Answer:
<u><em>note:</em></u>
<u><em>find the attached solution:</em></u>
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.
It would be, 1.000. Hope that helps :)
Answer:
Explanation:
Given that,
Mass of block
M = 2kg
Spring constant k = 300N/m
Velocity v = 12m/s
At t = 0, the spring is neither stretched nor compressed. Then, it amplitude is zero at t=0
xo = 0
It velocity is 12m/s at t=0
Then, it initial velocity is
Vo = 12m/s
Then, amplitude is given as
A = √[xo + (Vo²/ω²)]
Where
xo is the initial amplitude =0
Vo is the initial velocity =12m/s
ω is the angular frequency and it can be determine using
ω = √(k/m)
Where
k is spring constant = 300N/m
m is the mass of object = 2kg
Then,
ω = √300/2 = √150
ω = 12.25 rad/s²
Then,
A = √[xo + (Vo²/ω²)]
A = √[0 + (12²/12.5²)]
A = √[0 + 0.96]
A = √0.96
A = 0.98m
Answer:
I believe its a and c but my notes are all kinds of messed up so im sorry if its wrong
Explanation: