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PilotLPTM [1.2K]
1 year ago
10

So, RCF, or relative centrifugal force should be equal to RCF =

%5E2" id="TexFormula1" title="11,18*r*(RPM/1000)^2" alt="11,18*r*(RPM/1000)^2" align="absmiddle" class="latex-formula">
The radius is measured in centimeters and RPM is revolutions per minute.

What I don't know is, if I plug all the data in the formula, will the result equal to acceleration in meters per second squared, or it will be measured in g's so that if I multiply the g's by 9.8 meters per second square, it will give me the actual acceleration.
Physics
1 answer:
zysi [14]1 year ago
6 0

After plugging all the data into the equation, the result of the relative centrifugal force (RCF)  is measured in terms of g.

<h3>What is relative centrifugal force?</h3>

The relative centrifugal force (RCF) or the g force is the radial force generated by the spinning rotor as expressed relative to the earth's gravitational force.

RCF = ac/g

where;

  • ac is centripetal acceleration
  • g is acceleration due to gravity

RCF = \frac{\omega ^2 r}{g} = 1.118\times 10^{-5} \ (RPM)^2 r = 11.18r\ (RPM/1000)^2

where;

  • r is radius in cm

<h3>For example, </h3>

Find the maximum RCF of the JS-4.2 rotor can be obtained from its maximum speed (4200 rpm) and its rmax (250 mm);

RCF = 11.18 \times 25\ cm \times (\frac{4200 \ RPM}{1000} )^2 = 4,930.3 \times g

Thus, after plugging all the data into the equation, the result is measured in terms of g.

Learn more about relative centrifugal force here: brainly.com/question/26887699

#SPJ1

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Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could
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We can solve this with the following relation:

\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}

Where:

d=17.91 mm =17.91(10)^{-3}  m is the diameter of a dime

D is the diameter of the Sun

x_{sun-pinhole}=150,000,000 km=1.5(10)^{11}  m is the distance between the Sun and the pinhole

x_{sunball-pinhole}=100 d=1.791 m is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding D:

D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}

D=\frac{17.91(10)^{-3}  m}{1.791 m} 1.5(10)^{11}  m

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So, we have to divide this distance between D in order to find how many suns could it fit in this distance:

\frac{149,597,870,700 m}{1.5(10)^{9}  m}=99.73 suns \approx 100 suns

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