Answer:
a= 92. 13 m/s²
Explanation:
Given that
Amplitude ,A= 0.165 m
The maximum speed ,V(max) = 3.9 m/s
We know that maximum velocity in the SHM given as
V(max) = ω A
ω=Angular speed
A=Amplitude
ω=23.63 rad/s
The maximum acceleration given as
a = ω² A
a= (23.63)² x 0.165 m/s²
a= 92. 13 m/s²
Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².
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.