Тело массой 0.2 кг падает с высоты 1 м с ускорением 8 м/с^2. Найти изменение импульса тела за время полета
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Дано:
m=0.2 кг
h=1 м
a=8 м/с²
Найти: ∆p-?
р1 = mV1
p2 = mV2
∆p = р2-p1
∆p = mV2-mV1
т. к. V1=0, то формула принимает вид
∆p = mV2
S=at²/2
t= корень из (2S/a)
t= корень из (2м/8 м/с²)
t= 0.5 с
a = V2-V0/t
V0=0
=>
a = V2/t
V2= at
V2 = 0.8 м/с² × 0.5 с = 0.4 м/с
∆p = 0.4 м/с × 0.2 кг = 0.08 кг × м/с
Объяснение:
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