R1=1 Ом R2= 2 Ом
R3= 3 Ом
R4= 4 Ом
ЭДС = 10 В
найдите напряжение цепи
в центре находится конденсатор

please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please please

Площадь сечения равна: S=πR^2
R=0.6 м => S=3.14*(0.6)^2=1.13 м^2
Найдём массу воды в трубе: m=S*h*p
Здесь h-длина трубы
m=1.13*150*1000=169500 кг
Далее закон сохранения энергии:
Ek1+Ep1=Ek2+Ep2
В начале скорость нулевая, в конце он достигает нужной высоты т.е. уже нулевой. И так получается:
Ep1=Ek2
m*g*h/2=m*v^2/2
здесь h- перепад высоты.
h/2-это потому, что Ep- связанно с движением центра масс
После сокращений получаем: v=√gh = √10*19=13.8 м/с
Энергия, которую можно получить равна: Ep=mgh/2=16102500 Дж
Переводим в кВт-ч, получается Ep≈4,473 кВт-ч или Ep=16102.5 кВт-с