Imagine you have a collection of identical flat-bottomed coffee filters that can be nested (stacked inside of each other) so that the resulting area is roughly unchanged. The terminal velocity of one coffee filter is measured to be 0.856 m/s. What is the terminal velocity of 14 nested coffee filters

the terminal velocity of 14 nested coffee filters is 3.2 m/s

Explanation:

Given the data in the question;

we know that;

The terminal velocity is proportional to the square root of weight.

v ∝ √W

v = k√W

the proportionality constant depends upon the surface area and the density of the medium (like air). The coffee filters can be stacked such that the resulting area is roughly unchanged. So, the constant of proportionality k is also unchanged

v/√W = constant

v₂/√W₂ = v₁/√W₁

v₂ = v₁√(W₂ / W₁ )

given that;

v₁ = 0.856 m/s,

W₂ = 14W₁; meaning 14 coffee filters have 14 times the weight of a single coffee filter

so we substitute

v₂ = 0.856 √(14W₁  / W₁ )

v₂  = 0.856 √( 14( W₁/W₁)

v₂  = 0.856 √( 14(1)

v₂  = 0.856 √( 14 )

v₂  = 0.856 × 3.741657

v₂  = 3.2 m/s

Therefore, the terminal velocity of 14 nested coffee filters is 3.2 m/s

(75°f − 32) × 5/9 = 23.889°c