WAEC 2021 Physics Obj And Essay Answer – Aug/Sept Expo

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PHYSICS OBJ:

01-10: AAABACCDDB

11-20: ACCDCBCBAC

21-30: CCCBBDDBBB

31-40: ACBBCCCAAD

41-50: BBADCCBAAC

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Waec Physics Answers

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(1)

[img]https://i.imgur.com/spasrPD.jpg[/img]

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(2a)

(i)Navigation satellites e.g global position system - GPS

(ii) Communication satellites eg ANIK

(2b)

V = 2πR/T

For every second,

Speed, V = 2πR

V = 2 × 22/7 × 6300

V = 39,600km/s

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(3)

V=u cosθ -gt

V=40*cos30-10*1

V=34.64-10

V=24.64m/s

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(4a)

A safely separates the incoming AC voltage from the secondary side which is adjusted to control the output voltage of the rectifier 

(4b)

B smothers the pulsating output voltage

(4)

[img]https://i.imgur.com/JYxewWw.jpg[/img]

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(5a)

Wave particles duality is the concept in quantum mechanics that every particles or quantum entitle may be described as either a particle or wave.

(5b)

λ=h/MV

4.2*10^-¹¹=h/1.6*10^-²³

h=4.2*10^-¹¹ * 1.6*10^-²³

h=6.72*10^-³⁴

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(6)

(i)reflection

(ii)refraction

(iii)diffraction

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(8ai)

Hooke's law states that the extension e produced by an elastic material is directly proportional to the applied force provided the elastic unit is not exceeded.

(8aii)

(i) Given extension, e = I2 - I1 = 0.75 - 0.20 = 0.55m

Force applied, Fe = (1.95 - 0.30) × 10

= 1.65 × 10 = 16.5N

Force constant , K = F/R = 16.5/0.55 = 30N/m

(ii) Using F = K (I1 - Io)

m1g = K ( I1 - Io)

= 0.30 × 10 = 30(0.20 - Io)

= 0.10 = 0.20 - Io

Io = 0.20 - 0.10

Io = 0.10m

Length of spring when it was unloaded = 0.10m

(8bi)

Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration.

(8bii)

(i) Temperature

(ii) Size of particles

(8biii)

Rate of diffusion is inversely proportional to the square root of density under given conditions of temperature and pressure. ie R ∝ 1/√d

(8c)

T = circumference of the orbit/orbit velocity

T = 2πR/v

But V = √GM/R

T = 2π√R³/GM

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(11ai)

Root measurement value of an alternating current is the value of A.C current that has the same heating affect as a D.C current

(11aii)

Impedance is the affective resistance of an electric circuit arising from the combined affects of ohmic resistance and reactance .( ie resistors, indicators and capacitors)

(11b)

Current in circuit , I = 60/120 = 0.5A

Voltage across device , Vr = 120V

Voltage across capacitor , Vc = √V² - Vr

Vc = √240² - 120²

Vc = 207.846V

Ic = I = 0.5

Xc = Vc/Ic = 207.846/0.5 = 415.7Ω

Capacitance, C = Xc/2πf = 415.7/2*3.142*50 = 1.323F

(11ci)

The capacitance of a capacitor is the ratio of the amount of charge on its plates to the potential difference between them. ie C = ∑/V

(11cii)

(i) Charge in both capacitors are the same

∑1 = Q2

ie C1V1 = C2V2 .........(1)

Total capacitance ,C = C1C2/C1+C2

C = C2/1+C2/C1 = C2/1+V/V2

C = C2V2/V1+V2 = C2V2/2

C= 1/2C2V2

(ii)Voltage across, V1 = (C2/C1+C2) × V

V1 = (C1/C1+C2) × V

But C1/C2 = d2/d1 = 5/2

V1 = (1/(5/2+1)) × 2 = 2/7 ×2 = 4/7Volts

Voltage across, C2 : V2 = 2-4/7 = 10/7volts

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(7a)

The scientific principle underlying the operation of a fibre optics is the principle of total internal reflection by which light signals can be transmitted from one place to another with a negligible loss of energy.

(7bi)

Core: The core is the light transmission area of the fibre, either glass or plastic cladding. The function of the cladding is to provide a lower refractive index at the core interface in order to cause reflection within the core so as to enable light waves transmitted through the fibre.

(7bii)

Cladding: It functions as a gadget that provides a clear refractive index at the core interface in order to cause reflection within the core so that light waves are transmitted through the fibre.

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(9ai)

The dew point is the temperature to which air must be cooled to become saturated with water vapor.

(9aii)

Dew is a condensation phenomenon. The water vapour when cooled below its dew point and now comes in contact with a colder surface forms a dew. The metal part of the bicycle tends to be colder than rubber part of the bicycle because the metal surface get cold easily.

(9bi)

The specific heat capacity of copper is 400 JKg−¹K−¹ means that 400 J of heat energy is required to raise or lower the temperature of 1kg of a piece of copper by 1 kelvin.

(9bii)

Qp = Qq

Cp/Cq =3

Mp/Mq=¹/²

MpCpΔQp =MqCqΔQq

Δθp/Δθp=Mq/Mp Cq/Cq

=2*¹/³

Δθp ; Δθq =2:3

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