How many different types of atoms are present in one molecule of aluminum hydroxide, Al(OH)3?
2 answers:
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Answer:
U₂ = 20 J
KE₂ = 40 J
v= 12.64 m/s
Explanation:
Given that
H= 12 m
m = 0.5 kg
h= 4 m
The potential energy at position 1
U₁ = m g H
U₁ = 0.5 x 10 x 12 ( take g= 10 m/s²)
U₁ = 60 J
The potential energy at position 2
U₂ = m g h
U ₂= 0.5 x 10 x 4 ( take g= 10 m/s²)
U₂ = 20 J
The kinetic energy at position 1
KE= 0
The kinetic energy at position 2
KE= 1/2 m V²
From energy conservation
U₁+KE₁=U₂+KE₂
By putting the values
60 - 20 = KE₂
KE₂ = 40 J
lets take final velocity is v m/s
KE₂= 1/2 m v²
By putting the values
40 = 1/2 x 0.5 x v²
160 = v²
v= 12.64 m/s
