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topjm [15]
3 years ago
7

A 72 kg skydiver can be modeled as a rectangular "box" with dimensions 21 cm × 41 cm × 170 cm . if he falls feet first, his drag

coefficient is 0.80. part a what is his terminal speed if he falls feet first? use ρ = 1.2 kg/m3 for the density of air at room temperature.
Physics
1 answer:
LuckyWell [14K]3 years ago
4 0

Formula for terminal velocity is:


Vt = √(2mg/ρACd) 
<span>Vt = terminal velocity = ?
<span>m = mass of the falling object = 72 kg
<span>g = gravitational acceleration = 9.81 m/s^2
<span>Cd = drag coefficient = 0.80
<span>ρ = density of the fluid/gas = 1.2 kg/m^3</span>
<span>A = projected area of the object (feet first) = 0.21 m * 0.41 m = 0.0861 m^2

Therefore:</span></span></span></span></span>

Vt =  √(2 * 72 * 9.81 / 1.2 * 0.0861 * 0.80) 

<span>Vt = 130.73 m/s</span>

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2 years ago
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Answer:

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        3W_{d} = W_{do} \sqrt{1 +\frac{V_{R} }{Qj} }

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        Substituting value of φj

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   The solution for the b part of the question is uploaded on first image

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slavikrds [6]

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