PHYSICS OBJ:
1-10: CBBBCACBCB
11-20: CCCABBBDCD
21-30: CBDADCDDDA
31-40: BDDBBCAABC
41-50: ABDABDBCBA
1-10: CBBBCACBCB
11-20: CCCABBBDCD
21-30: CBDADCDDDA
31-40: BDDBBCAABC
41-50: ABDABDBCBA
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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

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 is initial velocity in m/s
β is angle of projection
H is maximum height
R is Range
P is initial velocity in m/s
β is angle of projection
H is maximum height
R is Range
(2b)
R = P²sin2β/g
at maximum range, sin2β = 1
=> Rmax = p²/g
This happens at β = 45degree

R = P²sin2β/g
at maximum range, sin2β = 1
=> Rmax = p²/g
This happens at β = 45degree

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(3)
(i) Photoelectricity
(ii) Thermonic equation
(iii) Compton effect
(i) Photoelectricity
(ii) Thermonic equation
(iii) Compton effect
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(4)
(i) magnetic declination
(ii) magnetic inclination
(iii) horizontal components
(i) magnetic declination
(ii) magnetic inclination
(iii) horizontal components
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(5a)
Free eletcron: These are electrons which are not attached to an atom. They are mobile electrons which are free for bonding.
Free eletcron: These are electrons which are not attached to an atom. They are mobile electrons which are free for bonding.
(5b)
Holes are spaces left by electrons in an atom in a crystal lattice. They are positively charged.
Holes are spaces left by electrons in an atom in a crystal lattice. They are positively charged.
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(6a)
Principle of operation of fiber optics state that the density of transparent glass and the refractive index is greater than the density of the two mirrors. It makes use of the principle of the total internal reflection.
Principle of operation of fiber optics state that the density of transparent glass and the refractive index is greater than the density of the two mirrors. It makes use of the principle of the total internal reflection.
(6b)
(i) It is used to view deep down the throat of a patient
(ii) It is used to conveying information
(i) It is used to view deep down the throat of a patient
(ii) It is used to conveying information
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(8)


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(9ai)
Latent heat is energy released or absorbed, by a body or a thermodynamic system, during a constant-temperature process.
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
(i)surface area in contact with atmosphere
(ii)nature of the liquid
(9bi)
The rate of absorption of heat by clay pot is less than the rate of absorption of heat by plastics. Also, water gets evaporated in a clay pot which the evaporation causes cooling; so water in a clay pot remains cool
The rate of absorption of heat by clay pot is less than the rate of absorption of heat by plastics. Also, water gets evaporated in a clay pot which the evaporation causes cooling; so water in a clay pot remains cool
(9bii)
An increase in pressure in the cooker increase the ambient temperature in the cooker hence the food cooks faster
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
(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
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

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

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(10)


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(11ai)
Force is anything that causes motion.Its S.I unit is Newton(N) or Kgm/s^2
Force is anything that causes motion.Its S.I unit is Newton(N) or Kgm/s^2
(11aii)
(i) Static friction
(ii) Dynamic friction
(i) Static friction
(ii) Dynamic friction
(11bi)
DRAW THE GRAPH
DRAW THE GRAPH
(11bii)
Deceleration
during last 5s=18/5=3.6m/s^2
Deceleration
during last 5s=18/5=3.6m/s^2
(11biii)
Total distance=S1+S2+S3+S4
=(30*20)+1/2(18+30)(6)+(18*10)=(1/2*5*18)
=600+144+180+45
=969metres

Total distance=S1+S2+S3+S4
=(30*20)+1/2(18+30)(6)+(18*10)=(1/2*5*18)
=600+144+180+45
=969metres

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