Предмет: Физика, автор: timer45677

какой массы алюминиевый груз следует привязать к деревянному бруску массой 680 г, чтобы, будучи погруженными в воду, они оставались в воде во взвешенном состоянии?

Ответы

Автор ответа: pavlushaarchi

1

Пусть плотность дерева: ρ(д) = 0,60 г/см^3.
Воды: ρ(в) = 1,00 г/см^3.
Алюминия: ρ(а) = 2,70 г/см^3.
m(д) = 680 г.
m(д) /(ρ(д) = V(д)..
Условие «взвешенности» — вес суммы тел равен весу воды в их суммарном объеме (запишем без g):
m(д) + m(a) = ρ(в) *(V(д) + V(a)) ==>
680 + 2.70*V(a) = 1,00*(680/0.6 + V(a)).
Получили уравнение для V(a): его решение — V(a) = 266,7 см^3.
Значит, масса алюминия равна: m(a) = 2,70*266,7 = 718.2 г.
Итак, искомая масса: m(a) = 718,2 г.

pavlushaarchi: Можно лучший ответ?

timer45677: Я не знаю как делать это

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