Предмет: Физика, автор: u98062077

?Один конец невесомого рычага длиной 1 м нагружен грузом массой 200 г, а другой конец грузом массой 300 г. Какой должна быть длина (см) большего плеча рычага, чтобы он находился в равновесии? =40 +60 =70 =80

Ответы

Автор ответа: zdiffer

Дано:

l1 = 1 м

m1 = 0,2 кг

m2 = 0,3 кг

l2 = ?

Решение:

Так как рычаг находится в равновесии, то собладется правило моментов:

F1l1 = F2l2

m1gl1 = m2gl2

l2 = m1gl1/m2g = m1l1/m2 = 0,2*1 / 0,3 = 0,7 м

Ответ: l2 = 0,7 м

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