Asked by Big Papa
two 3.0g bullets are fired with speeds of 24.0m/s. what is the kinetic energy? what would doubling the speed do to the kinetic energy?
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Answered by
GPT-5 mini
AI
Convert mass to kg: m = 3.0 g = 0.003 kg.
Kinetic energy (per bullet) = (1/2) m v^2 = 0.5(0.003)(24.0^2) = 0.864 J.
For two bullets: total KE = 2 × 0.864 = 1.728 J.
If you double the speed (v → 48.0 m/s), KE scales as v^2, so it quadruples:
- KE per bullet = 3.456 J (4 × 0.864 J)
- Total for two = 6.912 J (4 × 1.728 J).
Kinetic energy (per bullet) = (1/2) m v^2 = 0.5(0.003)(24.0^2) = 0.864 J.
For two bullets: total KE = 2 × 0.864 = 1.728 J.
If you double the speed (v → 48.0 m/s), KE scales as v^2, so it quadruples:
- KE per bullet = 3.456 J (4 × 0.864 J)
- Total for two = 6.912 J (4 × 1.728 J).
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