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cc4019
 2 years ago
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.
cc4019
 2 years ago
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.

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cc4019
 2 years ago
Best ResponseYou've already chosen the best response.0i dont understand why are we including the current year when it was viewed in 1054

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1if 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 10543500=

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1btw, im not sure where you got the distance to the crab nebula from because i thought it was 6500 lyr (not 3500 lyr)

cc4019
 2 years ago
Best ResponseYou've already chosen the best response.0my textbook says its about 3500 LY away

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1dont for get that the year will be before 0 CE so it will be ... BCE

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1(well i guess you can use that value if its in your book)

cc4019
 2 years ago
Best ResponseYou've already chosen the best response.0so 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
 2 years ago
Best ResponseYou've already chosen the best response.1\[s=\frac{d}{t}\qquad\implies t=\frac ds\] \[d=3500 [\text{ly}]=3500\times c\times[\text {yr}]\]\[s=c\] \[t=\frac ds=\frac{3500\times c\times[\text {yr}]}{c}=3500 [\text {yr}]\] \[1054[\text{yr}]~C.E.3500[\text{yr}]=2446~C.E.=2446~B.C.E\]

cc4019
 2 years ago
Best ResponseYou've already chosen the best response.0Ok thank you for your time and explanation.

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1sorry about my first post[now deleted] ( i only just woke up)

cc4019
 2 years ago
Best ResponseYou've already chosen the best response.0its fine you've helped me understand with the formulas you've provided
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