I don't understand how you're getting the Ax

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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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:
\[ \boxed{A_y = 1.06 \, \text{km}} \]
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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