Suppose Students A and B stand 50 m from Student C. Students B and C each hold a copper cup, and the two cups are attached by a copper wire measuring 50 m . If Student A shouts at Student C and Student B shouts into the copper cup, who will Student C hear first?

To determine who Student C will hear first, we need to consider the speed of sound in air and the speed of electrical signals in a copper wire. 1. **Speed of Sound in Air**: The speed of sound in air is approximately 343 meters per second (m/s) at room temperature. 2. **Speed of Electrical Signals in Copper Wire**: The speed of electrical signals in a copper wire is about two-thirds the speed of light, which is approximately \(2 \times 10^8\) meters per second (m/s). The speed of light in a vacuum is about \(3 \times 10^8\) m/s, so in copper, it would be around \(2 \times 10^8\) m/s. Now, let's calculate the time it takes for each sound to reach Student C: - **Time for Student A's shout to reach Student C**: \[ \text{Distance} = 50 \text{ m} \] \[ \text{Speed of sound} = 343 \text{ m/s} \] \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{50 \text{ m}}{343 \text{ m/s}} \approx 0.146 \text{ seconds} \] - **Time for Student B's shout through the copper wire to reach Student C**: \[ \text{Distance} = 50 \text{ m} \] \[ \text{Speed of electrical signals} \approx 2 \times 10^8 \text{ m/s} \] \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{50 \text{ m}}{2 \times 10^8 \text{ m/s}} = 2.5 \times 10^{-7} \text{ seconds} \approx 0.00000025 \text{ seconds} \] Comparing the two times: - Time for Student A's shout: approximately 0.146 seconds - Time for Student B's shout through the copper wire: approximately 0.00000025 seconds **Conclusion**: Student C will hear Student B's shout through the copper cup first, as the electrical signal travels much faster than sound in air.