While taking off from an aircraft carrier, a jet starting from rest accelerates uniformly to a final speed of 40. meters per sec
1 answer:
The acceleration of the jet is
Explanation:
Since the jet motion is a uniformly accelerated motion (=constant acceleration), we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
For the jet in this problem, we have
u = 0 (it starts from rest)
v = 40 m/s (final velocity)
s = 70 m (length of the runway)
Solving for a, we find the acceleration:
Learn more about acceleration:
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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