punkt znajdujacy sie na obzerzu wirujacej tarczy o promieniu 35 cm porusza sie z szybkoscia 3.14m/s oblicz okres tarczy.
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dane:
r = 35 cm = 0,35 m
v = 3,14 m/s
szukane:
T = ?
v = 2πr/T I*T
v·T = 2πr /:v
T = 2πr/v = 2·3,14·0,35m/3,14m/s = 0,7 s
T = 0,7 s
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