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The terminal speed for a person falling spread eagle is given by:

(a) v = √(2mg/(rhoACd))
(b) v = (2mg/(rhoACd))^2
(c) v = 2mg/(rhoACd)
(d) v = (rhoACd)/(2mg)

User James Wood
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Final answer:

The terminal speed for a person falling spread-eagle can be calculated using the formula v = √(2mg/(ρACd)), considering the weight and area of the person facing the fluid. Example calculations can determine the terminal velocity for a skydiver falling spread-eagle.

Step-by-step explanation:

The terminal speed for a person falling spread eagle is given by the formula v = √(2mg/(ρACd)), where v is the terminal velocity, m is the person's weight, ρ is the density of the medium (air), A is the frontal cross-sectional area facing the fluid, and Cd is the drag coefficient. This formula takes into account the weight and area of the person facing the fluid.



For example, a skydiver with a weight of 85 kg falling spread eagle can calculate their terminal velocity using the formula v = √(2 * 9.8 m/s^2 * 85 kg / (1.225 kg/m^3 * 0.7 * 0.85)). Plugging in the values would give the terminal velocity in m/s.



Keywords: terminal speed, spread eagle, formula, person's weight, terminal velocity, density, frontal cross-sectional area, drag coefficient, skydiver

User Mludd
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