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The problem involves sound traveling down a well and echoing back. Here's how to solve it:

### Given:
- Time for the echo to return: \( t = 7.7 \, \text{seconds} \)
- Speed of sound: \( v = 340.4 \, \text{m/s} \)

### Concept:
The sound travels down to the bottom of the well and back up, covering twice the depth of the well. Therefore, the total distance covered by the sound is \( 2d \), where \( d \) is the depth of the well.

The relationship between distance, speed, and time is given by:
\[
\text{Distance} = \text{Speed} \times \text{Time}
\]
So,
\[
2d = v \times t
\]
\[
d = \frac{v \times t}{2}
\]

### Substituting the values:
\[
d = \frac{340.4 \times 7.7}{2}
\]

Let me calculate that.The depth of the well is approximately **1310.54 meters**.

Thus, the correct answer is **b. 1310.54 meters**.

Would you like more details on the solution or have any other questions?

Here are 5 related questions to expand your understanding:
1. How would the depth change if the speed of sound were different?
2. How does temperature affect the speed of sound and, consequently, the depth calculation?
3. What is the formula for the speed of sound in air at different temperatures?
4. If Jacob had heard the echo after 15 seconds, what would the depth of the well be?
5. How do sound waves reflect differently from various surfaces?

**Tip:** The speed of sound in air changes with temperature. At higher temperatures, sound travels faster!

Echoes occur when sound reflects off a hard surface and returns to their point of origin. Jacob screams his name down to the bottom of a really deep well and hears his echo 7.7 seconds later. Knowing the speed of sound at this location is a constant 340.4 m/s, how deep is the well?

Learn how to calculate the depth of a well using echo time and the speed of sound. This problem uses the formula d = (v × t) / 2, where sound travels down and reflects back. Suitable for high school physics students.

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