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3 years ago

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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.

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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3 years ago

Read 2 more answers

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Answer:

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$q_{1}$ = 3 nC = 3 x $10^{-9}$ C

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k = constant (8.99 x $10^{9} Nm^{2} /C^{2}$ )

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since it is measured at the midpoint,

r = $\frac{0.5}{2}$

= 0.25 m

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F = $F_{1}+ F_{2}$

= k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $\frac{kq_{e} }{r}$ ( $q_{1} +q_{2}$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ +2 x $10^{-9}$ )/0.25

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the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

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= $kq_{e}$ ( $\frac{q_{1} }{r_{1} }$ + $\frac{q_{2} }{r_{2} }$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ /0.1+2 x $10^{-9}$ /0.4)

= - 5.04 x $10^{-17}$ J

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