3. In the above problem, what is, ĮD|, the magnitude of vector Ď = Ä – B ? (A) 1.0 cm (B) 1.1 cm (C) 1.2 cm (D) 1.4 cm

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ChapterP: Prerequisites: Fundamental Concepts Of Algebra
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Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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### Physics Problems - Vector Magnitudes and Relative Velocity

#### Problem 3:
In the above problem, what is \(|\vec{D}|\), the magnitude of vector \(\vec{D} = \vec{A} - \vec{B}\)?
- (A) 1.0 cm
- (B) 1.1 cm
- (C) 1.2 cm
- (D) 1.4 cm

#### Problem 4:
A motorboat heads due east at 5.0 m/s across a river that flows toward the south at a speed of 5.0 m/s. What is the velocity relative to an observer on the shore?
- (A) 3.2 m/s to the southeast
- (B) 5.0 m/s to the southeast
- (C) 7.1 m/s to the southeast
- (D) 10.0 m/s to the southeast

### Explanation:

**For Problem 3:**
The problem requires determining the magnitude of the resultant vector \(\vec{D}\) when vector \(\vec{B}\) is subtracted from vector \(\vec{A}\). The correct answer can be chosen from options (A) to (D) based on the given numerical values.

**For Problem 4:**
The problem states that a motorboat moves due east at 5.0 m/s while a river flows south at 5.0 m/s. To find the velocity of the boat relative to an observer on the shore, we need to consider both velocities as perpendicular components of the resultant velocity.
- The eastward velocity component: \(\vec{v}_\text{boat} = 5.0 \, \text{m/s}\)
- The southward velocity component: \(\vec{v}_\text{river} = 5.0 \, \text{m/s}\)

Using the Pythagorean theorem:
\[ v_{\text{resultant}} = \sqrt{(v_{\text{boat}})^2 + (v_{\text{river}})^2} \]

\[ v_{\text{resultant}} = \sqrt{(5.0 \, \text{m/s})^2 + (5.0 \, \text{m/s})^2} \]

\[ v_{\text{resultant}} = \sqrt{25 +
Transcribed Image Text:### Physics Problems - Vector Magnitudes and Relative Velocity #### Problem 3: In the above problem, what is \(|\vec{D}|\), the magnitude of vector \(\vec{D} = \vec{A} - \vec{B}\)? - (A) 1.0 cm - (B) 1.1 cm - (C) 1.2 cm - (D) 1.4 cm #### Problem 4: A motorboat heads due east at 5.0 m/s across a river that flows toward the south at a speed of 5.0 m/s. What is the velocity relative to an observer on the shore? - (A) 3.2 m/s to the southeast - (B) 5.0 m/s to the southeast - (C) 7.1 m/s to the southeast - (D) 10.0 m/s to the southeast ### Explanation: **For Problem 3:** The problem requires determining the magnitude of the resultant vector \(\vec{D}\) when vector \(\vec{B}\) is subtracted from vector \(\vec{A}\). The correct answer can be chosen from options (A) to (D) based on the given numerical values. **For Problem 4:** The problem states that a motorboat moves due east at 5.0 m/s while a river flows south at 5.0 m/s. To find the velocity of the boat relative to an observer on the shore, we need to consider both velocities as perpendicular components of the resultant velocity. - The eastward velocity component: \(\vec{v}_\text{boat} = 5.0 \, \text{m/s}\) - The southward velocity component: \(\vec{v}_\text{river} = 5.0 \, \text{m/s}\) Using the Pythagorean theorem: \[ v_{\text{resultant}} = \sqrt{(v_{\text{boat}})^2 + (v_{\text{river}})^2} \] \[ v_{\text{resultant}} = \sqrt{(5.0 \, \text{m/s})^2 + (5.0 \, \text{m/s})^2} \] \[ v_{\text{resultant}} = \sqrt{25 +
The image displays a coordinate system with the \(x\)-axis and \(y\)-axis. 

Two vectors are represented in this system:
1. Vector \(\vec{A}\) is positioned along the positive \(x\)-axis and has a magnitude of 2.0 cm.
2. Vector \(\vec{B}\) is oriented at an angle \(\alpha = 30^\circ\) relative to the positive \(x\)-axis. This vector also has a magnitude of 2.0 cm.

The image shows that the tail of each vector is positioned at the origin (0,0) of the coordinate system. Vector \(\vec{A}\) extends horizontally to the right, while vector \(\vec{B}\) extends diagonally upward to the right, forming a 30-degree angle with vector \(\vec{A}\).

This diagram can be useful for understanding the representation of vectors in a coordinate system, the concept of vector magnitude and direction, and how trigonometric functions can be applied to decompose vectors into their components.
Transcribed Image Text:The image displays a coordinate system with the \(x\)-axis and \(y\)-axis. Two vectors are represented in this system: 1. Vector \(\vec{A}\) is positioned along the positive \(x\)-axis and has a magnitude of 2.0 cm. 2. Vector \(\vec{B}\) is oriented at an angle \(\alpha = 30^\circ\) relative to the positive \(x\)-axis. This vector also has a magnitude of 2.0 cm. The image shows that the tail of each vector is positioned at the origin (0,0) of the coordinate system. Vector \(\vec{A}\) extends horizontally to the right, while vector \(\vec{B}\) extends diagonally upward to the right, forming a 30-degree angle with vector \(\vec{A}\). This diagram can be useful for understanding the representation of vectors in a coordinate system, the concept of vector magnitude and direction, and how trigonometric functions can be applied to decompose vectors into their components.
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