I Review | ConstantsOne problem for humans living in outer space is thatthey are apparently weightless. One way around thisproblem is to design a space station that spins aboutits center at a constant rate. This creates "artificialgravity" at the outside rim of the station.You may want to review (Page)Part AIf the diameter of the space station is 770 m , how many revolutions per minute are needed for the "artificial gravity" accelerationFor related problemsolving tips and strategies, youto be 9.80 m/s??may want to view a Video Tutor Solution of Force inExpress your answer in revolutions per minute.uniform circular motion.?f =rev/minSubmitRequest AnswerPart BIf the space station is a waiting area for travelers going to Mars, it might be desirable to simulate the acceleration due to gravityon the Martian surface (3.70 m/s2). How many revolutions per minute are needed in this case?Express your answer in revolutions per minute.f =rev/min

Given: The diameter of the is d=770 m. The acceleration is a=9.8 m/s2. Part (a) The expression for…

Answered: One problem for humans living in outer…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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I Review | Constants
One problem for humans living in outer space is that
they are apparently weightless. One way around this
problem is to design a space station that spins about
its center at a constant rate. This creates "artificial
gravity" at the outside rim of the station.
You may want to review (Page)
Part A
If the diameter of the space station is 770 m , how many revolutions per minute are needed for the "artificial gravity" acceleration
For related problemsolving tips and strategies, you
to be 9.80 m/s??
may want to view a Video Tutor Solution of Force in
Express your answer in revolutions per minute.
uniform circular motion.
?
f =
rev/min
Submit
Request Answer
Part B
If the space station is a waiting area for travelers going to Mars, it might be desirable to simulate the acceleration due to gravity
on the Martian surface (3.70 m/s2). How many revolutions per minute are needed in this case?
Express your answer in revolutions per minute.
f =
rev/min

Expert Solution

Step 1

Given:

The diameter of the is d=770 m.

The acceleration is a=9.8 m/s².

Part (a)

The expression for revolution per minute is:

$ω = \sqrt{\frac{2 a}{d}} ω = \sqrt{\frac{2 (9.8 \frac{m}{s^{2}})}{(770 m)}} ω = \frac{0.159 rad}{s} \times \frac{\frac{60}{2 π} \frac{rev}{min}}{1 \frac{rad}{s}} ω = 1.52 \frac{rev}{min}$

Thus, the revolution per minute is 1.52 rev/min.

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