Physics, 08.07.2019 11:00 lolirdc

Astandard 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.

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Answer from: drewje12

Mass of base ball $m_b=3.992*10^{14}kg$

Explanation:

Circumference of baseball = 2πr = 23 cm

So radius of baseball = 3.66 cm = $3.66*10^{-2}$ m

Mass per unit volume of baseball = Mass per unit volume of neutron or proton.

Mass of proton = $10^{-27}$ kg

Diameter of proton = $10^{-15}$ m

Radius of proton = $5*10^{-16}$ m

Volume of ball = $\frac{4}{3} \pi r^3$

Now substituting all values in Mass per unit volume of baseball = Mass per unit volume of neutron or proton.

$\frac{m_b}{\frac{4}{3}\pi *(3.66*10^{-2})^3} =\frac{10^{-27}}{\frac{4}{3}\pi *(5*10^{-16})^3}$

$\frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}$

$m_b=3.992*10^{14}kg$

So mass of base ball $m_b=3.992*10^{14}kg$

Answer from: mandi795

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 .

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.

Answer from: Quest

i'm just gonna leave this comment here, so i can get a challange done-

explanation:

Answer from: Quest

the device is called “ rocket engine”

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