Али20061111
23.07.2020 06:54

5. Жоғарғы ұшы бекітілген серіппеге 18 кг жүк ілгенде оның ұзында iппенің қатаңдығы 10 000 Н/м.
10 см, ал 30 кг жүк ілгенде 12 см болды. Серіппені 10 см-ден 15
дейін созу үшін қандай жұмыс істелді?​

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Ответ:
Nr2006
07.08.2021 09:42

pererіau 2.8 mm

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Ответ:
Вика1609000
18.01.2023 19:17

F = 0,05 H

Объяснение:

Дано:

q₁ = - 8·q

q₂ = + 4·q

F₁₂ = 0,4 Н

F₃₄ - ?

Поскольку заряды противоположных знаков, то они притягиваются с силой, равной по модулю:

F₁₂ = k·|q₁|·q₂ / r²

Отсюда:

r² = k·|q₁|·q₂ / F₁₂ = k·8·q·4·q / 0,4 = 80·k·q²     (1)

После соединения и развода зарядов на прежнее расстояние, каждый из зарядов (по закону сохранения):

q₃ = q₄ = (-8+4)·q / 2 = - 2·q

Сила отталкивания по модулю будет:

F₃₄ = k·2·q·2·q / r² =  4·k·q² / r²       (2)

И теперь, с учетом формулы (1), имеем:

F₃₄ =  F = 4·k·q² / r² = 4 / 80 =  0,05  Н

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