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Yanka [14]
3 years ago
9

A 60 kg Gila monster on a merry-go-round is traveling in a circle with a radius of 3 m, rotating at a rate of 9 revolutions/minu

te.
(A) What acceleration does the monster experience?



(B) What is the net force?



(C) If the Gila monster crawls inwards to a distance of 1 m from the center, what is the new value of the force the Gila monster experiences?
Physics
1 answer:
Yanka [14]3 years ago
5 0

PART A)

mass of the monster = 60 kg

radius = 3 m

frequency = 9 rev/min = 9/60 rev/s

now we can find the angular speed as

\omega = 2 \pi f

\omega = 2 \pi (9/60) = 0.94 rad/s

now the acceleration is given by

a_c = \omega^2 R

a_c = (0.94)^2(3) = 2.66 m/s^2

PART B)

now as per Newton's law we have

F_c = ma_c

from above all values we have

F_c = 60(2.66) = 159.9 N

Part C)

now monster moves inside at radius R = 1 m

so now centripetal acceleration is given as

a_C = \omega^2 R

a_C = 0.94^2(1) = 0.88 m/s^2

now the force experience by monster

F_C = ma_C

F_c = 60(0.88) = 52.8 N

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<h2>Answer:</h2>

4E

<h2>Explanation:</h2>

The elastic potential energy of an elastic material (e.g a spring, a wire), is the energy stored when the material is stretched or compressed. It is given by

U = \frac{1}{2}kx^2               --------------------(i)

Where;

U = potential energy stored

k = spring constant of the material

x = elongation (extension or compression of the material).

<em>From the first statement;</em>

<em>when elongation (x) is 4cm, energy stored (U) is E</em>

<em>Substitute these values into equation (i) as follows;</em>

E = \frac{1}{2}k(4)^2

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<em>Make k subject of the formula</em>    

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<em>From the second statement;</em>

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U = \frac{1}{2}\frac{E}{8} (8)^2  

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U = {4E}

Therefore, the potential energy stored will now be 4 times the original one.

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