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An object moves downward through still air. Which statement about the drag force is correct?
Drag acts downward because the object is moving downward Drag acts upward because it opposes the velocity Drag is zero once the object reaches terminal speed Drag acts upward only while the object is speeding up In the linear-drag model $F_{\text{drag}}=-kv$, what are the SI units of $k$?
$\text{kg s}^{-1}$ $\text{kg m}^{-1}$ $\text{N}$ $\text{s}^{-1}$ For linear drag with downward-positive convention, $m\,dv/dt=mg-kv$. What is the terminal speed $v_t$?
$v_t=mgk$ $v_t=mg/k$ $v_t=\sqrt{mg/k}$ $v_t=k/(mg)$ For linear drag with $v(0)=0$, the solution is $v(t)=v_t(1-e^{-t/\tau})$. What is the time constant $\tau$?
$\tau=k/m$ $\tau=m/k$ $\tau=mg/k$ $\tau=k/(mg)$ An object is released from rest at $t=0$ and falls under gravity with linear drag. With terminal speed $v_t$ and time constant $\tau$, how long does it take to reach $0.90\,v_t$?
$0.90\,\tau$ $\tau\ln 10\approx 2.30\,\tau$ $\tau\ln 2\approx 0.69\,\tau$ $10\,\tau$ In the quadratic-drag model $m\,dv/dt=mg-bv^2$, what is the terminal speed?
$v_t=mg/b$ $v_t=\sqrt{mg/b}$ $v_t=mg\,b$ $v_t=\sqrt{b/(mg)}$ For quadratic drag from rest, $v(t)=v_t\tanh(gt/v_t)$. What is $v(t)$ for very small $t$?
$v\approx v_t$ $v\approx gt$ $v\approx g^2t^2$ $v\approx 0$ for small $t$ Which expression matches the common aerodynamic drag form $F_{\text{drag}}=\tfrac12\rho C_D A v^2$ to $F_{\text{drag}}=bv^2$?
$b=\rho C_D A$ $b=\tfrac12\rho C_D A$ $b=\tfrac12\rho C_D /A$ $b=\tfrac12 C_D A/\rho$ In quadratic drag, if the cross-sectional area $A$ doubles (all else fixed), what happens to terminal speed $v_t$?
Doubles Halves Decreases by a factor of $\sqrt{2}$ Increases by a factor of $\sqrt{2}$ At terminal speed under drag, which statement is correct?
Net force is zero and acceleration is zero Net force is zero but acceleration is non-zero Net force is non-zero but acceleration is zero Both net force and acceleration are non-zero