A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 . Th
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Answer:
relative vorticity is -0.868 × s-1
Explanation:
given data
air parcel at latitude = 18◦N
relative vorticity = 5.8 × 10^−5 s−1
to find out
relative vorticity
solution
we will apply here conservation of vorticity that is
ζ + f = constant
we know ζ initial = 2Ωsin(5π/18)
and f initial = 5.8 ×
and f final = 2Ωsin(π/18)
and here angular frequency Ω = 0.7272 × s-1
its mean
ζ initial - ζ final = f final - f initial
so
ζ final = ζ initial - f final + f initial
ζ final = - 2Ωsin(5π/18) + ( 2Ωsin(π/18) + 5.8 × )
ζ final = - 2(0.7272 × ) sin(5π/18) + ( 2(0.7272 ×
)sin(π/18) + 5.8 ×
)
ζ final = -0.868 × s-1
so relative vorticity is -0.868 × s-1