Answer: 100 suns
Explanation:
We can solve this with the following relation:
[tex]\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}[/tex]
Where:
[tex]d=17.91 mm =17.91(10)^{-3} m[/tex] is the diameter of a dime
[tex]D[/tex] is the diameter of the Sun
[tex]x_{sun-pinhole}=150,000,000 km=1.5(10)^{11} m[/tex] is the distance between the Sun and the pinhole
[tex]x_{sunball-pinhole}=100 d=1.791 m[/tex] is the amount of dimes that fit in a distance between the sunball and the pinhole
Finding [tex]D[/tex]:
[tex]D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}[/tex]
[tex]D=\frac{17.91(10)^{-3} m}{1.791 m} 1.5(10)^{11} m[/tex]
[tex]D=1.5(10)^{9} m[/tex] This is roughly the diameter of the Sun
Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:
[tex]1 AU=149,597,870,700 m[/tex]
So, we have to divide this distance between [tex]D[/tex] in order to find how many suns could it fit in this distance:
[tex]\frac{149,597,870,700 m}{1.5(10)^{9} m}=99.73 suns \approx 100 suns[/tex]
