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       #Post#: 262--------------------------------------------------
       100% CORRECT FREE WAEC GCE PHYSICS OBJECTIVE AND ESSAY ANSWERS
       By: Ebenezer Date: September 12, 2013, 7:02 am
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       #Post#: 263--------------------------------------------------
       Re: 100% CORRECT FREE WAEC GCE PHYSICS OBJECTIVE AND ESSAY ANSWE
       RS
       By: Ebenezer Date: September 12, 2013, 9:20 am
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       Verified PHYSICS OBJ
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       THEORY
       (6a)
       - density of the substance
       -temperature of the substance
       (6b)
       -absorbtion of ink with blotting paper
       -rising of oil in lamp wick
       (1)
       Given, tita = 30degree , speed = 25cm/s
       (a)
       Time of flight (T) = 2Usin(tita)/g
       = 2*250*sin30/10
       = 250/10
       =25s
       (b)
       Max - lenght =U^2 sin^2 tita/ 2g
       Recall V^2=U^2 - 2gh
       The speed is zero at max height
       (4)
       Given, current(I) = 10A
       Time (T) = 1hr = 1*3600 = 3600s
       Z= 1.2*10^-7kgc^-1
       m = Zit
       m = 1.2*10^-7*10*3600
       m= 4.32*10^-3kg
       (7)
       i. Brownian motion
       ii. Tidal movement
       (10)
       i. Electron Diffraction
       ii. Inference
       iii. Reflection
       #Post#: 264--------------------------------------------------
       Re: 100% CORRECT FREE WAEC GCE PHYSICS OBJECTIVE AND ESSAY ANSWE
       RS
       By: Ebenezer Date: September 12, 2013, 9:35 am
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       WAEC GCE PHYSICS;
       Verified PHYSICS OBJ:
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       21-30: bcddadcada
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       41-50: abbcdaabac
       NOTE: tita is tita sign.
       ^ Means RAIST TO POWER.
       Degree means degree sign.
       (1) Given, tita = 30degree , speed = 25cm/s
       (a)
       Time of flight (T) = 2Usin(tita)/g
       = 2*250*sin30/10
       = 250/10
       =25s
       (b)
       Max - lenght =U^2 sin^2 tita/ 2g
       Recall V^2=U^2 - 2gh
       The speed is zero at max height
       (4)
       Given, current(I) = 10A
       Time (T) = 1hr = 1*3600 = 3600s
       Z= 1.2*10^-7kgc^-1
       m = Zit
       m = 1.2*10^-7*10*3600
       m= 4.32*10^-3kg
       (6a)
       - density of the substance
       -temperature of the substance
       (6b)
       -absorbtion of ink with blotting paper
       -rising of oil in lamp wick
       (12ai)it causes change of state of matter.
       (ii)it increases the temperature of a matter.
       (iii)it increases the average kinetic theory of the molecules in
       the matter
       (12bi)when a boiling water is poured into a thick glass,the
       inner of the glass suddenly become hotter than the outer layer
       causing irregular expansion hence creating a non-uniform
       pressure in the structure of the glass material hence it cracks.
       (12bii)the claws of a cat is naturally designed to create the
       required frictional force necessary for locomotion as a result
       of grip.but on a polished floor where gripping is
       impossible,hence,frictional force is minimal.
       #Post#: 265--------------------------------------------------
       Re: 100% CORRECT FREE WAEC GCE PHYSICS OBJECTIVE AND ESSAY ANSWE
       RS
       By: Ebenezer Date: September 12, 2013, 9:39 am
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       (4)
       Given, current(I) = 10A
       Time (T) = 1hr = 1*3600 = 3600s
       Z= 1.2*10^-7kgc^-1
       m = Zit
       m = 1.2*10^-7*10*3600
       m= 4.32*10^-3kg
       (7)
       i. Brownian motion
       ii. Tidal movement
       (10)
       i. Electron Diffraction
       ii. Inference
       iii. Reflection
       (12ai)it causes change of state of matter.
       (ii)it increases the temperature of a matter.
       (iii)it increases the average kinetic theory of the
       molecules in the matter
       the pressuere law states that the pressuere of a
       fixed mass of a gas at constant volume is directly
       proportional to the temperature.
       (12bi)
       this is because of expansion of the glass cup because
       the temperature of boiling water is greater than the
       glass cup.
       (12bii)
       this is because of the absence of frictional force on
       highly polished floor.
       (12cii)
       given T1+20degrees=20+273=293k
       T2=?
       P1=76cmHg
       p2=69cmHg
       p1/T1:-p2/T2
       76/293=69/T2
       76*T2=69*293
       T2=69*293=266k
       T2=-7degree
       (13a)
       (i)
       Object distance(u)= a
       Image distance(v) = b
       f^2 = ab
       (ii)
       1/a+1/b = 1/f
       1/16 - 1/25 = 1/f
       1/f = 25-16/ 400
       1/f = 9/400
       f = 400/9 = 44.4cm
       (iii)
       Magnification(M) = v/u =b/a
       M= 25/16 = 1.5625
       M=1.6
       (13b)
       (i) Principal focus is the point mid-way between the
       pole of a curved mirror and the centre of curvature in
       which all rays that are parallel and close to the
       principal axis appear to converge or diverge
       (ii) Optical Centre: this is the centre of the lens
       (iii) Focal lenght: This is the distance between the pole
       and the principal focus of mirror or lens
       (15a) This is because neutron is very light when
       compared with alpha particle
       (15b)
       i. They are negatively charged
       ii. They can cause florescent
       iii. They can ionize gases at high tempature
       iv. They are fast moving electron
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