Mechanics11/Seite14/Loesung5.1
Lösung
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)
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)