Recent content by b_o3

I'm sorry I'm not an expert on this I don't usually take science courses... but are u saying I'm gna take a(t)= k * t^(2/3) . and that gives me the answer?

cant i just equate 1/a(t)^4=1/a(t)^3 . then I would get the value of a(t) when they are equal

okay so after i do that... hows that going to help

to the power of 4

okay cool then .. I got 7.85 exp 10 .. which seems about right

so how would that work then?

sorry but I am not following

how would it run out of hydrogen in 2015... i said it was 3.93 exp 12. But now the problem is I am not sure whether to use the total energy of the sun to divided by rate of conversion..or should i use the total mass of Hydrogen

someone help me with it please

so do I have all the values i need?

but wait hows this right... the sun shines when there's energy, so I should probably use the total energy available to compute how long it will shine .. right? When I do that I get 3.93 exp 12 thanks,

so I found mass of H in sun = 1.5 exp30 kg, I divided that by rate of Hydrogen conversion which is equal to 1.91 exp 19 kg per year, and I got 7.85 years is this the correct answer?

so I found mass of H in sun = 1.5 exp30 kg, I divided that by rate of Hydrogen conversion which is equal to 1.91 exp 19 kg per year, and I got 7.85 years is this the correct answer?

yest mentz its that one

how do i get that? do i have to integrate first