Answered

What are:
a.) the angular speed
b.) the radial acceleration and
c.) the tangential acceleration of a spaceship negotiating
acircular turn of radius 3220 km at a speed of 29,000 km/h?

Answer :

MechEngineer

Answer:

a) 0.0025 rad/s

b) 20.15 m/s2

c) 0 m/s2

Explanation:

Unit conversion:

3220 km = 3220000m

29000km/h = 29000 km/h * 1000m/km * 1/3600h/s = 8056 m/s

a) The angular speed:

[tex]\omega = \frac{v}{R} = \frac{8056}{3220000} = 0.0025 rad/s[/tex]

b) The radical acceleration:

[tex]a_r = \frac{v^2}{R} = \frac{8056^2}{3220000} = 20.15 m/s^2[/tex]

c) Assume constant velocity, the tangential acceleration of the spaceship would be 0.

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