11)lambda=ln2/8.02*24*3600=1*10^-6s^-1, N=1860*10^6/1*10^-6=1.86*10^15 atoms

12)a)N=1.35*10^5*2*10^-3/1.38*10^-23*295=6.63*10^22 molecules

12)b)U proportional to T therefore 2*T=2*(273+12)=570

13)a)P=VI=230*6.75=1550W

13)b)i)T[f]=(1550/0.048*1.01*10^3)+18=50.0+273=323k

13)b)ii)As there are heat losses to the surroundings

14)a)1/11=0.091, 1/4.4=0.227 a graph is drawn with a 1:5 scale and takes 2/3 of the space of the graph it has axis labels and a line graph drawn with a line of best fit.

14)b)As mass of gas and pressure are directly proportional the pressure is inversely proportional to volume occupied by gas therefore the graph supports boyle's law

15)a)Acceleration is directly proportional to displacement from the equilibrium position and acts towards the equilibrium position

15)b)i)A=9m and T=12*3600=43200s

15)b)ii)w=2pi/12.5=0.503hr^-1, v=Aw=3.7*0.503=1.86mhr^-1

15)b)iii)A sin graph drawn with same period and leading by 2pi/4 increase in phase

16)High density and high pressure is required for frequent collisions to take place and high energy is required for powerful collisions to take place and transfer the energy to mass as shown by the equation /\E=mc^2 and also overcome the electrostatic repulsion of the particles. A strong magnetic field so plasma does not stick to the walls of the reactor

17)a)As alpha radiation has high ionization and low penetration power and it only able to travel a few cms in air

17)b)Before finding the count rate the student should substract the backgroud count for each count and then find the count rate. The student should average all the counts found by her.

18a)number of peaks in 15 seconds 11/15=0.733Hz

18)b)i)w=2*pi*0.8=5.03rads^-1, v=rw=1.4*10^9/2*5.03=3.52*10^9ms^-1

18)b)ii)As this would cause great differences in heat given to the earth leading to dangerous climatic conditions

18)c)i)This is due to the doppler effect where The A points will show different apparent wavelength which can be less or more depending on if the sun is moving closer to the earth or away from the earth

18)c)ii)/\lambda=(4*10^3/3*10^8)*5.90*10^-7=7.87*10^-12m

19)a)When a nearby star is seen from two positions the extending of the lines gives us a small angle formed near the back of the nearby star called and this is used in equations to calculate how far distant stars are.

19)b)Star A will have a parallax twich as small as star B when the stars are viewed from two positions at a interval of 6 months this change in angular position would be shown

19)c)i)Satellite communication would be easier and faster as the satellite would be at the same place above the orbit

19)c)ii)mw^2r=GMm/r^2, therefore (2pi/T)^2=GM/r^2 therefore, r=/GMT^2/4pi^2=/3/6.67*10^-11*6*10^24*(24*3600)^2/4pi^2, r=4.23*10^7m, h=r-R[E]=4.23*10^7-6.4*10^6=3.59*10^7m