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crawfishcollins
@crawfishcollins
January 2021
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Aidez moi SVP la semaine prochaine j'ai un contrôle en physique
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scoladan
Verified answer
Bonjour,
1) Soit F la force exercée par le ressort sur S d'intensité F = kx
P + F = 0 (en vecteurs)
⇒ P - F = 0 (en normes)
⇔ mg - kx₁ = 0
⇒ x₁ = mg/k
2) P + F + Pa = 0 (avec Pa Poussée d'Archimède)
⇒ P - F - Pa = 0 (en normes)
⇔ mg - kx₂ - (m₂ - m₁)g = 0
⇒ x₂ = [m - (m₂ - m₁)]g/k
donc x₂ < x₁
3) (m₂ - m₁) = m - kx₂/g
1 votes
Thanks 1
crawfishcollins
Merci beaucoup vous me sauvé la vie
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Lista de comentários
Verified answer
Bonjour,1) Soit F la force exercée par le ressort sur S d'intensité F = kx
P + F = 0 (en vecteurs)
⇒ P - F = 0 (en normes)
⇔ mg - kx₁ = 0
⇒ x₁ = mg/k
2) P + F + Pa = 0 (avec Pa Poussée d'Archimède)
⇒ P - F - Pa = 0 (en normes)
⇔ mg - kx₂ - (m₂ - m₁)g = 0
⇒ x₂ = [m - (m₂ - m₁)]g/k
donc x₂ < x₁
3) (m₂ - m₁) = m - kx₂/g