WAEC 2019 PHYSICS THEORY-OBJ EXPO

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(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

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(2a)

P - Initial velocity of the projectile

B - Angle of projection

H - Maximum height reached by the projectile

R - Range or the total horizontal distance

covered by the

projectile

(2b)

R=P^2 sin 2β/g

at maximum range sin2β=1

Rmax=P^2/g

This happens at β = 45degree

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(3)

(i)reflection

(ii)refraction

(iii)diffractio

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(4)

(i)Magnetic Declination

(ii)Horizontal Component of Earth’s Magnetic

Field

(iii)The angle of Dip or Magnetic Inclination

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8ai Terminal velocity is the highest velocity attainable by an object as it falls through a fluid (air is the most common example). It occurs when the sum of the drag force ( Fd ) and the buoyancy is equal to the downward force of gravity (F G ) acting on the object. Since the net force on the object is zero, the object has zero

acceleration . [1]

In fluid dynamics , an object is moving at its termin

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 (9a) Latent heat is energy released or absorbed, by a body or a thermodynamic system, during a constant-temperature process. 

(9aii)

 (i)surface area in contact with atmosphere 

(ii)nature of the liquid

 (9bi) 

Clay is poorer conductor of heat than plastic 

(9bii) An increase in pressure in the cooker increase the ambient temperature in the cooker hence the food cooks faster

 (9c)

 (i)Change in the state of a substance (ii)Increase in temperature

(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/

(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°

(5b)

Holes are spaces left by electrons in an atom in a crystal lattice. They are positively charged.

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11) u= 5ms-1                                   u= 6ms-1       A= 1ms-2                                     a= 3ms-2      p→                                                        ←a  M—————————————N                                 51m    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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