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the space station in the drawing is rotating to create artificial gravity. the speed of the inner ring is one - half that of the outer ring. as an astronaut walks from the inner to the outer ring, what happens to her apparent weight?
her apparent weight does not change.
her apparent weight becomes twice as great.
her apparent weight becomes half as great.
her apparent weight becomes one - fourth as great.
her apparent weight becomes four times as great.

Explanation:

Step1: Recall centripetal - force formula for apparent weight

The apparent weight \(W\) (centripetal force \(F_c\)) in a rotating frame is given by \(F_c = m\frac{v^{2}}{r}\), where \(m\) is the mass of the astronaut, \(v\) is the tangential speed, and \(r\) is the radius of the circular path. Let the speed of the outer - ring be \(v_{out}\) and its radius be \(r_{out}\), and for the inner - ring be \(v_{in}\) and \(r_{in}\). Given that \(v_{in}=\frac{1}{2}v_{out}\) and assume \(r_{in}=\frac{1}{2}r_{out}\) (since for a rotating space - station with a consistent angular velocity \(\omega\), \(v = r\omega\)).
The apparent weight at the outer ring \(W_{out}=m\frac{v_{out}^{2}}{r_{out}}\), and the apparent weight at the inner ring \(W_{in}=m\frac{v_{in}^{2}}{r_{in}}\).

Step2: Substitute \(v_{in}=\frac{1}{2}v_{out}\) and \(r_{in}=\frac{1}{2}r_{out}\) into the inner - ring weight formula

\[

$$\begin{align*} W_{in}&=m\frac{(\frac{1}{2}v_{out})^{2}}{\frac{1}{2}r_{out}}\\ &=m\frac{\frac{1}{4}v_{out}^{2}}{\frac{1}{2}r_{out}}\\ &=\frac{1}{2}m\frac{v_{out}^{2}}{r_{out}} \end{align*}$$

\]
Since \(W_{out}=m\frac{v_{out}^{2}}{r_{out}}\), we have \(W_{in}=\frac{1}{2}W_{out}\). So when the astronaut moves from the inner to the outer ring, her apparent weight becomes twice as great.

Answer:

Her apparent weight becomes twice as great.