### Problem StatementSuppose the particles shown below have masses \( m_1 = 0.18 \, \text{kg} \), \( m_2 = 0.36 \, \text{kg} \), \( m_3 = 0.54 \, \text{kg} \), \( m_4 = 0.72 \, \text{kg} \). The velocities of the particles are \( \mathbf{v}_1 = 3.5 \hat{\mathbf{i}} \, \text{m/s} \), \( \mathbf{v}_2 = (5.3 \hat{\mathbf{i}} - 5.3 \hat{\mathbf{j}}) \, \text{m/s} \), \( \mathbf{v}_3 = -2.6 \hat{\mathbf{j}} \, \text{m/s} \), \( \mathbf{v}_4 = -7.0 \hat{\mathbf{i}} \, \text{m/s} \). (Express your answers in vector form.)### Diagram ExplanationThe diagram consists of a coordinate system with the x and y axes marked in meters (m). Each particle is represented by a point, with labeled masses and velocity vectors:- **Particle \( m_1 \):** Positioned at \( (0, 2.0) \, \text{m} \), moving with velocity \( \mathbf{v}_1 = 3.5 \hat{\mathbf{i}} \).- **Particle \( m_2 \):** Positioned at \( (4.0, -2.0) \, \text{m} \), moving with velocity \( \mathbf{v}_2 = (5.3 \hat{\mathbf{i}} - 5.3 \hat{\mathbf{j}}) \).- **Particle \( m_3 \):** Positioned at \( (-2.0, -4.0) \, \text{m} \), moving with velocity \( \mathbf{v}_3 = -2.6 \hat{\mathbf{j}} \).- **Particle \( m_4 \):** Positioned at \( (4.0, 0) \, \text{m} \), moving with velocity \( \mathbf{v}_4 = -7.0 \hat{\mathbf{i}} \).### Questions(a) Calculate the angular **Title: Determining Angular Momentum Using the Right-Hand Rule****Instructions:**Use the right-hand rule to determine the directions of the angular momenta about the origin of the particles as shown below. The +z-axis is out of the page.**Diagram Explanation:**A two-dimensional plot displays four masses, \( m_1, m_2, m_3, \) and \( m_4 \), positioned on the xy-plane.- **Mass \( m_1 \):** Located at (2, 2) with velocity vector \( \vec{v_1} \) pointing in the positive x-direction.- **Mass \( m_2 \):** Placed at (2, -3) with velocity vector \( \vec{v_2} \) pointing upwards in the third quadrant.- **Mass \( m_3 \):** Positioned at (-3, 2) with velocity vector \( \vec{v_3} \) directed downwards in the fourth quadrant.- **Mass \( m_4 \):** Situated at (4, 0) with velocity vector \( \vec{v_4} \) extending to the right along the x-axis.**Legend:**- \( \vec{L_1} \) direction: [Dropdown] (Selected: -x) ❌- \( \vec{L_2} \) direction: [Dropdown] (Selected: -y) ❌- \( \vec{L_3} \) direction: [Dropdown] (Selected: -y) ❌- \( \vec{L_4} \) direction: [Dropdown] (Selected: -z) ❌**Method:**To determine the direction of angular momentum (\( \vec{L} \)), use the right-hand rule: point your fingers in the direction of the radius vector (\( \vec{r} \)) from the origin to the point mass, and curl them towards the velocity vector (\( \vec{v} \)). Your thumb will point in the direction of the angular momentum vector (\( \vec{L} \)).

Name: ### Problem StatementSuppose the particles shown below have masses \(
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Description: ### Problem StatementSuppose the particles shown below have masses \( m_1 = 0.18 \, \text{kg} \), \( m_2 = 0.36 \, \text{kg} \), \( m_3 = 0.54 \, \text{kg} \), \( m_4 = 0.72 \, \text{kg} \). The velocities of the particles are \( \mathbf{v}_1 = 3.5

The given masses and there respective velocities are m1=0.18 kg; m2=0.36 kg; m3=0.54 kg; m4=0.72 kgv1=3.5 i^ m/s; v2=5.3i^-5.3j^ m/s; v3=-26 j^ m/s; v4=-7 i^ m/s Angular momentum about the given point is defined as the product of the mass, velocity and the perpendicular distance form the point to the velocity vector. l =mvr*sin(θ)l = angular momentumm = massv =velocityr = radius andθ = angle between the velocity and the radius vector l1 = m1v1r1sin(θ1) =(0.18 kg)(3.5 m/s)(2 m)(sin(90°)) =1.26 kgm2/sl2 = m2v2r2sin(θ2) =(0.36 kg)(7.5 m/s)(2.82 m)(sin(0°)) =0l3 = m3v3r3sin(θ3) =(0.54 kg)(-2.6 m/s)(3 m)(sin(90°)) {from the figure: 3 is the perpendicular distance with v3} =-4.21 kgm2/sl4 = m4v4r4sin(θ4) =(0.72 kg)(-7 m/s)(4 m)(sin(0°)) =0b) The total angular momentum is the sum of all angular momentum l =l1+ l2+ l3+ l4 =1.26 + 0 + (-4.21) + 0 =-2.95 khm2/s hence the total momentum is l =-2.95 khm2/s 2) DIRECTION Right hand rule: Stretch the fore finger, middle finger and the thumb of your right hand perpendicular to each other and point fore finger along vector a and the middle finger along vector b then the thumb will be pointing towards the cross product of the two as: Fore finger ⇒ a→Middle finger ⇒ b→Thumb ⇒a→× b→ Now for the given situation Fore finger ⇒ v→Middle finger ⇒ r→Thumb ⇒l→ l1 ⇒+Z directionl2 ⇒0l1 ⇒-Z directionl1 ⇒0 Thank you!

