College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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I don't understand how you're getting the Ax
![## Diagram Explanation and Calculation
### Diagram Description
The provided diagram depicts a right-angled triangle where the following components are identified:
- The hypotenuse (A) measures 1.5 km.
- The opposite side (A_y) and the adjacent side (A_x) are the sides of the triangle with respect to the 45° angle.
- The adjacent side (A_x) is given as -√2 km (approximately -20 km).
- The opposite side (A_y) is to be determined.
### Point Coordinates
The coordinates of the point are (x=0 km, y=1.06 km).
### Given
\[ A = 1.5 \, \text{km} \]
\[ A_x = -20 \, \text{km} \]
\[ \theta = 45^\circ \]
### Problem
1. To find the component \( A_y \).
### Calculation
To determine \( A_y \):
Using the sine function,
\[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \]
\[ \sin(45^\circ) = \frac{A_y}{1.5 \, \text{km}} \]
\[ \frac{\sqrt{2}}{2} = \frac{A_y}{1.5 \, \text{km}} \]
Solving for \( A_y \):
\[ A_y = \left( \frac{\sqrt{2}}{2} \right) \times 1.5\, \text{km} \]
\[ A_y = \frac{1.5 \sqrt{2}}{2} \]
\[ A_y \approx 1.06 \, \text{km} \]
Thus,
\[ A_y = 1.06 \, \text{km} \]
This calculation shows that the vertical component \( A_y \) of the vector is approximately 1.06 km.
### Summary
The length of \( A_y \) obtained from the given diagram and calculations is:
\[ \boxed{A_y = 1.06 \, \text{km}} \]](https://content.bartleby.com/qna-images/question/9d4854d9-aadf-4fe0-b1f6-bd622f90aeab/342d9a08-f28d-49ac-a759-9e4f497b8a15/prhblnr_processed.jpeg)
Transcribed Image Text:## Diagram Explanation and Calculation
### Diagram Description
The provided diagram depicts a right-angled triangle where the following components are identified:
- The hypotenuse (A) measures 1.5 km.
- The opposite side (A_y) and the adjacent side (A_x) are the sides of the triangle with respect to the 45° angle.
- The adjacent side (A_x) is given as -√2 km (approximately -20 km).
- The opposite side (A_y) is to be determined.
### Point Coordinates
The coordinates of the point are (x=0 km, y=1.06 km).
### Given
\[ A = 1.5 \, \text{km} \]
\[ A_x = -20 \, \text{km} \]
\[ \theta = 45^\circ \]
### Problem
1. To find the component \( A_y \).
### Calculation
To determine \( A_y \):
Using the sine function,
\[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \]
\[ \sin(45^\circ) = \frac{A_y}{1.5 \, \text{km}} \]
\[ \frac{\sqrt{2}}{2} = \frac{A_y}{1.5 \, \text{km}} \]
Solving for \( A_y \):
\[ A_y = \left( \frac{\sqrt{2}}{2} \right) \times 1.5\, \text{km} \]
\[ A_y = \frac{1.5 \sqrt{2}}{2} \]
\[ A_y \approx 1.06 \, \text{km} \]
Thus,
\[ A_y = 1.06 \, \text{km} \]
This calculation shows that the vertical component \( A_y \) of the vector is approximately 1.06 km.
### Summary
The length of \( A_y \) obtained from the given diagram and calculations is:
\[ \boxed{A_y = 1.06 \, \text{km}} \]
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