No Regulus or Alpha Leonis is 23.8 parsecs from the Earth.
A star with an absolute magnitude of -20 would need to be about 100 times brighter than the sun. Since brightness decreases with the square of the distance from the observer, the supernova would need to be roughly 10 times closer than the sun, so at a distance of about 15 million kilometers.
There is no point in doing this. This would be even sillier than measuring the distance from New York to Los Angeles by the thickness of a piece of paper.Use the right tool for the job.
The "north" and "south" latitude references begin at the equator. So "20 degrees" north or south are both 20 degrees from the equator, and "10 degrees" north or south are both 10 degrees from the equator. 10 is closer to the equator than 20.
The apparent magnitude of an RR Lyrae star at a distance of 10 parsecs would be 0, given its absolute magnitude of 0. To find the distance at which it would appear as faint as 20th magnitude, we would use the distance modulus formula, which results in a distance of 10,000 parsecs or 32,600 light-years.
Energy output, as absolute brightness (magnitude) is taken at a standard distance of 10 parsecs.
9.99 is closer to 10 than 9.9 is.
Basically, that means that the number of tires (2) is closer to 1 than to 10.Basically, that means that the number of tires (2) is closer to 1 than to 10.Basically, that means that the number of tires (2) is closer to 1 than to 10.Basically, that means that the number of tires (2) is closer to 1 than to 10.
|-20 - (-40)| = 20 |-20 - 10| = 30 -20 is closer to -40 than 10.
10
-5 is larger than -10 because the closer a number is to zero, the larger it is. In this case, -5 is closer to zero than -10.
Your 1 kilometer equals 3.24077649 x 10-14 parsecs. And your second equals 3.16887646 x 10-8 years. Converting kilometers/second to parsecs/year involves multiplying by 3.24077649 x 10-14 and dividing by 3.16887646 x 10-8 to get your answer. Or do that bit of math in advance and then just multiply by 1.022689439 x 10-6 to get your answer. Make sure you have your beginning figure in km per one second. Your answer will appear in parsecs per one year, or just parsecs per year.
300mm is 9.72 x 10-18 parsecs.
The square root of 10 is 3.1622776601683793319988935444327, so it is closer to 3 than 4
10
A star with an absolute magnitude of -20 would need to be about 100 times brighter than the sun. Since brightness decreases with the square of the distance from the observer, the supernova would need to be roughly 10 times closer than the sun, so at a distance of about 15 million kilometers.
No. Brighter distant stars can have the same apparent magnitude as fainter stars that are closer.(Absolute magnitude does not refer to actual brightness, but rather to what the brightness of a star would likely be at an arbitrary distance of 10 parsecs, rather than its actual distance.)
10.05 is closer to 10 than 10.5 is.