12. A particle of mass \( m \) falls against a resistance, per unit mass, proportional to its velocity \( v \). Given that \( k>0 \), the terminal velocity is A \( k v \) B \( \frac{1}{s} \) C \( \frac{ \pi}{k} \) D kg
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When a particle falls, it experiences two forces: its weight downward (\( mg \)) and the resistive force upward, which is proportional to its velocity (\( kv \)). At terminal velocity, these forces balance out, so we have \( mg = kv_t \), where \( v_t \) is the terminal velocity. Solving for \( v_t \) gives us \( v_t = \frac{mg}{k} \). Thus, with some simplification based on the context and variables, the answer ties back to choices that could refer to mass or dimensions. Let’s continually explore how forces interact; it’s fascinating how simple physics governs such phenomena! Understanding terminal velocity not only helps in physics but can also be insightful for projects like skydiving or designing skydiving equipment. For those looking to delve deeper into motion under gravity and resistive forces, consider classic texts like "Physics for Scientists and Engineers" by Serway and Jewett. It offers a thorough treatment packed with engaging problems and illustrations!