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jk_16
 3 years ago
in air resistance (drag) is acceleration constant??
jk_16
 3 years ago
in air resistance (drag) is acceleration constant??

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alexray19
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, it is a force. But its magnitude, direction, behavior, etc is a function of several other variables related to the nature of the object experiencing the resistance (e.g. its shape, velocity, its environment, the properties of the air, etc).

vf321
 3 years ago
Best ResponseYou've already chosen the best response.0What do you mean? Acceleration has no influence on drag force, so \(\frac{dF_{drag}}{da}=0\) , but it has an influence on the acceleration since it's a force.

vf321
 3 years ago
Best ResponseYou've already chosen the best response.0***Instantaneous acc has no influence on drag

alexray19
 3 years ago
Best ResponseYou've already chosen the best response.0Assuming ceteris paribus, as an object travels at a constant velocity in a straight line through nonturbulent air, it experiences a constant acceleration backwards caused by the air.

jk_16
 3 years ago
Best ResponseYou've already chosen the best response.0the grap of a vt system under the influence of air resistance..is not a constant v(t)

alexray19
 3 years ago
Best ResponseYou've already chosen the best response.0That is, in order to maintain constant velocity, the object must constantly be accelerating forward with a force equal but opposite to the force caused by the air resistance. It's like driving your car at high speeds; you have to constantly give it some gas to maintain your velocity, which is goes against Newton's law that an object in motion remains in motion. The resistance caused by air (and internal friction) accounts for this need to constantly accelerate the car.

vf321
 3 years ago
Best ResponseYou've already chosen the best response.0If an object is ONLY encountering drag, \[F = \frac{1}{2}cAv^2\]\[m a = \frac{1}{2}cAv^2\]\[k=.5cA/m; \frac{d v}{dt}=kv^2\]

jk_16
 3 years ago
Best ResponseYou've already chosen the best response.0@vf321 where did you get the initial F=1/2 cAv^2 from

vf321
 3 years ago
Best ResponseYou've already chosen the best response.0That's the formula for drag, if I remember correctly.

jk_16
 3 years ago
Best ResponseYou've already chosen the best response.0F=12cAv2 ma=12cAv2 k=.5cA/m;dvdt=kv2
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