User:Nikih

(: I am impressed at what you are doing. Keep up the good work. It would be nice if you could upload a photo as well. For this purpose I have inserted an infobox. Warm regards, Patricia (L4C coordinator, WikiEducator))

Niki
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Occupation:	Student
Nationality:	German
Country:	Germany

[math]Insert formula here[/math]Hi outside there,

my name is niki i'm 17 jears old and i come from munich a city in germany

i'm a pupil at the Gisela gramma school and in case of a physic project i created my account on wikieducator

a special thanks to white eagle for this nice project.

sandbox

1.1

Zeichen ∫ ∑ ∏ √ − ± ∞ ≈ ∝ ≡ � ≤ ≥ → ⋅ × · ÷ ∂ ′ ″ ∇ ‰ ° ∴ ℵ ø ∈ ∉ ∩ ∪ ⊂ ⊃ ⊆ ⊇ ¬ ∧ ∨ ∃ ∀ ⇒ ⇔→ ↔ fett gedruckt kursiver text There was a young lady from Riga, who smiled as she rode on a tiger. They returned from the ride with the lady inside and the smile on the face of the tiger --Nikih 12:50, 5 October 2007 (CEST) x(t)= [math]x_0[/math] + v*t [math]\Delta x[/math] = v* [math]\Delta t[/math] v= [math]\frac {x}{t} [/math] v= [math]\frac {x_1-x_0}{\Delta t}[/math] [math]x_1[/math](t)= [math]x_2[/math](t) [math]x_1+v_1*t = x_2+v_2*t[/math] [math]x_1-x_2 = (v_2-v_1)*t[/math] [math]t = \frac {x_1-x_2}{v_2-v_1}[/math] = [math]\frac {173m-25m}{28km/h-11km/h}[/math] Lösung Hausaufgabe 1.10 [math]x_1[/math] =[math]x_0+v_1t[/math] x=47,7km + 45m/s*4870s x=47.7km + 219,2km = 266,9km [math]v_2 = \frac{\Delta x}{\Delta t}[/math] = [math]\frac{266,9km - 15,2km}{4800s}[/math] =[math]\frac{251,7km}{4800s}[/math] = 52,4m/s Schön, dass Sie an der Lösung dieser Aufgabe gearbeitet haben!--White Eagle 12:12, 22 October 2007 (CEST) Translation Mechanics11/Page2 To detect a movement we need one · Abscissa of the position · Coordinate of time t The position difference is accordent delta x: xxxxxxxxxxxxxxxxxxxxxxxformelxxxxxxxxxxxxxxxxxxxxxxx (Abscissa of the position at the end minus abscissa of the position at the start of the time slice) According to the time slice xxxxxxxxxx delta t xxxxxxxxxxxxx: xxxxxxxxxxxxxxx Formel xxxxxxxxxxxxxxxx Definiton: xxxxxxxxxxxx formel xxxxxxxxxxxxxx (Covered distance divided by needed time) Advice: Here it is a matter of a quotient out of differences, a differencequotient. If we take two neighbouring measuring points, then we get the momentary speed of the cart to time t1 and/or, we take first and the 51 to t2. , We get the average speed in the time interval to measuring point Formeln werden nachträglich eingetragen ! Kilian & Niklas Aufgabenlösung 3.9 geg.: m = 330000kg ; Fmax = 9,0 *10^5N  ; F-start = 8,2*10^5N Fr= 2,7*10^5N ; v-start = 300km/h