Answer:
Explanation:
Given
diameter of spacecraft [tex]d=148\ m[/tex]
radius [tex]r=74\ m[/tex]
Force of gravity [tex]F_g[/tex]=mg
where m =mass of object
g=acceleration due to gravity on earth
Suppose v is the speed at which spacecraft is rotating so a net centripetal acceleration is acting on spacecraft which is given by
[tex]F_c=\frac{mv^2}{r}[/tex]
[tex]F_c=F_g[/tex]
[tex]\frac{mv^2}{r}=mg[/tex]
[tex]\frac{v^2}{r}=g[/tex]
[tex]v=\sqrt{gr}[/tex]
[tex]v=\sqrt{1450.4}[/tex]
[tex]v=38.08\ m/s[/tex]
The speed of the spacecraft at its outer edge is 26.93 m/s.
The given parameters;
The speed of the spacecraft at its outer edge is calculated as follows;
[tex]F_g = F_c\\\\mg = \frac{mv^2}{r} \\\\v^2 = rg\\\\v = \sqrt{rg} \\\\v = \sqrt{(74)(9.8)} \\\\v = 26.93 \ m/s[/tex]
Thus, the speed of the spacecraft at its outer edge is 26.93 m/s.
Learn more here:https://brainly.com/question/20905151
