A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or

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A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or

a proton, about what would its mass be? suppose that a neutron or a proton is a ball that has a diameter of 10-15 m , and a mass of 10-27 kg.

Physics

2 answers:

inna [3.1K] · Answer 1 · 2026-03-10T04:45:56+03:00

The mass of the baseball is $\fbox{\begin\\3.922 \times {10^{14}}\,{\text{Kg}}\end{minispace}}$ .

Further explanation:

The mass-to-volume ratio of the baseball aligns with that of either neutrons or protons.

Given:

The circumference of the baseball measures $23\,{\text{cm}}$ .

The proton's diameter is ${10^{ - 15}}\,{\text{m}}$ .

The mass of a proton is ${10^{ - 27}}\,{\text{Kg}}$ .

Concept applied:

The circumference is defined as the perimeter around the sphere.

The formula for the sphere's circumference is expressed as.

$C = 2\pi r$

We can rearrange the above formula to derive the radius of the ball.

$\fbox{\begin\\r=\dfrac{C}{{2\pi }}\end{minispace}}$ …… (1)

In this context, r signifies the radius, C refers to the circumference of the sphere.

The formula for the volume of a sphere is expressed as.

$V = \dfrac{4}{3}\pi {r^3}$

The mass-to-volume ratios for the baseball and proton are identical.

This statement is depicted by the formula.

$\dfrac{{{m_b}}}{{{V_b}}}=\dfrac{{{m_p}}}{{{V_p}}}$

Rearranging this formula gives us the mass in terms of the ball.

${m_b}=\dfrac{{{m_p}}}{{{V_p}}}{V_b}$

Where ${m_b}$ represents the ball's mass, ${m_p}$ symbolizes the proton's mass, ${V_p}$ corresponds to the proton's volume, and ${V_b}$ denotes the ball's volume.

The equation for the mass of the ball, expressed in terms of the radius, is given as.

$\fbox{\begin\\{m_b}={\left( {\dfrac{{{r_b}}}{{{r_p}}}}\right)^3}{m_p}\end{minispace}}$ ...... (3)

Substituting $23\,{\text{cm}}$ for C in equation (1).

$\begin{aligned}r&=\frac{{23\,{\text{cm}}}}{{2\pi }}\\&=\frac{{23\,{\text{cm}}\left( {\frac{{1\,{\text{m}}}}{{100\,{\text{cm}}}}} \right)}}{{2\pi }}\\&=3.66\times {10^{ - 2}}\,{\text{m}}\\\end{aligned}$

Then replacing $3.66 \times {10^{ - 2}}\,{\text{m}}$ with ${r_b}$ , $\left( {\dfrac{{{{10}^{ - 15}}\,{\text{m}}}}{2}} \right)$ with ${r_p}$ , and ${10^{ - 27}}\,{\text{Kg}}$ with ${m_p}$ in equation (3).

$\begin{aligned}{m_b}&={\left( {\frac{{3.66 \times {{10}^{ - 2}}\,{\text{m}}}}{{\left( {\frac{{{{10}^{- 15}}\,{\text{m}}}}{2}} \right)}}} \right)^3}{10^{ - 27}}\,{\text{Kg}}\\&=3.922 \times {10^{41}} \times {10^{ - 27}}\,{\text{Kg}}\\&=3.922 \times {10^{14}}\,{\text{Kg}}\\\end{aligned}$

Consequently, the mass of the baseball is $\fbox{\begin\\3.922 \times {10^{14}}\,{\text{Kg}}\end{minispace}}$ .

Learn more:

1. Motion under friction .

2. Conservation of momentum .

3. Motion under force .

Answer Details:

Grade: College

Subject: Physics

Chapter: Modern Physics

Keywords:

Baseball, proton, neutron, mass, volume, radius, circumference, sphere, density, ratio, 3.922*10^14Kg, 3.9*10^14Kg, 3.66*10^-2m, mass to volume ratio.