# User:Olaff

Seas my Name is Florian *HEADBÄNG*. I'm a pupil @ Gisela Gymnasium in Germany.

I'm one of the Munich Freesytle Slalomskater. Some times I'm skatin' till 12 hours a day. I skate since 2009 and I'm training very hard. But it's very funny because most of the passants in the Olympiapark watch you while skating. Skating is... yeah let me say it's like a blessing. But I'm not only a slalomskater. @ the Munich Bladenight I do also freerides. It's not the same with rocked Skates, because you have not the same grip than with nonrocked skates but you are much more flexible with rocked skates.

Skating is, next to iceskating, the best kind of sports you will find. You can do freerides or do slalomskating but if you want to learn slalomskating than beginn skating with it because if you are a freerider, you normally are a speedskater, and as a speedskater you would have problems. You have to learn slalomskating very slow and than rise your speed. What you need is: Interest, skates, and funny legs =) *g* No, all you can see are turnings and it's only routine.

I'm a more or less good iceskater. Freestyle iceskating 4 eva. Some of my favourit videos:

# Sandbox

Aufgabe 1.7
a) x(t) = x0+ v $\Delta$t

$\frac {x(t)-x_0}{v}$ = $\Delta$t
Falke: x(t) = 0km + 153$\frac {km}{h}$$\Delta$t
Bussard: x(t) = 14km + 95$\frac {km}{h}$$\Delta$t
(153t)h=(14+95t)h
t=0,24h -->Einsetzten--> x=37km

Created by Me

Ein Läufer läuft mit einer konstatnten Geschwindigkeit von 5 $\frac {km}{h}$ 30 Minuten lang.

Wie schnell müsste ein Fahrradfahre fahren, wenn er die gleiche Strecke in 5 Minuten zurücklegen möchte?

Wann würde der Fahrradfahrer den Läufer einholen, wenn der läufer 12 Minuten Vorsprung hätte?

Lösung:

v(Läufer) = 5 $\frac {km}{h}$

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

=> s = v * t

s(Läufer) = 5 $\frac {km}{h}$ * 30 Minuten s(Läufer) = 5 $\frac {km}{h}$ * $\frac {1}{2}h$ s(Läufer) = 2,5 km

3.10

v = 7,5 $\frac {m}{s}$
v = a * t
t = $\frac {v}{a}$ -> t = $\frac {7,5}{9,81}$ = 0,764 s

h = $\frac {1}{2}$ * g* t = $\frac {1}{2}$ * 9,81 * 0,764² = 2,86m


3.11

h = 3,45m *4
t = $\sqrt{\frac {2 * h}{g}}$ -> t = 1,67s

v = a * t -> v = 1,64 m/s
v' = $\frac {1}{7}$*a*t -> v = 6,2 $\frac {m}{s}$
v = $\frac {x}{t}$ -> t' = 4,4s


3.12

v = 7,5 $\frac {m}{s}$
x(t) = $\frac {1}{2}$a * t -> $\frac {1}{2}$ * 9,81 * 5 -> x = 0,12km
v = $\frac {x}{t}$ -> x' = v * t -> x' = 1537,5m
vx = 1537,5 + 122,6m = 1,66km


Übungsaufgabe:

Pfeil an die Decke!

Augabenstellung:
Mit einer Schraubenfeder soll ein Pfeil an die Decke katapultiert werden, dass er die Decke mit der Spitze gerade
nur berührt.

Wir messen: Die Federhärte        D
Die Steighöhe         h
Die Masse des Pfeils  m

Theorie:    Spannarbeit Wsp = $\frac {1}{2}$Ds²
Dehnung s
Hubarbeit   Whub = m * g * h

Messungen:  Masse des Pfeils m = 93,1g
Federhärte = $\frac {F}{s}$ = $\frac {8,0N}{0.25m}$ = 32$\frac {N}{m}$
Steighöhe h = 1.37m + s

Berechnung: Whub = m*g*h = 0,0931kg*9,81$\frac {N}{kg}$*1,37m = 1,25J
s = $\sqrt{\frac {Wsp*2}{D}}$
s = $\sqrt{\frac {Wsp*2}{D}}$
s = $\sqrt{\frac {1,25J*2}{32\frac{N}{m}}}$ = 0,279m (Näherungswert!!)

            h+s =hges.
1,37m + 0,279m = 1,649m
Whub = m * g * h = 0,0931kg* 9,81$\frac {N}{kg}$ * 1.649m = 1.50J
s = $\sqrt{\frac {1,50J*2}{32\frac {N}{m}}}$ = 31cm


5.7

a)

  geg.: r = 1,0 m
ges.: v
lsg.: v0= $\sqrt{g*r}$
v0= $\sqrt{9,81*1}\frac{m}{s}$
v0= 3,13$\frac {m}{s}$



b)

 ges.:v
lsg.:  v0= v
v1= $\sqrt{v^2+4*g*r}$
v1= $\sqrt{3,13^2+4*9,81*1}\frac{m}{s}$
v1= 7,00$\frac {m}{s}$