PHYSICS REAL ANSWERS
PHYSICS REAL ANSWERS
PHYSIC OBJ
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ile is a body undergoing a motion
under the combined effect of horizontal and
vertical forces
(1b) The horizontal components of the velocity remains the same because the body undergoes a uniform motion ==================================
(2) Angle of projection(tita)=30degrees velocity(u)=150m/s At greatest height, v=0 t=Usin*tita/g t=(150*sin30degrees)/10 =150*0.5/10 t=7.5s =================================
(3) Given original lenght(Lo)=2.00m radius(r)=1.0mm=1*10^-3m Extension(e)=25mm=2*10^-3m Force(f)=10^3N stress =force/area Area
(a)pie*R^2 =22/7*(10^-3)^2 3.142*10^16m^2 Stress=10^3/3.142*10^-6 =3.18*10^-1*10^3*10^6 =3.18*10^8Nm^-2 Strain=Extension/original length =25*10^-3/2 =1.25*10^-2 Young modulus(Y)=3.18*10^8/1.25*10^-2 Y-2.54*10^10Nm^-2 =================================
(4) -By wind -By increasing the cup surface area -By reducing the temperature ===================================
(7a) De brogile equation states that landa=h/mv where landa=wavelenght of light mv=p=momentum h=planck's constant
(7b) The equation is used to explain the wave- nature of matter ==============================
(8a
i) Uniform acceleration occurs when there is an equal changein n velocity with equal time interval no matter how small the time interval may be (8aii) V=wr V=linear velocity w=angular velocity r=radius
(8b) R= resultant R=root(F1^2+F2^2) R=root(30^2+40^2) R=root(2500) R=50N
(8c) This is done to increase the surface friction in water (8di) Given;mass(m)=20kg angle of inclineation(tita)=30degrees F=UR F=Umgcos*tita ma*sin*tita=Umg cos tita U=tan*tita U=tan30 U=0.51
(8dii) Frictional force(Fr) Fr=Umgcos*tita Fr=0.51*20*10*c0s30 Fr=90.9N
(10ai) Resonance ia a phenomenon whereby a vibrating body sets another bodies vibrating all oscillating at d same neutal frequency
(10aii) (1)Radio-recieving circuit (2)Sets of bob of pendulum
10aiii) Loudness is the sensation of sound reaching d ear while intensity is d ratio of d power to d area
(10b) Wavelenght & speed
(10d) Given; focal lenght(f)=18.0m Object distance(u)=90.0m Image height(hi)=1.0m (10di) 1/u+1/v=1/f 1/90+1/v=-1/18 1/v=-1/18-1/90 1/v=(-5-1)/90 1/v=-6/90 v=-90/6 =-15cm
(10dii) v/u=hi/ho 15/90=1.0/h0 ho=90/15 =6cm Lenght of shark is 6cm
(12a) Nucleon number is the sum of proton number and neutron number
(12b) 95=c+a 95=c+2 c=95-2 c=93
(12bi) Number of neutrons in americium-241 is 237-93=144
(12ii) a=2, b=4, c=93
(12ci) Gamma ray is not deflected because it has no charge
(12cii) -It has the highest penetrating power -it has low ionizing power
(12d) N/Nt=(1/2)^t1/2/T 1/32=(1/2)^(110/t1/2) 5=110/t1/2 5t1/2=110 t1/2=110/5 t1/2=22s t1/2=0.693/landa 22=0.693/landa landa=0.693/22 landa=0.0315 landa=0.0315 per year
9a.
