A rod of length L and mass M has a nonuniform mass distribution. The linear mass density (mass per length) is λ=cx2, where x is
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Answer:
Maximum height attained by the model rocket is 2172.87 m
Explanation:
Given,
- Initial speed of the model rocket = u = 0
- acceleration of the model rocket =
- time during the acceleration = t = 2.30 s
We have to consider the whole motion into two parts
In first part the rocket is moving with an acceleration of a = 85.0 for the time t = 2.30 s before the fuel abruptly runs out.
Let be the height attained by the rocket during this time intervel,
And Final velocity at that point be v
Now, in second part, after reaching the altitude of 224.825 m the fuel abruptly runs out. Therefore rocket is moving upward under the effect of gravitational acceleration,
Let '' be the altitude attained by the rocket to reach at the maximum point after the rocket's fuel runs out,
At that insitant,
- initial velocity of the rocket = v = 195.5 m/s.
- a =
- Final velocity of the rocket at the maximum altitude =
From the kinematics,
Hence the maximum altitude attained by the rocket from the ground is
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.