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

where;
<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);

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
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"Acceleration" means any change in the speed or direction of motion ... speeding up, slowing down, or turning. So . . .
<span>-- </span><span>The distance traveled in a certain time may increase or decrease.
-- The displacement covered in a certain time may increase or decrease.
-- The speed of the object may increase or decrease.
-- The velocity of the object (speed/direction) will change.
</span>
Hope this helps..
W = F x D
The work you put into a machine -- the input force -- is the force you apply times the distance you apply it. The work done by the machine equals the resisting weight times the distance it moves when you perform the work.
Answer:
a) f ’’ = f₀
, b) Δf = 2 f₀ 
Explanation:
a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.
Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source
f ’= fo
This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.
f ’’ = f’
where c represents the sound velocity in stationary blood
therefore the received frequency is
f ’’ = f₀
let's simplify the expression
f ’’ = f₀ \frac{c+v}{c-v}
f ’’ = f₀
b) At the low speed limit v <c, we can expand the quantity
(1 -x)ⁿ = 1 - x + n (n-1) x² + ...
f ’’ = fo
f ’’ = fo 
leave the linear term
f ’’ = f₀ + f₀ 2
the sound difference
f ’’ -f₀ = 2f₀ v/c
Δf = 2 f₀ 