If the Sun’s 1,395,000 km diameter is shrunken down to 12 inches, and the next closest star is 4.20 × 1013 km from the Sun, how far apart should the two be placed in a properly scaled model of the Sun and its nearest neighbor, Proxima Centauri? (For your answer, convert the scale model distance into miles, since the separation in inches will be huge)

30,179,400 miles
340 miles
362,152,200 miles
2.295×1013 miles
5720 miles
30,107,500 miles
68,590 miles
Human beings have not evolved to the point where inches can be converted into miles

5702 miles

Explanation:

First, let us find the rate at which the shrinking down is occurring.

If 1,395,000 km has been shrunk down to 12 inches, that means that 1 km will be:

1 km = =

Therefore, the  distance between the sun and the next closest star, in inches, will be:

inches

Finally, we have to convert this distance to miles.

1 inch = miles

Therefore, 361284000 inches will be:

= miles

The nearest neighbor of the sun is 5702 miles away from the sun.

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day