23) A spring has a length of 1.0 meter when there is no tension on it. The spring is then stretched between two points 10 meters apart. A wave pulse travels between the two end points in the spring in a time of 1.0 seconds. The spring is now stretched between two points that are 20 meters apart. The new time it takes for a wave pulse to travel between the ends of the spring is closest to which of the following?

A) 0.5 seconds
B) 0.7 seconds
C) 1 second
D) 1.4 seconds
E) 2 seconds
1. **Determine the wave speed in the spring:**
   The wave speed \( v \) in a medium under tension \( T \), with mass \( M \) and length \( L \), is given by:
   \[
   v = \sqrt{\frac{T}{M/L}}
   \]
   Here, \( T \) is the tension, \( M \) is the mass, and \( L \) is the length of the spring.

2. **Express the tension in terms of the spring constant:**
   For a spring, the tension \( T \) is related to the spring constant \( k \) and the length \( L \) by Hooke's Law:
   \[
   T = kL
   \]
   Substituting this into the wave speed formula, we get:
   \[
   v = \sqrt{\frac{kL}{M/L}} = \sqrt{\frac{kL^2}{M}}
   \]

3. **Simplify the wave speed expression:**
   Since \( k \) and \( M \) are constants, the wave speed \( v \) is proportional to \( L \):
   \[
   v \propto L\sqrt{\frac{k}{M}}
   \]
   Therefore, the wave speed \( v \) can be written as:
   \[
   v = C \cdot L
   \]
   where \( C \) is a constant that includes \( \sqrt{\frac{k}{M}} \).

4. **Calculate the initial wave speed:**
   Initially, the spring is stretched to 10 meters, and the wave pulse travels this distance in 1.0 seconds. Thus, the initial wave speed \( v_1 \) is:
   \[
   v_1 = \frac{10 \text{ meters}}{1.0 \text{ seconds}} = 10 \text{ m/s}
   \]

5. **Determine the new wave speed:**
   When the spring is stretched to 20 meters, the new wave speed \( v_2 \) is:
   \[
   v_2 = C \cdot 20
   \]
   Since \( v_1 = C \cdot 10 \), we have:
   \[
   10 = C \cdot 10 \implies C = 1
   \]
   Therefore:
   \[
   v_2 = 1 \cdot 20 = 20 \text{ m/s}
   \]

6. **Calculate the new travel time:**
   The new time \( t_2 \) it takes for a wave pulse to travel between the ends of the spring is:
   \[
   t_2 = \frac{\text{distance}}{\text{speed}} = \frac{20 \text{ meters}}{20 \text{ m/s}} = 1.0 \text{ seconds}
   \]

Conclusion:
The time it takes for the wave pulse to travel between the ends of the spring remains the same, which is 1.0 seconds.

The final answer is \( \boxed{1 \text{ second}} \)