The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. Given that Earth’s mass is 5.97 × 1024 kg and its radius i
s 6.38 × 106 m, what is HST’s tangential speed?
2 answers:
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Answer:
v = 9.936 m/s
Explanation:
given,
height of cliff = 40 m
speed of sound = 343 m/s
assuming that time to reach the sound to the player = 3 s
now,
time taken to fall of ball
t = 2.857 s
distance
d = v x t
d = v x 2.875
time traveled by the sound before reaching the player
distance traveled by the wave in this time'
r = 0.143 x 343
r= 49.05 m
now,
we know.
d² + h² = r²
d² + 40² = 49.05²
d =28.387 m
v x 2.875=28.387 m
v = 9.936 m/s