Mechanics11/Seite14/Loesung5.1

5.1 a.

[math]w=\frac {\Delta \phi}{\Delta t}[/math]

[math]w=\frac {26\pi}{300s}= \frac {13\pi}{150}*s^{-1}[/math]

Umlaufdauer: [math]5min=300s [/math]

        [math]300s:13=23,1s[/math]

--Lenny 10:57, 18 April 2008 (UTC) --Benni.m 10:59, 18 April 2008 (UTC)

2.Lösung

a) Umlaufdauer ca. 23,1 s

  [math]\;w = frac{\Delta \phi }{\Delta t}[/math]  = [math]\frac{26 \pi}{300s}[/math] =  [math]\frac{13 \pi}{150s^-1}[/math]

--Nikih 11:01, 18 April 2008 (UTC) --Kiprilo 11:03, 18 April 2008 (UTC)

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5.1. Drehwurm:

a)

ges: Frequenz, Umlaufdauer, Winkelgeschwindigkeit

geg: [math]\Delta t[/math] = 5min = 5 * 60s = 300s

[math]\Delta \phi[/math] = 13 * [math]2 \pi[/math] = 26[math]\pi[/math]

-> Umlaufdauer pro Runde: 300s : 13 = 23,1 s

-> Winkelgeschwindigkeit: [math]\omega = \frac{\Delta \phi}{\Delta t} = \frac{26 \pi}{300 s } = \frac{13 \pi}{150 } s^{-1}[/math]

-> Frequenz: 13 : 300s = 0,043 [math]\frac{1}{s}[/math]

b)

ges: [math]\Delta \phi[/math]

geg: [math]\Delta t[/math] = 10s

[math]\omega = \frac{13 \pi}{150 } s^{-1}[/math]

[math]\omega = \frac{\Delta \phi}{\Delta t}[/math]

 -> [math]\Delta \phi[/math] =  [math]\omega *  t[/math]

= [math]\frac{13 \pi}{150 } s^{-1}[/math] * 10s = [math]\frac{130 \pi}{150 }[/math] = [math]\frac{13 \pi}{15 }[/math] = 2,7227

c)

ges: [math]\Delta t[/math]

geg: [math]\Delta \phi[/math] = [math]5 \pi[/math]

 [math]\omega [/math]  = [math]\omega [/math]1

[math]\frac{5 \pi}{\Delta t}[/math]= [math]\frac{26 \pi}{300s}[/math]

-> [math]\Delta t[/math] =  [math]5 \pi[/math] * 300s : [math]26 \pi[/math] 

= 57,69 s

d)

ges: b

geg: [math]\Delta t[/math] = 2min = 120s

r = 5,0m

[math]\omega = \frac{13 \pi}{150 } s^{-1}[/math]

[math]\frac{26 \pi}{300s}[/math] = [math]\frac{x2 \pi}{120s}[/math]

-> x = 5,2

b = x * U = x * [math]2 \pi[/math]r

= 5,2 * [math]2 \pi[/math] * 5,0m

= [math]52 \pi[/math]m = 163,36m

--Anki 12:39, 20 April 2008 (UTC)