11)a)/\E=mc/\*theta=E=Pt --> t=110*720*12/85*44=254s

11)b)Due to conduction which takes place through the train carriage

12)Dark matter is theorized to be the excess mass in the universe which enables the galaxies to spin at the speeds they do and the orbits of different celestial objects to be the way they are. The absence of dark matter in the universe would decrease the mass of the universe which would cause the mass to be less than the critical density and this would make the fate of the universe to expand forever.

13)a)i)P[1]/T[1]=P[2]/T[2]--> T[2]=P[2]*T[1]/P[1]=(5.6*10^5)*(273+22)/(5.52*10^5)=299K Temp rise=299-273=26K

13)a)ii)As air moves out heat energy will be lost by radiation which will cause the temperature of the air to decrease

13)b)As temperature increases the average kinetic energy of the particles and the momentum of the particles and velocity of the particles also increases leading to greater number of collisions per unit time to greater force exerted on the walls of the container causing pressure to increase.

14)a)137^[54]Ba+0^[-1]B-

14)b)1/2 life=30.2 years Cs-137, A[o]=-lambda*N[o], lambda=ln2/t[1/2]=ln2/30.2*3600*24*365=(7.28*10^-10)s^-1, A[o]=(-7.28*10^-10)*(1.36*10^24)=(9.9*10^14), A=A[o]*e^-lambda*t=(9.9*10^14)*(e^-7.28*10^-10*20*3600*24*365)=(6.25*10^14)

14)c)gamma radiation has higher penetration power than beta radiation therefore it can damage the cells of the body, gamma radiation requires more shielding

15)a)i)T=2pi/w, F=mrw^2=GMm/r^2,w^2=GM/r^3, pi^2/T^2=GM/r^3, T^2=4*pi^2*r^3/GM

15a)ii)T^2=4*pi^2*r^3/GM=4*pi^2*(2.66*10^7)3/6.67*10^-11*6*10^24, T=/|1.86*10^9=43127s, 43127/24*3600=0.50, 1/0.5=2 times

15)b)The actual value would be lower as altitude increase the effective value of g decreases therefore /\E[grav] would be less than the calculated value

16)a)i)Fe-56 has the highest binding energy per nucleon which shows that it cannot do any fusion or fission reactions to increase binding energy therefore its the most stable

16)a)ii)/\m=(1.673*10^-27*26+1.675*10^-27*30)-9.287*10^-26=8.78*10^28, /\E=/\m*c^2=8.78*10^-28*(3*10^8)^2*10^-6/1.6*10^-19=494 MeV, 494/56=8.82MeV

16)a)iii)/\E[[2]He]=1 unit of graph of 2^He/9 units of graph of Fe*8.82=0.98MeV

16)b)As in a fission reaction in the conversion the mass decreases which due to the equation /\E=m*c^2 the lost mass is converted to energy

17)a)Simple harmonic motion occurs when acceleration of an object is directly proportional to displacement from equilibrium position and acceleration acts towards the equilibrium position

17)b)i)Place a fiducial marker at the equilibrium position then time 25-30 complete oscillations and count 1 oscillation when the test tube passes the equilibrium position and the next oscillation when it passes through the equilibrium position again and then divide the time taken for all these oscillations and divide by he number of complete oscillations and repeat this time once again and average the periods

17)b)ii)w=2*pi/T=2*pi/0.57=11.02rads^-1, A=wA=11.02*2*10^-2m=0.22ms^-1

17)b)iii)A graph drawn which is 90 degrees shifted to the left and has higher amplitude and has the same frequency or period

17)b)iv)As the oscillation will become damped and lose energy to resistive forces like friction

18)a)i)lambda[max]*T=2.898*10^-3--> lambda[max]=2.898*10^-3/T=2.898*10^-3/3150=9.2*10^-7m

18)a)ii)L[L]/L[S]=(sigma/sigma)*(4*pi/4*pi)*(35%/1)*(3150/5800)^2=0.0107

18)a)ii)K.E=3/2kT=3/2*1.38*10^-23*3150=6.52*10^-20J

18)b)i)This is due to doppler effect in which the observer sees a decrease in wavelength as the star is moving away from the earth and there are a range of wavelength as due to z=v/c=/\lambda/lambda and different isotopes of hydrogen have different velocities

18)b)ii)/\lambda/lambda=/|kT/mc^2 = v/c, v=(/|1.38*10^-23 *3150/1.67*10^-27 * (3*10^8)^2 *3*10^8 =5102ms^-1

18)c)Due to the Sun having a larger mass than luyten star due to higher radius there is more hydrogen nuclei in the sun therefore there is more fusion reactions therefore the Sun will do fusion reactions increase in heat energy faster than the luyten and therefore it will reach red giant faster than the luyten star.