Question

Verify that a 2.3 x 10¹⁷ kg mass of water at normal density would make a cube 60 km on a side. (This mass at nuclear density would make a cube 1.0 m on a side.)

How Small and Dense Is a Nucleus?

(a) Find the radius of an iron-56 nucleus.

Answer 1

To verify that a 2.3 × 10¹⁷ kg mass of water at normal density would make a cube 60 km on a side, we can use the formula for the volume of a cube. A cube with a mass of 1.0 kg and a density of nuclear matter (2.3 × 10¹⁷ kg/m³) would have a length of approximately 0.0367 meters.

Step-by-step explanation:

To verify that a 2.3 × 10¹⁷ kg mass of water at normal density would make a cube 60 km on a side, we can use the formula for the volume of a cube. The volume of a cube is given by V = s^3, where s is the length of one side of the cube. In this case, the mass of the water is given, so we can use the density of water (1000 kg/m³) to find the volume and then solve for the length of one side:

V = m / ρ, where V is the volume, m is the mass, and ρ is the density.

Substituting the given values:

V = (2.3 × 10¹⁷ kg) / (1000 kg/m³)

V ≈ 2.3 × 10¹⁴ m³

Since the volume of a cube is V = s^3, we can solve for s:

s = V^(1/3)

s ≈ (2.3 × 10¹⁴)^(1/3) m

s ≈ 60959 m

Therefore, a 2.3 × 10¹⁷ kg mass of water at normal density would make a cube approximately 60959 meters on a side.

To find the length of a side of a cube having a mass of 1.0 kg and a density of nuclear matter (2.3 × 10¹⁷ kg/m³), we can use the same formula for the volume of a cube:

V = m / ρ

Substituting the given values:

V = (1.0 kg) / (2.3 × 10¹⁷ kg/m³)

V ≈ 4.35 × 10^-18 m³

s = V^(1/3)

s ≈ (4.35 × 10^-18)^(1/3) m

s ≈ 0.0367 m

Therefore, a cube with a mass of 1.0 kg and a density of nuclear matter (2.3 × 10¹⁷ kg/m³) would have a length of approximately 0.0367 meters.