Предмет: Физика, автор: alan2001

скорость шлифовального круга 100 м/с. Найдите частоту вращения круга,если его диаметр 40см

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Автор ответа: dfkth

0

V=100 м/с
R=0.2 м
n-?
n=1/T
T=2πR/V n=V/2πR n=V/πD n=100/3.14*0.4=79.6 Гц

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