Answer:
[tex](a)w=2.485*10^{-3}m/s\\(b)a=26.92m/s^2\\(c)\alpha =0rad/s^2[/tex]
Explanation:
Given data
Linear speed of spaceship V=39000 km/h =10833.3 m/s
The radius of circular motion r=4360 km=4360000 m
For Part (a) angular velocity
The angular velocity is given by:
[tex]w=\frac{v_{speed}}{r_{radius}}[/tex]
Substitute the given values
So
[tex]w=\frac{10833.3m/s}{4360000m} \\w=2.485*10^{-3}m/s[/tex]
For Part(b) The radial acceleration of spaceship
The radial acceleration is given by:
[tex]a_{r}=w^2 r[/tex]
Substitute given values and value of ω
So
[tex]a_{r}=(2.485*10^{-3}m/s)^2(4360000m)\\a_{r}=26.92m/s^2[/tex]
For Part (c) The tangential acceleration
The spaceship is rotating with constant angular velocity,then the tangential acceleration of spaceship is zero
So
[tex]\alpha =0rad/s^2[/tex]
