Answer:
0.68 s
Explanation:
We are given that
Initial velocity of box=
Final velocity of box=v=11.5 m/s
Distance=d=8.5 m
We have to find the time taken by box to slow by this amount.
We know that

Substitute the values




We know that
Acceleration=
Substitute the values



Hence, the time taken by box to slow by this amount=0.68 s
Explanation:
Given that,
Distance 1, r = 100 m
Intensity, 
If distance 2, r' = 25 m
We need to find the intensity and the intensity level at 25 meters. Intensity and a distance r is given by :
.........(1)
Let I' is the intensity at r'. So,
............(2)
From equation (1) and (2) :



Intensity level is given by :
, 

dB = 32.96 dB
Hence, this is the required solution.
surface wave is a wave that travels along the surface of a medium. The medium is the matter through which the wave travels. Ocean waves are the best-known examples of surface waves. They travel on the surface of the water between the ocean and the air.
HOPE IT HELPS