Answer:
9.89 m/s
Step-by-step explanation:
d = diameter of the space station = 20.0 m
r = radius of the space station
radius of the space station is given as
r = (0.5) d
r = (0.5) (20.0)
r = 10 m
a = acceleration produced at outer rim = 9.80 m/s²
v = speed at which it rotates
acceleration is given as
![a = (v^(2))/(r)](https://img.qammunity.org/2020/formulas/physics/college/vyem1neufwvanbz0esq94r0ku9ekfabbo9.png)
![9.80 = (v^(2))/(10)](https://img.qammunity.org/2020/formulas/physics/college/nncpgf0wucrhaexao1t2rilroasggtnjyn.png)
v = 9.89 m/s