<span>The angular momentum of a particle in orbit is
l = m v r
Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"
m_1 v_1 r_1 = m_2 v_2 r_2
Assuming that the mass did not change, conservation of angular momentum demands that
v_1 r_1 = v_2 r_2
or
v1 = v_2 (r_2/r_1)
Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have
v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s
Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
Answer:
The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency is 15.7 m/s
Explanation:
The Doppler shift equation is given as follows;
Where:
f' = Required observed frequency = 20.0 kHz
f = Real frequency = 21.0 kHz
v = Sound wave velocity = 330 m/s
= Observer velocity = X m/s
= Source velocity = 0 m/s (Assuming the source is stationary)
Which gives;
330 - = (20/21)*330
= 330 - (20/21)*330 = 15.7 m/s
The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency = 15.7 m/s.
