Let's solve the problem step by step: --- ### **Step 1: Calculate the speed of the wave in the wire** The speed \( v \) of a wave on a stretched string is given by: \[ v = \sqrt{\frac{T}{\mu}} \] where - \( T \) = tension in the wire = **28.5 N** - \( \mu \) = linear density = **6.49 g/m = .00649 \, \text{kg/m}** (convert to kg/m) Plug in the values: \[ v = \sqrt{\frac{28.5}{.00649}} \approx \sqrt{4391.37} \approx 66.3 \, \text{m/s} \] **Explanation:** The speed of the wave in the wire is calculated using the tension and the mass per unit length. --- ### **Step 2: Calculate travel times for both waves and find the difference** **Distance:** \( d = 20. \) m - **Time for wave in wire (wave 1):** \[ t_1 = \frac{d}{v_{\text{wire}}} = \frac{20.}{66.3} \approx .302 \, \text{s} \] - **Time for wave in air (wave 2):** \[ t_2 = \frac{d}{v_{\text{air}}} = \frac{20.}{343} \approx .0584 \, \text{s} \] - **Time difference:** \[ \Delta t = t_1 - t_2 = .302 - .0584 = .244 \, \text{s} \] **Explanation:** The travel time through the wire is much longer than through the air. Wave 2 (air) arrives first. --- ## **Final Answer** \[ \boxed{-.244} \] (Wave 2 arrives **sooner** by .244 seconds; negative because wave 2 arrives first.)