Answer:
\(2.72\cdot 10^{-4} kg/m\)
Explanation:
The terminal speed is reached by an object in free fall when the force of gravity becomes equal to the air resistance:
\(W=R\)
This can be rewritten as
\(mg=bv^2\)
where
m is the mass of the object
g is the gravitational acceleration
b is the coefficient of the air resistance
v is the speed of the object
In this problem, we have m = 2.3 g = 0.0023 kg, g=9.8 m/s^2 and v=9.1 m/s. Substituting and re-arranging the equation we find
\(b=\frac{mg}{v^2}=\frac{(0.0023 kg)(9.8 m/s^2)}{(9.1 m/s)^2}=2.72\cdot 10^{-4} kg/m\)