Specific latent heart of vaporization can be defined as the quantity of heat required to change the state of a liquid (wata) to gas (vapour) at constant temperature and pressure. 9bi. Renewable energy sources are:those that can be replaced as they are used up. They are continually replenished 9bii. (i) Biomass
(ii) water energy
(iii) wind energy and
(iv) solar enemy. 9c. Tomatoes kept in a moist jute bag stay longer because the quantity od heat required to change the state of solid to liquid is very low as a result of the moisture content of the jute bag. 9d. Given that V=0.28 M3 (i.e m square), (tita)= 25'c (decgree c), Density (p)= 8.00x10 square kg|m sqaure Specific latent heat of fudion of iron=1.38x10 raise to power 5 j|kg, SHC= 460 j|kg|k, tita= 1500'c (degree )
9di. Total mass of the iron filling=density x volume from Density=mass/volume, p= 8.00x10 raise to power 3kg|m raise to power 3, v=0.28m raise to power 3 Therefore: m= 8.00 x10 raise to power 3 kg/ m raise to poer 3 M= 2240kg : 70% of m= 70/100x2240/1 =1568kg answer
9dii. Q1= ml tita,Q2.=MC tita. Qr=Q1+Q2 But 10% of the iron filling =10/100x2240/1=224kg therefore Q1=2241.38x10 raise to power 5x (1500-25)+224x460x(1500) =3091200x1475)+(103040x1475) 4.5592x10 raise to power 10+151984000 =4.575x10 raise to power 10j
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a.project-
(1b) The horizontal components of the velocity remains the same because the body undergoes a uniform motion ==================================
(2) Angle of projection(tita)=30degrees velocity(u)=150m/s At greatest height, v=0 t=Usin*tita/g t=(150*sin30degrees)/10 =150*0.5/10 t=7.5s =================================
(3) Given original lenght(Lo)=2.00m radius(r)=1.0mm=1*10^-3m Extension(e)=25mm=2*10^-3m Force(f)=10^3N stress =force/area Area
(a)pie*R^2 =22/7*(10^-3)^2 3.142*10^16m^2 Stress=10^3/3.142*10^-6 =3.18*10^-1*10^3*10^6 =3.18*10^8Nm^-2 Strain=Extension/original length =25*10^-3/2 =1.25*10^-2 Young modulus(Y)=3.18*10^8/1.25*10^-2 Y-2.54*10^10Nm^-2 =================================
(4) -By wind -By increasing the cup surface area -By reducing the temperature ===================================
(7a) De brogile equation states that landa=h/mv where landa=wavelenght of light mv=p=momentum h=planck's constant
(7b) The equation is used to explain the wave- nature of matter ==============================
(8a
i) Uniform acceleration occurs when there is an equal changein n velocity with equal time interval no matter how small the time interval may be (8aii) V=wr V=linear velocity w=angular velocity r=radius
(8b) R= resultant R=root(F1^2+F2^2) R=root(30^2+40^2) R=root(2500) R=50N
(8c) This is done to increase the surface friction in water (8di) Given;mass(m)=20kg angle of inclineation(tita)=30degrees F=UR F=Umgcos*tita ma*sin*tita=Umg cos tita U=tan*tita U=tan30 U=0.51
(8dii) Frictional force(Fr) Fr=Umgcos*tita Fr=0.51*20*10*c0s30 Fr=90.9N
(10ai) Resonance ia a phenomenon whereby a vibrating body sets another bodies vibrating all oscillating at d same neutal frequency
(10aii) (1)Radio-recieving circuit (2)Sets of bob of pendulum
10aiii) Loudness is the sensation of sound reaching d ear while intensity is d ratio of d power to d area
(10b) Wavelenght & speed
(10d) Given; focal lenght(f)=18.0m Object distance(u)=90.0m Image height(hi)=1.0m (10di) 1/u+1/v=1/f 1/90+1/v=-1/18 1/v=-1/18-1/90 1/v=(-5-1)/90 1/v=-6/90 v=-90/6 =-15cm
(10dii) v/u=hi/ho 15/90=1.0/h0 ho=90/15 =6cm Lenght of shark is 6cm
(12a) Nucleon number is the sum of proton number and neutron number
(12b) 95=c+a 95=c+2 c=95-2 c=93
(12bi) Number of neutrons in americium-241 is 237-93=144
(12ii) a=2, b=4, c=93
(12ci) Gamma ray is not deflected because it has no charge
(12cii) -It has the highest penetrating power -it has low ionizing power
(12d) N/Nt=(1/2)^t1/2/T 1/32=(1/2)^(110/t1/2) 5=110/t1/2 5t1/2=110 t1/2=110/5 t1/2=22s t1/2=0.693/landa 22=0.693/landa landa=0.693/22 landa=0.0315 landa=0.0315 per year
9a.
Specific latent heart of vaporization can be defined as the quantity of heat required to change the state of a liquid (wata) to gas (vapour) at constant temperature and pressure. 9bi. Renewable energy sources are:those that can be replaced as they are used up. They are continually replenished 9bii. (i) Biomass
(ii) water energy
(iii) wind energy and
(iv) solar enemy. 9c. Tomatoes kept in a moist jute bag stay longer because the quantity od heat required to change the state of solid to liquid is very low as a result of the moisture content of the jute bag. 9d. Given that V=0.28 M3 (i.e m square), (tita)= 25'c (decgree c), Density (p)= 8.00x10 square kg|m sqaure Specific latent heat of fudion of iron=1.38x10 raise to power 5 j|kg, SHC= 460 j|kg|k, tita= 1500'c (degree )
9di. Total mass of the iron filling=density x volume from Density=mass/volume, p= 8.00x10 raise to power 3kg|m raise to power 3, v=0.28m raise to power 3 Therefore: m= 8.00 x10 raise to power 3 kg/ m raise to poer 3 M= 2240kg : 70% of m= 70/100x2240/1 =1568kg answer
9dii. Q1= ml tita,Q2.=MC tita. Qr=Q1+Q2 But 10% of the iron filling =10/100x2240/1=224kg therefore Q1=2241.38x10 raise to power 5x (1500-25)+224x460x(1500) =3091200x1475)+(103040x1475) 4.5592x10 raise to power 10+151984000 =4.575x10 raise to power 10j
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