Suppose a star has a luminosity of 6.0×1026 watts and an apparent brightness of 1.5×10−12 watt/m2 . How far away is it? Give your answer in both kilometers and light-years.
Use the inverse square law for light to answer each of the following question.

The distance of this star in kilometers is equal to 5.64 × 10¹⁸ kilometers. Also, the distance of this star in light-years is equal to 626.67 light-years.How to calculate the distance of light?Mathematically, the inverse square law for light can be used for calculating its luminosity at a distance by using this mathematical expression:d = √(L/4πF)Where:d represents the distance of light.L represents the luminosity of light.F represents the apparent brightness of light.Substituting the given parameters into the formula, we have;Distance in kilometers, d = √(6.0 × 10²⁶/4 × 3.142 × 1.5 × 10⁻¹²)Distance in kilometers, d = √(6.0 × 10²⁶/1.885 × 10⁻¹¹)Distance in kilometers, d = √3.183 × 10³⁷Distance in kilometers, d = 5.64 × 10¹⁸ kilometers.For the distance in light-years, we would divide the distance in kilometers by 9.0 × 10⁻¹⁵:Distance in light-years = 5.64 × 10¹⁸/9.0 × 10¹⁵Distance in light-years = 626.67 light-years.Read more on distance in light-years here: https://brainly.com/question/1302132#SPJ1