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2 answers:
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Answer:
27.82 m/s
Explanation:
The radius of the hose is half of its diameter
So its area must be
The speed of water coming out of the hose is its flow rate divided by the cross-section area of the hose
Answer:
In first case speed of fish will be m/s.
In the second case the speed of fish m/s
Explanation:
Given data :-
Mass of bigger fish ( m₁ ) = 5 kg.
Mass of small fish ( m₂ ) = 1 kg.
Speed of large fish ( v₁ ) = 1 m/s
Mass of bigger fish after eating smaller one = 5 + 1 = 6 kg.
Case - 1
Momentum of bigger fish before eating the smaller fish = m₁* v₁ = 5 * 1 = 5 kg.m/s
Momentum of bigger fish after eating the larger fish = ( m₁ + m₂)*v
v = speed of bigger fish immediately after lunch.
Using the conservation of momentum.
m₁* v₁ = ( m₁ + m₂)*v
5 = 6 * v
v = m/s.
Case -2
Speed of small fish = 4 m/s
Momentum of bigger fish before lunch = 5 kg.m/s
Momentum of smaller fish before lunch = 4*1 = 4 kg.m/s
Net momentum before lunch = 5 - 4 = 1 kg.m/s
Momentum of bigger fish after eating the larger fish = 6 * V
Using the conservation of momentum.
1 = 6 * V
V = m/s.