a) ges.: t Lsg.: F-start - Fr = 8,2*10^5N - 2,7*10^5N = 5,5*10^5N F=m*a a=[math]\frac{F}{m}[/math] a=[math]\frac{5,5*10^5N}{3,3*10^5kg}[/math]=1,66[math]\frac{m}{s^2}[/math] a=[math]\frac{v}{t}[/math] t=[math]\frac{v}{a}[/math] = [math]\frac{83,33\frac{m}{s}}{1,66\frac{m}{s^2}}[/math] = 50,1s b) ges.: x Lsg.: x = 0.5 * a * t^2 = 0.5 * 1.66[math]\frac{m}{s^2}[/math] * (50s)^2 = 2075m c) zusätzlich geg.: x = 3000m ges.: F Lsg.: v^2 = 2ax a = [math]\frac{v^2}{2x}[/math] = [math]\frac{(83,33\frac{m}{s})^2}{2 * 3000m}[/math] = 1,15[math]\frac{m}{s^2}[/math] F = m * a = 3,3*10^5kg * 1,15[math]\frac{m}{s^2}[/math] = 3,795 * 10^5N = 3,8*10^5N Lösung 3.10 geg.: v = 7.5 [math]\frac{m}{s}[/math] ; g = 9,81 [math]\frac{N}{kg}[/math] ges.: h Lsg.: v2 = 2ax x = [math]\frac{v^2}{2a}[/math] x = [math]\frac{(7,5 m/s)^2}{2*9,81 N/kg}[/math] x = 2,87m Lösung 5.5 a) w = f/T = (2 pi * 15) / 120s = 1/4 pi aZ = (1/4 pi)2 * 5,3m = 3,26m/s2 3,26m/s2 / 9,81m/s2 = 0,33 = 33% b) FR = mue * FN => FG = mue * FZ => g = mue * aZ g = mue * w2 * r => w = [math]g/(mue * r)[/math] = [math]9,81/(0,1 * 4,2)[/math] s-1 = 4,8s-1 w = 2 * pi * f => f = w/2pi = 0,77s-1 T = 1,3s Lösung 5.6 a) geg.: Masse m = 0,80 kg Radius r = 50 cm Geschwindigkeit v = 3,4 m/s ges.: (Umkreis u) (Kraft FZ) (Kraft mit Erdanziehungskraft FZ.grav) Kraft unten am Kreis FZ.unten Kraft oben am Kreis FZ.oben [math]u = 2r\pi = 3,14 m \frac{}{}[/math] [math]T = \frac{u}{v}[/math] [math]T = \frac{3,14 m}{3,4 m}[/math] [math]T = 0,924 s \frac{}{}[/math] [math]\omega = \frac{2\pi}{T}[/math] [math]F_Z = m\omega^2r \frac{}{}[/math] [math]F_Z = 0,80 kg * (\frac{2\pi}{0,924s})^2 * 0,50m[/math] [math]F_Z = 18,49 N \frac{}{}[/math] [math]g = 9,81 \frac{N}{kg}[/math] [math]F_{Z.grav} = g * 0,8 kg \frac{}{}[/math] [math]F_{Z.grav} = 9,81 * \frac{N}{kg} * 0,8 kg[/math] [math]F_{Z.grav} = 7,84 N \frac{}{}[/math] [math]F_{Z.oben} = F_Z - F_{Z.grav} \frac{}{}[/math] [math]F_{Z.oben} = 10,65 N \frac{}{}[/math] [math]F_{Z.unten} = F_Z + F_{Z.grav} \frac{}{}[/math] [math]F_{Z.unten} = 26,35 N \frac{}{}[/math] b) geg.: FZ = 50 N ges.: [math]\omega_2[/math] FZ = m * w2 * r ->w = [math]F\ltsub\gtZ\lt/sub\gt : m*r[/math] Aufgabe: Fußballfans Während der EM läuft ein frustrierter betrunkener Fußballfan(90kg) konstant mit 4km/h gegen einen Leidensgenossen(100kg). Nach dem Zusammenprall torkeln sie Arm in Arm mit 3km/h weiter. a) Wie schnell war der Leidensgenosse vor dem Zusammenstoß? b) In welche Richtung ist der Leidensgenosse ursprünglich gelaufen? c) Die menschliche Schmerzgrenze liegt bei 400N. Ist der Zusammenprall schmerzhaft? d) Von welcher EM-Mannschaft stammen die beiden Fans?