Предмет: Физика, автор: nrswift49

под действием силы 4h пружина удлинилась на 0.02м чему равна жесткость пружины?
А.2нм
б.0.5нм
в.0.02нм
г.500нм
д.200нм

Ответы

Автор ответа: kamei34

1

Ответ:

k=F/x 4/0,02=200нм .......

Автор ответа: elena20092

1

Ответ:

д. 200 Н/м

Объяснение:

F = 4H

ΔL = 0.02 м

--------------------

k - ?

---------------------------

Жёсткость пружины

k = F : ΔL = 4 : 0.02 = 200 (Н/м)

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