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Neue Datenschutzrichtlinie, Mai 2018
Hallo,
Diese Webseite wurde aktualisiert mit kleineren Anpassungen an die DS-GVO.
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विषय से परेे
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admin
27 Mei 2018
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New Privacy Policy, May 2018
Hello,
this website was updated today because of some legal changes here in Europe. ("GDPR").
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विषय से परेे
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admin
27 Mei 2018
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New version is online for two weeks now, any problems?
Hi,
Any problems? You can use the "Report a Problem" link in the footer, or send me an email to britnex+bugs@gmail.com
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admin
12 Sep 2015
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New website is online.
Hi,
Any problems? You can use the "Report a Problem" link in the footer, or send me an email to britnex+bugs@gmail.com , thank you!
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Melody
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admin
30 Agu 2015
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Neue Version der Webseite ist online
Hi,
Probleme bitte per Email an britnex+bugs@gmail.com melden, Danke!
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admin
30 Agu 2015
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three times ten to the power of negative 2
help what is 3x10to the power of negative 2
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admin
26 Jan 2012
«
नवीनतम
0
-1
2
नवीनतम
»
#1
+3146
0
base "10":
[input]log( 10^3 )[/input]
base "e":
[input]ln( e^3 )[/input]
admin
13 Des 2012
#1
+3146
0
[input]-2x^2+4x+6=6[/input]
admin
12 Des 2012
#1
+3146
0
[input]33750=-2/(3x^2)+300x[/input]
admin
11 Des 2012
#1
+3146
0
[input]factorize( 512 )[/input]
admin
10 Des 2012
#1
+3146
0
press enter..
[input]12300000000000000[/input]
admin
7 Des 2012
#2
+3146
0
3169 cases sold 2011
750 cases sold 2012
[input]100-100*750/3169[/input]
[input]3169-76.3332281476806563584727043%[/input]
admin
7 Des 2012
#2
+3146
0
1/4 + a > -3/4
1/4 + a - 1/4 > -3/4 -1/4
a > -1
[input]plot( 1/4 + x, -3/4, x=-2..2)[/input]
admin
6 Des 2012
#1
+3146
0
a=8,0*10^5
b=2,3
c=631
LauweJoster:
p=b+c=
[input]a=8.0*10^5; b=2.3; c=631; b+c[/input]
LauweJoster:
y=p*a=
[input]a=8.0*10^5; b=2.3; c=631; p=b+c; p*a[/input]
[input]506640000[/input]
admin
5 Des 2012
#2
+3146
0
v0 = 36km/h in m/s = 10m/s (Startgeschwindigkeit)
Gleichung 1:
Geschwindigkeits-Zeit-Gesetz der gleichmäßig beschleunigten Bewegung:
v=v0+a*t
v=v0+(F/m)*t
Kraft = Masse * Beschleunigung:
F = m*a, somit a = F/m
somit:
0 = 10m/s - (F/60kg)*t
0 = 10 - (F/60)*t
Gleichung 2:
Weg-Zeit-Gesetz der gleichmäßig beschleunigten Bewegung:
s = v0*t+(1/2)*a*t^2
50m = t*10m/s -(1/2)*(F/60kg) *t^2
(minus wegen Abbremsvorgang)
50 = t*10 -(1/2)*(F/60) *t^2
Gleichungssystem mit 2 Gleichungen und 2 Unbekannten (F, t) lösen:
[input]solve( 0 = 10 - (F/60)*t, 50 = t*10 -(1/2)*(F/60) *t^2 )[/input]
per Hand:
0 = 10 - (F/60)*t
-10 = - (F/60)*t
10 = (F/60)*t
10/(F/60) = t
t = 600/F
[input]solve( 0 = 10 - (F/60)*t, t)[/input]
t einsetzen:
(oder in Rechte Seite das Ergebnis F=60N einsetzen und aussrechen und mit 50 vergleichen..)
50 = (600/F)*10 -(1/2)*(F/60) *(600/F)^2
50 = (6000/F) - (F/120) *(600/F)^2
50 = (6000/F) - (F/120) *(600/F)*(600/F)
50 = (6000/F) - 360000F / (120F^2)
50 - (6000/F) = - 360000F / (120F^2)
( 50 - (6000/F) ) * (120F^2) = - 360000F
((50F-6000) / F ) * (120F^2) = - 360000F
(50F-6000) * 120F = - 360000F
6000F^2-720000F = - 360000F
6000F^2-720000F + 360000F = 0
6000F^2 - 360000F = 0
p-q-Formel:
F= 60 (Newton)
[input]solve( 50 = (600/F)*10 -(1/2)*(F/60) *(600/F)^2, F)[/input]
somit:
F == 60N (Reibungskraft ) q.e.d.
admin
5 Des 2012
#1
+3146
0
[input]((2/5)/(2/7)) / (1/2)[/input]
((2/5)*(7/2)) / (1/2) = ((2/5)*(7/2)) * (2/1) = (14/10) * 2 = 28/10 = 14/5
admin
4 Des 2012
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