# 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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# 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)= $x_0$ + v*t

$\Delta x$ = v* $\Delta t$

v= $\frac {x}{t}$

v= $\frac {x_1-x_0}{\Delta t}$

$x_1$(t)= $x_2$(t)

$x_1+v_1*t = x_2+v_2*t$

$x_1-x_2 = (v_2-v_1)*t$

$t = \frac {x_1-x_2}{v_2-v_1}$ = $\frac {173m-25m}{28km/h-11km/h}$

1.10 Eingeholt

b)

$x_1 = x_0 + v_1 t = 47,7km + 45m/s4870s = 47,7 km + 219,2 km = 266,9 km$

$v_2=\frac{\Delta x}{\Delta t}=\frac{266,9km-15,2km}{4800s}=\frac{251,7km}{4800s}=52,4m/s$

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: $70\frac{km}{h}=19,4\frac {m}{s} t=0,2s$

Lösung: $a=\frac {v}{t}=\frac{19,4\frac {m}{s}}{0,2s}=97,0\frac {m}{s^2}$

b.

Lösung: $\frac{F_b}{F_g}= \frac{m*a}{m*g}=\frac{97 \frac{m}{s^2}}{9,81 \frac{kg}{N}}= 9,9$

Antwort: Der Fahrer erfährt das 9,9 fache seiner Gewichtskraft.