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lord [1]
3 years ago
9

If the diameter of the space station is 730 m , how many revolutions per minute are needed for the "artificial gravity" accelera

tion to be 9.80m/s2?
Physics
1 answer:
nignag [31]3 years ago
4 0
For circular motion we know that

<span>F=ma=v^2/r </span>

<span>Therefore: </span>

<span>v = sqrt (rma) </span>

<span>Also, for cicular motion: </span>

<span>rev/min. = 60v/(2r*pi) </span>

<span>So your equation is: </span>

<span>rev./min = 60sqrt(rma)/(2pi*r) </span>

<span>For the mass (m) we can just use 1 kg. 
</span>
rev./min = 60sqrt(730*1*9.8)/(2pi*730) =60sqrt(7154)/(4584.4) 
rev./min = 60sqrt(7154)/(4584.4) =1.11 rev/min
<span>
the answer is </span>1.11 rev/min<span>

 </span>
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Answer:

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Explanation:

The definition of angular velocity is as follows:

\omega=2\pi f

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Frequency can also be represented as:

f=\frac{1}{T}

where T is the period, (the time it takes to conclude a cycle)

with this, the angular velocity is:

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The period T of rotation around the sun 365 days, thus, the angular velocity:

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if we want the angular velocity in rad/second, we need to convert the 365 days to seconds:

Firt conveting to hous

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then to minutes

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525,600minutes(\frac{60seconds}{1minute})=31,536x10^3seconds

thus, angular velocity in rad/second is:

\omega=\frac{2\pi}{31,536x10^3seconds}=1.99x10^{-7}rad/second

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