Answer:

Explanation:
Unit conversions:
1890 km/h = 1890 km/h * 1000m/km * 1/3600 h/s = 525 m/s
5.2 km = 5200 m
Assume that the jets is traveling in perfect circular motion, we can calculate the centripetal acceleration of the motion:

where v = 525m/s is the velocity of the jet and r = 5200 is the radius of the arc

Answer:

Explanation:
From the question we are told that
Angle of cable 2 
Weight of sculpture 
Generally the Tension from cable 2
is mathematically given by



Generally the Tension from Cable 1
is mathematically given by



The correct answer for this question is 6m/s. I hope this helps.
( 1.05 x 10¹⁵ km ) x ( 1 LY / 9.5 x 10¹² km ) x ( 1 psc / 3.262 LY ) =
(1.05) / (9.5 x 3.262) x (km · LY · psc) / (km · LY) x (10¹⁵⁻¹²) =
(0.03388) x (psc) x (10³) =
33.88 parsecs
(Hint: the time<span> to rise to the </span>peak<span>is one-half the </span>total hang-time<span>.).</span>