1)Медный шарик радиусом r=1 мм равномерно падает в жидкости со скоростью v=5 м/с . Определите n динамическую вязкость жидкости, если её плотность pж= 1.5*10^3 кг/м^3 2)Определите Q теплоту, необходимую для превращения льда массой m=5кг, взятого при температуре t1= -20 C в пар, температура которого t2=100 C
107 (возможно 106)
Объяснение:
Дано
l=2,3м
t=2,7мин=162с
g=9,8м/с²
π=3,14
Найти n
период математического маятника можно найти по формуле
T=2π√(l/g)=2*3,14√(2,3м/9,8м/с²)=6,28√(0,235с²)=6,28*0,49с=3,04с
За время t маятник совершит t/T=162c/3,04с =53,25 полных колебаний
За один период маятник проходит точки, где его кинетическая энергия масимальна дважды
за 53 периода это случиться 53*2=106 раз, за еще 0,25 периода маятник придет в эту точку еще раз
итого n=107
на самом деле более точное зачение не 53,25 а 53,2481, поэтому возможно, что в задаче подразумевается, что за 162 с маятник всеже не дойти до этой точки 107 раз и тогда правильный ответ 106
хз
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Объяснение: