Answer:
A. 51.42 m.
B. 17.14 s.
Explanation:
Using equations of motion:
vf^2 = vi^2 + 2 * aS
Where,
vf = final velocity
a = acceleration
S = distance to which swan traveled
vi = 0 m/s
6.00^2 = 2 * 0.350S
S = 36/0.7
= 51.42 m.
B.
vf = vi + at
6 = 0 + 0.35t
t = 6/0.35
= 17.14 s.
1.28m/s the velocity is found by distance/time.
Answer:
0.21 lunar month
Explanation:
the radius of moon = r₁
time period of the moon = T₁ = 1 lunar month
The radius of the satellite = 0.35 r₁
Time period of satellite
The relation between time period and radius

now,



T₂ = 0.21 lunar month
hence, the time period of revolution of satellite is equal to 0.21 lunar month