anonymous
  • anonymous
The astronomical object known as Crab Nebula is thhe remnant of an exploded star. The explosion was seen by the Chinese in 1054 C.E. However, the Crab Nebula is about 3500 LY distance from the Earth In what year did the star actually explode.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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UnkleRhaukus
  • UnkleRhaukus
http://apod.nasa.gov/apod/image/0802/crabmosaic_hst_big.jpg
anonymous
  • anonymous
i dont understand why are we including the current year when it was viewed in 1054
UnkleRhaukus
  • UnkleRhaukus
hmm,

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More answers

UnkleRhaukus
  • UnkleRhaukus
if the crab nebula is 3500 [lyr] away, then light traveling from the crab nebula will take 3500 [yr] to reach earth so to find the year that the light left the crab nebula just take away the travel time from the year it arrived at earth 1054 CE 1054-3500=
anonymous
  • anonymous
im here
anonymous
  • anonymous
2466
UnkleRhaukus
  • UnkleRhaukus
btw, im not sure where you got the distance to the crab nebula from because i thought it was 6500 lyr (not 3500 lyr)
anonymous
  • anonymous
my textbook says its about 3500 LY away
UnkleRhaukus
  • UnkleRhaukus
dont for get that the year will be before 0 CE so it will be ... BCE
UnkleRhaukus
  • UnkleRhaukus
(well i guess you can use that value if its in your book)
anonymous
  • anonymous
so what was the first answer you gave because that seemed to make a lot of sense if the speed of light is pretty fast
UnkleRhaukus
  • UnkleRhaukus
\[s=\frac{d}{t}\qquad\implies t=\frac ds\] \[d=3500 [\text{ly}]=3500\times c\times[\text {yr}]\]\[s=c\] \[t=\frac ds=\frac{3500\times c\times[\text {yr}]}{c}=3500 [\text {yr}]\] \[1054[\text{yr}]~C.E.-3500[\text{yr}]=-2446~C.E.=2446~B.C.E\]
anonymous
  • anonymous
Ok thank you for your time and explanation.
UnkleRhaukus
  • UnkleRhaukus
sorry about my first post[now deleted] ( i only just woke up)
anonymous
  • anonymous
its fine you've helped me understand with the formulas you've provided

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