When a crest-trough meet the interference produced will be destructive in nature hence they both will cancel out and the amplitude produced will be equal to zero hence the loudness will reduce to zero.
Answer:
Minimum work = 5060 J
Explanation:
Given:
Mass of the bucket (m) = 20.0 kg
Initial speed of the bucket (u) = 0 m/s
Final speed of the bucket (v) = 4.0 m/s
Displacement of the bucket (h) = 25.0 m
Let 'W' be the work done by the worker in lifting the bucket.
So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.
Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

Therefore, the work done by the worker in lifting the bucket is given as:

Now, plug in the values given and solve for 'W'. This gives,

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.
Answer:
1.2 x 10¹¹ kgm²/s
Explanation:
m = mass of the airplane = 39043.01
r = altitude of the airplane = 9.2 km = 9.2 x 1000 m = 9200 m
v = speed of airplane = 335 m/s
L = Angular momentum of airplane
Angular momentum of airplane is given as
L = m v r
Inserting the values
L = (39043.01 ) (335) (9200)
L = (39043.01 ) (3082000)
L = 1.2 x 10¹¹ kgm²/s