**13. You are on a train that is traveling at 3.0 m/s along a level straight track. Very near and parallel to the track is a wall that slopes upward at a 12° angle with the horizontal. As you face the window (0.90 m high, 2.0 m wide) in your compartment, the train is moving to the left, as the drawing indicates. The top edge of the wall first appears at window cor- ner A and eventually disappears at window corner B. How much time passes between appearance and disappearance of the upper edge of the wall? A km) mangle Problem 13 A 3.0 m/s 12° B

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### Problem 13

#### Scenario:
You are on a train that is traveling at 3.0 m/s along a level straight track. Very near and parallel to the track is a wall that slopes upward at a 12° angle with the horizontal. As you face the window (0.90 m high, 2.0 m wide) in your compartment, the train is moving to the left, as the drawing indicates. The top edge of the wall first appears at window corner A and eventually disappears at window corner B.

#### Question:
How much time passes between the appearance and disappearance of the upper edge of the wall?

#### Diagram Explanation:
1. The diagram shows the window of the train compartment (0.90 m high and 2.0 m wide) and the wall sloping upward at an angle of 12°.
2. The upper edge of the wall first appears at point A (lower-left corner of the window).
3. The top edge of the wall moves across the window to point B (upper-right corner of the window) as the train moves.
4. The train is traveling to the left at a speed of 3.0 m/s.

To find the time that passes between the appearance and disappearance of the upper edge of the wall, we need to analyze the relative motion of the train and the wall's slope. Here is the detailed approach:

#### Calculations: 

1. The change in height of the top edge of the wall as it moves across.
   - Window height = 0.90 m
   - Window width = 2.0 m

2. Determine the horizontal distance for the top of the wall to appear at point B:
   - \[ \tan(12^\circ) = \frac{0.90 \, \text{m}}{x} \]
   - Solve for x: \[ x = \frac{0.90 \, \text{m}}{\tan(12^\circ)} \approx 4.27 \, \text{m} \]

3. The time it takes for the top of the wall to move across the window:
   - Distance = 2.0 m (width of the window)
   - Speed of the train = 3.0 m/s

The horizontal speed of the train relative to the wall is:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac
Transcribed Image Text:### Problem 13 #### Scenario: You are on a train that is traveling at 3.0 m/s along a level straight track. Very near and parallel to the track is a wall that slopes upward at a 12° angle with the horizontal. As you face the window (0.90 m high, 2.0 m wide) in your compartment, the train is moving to the left, as the drawing indicates. The top edge of the wall first appears at window corner A and eventually disappears at window corner B. #### Question: How much time passes between the appearance and disappearance of the upper edge of the wall? #### Diagram Explanation: 1. The diagram shows the window of the train compartment (0.90 m high and 2.0 m wide) and the wall sloping upward at an angle of 12°. 2. The upper edge of the wall first appears at point A (lower-left corner of the window). 3. The top edge of the wall moves across the window to point B (upper-right corner of the window) as the train moves. 4. The train is traveling to the left at a speed of 3.0 m/s. To find the time that passes between the appearance and disappearance of the upper edge of the wall, we need to analyze the relative motion of the train and the wall's slope. Here is the detailed approach: #### Calculations: 1. The change in height of the top edge of the wall as it moves across. - Window height = 0.90 m - Window width = 2.0 m 2. Determine the horizontal distance for the top of the wall to appear at point B: - \[ \tan(12^\circ) = \frac{0.90 \, \text{m}}{x} \] - Solve for x: \[ x = \frac{0.90 \, \text{m}}{\tan(12^\circ)} \approx 4.27 \, \text{m} \] 3. The time it takes for the top of the wall to move across the window: - Distance = 2.0 m (width of the window) - Speed of the train = 3.0 m/s The horizontal speed of the train relative to the wall is: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac
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