Particles at the very outer edge of Saturn’s A Ring are in a 7:6 orbital resonance with the moon Janus. If the orbital period of
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Answer:
The rate of change of distance between the two ships is 18.63 km/h
Explanation:
Given;
distance between the two ships, d = 140 km
speed of ship A = 30 km/h
speed of ship B = 25 km/h
between noon (12 pm) to 4 pm = 4 hours
The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =
140 km - 120 km = 20 km
(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)
The displacement of ship B at 4pm = 25 km/h x 4h = 100 km
Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;
r² = a² + b²
r² = 20² + 100²
r = √10,400
r = 101.98 km
The rate of change of this distance is calculated as;
r² = a² + b²
r = 101.98 km, a = 20 km, b = 100 km
Answer:
20 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
r = Radius of robot = 0.5 m
= Angular acceleration = 2 rad/s²
Linear acceleration is given by
While speeding up time taken is 10 seconds
The time taken to slow down is 10 seconds
Total time the trial is rum is 10+10 = 20 seconds