User:Kiprilo

Hello there,

my name is Kilian Prikryl I'm nearly 20 years old and come from Germany - Bavaria - Munich.

I visit the Gisela Gymnasium as a pupil, my favorite lesson is sports and geographic. In my freetime I am activ in my fitnesscenter and meeting with my friends every free second.

White Eagle is called my physics teacher who has created the Ted !

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Zeichen √ − ± ∞ ≈ ∝ ≤ ≥ → × · ÷ ∇ ∈ ¬ ∧ ∨ ∃ ⇒ ⇔→ ↔ GRAPHENTESTINGAREA Testing Area Kursiv & Fett: Die Wies'n is bald rum, sie war net grad g'sund, O'zapft is !!! Limeric Lieber den A*sch voll Zecken als eine Nacht in Wildflecken! Gleichungen 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] 1.10 Eingeholt b) [math]x_1 = x_0 + v_1 t = 47,7km + 45m/s4870s = 47,7 km + 219,2 km = 266,9 km[/math] [math]v_2=\frac{\Delta x}{\Delta t}=\frac{266,9km-15,2km}{4800s}=\frac{251,7km}{4800s}=52,4m/s[/math] c) LKW: 219 km PKW: 252 km Schön, dass Sie an der Lösung dieser Aufgabe gearbeitet haben!--White Eagle 12:13, 22 October 2007 (CEST) Transaltion 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 Lösung der Aufgabe 3.8 am 7.12.07 a. Gegeben: [math]70\frac{km}{h}=19,4\frac {m}{s} t=0,2s[/math] Lösung: [math]a=\frac {v}{t}=\frac{19,4\frac {m}{s}}{0,2s}=97,0\frac {m}{s^2}[/math] b. Lösung: [math]\frac{F_b}{F_g}= \frac{m*a}{m*g}=\frac{97 \frac{m}{s^2}}{9,81 \frac{kg}{N}}= 9,9[/math] Antwort: Der Fahrer erfährt das 9,9 fache seiner Gewichtskraft.