Easha

1) Every second the Sun outputs 3.90* 10^26 Joules of energy. Power is energy output per unit time and is given in units of Watts (W) where 1 W = 1 Joule per second. The Sun's power output is therefore 3.90*10^26 Watts. This is called the solar luminosity...

hey..i have this same question problem and i also posted this same question here in this site..and as far as i understand u need to get each in m^3/year...!!

i have this same assignment..and i am stuck as well...as far as i understand..u need to get each individual parts in m^3/year.

no..i meant that input shud all b in m^3/year...den output also..and the other individual parts as well..so change in storage will also b in m^3/year..and change in storage must b zero..!! i was wondering...by any chance do u have a question number 1a.b.c...

no need for 1a b c d e..i think my one is already right!!:)

no need for any answer..ty..i solved it...but i don't no if its right or wrong..but i'll go with it anyways..:)

ty so much for the help..for the 1st 1 i got the same answer...but y did u use pir^2 for the second 1 and 4pir^2 for the 1st 1??

yeah...i got it..ty sooo much...i can do the rest i guess..but i have any problem with any other questions...i'll post it..!...ty once agn!!:D

sry to disturb u 1ce agn..but i had another problem related to galilean moons...i thot i would just post the question in case u can help me...i have done it...i just want to check my answer!!.. What is the largest angle that can separate Ganymede from Eur...

what i did was.. distance between earth and jupiter= 4.20 AU= 6.30*10^8 km distance between jupiter and europa= 6710000 km distance between jupiter and ganymede= 1070000 km then i found the angles subtended by jupiter and europa, x (from earth viewpoint)...

yea...i figured that out earlier..my answer was in degree (as the calculator was in degree mode)...but i took the answer to be in radian then i converted that to degree...which is completely wrong...its 2.18arc mins now...ty sooo much for all ur help!!