If the horizontal circular path the riders follow has a 7.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 2.25 times that due to gravity?

A fairgrounds ride spins its occupants inside a flying saucer-shaped container. rev/min

Answer 1
centripetal acceleratrion can be expressed as: #a=romega^2#
Where #omega# is angular velocity . Equating this to the question: #romega^2=2.25timesg#
Substituting in values: #7.00timesomega^2=2.25times9.81#
solving for #omega# we get: #omega = 1.78#
Angular velocity is the rate of change of angle #omega=(2*pi)/T#

where T is the time interval (one revolution's worth)

Thus, we can figure out T.

#T=(2*pi)/omega=3.5s#

In one minute, it will have completed one revolution every 3.5 seconds.

#60/3.5=17.1# revolutions
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Answer 2

The riders will experience a centripetal acceleration equivalent to 22.5 times the acceleration due to gravity. They will undergo approximately 4.77 revolutions per minute.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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