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2πr = 2 * 22/7 * 0,5 m = 22/7 = 3,14 m
1000 : 22/7 ≠ 318 obrotów
d=50cm
2p=50π
1km=100000cm
100000:50π =2000π
2000*22/7≈ 6285,7142..
Odp:.Koło o średnicy 50cm wykona 6285 pełnych obrotów na drodze 1km.
2πr = 2 x 0,5 m x 22/7 = 1 m x 22/7 = 3,14 m
1 km = 1000 m
1000 m : 22/7 ≈ 318 obrotów