Suppose you want to make a scale model of a hydrogen atom. you choose, for the nucleus, a small ball bearing with a radius of 1.5 mm. the radius of the hydrogen atom is 0.529 ã 10â10 m and the radius of the nucleus is 1.2 ã 10â15 m. (a) what would be the radius (m) of the model? (b) suppose that now you want to make a scale model of the solar system using the same ball bearing as in part (a) to represent the sun. how far from it (mm) would you place a sphere representing the earth? (center to center distance .) (see the inside cover of your textbook for data.) (c) what would be the radius (mm) of the sphere representing the earth in part (b)?

The radius of the models will be:

Radius of small ball bearing = 1.5 mm = 1.5 x 10⁻³ m

Radius of the hydrogen atom 0.529 x 10⁻¹⁰ m

Radius of the nucleus is 1.20 x 10⁻¹⁵ m.

Radius of the sun = 6.96 x 10⁸ m

Radius of the Earth = 6.37 x 10⁶ m

Distance of earth from sun = 1.50 x 10¹¹ m

A. Ratio of radius of ball bearing to that of the hydrogen nucleus, R₁, is first determined.

R₁ = 1.5 x 10⁻³ m/1.20 x 10⁻¹⁵ m

R₁ = 1.25 x 10¹²

Therefore, radius of model = radius of hydrogen atom x R₁

Radius of model hydrogen atom = 0.529 x 10⁻¹⁰ m x  1.25 x 10¹²

Radius of model hydrogen atom = 0.661 x 10² m

Radius of model hydrogen atom = 66.1 mm

B. Ratio of radius of ball bearing to that of the sun R₂, is first determined.

R₂ = 1.5 x 10⁻³ m/6.96 x 10⁸ m

R₂ = 0.215 x 10⁻¹¹

Therefore, distance of model earth from the sun = Distance of earth from sun x R₂

distance of model earth from the sun = 1.50 x 10¹¹ m x  0.215 x 10⁻¹¹

distance of model earth from the sun = 0.3225 m

Distance of model earth from the sun = 32.35 mm

C. Radius of model earth = radius of earth x R₂

Radius of model earth = 6.37 x 10⁶ m x  0.215 x 10⁻¹¹

Radius of model earth = 1.369 x 10⁻⁵ m

Radius of model earth = 0.137 mm

Therefore, the radius of the models will be:

Learn more about the solar system, atoms and models at: link

A)  x _electron = 0.66 10² m , B)   x _Eart = 1.13 10² m , C)  d_sphere = 1.37 10⁻² mm

Explanation:

A) Let's use a ball for the nucleus, the electron is at a farther distance the sphere for the electron must be at a distance of

Let's use proportions rule

                x_ electron = 0.529 10⁻¹⁰ /1.2 10⁻¹⁵ 1.5

               x _electron = 0.66 10⁵ mm = 0.66 10² m

B) the radii of the Earth and the sun are

                = 6.37 10⁶ m

                tex]R_{Sum}[/tex] = 6.96 10⁸ m

                Distance = 1.5 10¹¹ m

                x_Earth = 1.5 10¹¹ / 6.96 10⁸  1.5

                x _Eart = 1.13 10² m

C) The radius of a sphere that represents the earth, if the sphere that represents the sun is 1.5 mm, let's use another rule of proportions

            d_sphere = 1.5 / 6.96 10⁸  6.37 10⁶

            d_sphere = 1.37 10⁻² mm

the force required to move the cart is divided by three or 33.33 n.

there is no correct list among the choices you've provided.

the list might look something like this:

-- baby human

-- piccolo

-- flute

-- soprano singer

-- trumpet

-- tenor singer

-- trombone

-- baritone singer

-- baritone horn

-- bass singer

-- tuba

since i had to make up the list myself, i made it easy on myself, and i didn't try to shuffle the string instruments and woodwinds into it too.