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Lösning 3.1:1

FörberedandeFysik

(Skillnad mellan versioner)
Hoppa till: navigering, sök
(Ny sida: a) <math>v=v_0+at=10m/s+3 m/s^2\csot 6s= 28m/s</math><br\> b) OA:v2=v02+2aÁs=)(10(m=s)2=0+2(4m=s2)ÁOA(m) =)OA=12;5m AB:s=v+21Áat2=)AB=(10m=s)Á(6s)+21(3m=s2)(6s)2=114m OB=OA+AB=12...)
Nuvarande version (22 december 2009 kl. 15.37) (redigera) (ogör)
 
(3 mellanliggande versioner visas inte.)
Rad 1: Rad 1:
-
a) <math>v=v_0+at=10m/s+3 m/s^2\csot 6s= 28m/s</math><br\>
+
a) <math>v=v_0+at=10m/s+3 m/s^2\cdot 6s= 28m/s</math><br\>
-
b) OA:v2=v02+2aÁs=)(10(m=s)2=0+2(4m=s2)ÁOA(m) =)OA=12;5m
+
b) <math>OA:</math> <math>v^2=v_0^2+2a\cdot s \rightarrow (10(m/s)^2=0+2(4m/s^2)\cdot OA(m) \Rightarrow OA=12,5 m</math><br\>
-
AB:s=v+21Áat2=)AB=(10m=s)Á(6s)+21(3m=s2)(6s)2=114m
+
-
OB=OA+AB=126;5m
+
<math>AB:</math> <math>s=v+\frac{1}{2}\cdot at^2 \Rightarrow AB=(10m/s)\cdot (6s)+\frac{1}{2}(3m/s^2)(6s)^2= 114 m </math><br\>
 +
 
 +
<math>OB=OA+AB=126,5m</math>

Nuvarande version

a) v=v0+at=10ms+3ms26s=28ms


b) OA: v2=v02+2as(10(ms)2=0+2(4ms2)OA(m)OA=125m

AB: s=v+21at2AB=(10ms)(6s)+21(3ms2)(6s)2=114m

OB=OA+AB=1265m

Den här artikeln är hämtad från http://wiki.math.se/wikis/forberedandefysik/index.php/L%C3%B6sning_3.1:1