V1=10 , v2=16 D=7.49/6
M= (27.61-20.12)g =1.248g/cm^3
=7.49g

Number of particles/molecules increace reducing time between collision.

M1u1 +m2u2 = (m1+m2)v
400 *3)+0=(400+m2)2.5
1200 = 1000+2.5m²
M2=80Kg

Position of cog 1mk
Area of the base 1mk

Reduce the speed of flow of the liquid

sLower (decrease) the temperature;
Metals contract more than glass for same temperature decrease ;

Upthrust = δvg = 5.5 x 800 x 10 = 44N
δ = 𝑚m/v𝑚𝑣𝑣 = 4.4/20 = 0.22kg

W =2𝜋πf
= 2 x 3.142 x 43S^-1

= 270.2 S^-1

V = rw ✓1
= 10/100 m x 270.2S-1
= 27.02m/s ✓1

(a)
(b) The difference between the adhesive and cohesive force produce resistant force which can hold a given mass of water above the level in the beaker. For mass lifted to be equal level the level must be higher in the narrow tube.

Absorbs latent heat vaporisation from the hand

Pmin =Force/Amax =6.0/0.02 = 300N/M^2

(a) The pressure of a fixed mass of gas is directly proportional to its absolute temperature provided volume is kept constant.
(1 Mark)
(b) (i) Sulphuric acid index (1 Mark)
(ii) A drying agent (1 Mark)
Indicate the volume of gas
(iii)
− The apparatus are set up as shown and the water bath heated
− The temperature and volume /length of index is recorded at regular intervals of time
− A graph of volume of versus absolute temperature is drawn and graph analysed.
− A straight line cutting the temperature axis at about -273K is obtained; hence volume is directly proportional to absolute temperature. (3 Marks)
(c) P₁V₁/ T₁ = P₂V₂/ T₂ (2 Marks)
1.5 X 10⁵ X 1.6 / 28⁵ = 1.0 X 10³ X V₂ / 27³
V₂ = 1.5 X 10⁵ X 1.6 X 27³ /(2.85 X 1.0 X 10³)
= 229.89 m³

.

(a) For a helical spring or any other elastic material, the extension of a string is directly proportional to the force applied, so long as the elastic limit is not exceeded (1mark)
(b) (i) Gradient = (3– 1)/(12 – 4) (2 marks)
= 0.25Ncm^-1
(ii) In parallel arrangement, k = 25 X 2 = 50 Ncm^-1
F = k e = 50 X 0.1
= 5Ncm^-1

(a) A body remains in its state of rest or uniform motion unless acted upon by an external force. (1 Mark)
(b) a =(v − u)/t
𝑡𝑡 = (30−0)/10
= 3m/s² (2 Mark)
(c) (i) Ft = mv – mu (3 Marks)
= 800 (0 – 15)
= -12000
I = 12000kgm/s
(ii) F = (mv −𝑚mu)/t
𝑡𝑡
= 12000
0.4 = 30,000N

(a)
(i) Valve Q opens due to high atmospheric pressure on the water while valve P closes due to its weight and that of the water above it. ✓
(ii) Valve Q closes due to its weight and pressure of water above the piston while valve P opens due to the pressure of water below it. 2mk
(iii) Can raise water to height greater than 10m. ✓
Unlike the lift pump, the flow of water out of spout is continuous. ✓
(b) (i) Work done = 108 x 3.2✓
= 345.6J✓
(ii) M.A. = L/E
= 108/60✓

= 1.8✓
(iii) Due to:
(i) Friction between the moving parts of the machine or
ii) Weight of the moving parts of the machine. ✓ (award 1mk for either