : 0.6667 || 1.0000 || 1.2000 || 1.3310 || 1.4272

R-¹(Ω-¹): 1.0000 || 0.5000 || 0.3333 || 0.2500 || 0.2000

(3vi)
GRAPH

(3viii)
Slope, S = ΔZ/ΔR-¹
S = (0.84 – 1.28)/(0.6 – 0.29)
S = -0.44/0.31
S = -1.419
Intercept, c = 1.68

(3ix)
K = c-¹ = (1.68)-¹
K = 0.595

(3x)
(i) I avoided parallex error when reading the potentiometer
(ii) I ensured tight connections

(3bi)
A possible reason is wrong circuit connection
Solution: Check connections and ensure they are all tight.

(3bii)
I – Voltage
II – Current.

(3)

1bi)
The period of an oscillatory motion is the time taken for a body to complete one oscillation.

(1bii)
-TABULATE-
(i)Mass is a scalar quantity WHILE weight is a vector quantity.
(ii)The SI unit of mass is the kilogram (kg) WHILE the SI unit of weight is Newton (N).
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°′
(3av)
Tabulate

S/N ||1 || 2 || 3 || 4 || 5

d(cm) || 80.00|| 70.00 || 50.00 || 40.00 || 30.00

I(A) || 0.30||0.35 || 0.40|| 0.45||0.50 ||

(I)^-¹ ||3.333 ||2.857 ||2.500 || 2.222||2.000 ||

(3avii)
from the graph, Slope s, = ∆d/∆I^-¹
= 71 – 20 ÷ 3 – 1.75
= 51/1.25
= 40.8cmA
Or 0.408Am

(3aviii)
(i) I ensured tight connections
(ii) I avoided error due to parallax while reading the ammeter

(3bi)
(i) Straight at the magnetic field
(ii) The number of turns of it’s coils

(3bii)
Given:
Diameter d, = 0.6cm = 6×10^-³m
Resistivity, e = 1.0×10^-6 Ωm
Resistance R = 4Ω

Using R = eL/A
L=RA/e
= Rπd²/4e
= 4 × 22/7 × 36 × 10^-6 ÷ 4 × 1 × 10^-⁶
= 22/7 × 36
= 113.14m
:. L = 113.14metres

8ai) Work is the measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement.

(8aii) Energy is defined as the “ability to do work, which is the ability to exert a force causing displacement of an object.”

8d)At constant pressure, the number of collision per unit time is constant. An increase in temperature causes average Kinetic energy increases. The molecules then move faster. The volume therefore increases so as to maintain the number of collision per unit time.

(9ai) The momentum of a body is the product of its mass and its velocity.

(9aii) The total momentum of a system before collision equals to the total momentum after collision provided no external force acts. OR In a system of colliding bodies the total momentum is constant provided no external force acts.

9b) (b) Mp Up + Mq Uq = Mp Vp + Mq Vq

0.25 Up + 0 = 0.25 x 23 Up + 0.25 x 2; 0.25

(Up – 2/3 Up) = 0.50; 1/3

Up = 0.500.25 = 2m/s

Up = 2 x 3 = 6m/s

(9c) Newton’s second law of motion states that the time rate of change of linear momentum is directly proportional to the force causing the change and the change takes place in the direction of the force

(9d) Resultant force on the vehicle = 9600 – 200 = 7400N.

But F = ma

a = Fm=74003400 = 2.18ms−2 or 2.2ms−2

(10b)

Heat lost = 0.024 x 400 (230 – T).

Heat gained by water = 0.054 x 4200 (T – 31)

Heat gained by calorimeter = 0.060 x 400 (T – 31)

Heat lost = Heat gained = 0.024 x 400 (230 – T) = 0.054 x 4200

(T – 31) + 0.060 x 400 (T – 31) T = 38.34°C

PHYSICS OBJ:
1-10: DCCAACDCCC
11-20: BDACABDCCB
21-30: ACBACCCBAD
31-40: BDDBAACCCA
41-50: CAAABBCBBC

loading…

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PLEASE NOTE: asteric(*) MEANS multiplication(x).

(1)
Original length Lo = 3m
Extension e = 310 = 3*3 = 9m
F×e
F = ke where k = 982.3NM-²
F = 982.3*9 = 8,840.7N
The five that produce the extension F = 8.840.7N

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

(2a)
Metal flat plate: It is a collector of heat that converts the radiant solar energy from the sun into heat energy by the process of greenhouse effect.

(2b)
Thermal Insulator: It prevents loss of accumulated heat in the solar collector and thereby improves its absorption efficiency.

(2c)
Tubes: They are use to capture almost all the sun’s energy. It is the most effecient solar collector
====================

(4)

V∝ Tx μy
LT-∝(MLT-²)x(MLT-¹)y
LT-¹∝ m*L*T-²x x m y L-y
LT-¹∝ m(x+y)L(x-y)T-²x
Comparing
X-y=1
X+y=0
-2x=1
From equation 3
X=½
Putting x = ½ into (eq) 2
y=-½
Equation is
V∝ T½ μ-½
V∝ √T/μ
V= k√T/μ
Where k is constant
(6)
Angle at horizontal θ = 30°
Time of light , T = 25sec
Velocity of projection u = ?
But T=2using sinθ/g
25= 2u*sin30/10
25=0.5u/5
0.5u=125
U=125/0.5
=250ms-¹

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

(8a)
A torque or moment of a couple is the product of one of the forces and the perpendicular distance between the lines of action of the two forces.

(8aii)
(i)The two forces must be equal
(ii)The two forces must be opposite parallel
(iii)They must be equal distance between them

(8bi)
Free fall is defined as the motion of the body where gravity is the only force acting upon it.

(8bii)
Height of tower H = 40.0m
Velocity v = 10.0m/s, time t= ? , g=10.0m/s

Using the relation
S= ut + 1/2gt^2
H= ut + 1/2gt^2
40= 10t + 1/2*10t^2
4= t + t^2/2
= t^2 + 2t – 8=0 ( solving quadratically)
= (t+4)(t-2)=0
t+4=0 or t-2=0
t= -4 or t=2

Therefore time to fall on the ground t=2s

Conclusion

Waec Physics Answers 2023 for 23rd May 2023 (Objective and Essay)

f you want to succeed in your Physics exam, you need to prepare thoroughly and have access to accurate and reliable answers and questions. Our WAEC Physics Answers and Questions for 2023 offer just that. With our expertly crafted material and our unbeatable price of just N600, you can give yourself the best possible chance of success in your upcoming exam.

Examking.net, Examkey.net, Examplaza, Examgod.net, ceebook, noniwap, noniexpo, examfada, sureanswer, expotab, runsmaster, examgrand, runslink, expobite.

Don’t wait until it’s too late – Subscribe now and take advantage of this amazing opportunity to excel in your Physics exam. We are confident that our answers and questions will help you achieve your goals and secure your future.

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