### Angular Momentum and Neutron Star FormationIn this educational example, we explore the fascinating dynamics of a star's collapse into a neutron star and how its rotational characteristics change due to conservation of angular momentum. #### Scenario:- A star completes one revolution every 10 days.- Initial radius: \(7.00 \times 10^5 \) km.- The star collapses into a neutron star: - New radius: 20.0 km. - The star loses one half of its mass during the collapse.#### Question:**What is the new rate of revolutions per day?** (The moment of inertia of a star is given by \( \frac{2}{5} mr^2 \).)#### Multiple Choice Options:- \(3.13 \times 10^8 \) rev/day- \(3.20 \times 10^8 \) rev/day- \(3.32 \times 10^8 \) rev/day- \(1.70 \times 10^8 \) rev/day- \(2.45 \times 10^8 \) rev/day#### Explanation:To find the new revolutions per day, consider the conservation of angular momentum. Angular momentum \(L\) is conserved and is defined as \(L = I \omega\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.1. **Initial Moment of Inertia (\(I_{\text{initial}}\)):** \[ I_{\text{initial}} = \frac{2}{5} m r_{\text{initial}}^2 \] Given: - Initial radius \( r_{\text{initial}} = 7.00 \times 10^5 \) km.2. **New Moment of Inertia (\(I_{\text{final}}\)):** \[ I_{\text{final}} = \frac{2}{5} \left( \frac{m}{2} \right) r_{\text{final}}^2 \] Given: - Final radius \( r_{\text{final}} = 20.0 \) km. - The mass is halved: \( \frac{m}{2} \).3. **Calculate Initial Angular Velocity (\(\omega_{\text{initial}}\)):** - The period (\(T\

Answered: Image /qna-images/answer/606578be-66f8-4e8a-a5f7-df321ba33e1d.jpg

A star spins one revolution every 10 days and its radius is 7.00*105 km. It then collapses and becomes a neutron star with a radius of 20.0 km. It also loses one half its mass during the collapse. ). What is the new revolutions /day? (The moment of inertia of a star is 2/5 mr²) O 3.13*108 rev/day O 3.20*108 rev/day O 3.32*108 rev/day 1.70*108 rev/day O 2.45*108 rev/day

A star spins one revolution every 10 days and its radius is 7.00105 km. It then collapses and becomes a neutron star with a radius of 20.0 km. It also loses one half its mass during the collapse. ). What is the new revolutions /day? (The moment of inertia of a star is 2/5 mr²) O 3.13108 rev/day O 3.20108 rev/day O 3.32108 rev/day 1.70108 rev/day O 2.45108 rev/day

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Central Force

Central force is the force that acts along the center of mass of one body to the center of mass of another body. Its direction from or away from a point on a body to another is called the center of the force. Such forces include gravitational force between two massive objects or electrostatic forces between two charged particles. So from atoms to planets, the central forces play an important role in shaping our daily lives. One of the remarkable features of central forces is that they are conservative, i.e. the total energy of a central force system can always be expressed as a gradient of a potential energy function.

Question

### Angular Momentum and Neutron Star Formation

In this educational example, we explore the fascinating dynamics of a star's collapse into a neutron star and how its rotational characteristics change due to conservation of angular momentum.

#### Scenario:

- A star completes one revolution every 10 days.
- Initial radius: \(7.00 \times 10^5 \) km.
- The star collapses into a neutron star:
- New radius: 20.0 km.
- The star loses one half of its mass during the collapse.

#### Question:

**What is the new rate of revolutions per day?**
(The moment of inertia of a star is given by \( \frac{2}{5} mr^2 \).)

#### Multiple Choice Options:

- \(3.13 \times 10^8 \) rev/day
- \(3.20 \times 10^8 \) rev/day
- \(3.32 \times 10^8 \) rev/day
- \(1.70 \times 10^8 \) rev/day
- \(2.45 \times 10^8 \) rev/day

#### Explanation:

To find the new revolutions per day, consider the conservation of angular momentum. Angular momentum \(L\) is conserved and is defined as \(L = I \omega\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

1. **Initial Moment of Inertia (\(I_{\text{initial}}\)):**
\[
I_{\text{initial}} = \frac{2}{5} m r_{\text{initial}}^2
\]
Given:
- Initial radius \( r_{\text{initial}} = 7.00 \times 10^5 \) km.

2. **New Moment of Inertia (\(I_{\text{final}}\)):**
\[
I_{\text{final}} = \frac{2}{5} \left( \frac{m}{2} \right) r_{\text{final}}^2
\]
Given:
- Final radius \( r_{\text{final}} = 20.0 \) km.
- The mass is halved: \( \frac{m}{2} \).

3. **Calculate Initial Angular Velocity (\(\omega_{\text{initial}}\)):**
- The period (\(T\

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