1. Vpa = 180m/s. @ 0 deg.
Vag = 40m/s @ 120 deg,CCW.
<span>
Vpg = Vpa + Vag,
Vpg = (180 + 40cos120) + i40sin120,
Vpg = 160 + i34.64,
Vpg=sqrt((160)^2 + (34.64)^2)=163.7m/s.
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<span>2. tanA = Y / X = 34.64 / 160 = 0.2165,
A = 12.2 deg,CCW. = 12.2deg. North of
East. </span>
3. 1 hr = 3600s. <span>d = Vt = 163.7m/s * 3600s = 589,320m.
hope this helps</span>
Answer:
Explanation:
We know that the pressure can be calculated in the following way:
p = d·g·h
with d being the density of the water, g the gravitational acceleration and h the depth.
Also d of the water = 1000 kg/m^3 circa and g = 9.8 m/s^2 circa
117,500 Pa = 1000kg/m³ · 9.8m/s² · h
Therefore h = 11,9 m
Hey the answer to the question is
m = 0.40
r₁ = distance of the point from the source = 43 km = 43000 m
I₁ = intensity of earthquake wave at distance "r₁" = 2.5 x 10⁶ W/m²
r₂ = distance of the point from the source = 1.5 km = 1500 m
I₂ = intensity of earthquake wave at distance "r₂" = ?
we know that , for a constant power , the intensity of wave is inversely proportional to the distance from the source .
I α 1/r² where I = intensity of wave , r = distance from source
hence we can write
I₁/I₂ = r₂²/r₁²
inserting the values
(2.5 x 10⁶) /I₂ = (1500/43000)²
I₂ = 2.1 x 10⁹ W/m²