The force it would take to accelerate a 900-kg car at a rate of 3 m/s2 is
2 answers:
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Answer:
Part A
The angular speed of rotation of the plane is 8.11781 × 10⁻⁵ rad/s
Part B
The angular speed of orbit of the planet is 1.04938 × 10⁻⁶ rad/s
Explanation:
The parameters of the planet are;
The duration of a day on the distant planet = 21.5 hr.
The duration of a year on the distant planet = 69.3 Earth days
Part A
The duration of a day = The time to make one complete revolution of 2·π radians
∴ The average angular speed about its axis, = Angle turned/Time
∴ = 2·π/(21.5 × 60 × 60) s ≈ 8.11781 × 10⁻⁵ rad/s
The average angular speed of the planet about its own axis, = 8.11781 × 10⁻⁵ rad/s
The angular speed of rotation of the plane = 8.11781 × 10⁻⁵ rad/s
Part B
The time it takes the planet to revolve round the neighboring star once = 69.3 Earth days
Therefore, the average angular speed of the planet around its neighboring star, , is given as follows;
= 2·π/((69.3 × 24 × 60 × 60) s) = 1.04938 × 10⁻⁶ rad/s
The average angular speed of orbit, = 1.04938 × 10⁻⁶ rad/s
The angular speed of orbit of the planet, = 1.04938 × 10⁻⁶ rad/s.