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Physics

Suppose the particles shown below have masses m, = 0.18 kg, m, = 0.36 kg, m, = 0.54 kg, m, = 0.72 kg. The velocities of the particles are v, = 3.5î m/s, v, = (5.3î - 5.3j) m/s, v, = -2.6j m/s, V, = -7.0î m/s. (Express your answers in vector form.) y(m)4 4.0 m, 2.0 m4 -4.0 -2.0 2.0 4.0 x(m) -2.0- m2. -4.0- V2 (a) Calculate the angular momentum of each particle about the origin (in kg · m2/s). i, = kg · m2/s 1, = kg • m2/s kg • m2/s kg - m2/s (b) What is the total angular momentum of the four-particle system about the origin (in kg · m2/s)? kg • m2/s

Suppose the particles shown below have masses m, = 0.18 kg, m, = 0.36 kg, m, = 0.54 kg, m, = 0.72 kg. The velocities of the particles are v, = 3.5î m/s, v, = (5.3î - 5.3j) m/s, v, = -2.6j m/s, V, = -7.0î m/s. (Express your answers in vector form.) y(m)4 4.0 m, 2.0 m4 -4.0 -2.0 2.0 4.0 x(m) -2.0- m2. -4.0- V2 (a) Calculate the angular momentum of each particle about the origin (in kg · m2/s). i, = kg · m2/s 1, = kg • m2/s kg • m2/s kg - m2/s (b) What is the total angular momentum of the four-particle system about the origin (in kg · m2/s)? kg • m2/s

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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### Problem Statement

Suppose the particles shown below have masses \( m_1 = 0.18 \, \text{kg} \), \( m_2 = 0.36 \, \text{kg} \), \( m_3 = 0.54 \, \text{kg} \), \( m_4 = 0.72 \, \text{kg} \). The velocities of the particles are \( \mathbf{v}_1 = 3.5 \hat{\mathbf{i}} \, \text{m/s} \), \( \mathbf{v}_2 = (5.3 \hat{\mathbf{i}} - 5.3 \hat{\mathbf{j}}) \, \text{m/s} \), \( \mathbf{v}_3 = -2.6 \hat{\mathbf{j}} \, \text{m/s} \), \( \mathbf{v}_4 = -7.0 \hat{\mathbf{i}} \, \text{m/s} \). (Express your answers in vector form.)

### Diagram Explanation

The diagram consists of a coordinate system with the x and y axes marked in meters (m). Each particle is represented by a point, with labeled masses and velocity vectors:

- **Particle \( m_1 \):** Positioned at \( (0, 2.0) \, \text{m} \), moving with velocity \( \mathbf{v}_1 = 3.5 \hat{\mathbf{i}} \).
- **Particle \( m_2 \):** Positioned at \( (4.0, -2.0) \, \text{m} \), moving with velocity \( \mathbf{v}_2 = (5.3 \hat{\mathbf{i}} - 5.3 \hat{\mathbf{j}}) \).
- **Particle \( m_3 \):** Positioned at \( (-2.0, -4.0) \, \text{m} \), moving with velocity \( \mathbf{v}_3 = -2.6 \hat{\mathbf{j}} \).
- **Particle \( m_4 \):** Positioned at \( (4.0, 0) \, \text{m} \), moving with velocity \( \mathbf{v}_4 = -7.0 \hat{\mathbf{i}} \).

### Questions

(a) Calculate the angular

**Title: Determining Angular Momentum Using the Right-Hand Rule**

**Instructions:**
Use the right-hand rule to determine the directions of the angular momenta about the origin of the particles as shown below. The +z-axis is out of the page.

**Diagram Explanation:**
A two-dimensional plot displays four masses, \( m_1, m_2, m_3, \) and \( m_4 \), positioned on the xy-plane.

- **Mass \( m_1 \):** Located at (2, 2) with velocity vector \( \vec{v_1} \) pointing in the positive x-direction.
- **Mass \( m_2 \):** Placed at (2, -3) with velocity vector \( \vec{v_2} \) pointing upwards in the third quadrant.
- **Mass \( m_3 \):** Positioned at (-3, 2) with velocity vector \( \vec{v_3} \) directed downwards in the fourth quadrant.
- **Mass \( m_4 \):** Situated at (4, 0) with velocity vector \( \vec{v_4} \) extending to the right along the x-axis.

**Legend:**
- \( \vec{L_1} \) direction: [Dropdown] (Selected: -x) ❌
- \( \vec{L_2} \) direction: [Dropdown] (Selected: -y) ❌
- \( \vec{L_3} \) direction: [Dropdown] (Selected: -y) ❌
- \( \vec{L_4} \) direction: [Dropdown] (Selected: -z) ❌

**Method:**
To determine the direction of angular momentum (\( \vec{L} \)), use the right-hand rule: point your fingers in the direction of the radius vector (\( \vec{r} \)) from the origin to the point mass, and curl them towards the velocity vector (\( \vec{v} \)). Your thumb will point in the direction of the angular momentum vector (\( \vec{L} \)).

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