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Sagot :
Correct question is;
A tiny water droplet of radius 0.010 cm descends through air from a high building .Calculate its terminal velocity . Given that η of air = 19 × 10^(-6) kg/m.s and density of water ρ = 1000kg/ms
Answer:
1.146 m/s
Explanation:
We are given;
Radius; r = 0.010 cm = 0.01 × 10^(-2) m
η = 19 × 10^(-6) kg/m.s
ρ = 1000 kg/ms
The formula for the terminal velocity is given by;
V_t = 2r²ρg/9η
g is acceleration due to gravity = 9.8 m/s²
Thus;
V_t = (2 × (0.01 × 10^(-2))² × 1000 × 9.8)/(9 × 19 × 10^(-6))
V_t = 1.146 m/s
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