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force acting upon the object. thus, to minimize the force needed to sto…

Question

force acting upon the object. thus, to minimize the force needed to stop an object, the time must be
____minimized__ (minimized/maximized).

big t … little f: some applications
| bag
decreased | force
increased | minimize
time |

there are several real - world applications of this idea. one example is the use of air bags in cars. air bags are
used in cars because they _______ the force on an object involved in a collision. air bags accomplish this
by extending the _______ required to stop the momentum of the driver and passenger. if in a collision, a
driver and passenger hit the windshield, that results in a large _______ exerted over a short time in order to
stop their momentum. if instead of hitting the windshield, the driver and passenger hit the air _______, then
the time duration of the impact is _______. if the time is increased, the force on the passenger and driver is
_______.

Explanation:

Brief Explanations
  1. For the first blank (about air bags' effect on force): From the concept of impulse ($J = F\Delta t$) and momentum change, to reduce force, with air bags, they decrease the force.
  2. Second blank (what air bags extend): They extend the time (since $F=\frac{\Delta p}{\Delta t}$, increasing $\Delta t$ decreases $F$).
  3. Third blank (when hitting windshield): Hitting windshield gives a large force (because $\Delta t$ is small, so $F$ is large from $F=\frac{\Delta p}{\Delta t}$).
  4. Fourth blank: Air bag (the object they hit).
  5. Fifth blank (time duration when hitting air bag): The time duration is increased (air bags increase the time of impact).
  6. Sixth blank (force when time is increased): From $F=\frac{\Delta p}{\Delta t}$, if $\Delta t$ increases, force is decreased (or minimized, but "decreased" fits context here).

For the initial question (minimize force to stop an object, time must be...): From $F=\frac{\Delta p}{\Delta t}$, to minimize $F$, we need to maximize $\Delta t$ (since $F$ is inversely proportional to $\Delta t$ when $\Delta p$ is constant). Wait, the original filled was "minimized" which is incorrect. Wait, let's re - check: Impulse $J = F\Delta t=\Delta p$. So if we want to minimize $F$ (force to stop, so $\Delta p$ is fixed as it's the change in momentum to stop), then $F=\frac{\Delta p}{\Delta t}$. So to minimize $F$, we need to maximize $\Delta t$. So the correct answer for the first (top) blank should be "maximized", not "minimized" as filled. But for the air - bag blanks:

  1. Air bags decrease the force.
  2. Extend the time.
  3. Large force (when hitting windshield, short time, so large force).
  4. Air bag.
  5. Time duration is increased.
  6. Force is decreased (or minimized).

Answer:

  1. (First top blank correction) maximized
  2. (Air bag force blank) decreased
  3. (extended what) time
  4. (large what) force
  5. (air what) bag
  6. (time duration) increased
  7. (force when time increased) decreased

(Note: The initial filled "minimized" for the top blank is incorrect; the correct is "maximized" based on $F=\frac{\Delta p}{\Delta t}$.)