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

Question

4 years ago

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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:

Firlakuza [10] · Answer 1 · 2021-12-28T22:21:47+03:00

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

Further explanation:

Mass per unit volume of the baseball is equal to the mass per unit volume of the proton or neutron.

Given:

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

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

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

Concept used:

The distance around the outer line of the sphere is known as the circumference of the sphere.

The expression for the circumference of the ball is given as.

$C = 2\pi r$

Rearrange the above expression for the radius of the ball.

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

Here, r is the radius of the ball and C is the circumference of the ball.

The expression for the volume of the sphere is given as.

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

Mass to volume ratio of the baseball and the proton are same.

The expression for the above statement is given as.

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

Rearrange the above expression for the mass of ball.

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

…… (2)

Here, ${m_b}$ is the mass of the ball, ${m_p}$ is the mass of the proton, ${V_p}$ is the volume of proton and ${V_b}$ is the volume of ball.

The expression for the mass of ball in terms of radius is given as.

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

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

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

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

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

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

Learn more:

1. Motion under friction brainly.com/question/7031524.

2. Conservation of momentum brainly.com/question/9484203.

3. Motion under force brainly.com/question/4033012.

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.