The correct equation for terminal velocity \( V_t \) is:
\[ V_t = \sqrt{\frac{2mg}{\rho C_d A}} \]
where:
- \( m \) is the mass of the falling object,
- \( g \) is the acceleration due to gravity,
- \( \rho \) is the density of the fluid,
- \( C_d \) is the drag coefficient,
- \( A \) is the cross-sectional area of the object.
However, none of the provided options directly match this standard equation format. An equivalent formulation may appear depending on the variables used. Among the options you provided, the one closest to standard physics expressions—taking into account some possible variable substitutions relevant to different contexts—may not be explicit.
If we denote \( w \) as weight (which is \( mg \)), the first response can be rewritten as: \[ V_t = \sqrt{\frac{2w}{\rho C_d A}} \]
This is a valid rearrangement if we take \( w \) as the weight of the object. Therefore, the first option:
\[ V_t = \sqrt{\frac{2w}{d A C_d}} \]
could be interpreted based on context (considering \( d \) is the density \(\rho\)), as a valid terminal velocity equation under appropriate assumptions.
If it's typically represented, the key take from this is understanding that \( V_t \) depends fundamentally on the object's properties and those of the fluid it moves through, thus confirming that option provides a proper representation in context depends significantly on how symbols are conventionally defined in the scenario.
Hence - If \(d\) equals the fluid density \(\rho\), then the first option accurately conveys the relationship for terminal velocity where \( w \) corresponds to weight effectively in practice.