WAEC 2019 Physics Answers Now Available Here

(1)
Mass = 20g = (20/100)kg = 0.02kg
Extension = 5cm = (5/100)m = 0.05m
K = 200N/
V = ?
1/2mv² = 1/2ke²
V = √ke²/m = √200×0.05×0.05/0.02
V = √25
V = 5m/s

(3)
(i) magnetic declination
(ii) magnetic inclination
(iii) horizontal components

(5)
Draw the diagram

Using cosine law
Cos∅ = 16²+10²-14²/2(16)(10)

Cos∅ = 256 + 100 – 196/320

Cos∅ = 160/320
Cos∅ = 0.5
∅ = cos-¹(0.5)
∅ = 60°

Angle between 10N and 16N
= 180 – ∅ (sum of angles on a straight line)
= 180 – 60
=120°

8ai) Terminal Velocity: this is the constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is falling prevents further acceleration.

(9b)
(i) Clay is proper conductor of heat than plastic

(ii) An in pressure in the cooker increases the ambient temperature in the cooker, hence the food cooks faster

(9c)
Change in the state of a substance
Increase in temperature

9ai)
Latent heat , energy absorbed or released by a substance during a change in its physical state (phase) that occurs without changing its temperature . The latent heat associated with melting a solid or freezing a liquid is called the heat of fusion ; that associated with vaporizing a liquid or a solid or condensing a vapour is called the heat of vaporization . The latent heat is normally expressed as the amount of heat (in units of joules or calories ) per mole or unit mass of the substance undergoing a change of state.

(9di)
Heat supplied = m1
lvt = m1
12*40*1400=1.5*1
L=12*40*1400/1.5
L=672000/1.5= 448,000J/kg
Laten heat of fusion of the body = 4.48*10^5J/Kg
(9dii)
lvt=mc D tita
12*40*72=1.5*c*60
C=12*40*72/1.5*60
C=384J/Kg/K
Specific heat capacity of the body =384J/Kg/K

No 10a

Atomic spactra is the spectra of frequencies of electromagnetic radiation emitted or absorbed during transitions of electrons between energy levels within an atom.

11a)t when particles are 30 meters apart
S= ut + ½ at2
Sp= st + ½ t2
Sa = 6t + 1/2 (3)t2
Where sp and sa are distances covered by particles respectively at any time t
At time, t= 1.5s
Sp = 5(1.5) + ½(1.5)2 =8.625m
sa = 6(1.5) + ½ (3)(1.5)2 = 12.375m
sp + sa = 8.625 + 12.375
= 21m
Distance between particles = 51 – 21 = 30m
At time t = 1.5s, the particles will be 30 meters apart.

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