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ρ = m/V
= 2/40
= 0,05 g/L
v = √3P/ρ
v = √(3 • 0,24 x 10⁵)/0,05
v = √1,44 x 10⁶
v = 1200 m/s (Jawaban C)
15.)
Q = +3000 (Karena menerima kalor)
W = -1200 (Karena dikenai usaha)
Q = ΔU + W
+3000 = ΔU - 1200
ΔU = 3000 + 1200
ΔU = 4200 J (Jawaban D)
18.)
η₁= 40% = 0,4
η = 1 - Tr/Tt
0,4 = 1 - 480/Tt
480/Tt = 0,6
Tt = 480/0,6
Tt = 800⁰K
Suhu reservoir tinggi = tetap = 800⁰K
η₂ = 60% = 0,6
η = 1 - Tr/Tt
0,6 = 1 - Tr/800
Tr/800 = 0,4
Tr = 800 x 0,4
Tr = 320⁰K (Jawaban B)