I am having trouble with this question, part c) My visualization of the problem: First, if I imagine myself on a mountain and there is a star directly above me at a distance of 10000 counts, then if the star were to move by 60 degrees, I would have an air mass of 2. So, my detector would read 20000 counts. But this is where my doubt comes in because it says it is 60% less at sea level. I can't help but think I'm missing something. So, I then visualized myself climbing down the mountain while still being under the zenith where the star is. In this case, I should still be reading 10000 counts, correct? But now, as the star, I am looking through a smaller angle than I did when I was on top of the mountain. However, I can't see how the air pressure is affecting the count rate. Document Viewer @takehome_test_1_2018.pdF ATOMIC PHYSICS of 2 163.12% V Thumbnails Identical to the Sun Q2) A Type I delta-Cepheid star with a one-month (30-day) period is put at the distance of the star Vega. (a) [20 marks] Determine the apparent magnitude this star would have. (b) [10 marks] Determine how many times more luminous the delta-Cepheid star is compared to Vega. (c) [20 marks] Assume you observe the above delta-Cepheid star at an observatory on a mountain high enough that the typical air pressure is only 60% of the conditions at sea level. If the star is at an airmass of one, your detector registers 10000 counts from the star. Determine the number of counts your detector would register if the star was positioned at a zenith distance of 60 degrees - assume the star has not changed its apparent magnitude between the two observations.

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