Got it: reasoning and solution..
the normal force exerted by the wall on each astronaut is the centripetal force needed to keep him in the circular path
so, Fc=mv^2/r. rearranging and letting Fc=(1/3)mg yields
r=3v^2/g= 3(29.1m/s)^2/(9.8m/s^2)=259m
Suppose the surface (radius = r) of a cylindrical space station is rotating at 29.1 m/s to provide artificial gravity. What must be the value of r for the astronauts to weigh 1/3 of their earth weight?
Don't know how to start please help.
3 answers
hell yeah s thats what im talking about buddy. if you disliked this u probably didnt plug ur numbers in correctly
that was so simple. why are there 9 thumbs down. y'all are dumb fr
2 answers