Respuesta :
Explanation:
Let the diameter of the Sun in the model be = x
Scale used in the model = [tex]\frac{1}{10^{10}}=10^{-10}[/tex]
[tex]\frac{x}{D}=10^{-10}[/tex]
The real diameters of the Sun ,D= 1.4 million kilometers = [tex]1.4\times 10^6 km[/tex]
[tex]x=1.4\times 10^6 km\times 10^{-10}=1.4\times 10^{-4} km=14 cm[/tex]
[tex]1 km = 100,000 cm[/tex]
The diameter of the sun in the model is 14 cm.
The real diameters of the Earth,d = 12,800 km
Let the diameter of the Earth in the model be = y
Scale used in the model = [tex]\frac{1}{10^{10}}=10^{-10}[/tex]
[tex]\frac{y}{d}=10^{-10}[/tex]
[tex]y=12,800 km\times 10^{-10}=1.28\times 10^{-6} km=0.128 cm[/tex]
The diameter of the Earth in the model is 0.128 cm.
The Earth-Sun distance ,l = [tex]150\times 10^6 km=1.5\times 10^8 km[/tex]
Let the Earth-Sun distance in the model be = z
[tex]\frac{z}{l}=10^{-10}[/tex]
[tex]z=1.5\times 10^8 km\times 10^{-10}=1.5\times 10^{-2} km=1500 cm[/tex]
The distances between Earth-Sun in the model is 1500 cm.
