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The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid.
Find the terminal velocity (in meters per second and kilometers per hour) of an 80.0-kg skydiver falling in a headfirst position with a surface area of 0.140 m2

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Answer:

The terminal velocity of the diver is 115 m/s = 414 km/hr

Explanation:

At terminal velocity,

Fnet = mg - Fd = 0

Drag force, Fd = cρAv²/2

mg = cρAv²/2

Terminal Velocity of a body falling through a fluid as in a diver falling through air is given by

v = √(2mg/ρcA)

where m = mass of body falling through fluid = 80 kg

g = acceleration due to gravity = 9.8 m/s²

ρ = density fluid, density of air, as obtained from literature = 1.21 kg/m³

c = coefficient of drag friction of diver falling through air, as obtained from literature = 0.7

A = the area of the diver facing the fluid = 0.14 m²

v = √(2mg/ρcA) = √((2 × 80 × 9.8)/(1.21 × 0.7 × 0.14)) = 115 m/s = 115 × (3600/1000) km/hr = 414 km/hr

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