00:01
Hello everyone for the first part of the question.
00:04
Let the age of universe be t, then t is equal to h0 rest of the power minus 1, whole divided 1 plus 2, then here h0 is equal to 70 km per second m -p -e.
00:31
Therefore h0 is equal to 2 .2 multiplied by 10 -d -pour -1.
00:37
Minus 13 per second.
00:41
Hence t is equal to 8 .66 multiplied by 10 to power 16 second or 2 .75 multiplied by 10 to power 9 year or 2 .75 beer.
01:04
Now for the second part the proper distance given as dp t0 which is equal to c divided by h0 z which is z where t not is present time therefore c multiplied by three multiplied by tenders to power 8 all divided by 70 multiplied by 4 .5 which is equal to 1 .93 multiplied by tenders to power 7 mp now for the seat part the proper distance when the light was emitted is d .p.
01:46
Te.
01:47
Therefore, d .p.
01:50
T .0 divided by 1 plus z.
01:55
Hence, where t .e is the time of emission, therefore it is equals to 3 .67 multiplied by 10 to power 6 mpe.
02:05
Now for the fourth part, the luminosity distance dl is equal to d .p.
02:13
T .0.
02:15
Multiplied by 1 plus z.
02:17
Therefore, the value is 10 .13 multiplied by tendous to power 7 m p.
02:26
Now for the fifth part, the angular diameter of the nucleus is 5 seconds...