When rain drops fall under the influence of gravity, their terminal velocity is?

Question

A: same for the drops

B: inversely proportional to their size

C: directly proportional to their size

D: none of these

Please help me with this question. Answer it only if you know the answer and please explain your answer as well. I'll be very grateful to you......

Answer 1

A free falling object achieves it's terminal velocity when the downward force of gravity equals the upward force of drag.This causes the net force on the object to be zero, resulting in an acceleration of zero. Mathematically an object asymptotically approaches and can never reach it's terminal velocity.I believe that the correct answer is "D", (none of these).  GOOD LUCK!!!

Answer 2

say WHAT?????

Answer 3

minister of truth has got it right, if something is small, the force of gravity doesn't give it enough momentum to overcome the wind resistance, you can find a chart of velocities here.

http://madsci.org/posts/archives/2000-07...

Answer 4

If you assume a raindrop is a sphere, then its weight increases as the cube of its radius, so the force of gravity on the raindrop is proportional to r^3. The cross-sectional area (a circle) increases as the square of the radius. Wind resistance is proportional to the area and the square of the velocity, so the force of wind is proportional to r^2 * v^2.

r^3 = k * r^2 * v^2, where k is some constant

r = k * v^2

v = sqrt(r/k)

So the answer would be (D), none of these. Terminal velocity is directly proportional to the SQUARE ROOT of the drop size (radius or diameter)

Answer 5

The larger the drop the higher the velocity.  C

larger mass is needed to overcome the updraft.

Answer 6

The terminal velocity is directly proportional to their diameter.  "Size" is and ambiguous term here--it could mean volume, in which case the terminal velocity is directly proportional to the cube root of the volume.