Is Centripetal Acceleration Noticeable When Running in a Circle?

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Centripetal acceleration can be calculated using the formula a = v^2/r, resulting in 1.7 m/s² for a speed of 4.5 m/s around a circle with a 12m diameter. The term "perceptible" refers to whether one can feel this acceleration during circular motion. When running in a circle, the centripetal acceleration is indeed noticeable, as it creates a force that pushes the runner outward, similar to the sensation experienced while driving in a roundabout. This sensation occurs due to the constant inward acceleration directed toward the center of the circle. Thus, centripetal acceleration is perceptible when running in a circle.
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Homework Statement


You are running at 4.5 m/s around a circle 12m in diameter. Is the centripital acceleration perceptible?


Homework Equations


a=v^2/r


The Attempt at a Solution


I know that the centripital acceleration is 1.7m/s^2 (4.5^2/12). What I don't know is what it means for centripital acceleration to be perceptible. Can anyone help me/ explain what this means?
 
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Ey yo, waddap.

I'm not a native English speaker, and I'm learning physics in my own language, so the question is unfamiliar to me, BUT I think I can help.

Perceptible.. To be perceived, experienced, felt.

Do you feel the acceleration? Yeah, you sure do. The acceleration vector in circle motion, as long as the orbital acceleration is constant, which it is in your case, points towards the center of the circle. So the object (you) will feel it pushing you the other way, I.E. 180 degrees from the center of the circle. Much as when you're driving in a roundabout. Yeah, the centripetal acceleration is perceptible.

That's my interpretation of the question.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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