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.
The pressure at the depth 11 km below sea level can be
calculated using
P=ρgh
P is pressure, ρ is the density of the fluid; g is the
gravitational constant, h is the height from the surface, or depth that the
object is submerged.
P = ( 1000 kg/ m3) ( 9.81 m.s2)( 11 000m) + 1 atm
P = 107,910,000 pa ( 1 atm/ 101 325 Pa) + 1 atm = 1066 atm
Answer:
2.667m/s to the north and 3.333 m/s to the west
Explanation:
According to law of momentum conservation, the total momentum should be conserved before and after the explosion.
Before the explosion, the momentum was
0.5*2 = 1 kg m/s to the west
Therefore the total momentum after the explosion should be the same horizontally and vertically.
Vertically speaking, it was 0 before the explosion. After the explosion:
0.2*4 + 0.3v = 0
0.3v = -0.8
v = -0.8/0.3 = -2.667 m/s
So the vertical component of the 0.3kg piece is 2.667m/s to the north
Horizontally speaking, since the 0.2kg-piece doesn't move west or east post-explosion:
0.2*0 + 0.3V = 1
0.3V = 1
V = 1/0.3 = 3.333 m/s
So the horizontal component of the 0.3kg piece is 3.333 m/s to the west
