Explanation:
Average speed =total distance ÷ total time
speed = d ÷ t
d= 74400
t= 16
then,
speed = 74400÷16
=4650
Answer:
The maximum speed a subway train can attain between stations is 32.93 m/s.
Explanation:
Given;
maximum tolerable acceleration = 1.39 m/s²
distance between subway train, d = 780 m
The distance available to accelerate between stations = ¹/₂ x 780 m = 390 m
Apply the following kinematic equation to determine the maximum speed;
v² = u² + 2ad
v² = 0 + 2ad
v² = 2(1.39)(390)
v² = 1084.2
v = √1084.2
v = 32.93 m/s
Therefore, the maximum speed a subway train can attain between stations is 32.93 m/s.
Answer:
Explanation:
Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:
Where 
and 
are the orbital periods of Mercury and Earth respectively. We have 
and 
. Replacing this and solving for
