Nabteb 2017 Physics (Essay & Obj) Answers – May/June Expo
Nabteb 2017 Physics (Essay & Obj) Answers – May/June Expo
100% Real and Correct 2017 NABTEB SSCE Physics Obj Answer
01-10: DACDCBDCDA 11-20: AACDBADACA 21-30: ABDBAAABAC 31-40: BBBADBBDDB 41-50: AACCBACDDD100% Real and Correct 2017 NABTEB SSCE Physics Essay Answer
1ai) Force (F)=ma a=v/t =d/t/t =d/t^2 a=LT^-2 F=ma=MLT^-21aii) Momentum(M)=mv =M*LT^-1 =MLT^-1
1aiii) Density=mass/volume =M/L^2 =ML^-2 1bi) Relative density is the ratio of the density of substance to the density of water
1bii) DRAW THE DIAGRAM Let Wo represent weight in air Wl represent weight in water Relative density=weight in air/weight in water relative density=Wo/(Wl-Wo)
1c) A force(F)=60N Extension(e)=0.3mm=0.3*10^-3 Original lenght Lo=2m Diameter(d)=1.5mm=1.5*10^-3m Area=pie*d^2/4 =3.142*(1.5*10^-3)^2/4 =7.0695*10^-6/4 A=1.77*10^-6m^2
(i)Stress=F/A =60/1.77*10^-6 =33.9*10^6 =3.39*10^7Nm^-2
(ii)Strain=e/Lo =0.3*10^-3/2 =0.15*10^-3 Strain=1.5*10^-4
(iii)Young modulus=stress/strain =3.39*10^7/1.5*10^-4 =2.26*10^11Nm^-2 ============================== 2ai) -It changes the shape of the body -It increases the temperature -It increases the length of the body -It increases the volume of the body
2aii) -Volume -Potential difference -Pressure -Resistance
2bi) An increase in pressure of a substance increases the boiling point of the liquid
2bii) Addition of salts to boiling water increases the boiling point of the liquid
II) -Saturated vapour is a vapour in contact with its own liquid while unsaturated vapour is a vapour that is not in contact with its own liquid -Saturated vapour is a vapor that us about to condense while unsaturated vapour is not about to condense -Saturated vapour is affected by pressure whereas unsaturated vapour is not affected by pressure
2c) Note @=tita Mass of copper(m1)=100g Initial temperature(@1)=100C Mass of copper can(m2) =509 Mass of water(m3)=200g Initial temperature(@2)=10C Final temeprature(@3)=? S.H.C of water(C3)=4.2J/gK S.H.C of copper(C1)=4J/gK m1C1(@1-@3)=m2C2(@3-@2)+m3C3(@3-@2) 100*4(100-@3)=50*4(@3-10)+200*4.2(@3-10) 400(100-@3)=200(@3-10)+840(@#-10) DIVIDE THROUGH BY 10 and expand 4000-40@3=20@3-200+84@3+840 5040=144@3 @3=5040/144 =35degrees centigrade =35C
2d) density(e1)=4.8g/cm^3 e2=? @1=20C @2=80C Gamma(Y)=3alpha=3*1.2*10^-5 Y=3.6*10^-5K^-1 Y=(e2-e1)/e1(@2-@1) 3.6*10^-5=(e2-4.8)/4.8(80-20) 1.036*10^3*10^-5=e2-4.8 e2=0.01036+4.8 e2=4.81g/cm^3 ============================== 3a) DRAW THE DIAGRAM
3bi) Electric field potential is the amount of workdone in bringing a positive charge from that point to a point of infinity against the attractive forces of the field
3bii) Electric field intensity is the force per unit test charge in an electric field
3biii) E=f/q =[(1/4*pie*Eo*(q^2/r2)]/q E=1/(4pieEo)*q/r^2 Electric potential(V) V=1/4pieEo*q/r
3ci) d=5*10^-3 V=5*10^4 1cii) V=Ed E=V/d =(5*10^4)/(5*10^-3) E=10^7NC^-1
1ciii) E=F/q 10^7=F/q C=q/V F=1.6*10^-19*10^7 =1.6*10^-12 ============================== 4ai) Mechanical waves are waves that requires a material medium for their propagation eg water waves,waves generated in a spring
4aii) Electromagnetic waves are waves that do not require a material medium for their propagation
4b) Period(T)=2.5*10^-6s Velocity(v)=3.10^8m/s Wavelength(landa)=? f=1/T=1/(2.5*10^-6) f=4*10^5Hz V=f*landa landa=V/f landa=(3*10^8)/(4*10^5) =7.5*10^2m
4ci) -Light must be travelling from optical densed medium to optically less densed medium -The angle of incidence must be greater than critical angle 4cii) -Mirage -Reflecting telescope
4d) Given y=2sin*pie(0.5x-500) y=Asin2pie(ft-x/landa) 2pieft=500piet 2f=500/2 f=250Hz 2piex/landa=0.5xpie 2/landa=0.5 landa=2/0.5 =4m Wavelength=4m Speed(v)=flanda =250*4 =1000m/s
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