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Llana [10]
1 year ago
9

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

roton, about what woul its mass be?
Physics
1 answer:
BigorU [14]1 year ago
7 0

Baseball, neutrons, and protons are all perfect spheres.

Circumference of the baseball, $L_{\text {ball }}=23 \mathrm{~cm}$.

mass of the proton or neutron, $m_{\mathrm{p}}=10^{-27} \mathrm{~kg}$

diameter of the proton or neutron, $r_{0}=10^{-15} \mathrm{~m}$

Mass per unit volume is the definition of a material's density. The density equation is

$$\rho=\frac{M}{V} .$$

Here, M is the mass of the material, V is the volume, and $\rho$ is the density of the material.

We know that formula for the volume of a sphere having diameter d is V=\frac{\pi d^{3}}{6}$.

You may obtain this result by plugging this number into the proton/neutron density equation:

$$\rho_{0}=\frac{m_{0}}{\left(\frac{\pi d_{0}^{3}}{6}\right)}$$

By replacing the variables in the aforementioned equation with their values, you will obtain

$$\begin{aligned}\rho_{\circ} &=\frac{10^{-27}}{\left(\frac{\pi\left(10^{-15}\right)^{3}}{6}\right)} \\&=1.91 \times 10^{18} \mathrm{kgm}^{-3}\end{aligned}$$

The formula for calculating a circle's circumference with diameter d is L=\pi d$.

Using this, you get the diameter of the baseball:

$$\begin{aligned}d_{\text {ball }} &=\frac{L_{\text {ball }}}{\pi} \\&=\frac{23}{\pi} \\&=7.32 \mathrm{~cm} \\&=0.0732 \mathrm{~m}\end{aligned}$$

The volume of the baseball may be calculated using the formula for the volume of a sphere:

$$\begin{aligned}V_{\text {ball }} &=\frac{\pi d_{\text {ball }}^{3}}{6} \\&=\frac{\pi \times(0.0732)^{3}}{6} \\&=2.054 \times 10^{-4} \mathrm{~m}^{3}\end{aligned}$$

The mass of the baseball is determined by the concept of density if you suppose that its density is equal to that of a proton or neutron.

$m_{\text {ball }}=\rho_{\circ} \cdot V_{\text {ball }}$

Here, $\rho_{\circ}$ is the density of the proton/neutron.

By replacing the variables in the aforementioned equation with their values, you will obtain

$$\begin{aligned}m_{\text {ball }} &=\left(1.91 \times 10^{18}\right) \cdot\left(2.054 \times 10^{-4}\right) \\&=3.92 \times 10^{14} \mathrm{~kg}\end{aligned}$$

Therefore, if the density of the baseball was equal to that of the neutron/proton, then its mass would be 3.92 \times 10^{14} \mathrm{~kg}$.

Learn more about  density brainly.com/question/15164682

#SPJ9

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Explanation:

Hello,

Let's get the data for this question before proceeding to solve the problems.

Mass of flywheel = 40kg

Speed of flywheel = 590rpm

Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.

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During the power off, the flywheel made 230 complete revolutions.

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + ω₂) / 2] × 0.5

But ∇θ = 230 revolutions

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ω₂ = ω₁ + αt

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α = (330 - 590) / 0.5

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α = -520rev/min

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ω₂ = ω₁ + αt

0 = 590 +(-520)t

520t = 590

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t = 590 / 520

t = 1.13min

60 seconds = 1min

X seconds = 1.13min

x = (60 × 1.13) / 1

x = 68seconds

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