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A sheet of paper has a mass of 5 grams. The paper is dropped from a high building twice (it reaches terminal speed in both cases), once as a flat paper of area 600 cm2 and once crumpled as a ball of radius 2 cm. The first time it reaches terminal speed v1" and the second time the terminal speed is v2. What is the ratio v2/v1"assuming the density of air is 1.2kg/m?

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gkosworldreign

Answer:

6.91

Explanation:

For any given height, the terminal velocity Vt = [2W/ρACd]^1/2

where W represent the object's weight, ρ represent the density of the gas, A represent the cross sectional area of the object, and Cd represent the drag coefficient.

v2/v1 = [2W/ρACd]^1/2 / [2W/ρACd]^1/2

v2 and v1  has same weight, influenced by same density of air is 1.2kg/m

and same Cd

v2/v1 = [A2/A1]^1/2

A2 =600 cm2

A1 = πr^2 = π*2^2 = 12.57cm2

v2/v1 = [600/12.57]^1/2  = 6.91

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