A satellite’s velocity is 30000m/s. After 60 secs, it’s velocity shows to 15000m/a. What is the satellite’s acceleration?
1 answer:
You might be interested in
Answer:
The magnitude of the final velocity is approximately 0.526 m/s in approximately the direction of 8.746° East of South
Explanation:
The given collision parameters are;
The kind of collision experienced by the four velcro-lined air-hockey disk = Inelastic collision
The mass of the first disk, m₁ = 50.0 g
The velocity of the first disk, v₁ = 0.80 m/s West = -0.8·i
The mass of the second disk, m₂ = 60.0 g
The velocity of the second disk, v₂ = 2.50 m/s North = 2.5·j
The mass of the third disk, m₃ = 100.0 g
The velocity of the third disk, v₃ = 0.20 m/s East = 0.20·i
The mass of the fourth disk, m₄ = 40.0 g
The velocity of the fourth disk, v₄ = 0.50 m/s South = -0.50·j
Therefore, the total initial momentum of the four velcro-lined air-hockey disk, is given as follows;
= m₁·v₁ + m₂·v₂ + m₃·v₃ + m₄·v₄ = 50.0×(-0.80·i) + 60.0×(2.50·j) + 100 × (0.20·i) + 40.0 × (-0.50·j)
∴ = -40·i + 150·j + 20·i - 20·j = -20·i + 130·j
∴ = -20·i + 130·j
By the law of conservation of linear momentum, we have;
= -20·i + 130·j
Therefore, given that the collision is perfectly inelastic, the disks move as one after the collision and the four masses are added to form one mass, "m", m = m₁ + m₂ + m₃ + m₄ = 50.0 + 60.0 + 100.0 + 40.0 = 250.0
∴ m = 250.0 g
Let, "v" represent the final velocity of the four disks moving as one after the collision
We have;
= m × v = 250.0 × v = -20·i + 130·j
∴ v = -20·i/250 + 130·j/250 = -0.08·i + 0.52·j
The final velocity of the four disks after collision, v = -0.08·i + 0.52·j
The magnitude of the final velocity, = √((-0.08)² + (0.52)²) ≈ 0.526
≈ 0.526 m/s
The direction of the final velocity, θ = arctan(0.52/(-0.08)) ≈ -81.254°
The direction of the final velocity, θ ≈ -81.254° which is 8.746° East of South
Answer:
Time period for first satellites 24.46 days and for second satellites 37.67 days
Explanation:
Given :
Distance of first satellites m
Distance of second satellites m
Distance of charon m
Time period of charon days
From the kepler's third law,
Square of the time period is proportional to the cube of the semi major axis.
For first satellites,
days
For second satellites,
days
Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days