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 !

= Sandbox =

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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)= $$x_0$$ + 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.