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R456
R456
20.04.2021 16:53 •  Физика

Ввертикально поставленный цилиндр с площадью основание 40^2 см вставлен поршень,под которым находится столб воздуха 60см.на сколько опустится поршень,если на него поставить гирю массой 10кг? масса поршня 2кг,атмосферное давление нормальное

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Ответ:
SabrinaSirotina
SabrinaSirotina
20.04.2021
Давление воздуха под поршнем P=P0+F/S
Площадь поршня S=0,004 м²Без гири Р1= 100 000 Па + 2 кг * 9,8 м/с² / 0,004 м² = 104900 ПаИзбыточное давление, которое создает гиря:
ΔР=m гири * g / S = 10 * 9,8 / 0,004 ≈ 24500 ПаДавление Р2 = Р1+ΔР = 104900+24500 = 129400 ПаДля идеального газа P*V=R*T. Предположим Т газа не изменилось (процесс изотермический), R=const, следовательно P*V=const.P1*V1=P2*V2
Объем газа V1 = S*h1 = 0,004 * 0,6 = 0,0024 м³V2=P1*V1 / P2 = 104900*0,0024 / 129400 = 0,00195 м³h2 = V2 / S = 0,00195 / 0,004 = 0,4875 мΔh=h1-h2 = 0,6-0,4875 = 0,1125 м = 11,25 см.ответ: поршень опуститься на 11,25 см.
4,5(53 оценок)
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Ответ:
vika2499
vika2499
20.04.2021

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4,6(17 оценок)
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Evelina0889
Evelina0889
20.04.2021

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4,8(38 оценок)
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