Ask Question
5 April, 09:39

A sparrow is flying around in a circle at a constant speed and height. there is air resistance. In what direction is the net force of the sparrow on the air as the sparrow flies. 1. outward and downward 2. Inward and Downward 3. Downward and Backward 4. outward, downward and backward

+3
Answers (2)
  1. 5 April, 09:42
    0
    Option 1.

    Explanation:

    As the bird flies around in a circle at constant speed and height, It exerts to force to keep that flight and continue moving.

    The first force is the lift of the bird and the other force is the thrust.

    The bird, having weight experiences a force due to gravity thus it requires a counter force, i. e., lift in order to overcome the force of gravity due to its weight to maintain flight.

    The second is due to air resistance, it experiences drag which is counter by the force that the bird applies in opposite direction in the form of thrust.
  2. 5 April, 10:07
    0
    Option 3 = Downward and Backward

    Explanation:

    While moving in a circle, the bird will experience following forces:

    Inward force (Centripetal Force) Outward force (Centrifugal Force) Backward force (due to air resistance) Downward force (due to Gravitational Force)

    The first two forces are due to circular motion and these are equal in magnitude (mv^2/r) but opposite in direction hence nullifying each other's effect.

    Air resistance is forcing the bird backward while earth's gravitational field attracting it downward by a force weight = mg (where m = mass of body; g=gravitational acceleration = 9.8m/s^2).

    Bird is putting force to overcome these two forces and eventually maintaining a uniform speed and constant height
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A sparrow is flying around in a circle at a constant speed and height. there is air resistance. In what direction is the net force of the ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers