Answer:
L = m v r (The momentum remains constant)
Explanation:
Even in an ellipsoidal orbit, the law of conservation of angular momentum always apply. When the plant approached the perihelion, the radius of the orbit decreases and the speed of the star increases to conserve the momentum. Similarly, when the planet approaches the aphelion, the speed of the star decreases as the radius increases to conserve the momentum. So, the momentum at a particular instant can be calculated by L = m v r
Answer:
v = 54.2 m / s
Explanation:
Let's use energy conservation for this problem.
Starting point Higher
Em₀ = U = m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v² = 2gh
v = √ 2gh
Let's calculate
v = √ (2 9.8 150)
v = 54.2 m / s
Answer:
1. 60 m/s.
2. 3600 m.
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 0
Acceleration (a) = 0.5 m/s²
Time (t) = 2 mins
Final Velocity (v) =?
Distance travelled (s) =?
1. Determination of the velocity at the end of 2 minutes.
Initial velocity (u) = 0
Acceleration (a) = 0.5 m/s²
Time (t) = 2 mins = 2 x 60 = 120 secs
Final Velocity (v) =?
v = u + at
v = 0 + (0.5 x 120)
v = 60 m/s
Therefore, the velocity at the end of 2 minutes is 60 m/s.
2. Determination of the distance travelled.
Initial velocity (u) = 0
Acceleration (a) = 0.5 m/s²
Final velocity (v) = 60 m/s
Distance travelled (s) =..?
v² = u² + 2as
60² = 0 + 2 x 0.5 x s
3600 = 1 x s
s = 3600 m
Therefore, the distance travelled is 3600 m.
Answer:
0.75 second
Explanation:
The equation is
Let t be the time when it is at 73 feet. Substitute, h = 73 in the above equation and then find the value of t.
Take positive sign
t = 0.75 second
Answer:
Because they are moving to fast for Earth's gravity to pull it down.
