The figure below is a video that starts with a graph of y = x^1000 in a standard

Question

The figure below is a video that starts with a graph of y = x^1000 in a standard coordinate system. It then shows the vertical axis being stretched so that the top of the graph window shows a y-value of 6e-64 (which means 6 middot 10^-64, or 6 divided by 10^64). Use a ruler to measure the distance from 0 to 6 middot 10^64 on the vertical axis on your screen. Approximately how far away, in light years, is 1 from the horizontal axis after the stretching stopped? Interpreter your answer in units of (1) trips to the moon: (2) trips to the Sun. (3) trips across the Milk}- Way: and (4) trips from earth to the edge of the visible universe. y = x^1000

Explanation / Answer

given, y = x^(1000)

now, for, y = 6*10^-64

so from ruler

3.9 cm = 6*10^-64

so, 3.9 cm/6*10^-64 = 1 = 6.5*10^63 cm = 6.5*10^61 m

so, distnace of 1 from horixontal acisd, d = 6.5*10^61 m

but 1 L.Y = speed of light* seconds in 1 year = 3*10^8*365*24*60*60 = 9.4608*10^15 m

so d = 6.5*10^61/9.4608*10^15 = 6.87045*10^45 Light Years

distance of moon from earth, r = 3844*10^5 m

distance of sun from earth, R = 149.5*10^9 m

diameter of milky way, D = 100,000 LIGHT YEARS

distance of edge of visible universe from earth, l = 46.5*10^9 LY

so,

a. number of round triops from moon = 6.5*10^61 m / 2*3844*10^5 m = 8.454*10^52 trips

number of triops to moon = 6.5*10^61 m / 3844*10^5 m = 16.908*10^52 trips

b. number of round triops from sun = 6.5*10^61 m / 2*149.5*10^9 m = 2.173*10^50 trips

number of triops to sun = 6.5*10^61 m / 149.5*10^9 m = 4.346*10^50 trips

c. trips acrtoss milky way = 6.87*10^45 LY / 100,000 LY = 6.87*10^40 yrips

d. trips to the edge of the visible universe = 6.87*10^45/46.5*10^9 = 1.477*10^35 trips

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