Jawaban:
maaf sanggupnya segini
1. p = mv = 0,02x300 = 6 kg m/s
2. I = F Δt = 50x0,01 = 0,5 Ns
3. I = F Δt = 20x0,2 = 4 Ns
I = Δp = mv'-mv
4 = 0,2 x v' - 0,2 x 0 ( kecepatan awal 0)
4 =0,2 x v'
v' = 4/0,2 = 20 m/s
7.
h₁ = 6m
h₂ = 1,5m
a.
e² = h₂ / h₁ = 1,5/6 = 1/4
e = 1/2
b.
e² = h₃ / h₂
1/4 = h₃ / 1,5
h₃ = 1/4x1,5 = 0,375m
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Jawaban:
maaf sanggupnya segini
1. p = mv = 0,02x300 = 6 kg m/s
2. I = F Δt = 50x0,01 = 0,5 Ns
3. I = F Δt = 20x0,2 = 4 Ns
I = Δp = mv'-mv
4 = 0,2 x v' - 0,2 x 0 ( kecepatan awal 0)
4 =0,2 x v'
v' = 4/0,2 = 20 m/s
7.
h₁ = 6m
h₂ = 1,5m
a.
e² = h₂ / h₁ = 1,5/6 = 1/4
e = 1/2
b.
e² = h₃ / h₂
1/4 = h₃ / 1,5
h₃ = 1/4x1,5 = 0,375m
Penjelasan: