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jk_16

  • 2 years ago

in air resistance (drag) is acceleration constant??

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  1. alexray19
    • 2 years ago
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    Yes, 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).

  2. vf321
    • 2 years ago
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    What 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.

  3. vf321
    • 2 years ago
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    ***Instantaneous acc has no influence on drag

  4. alexray19
    • 2 years ago
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    Assuming ceteris paribus, as an object travels at a constant velocity in a straight line through non-turbulent air, it experiences a constant acceleration backwards caused by the air.

  5. jk_16
    • 2 years ago
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    the grap of a v-t system under the influence of air resistance..is not a constant v(t)

  6. vf321
    • 2 years ago
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    grap?

  7. alexray19
    • 2 years ago
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    That 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.

  8. vf321
    • 2 years ago
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    If an object is ONLY encountering drag, \[F = \frac{1}{2}cAv^2\]\[m a = \frac{1}{2}cAv^2\]\[k=.5cA/m; \frac{d v}{dt}=kv^2\]

  9. jk_16
    • 2 years ago
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    @vf321 where did you get the initial F=1/2 cAv^2 from

  10. vf321
    • 2 years ago
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    That's the formula for drag, if I remember correctly.

  11. jk_16
    • 2 years ago
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    how do i derive it?

  12. vf321
    • 2 years ago
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    derive what

  13. jk_16
    • 2 years ago
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    F=12cAv2 ma=12cAv2 k=.5cA/m;dvdt=kv2

  14. vf321
    • 2 years ago
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    Well u just did it...

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