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riadik2000 [5.3K]
3 years ago
15

A marble is placed at the bottom of a semi-spherical bowl, as shown in the figure. The marble is then displaced from the bottom

of the bowl to a position about halfway from the top of the bowl. The marble is then released from rest such that the marble always remains in contact with the bowl. Students observe that the marble rolls back and forth as it oscillates about the bottom of the bowl. Which of the following statements best explains why the marble undergoes oscillatory motion?
a)The sides of the bowl become steeper at positions farther from the bottom of the bowl.b) The net force exerted on the marble always has a component directed toward the bottom of the bowl. c) The normal force exerted on the marble decreases with increasing distance from the bottom of the bowl. D ) The gravitational force exerted on the marble is constant in magnitude and direction.​

Physics
1 answer:
Andrew [12]3 years ago
8 0

Answer:

It is D

Explanation:

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A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s
yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

E_{pe}\:(elastic\:potential\:energy) = 5184\:J

K\:(constant) = 16200\:N/m

x\:(displacement) =\:?

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

E_{pe} = \dfrac{k*x^2}{2}

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:  

E_{pe} = \dfrac{k*x^2}{2}

5184 = \dfrac{16200*x^2}{2}

5184*2 = 16200*x^2

10368 = 16200\:x^2

16200\:x^2 = 10368

x^{2} = \dfrac{10368}{16200}

x^{2} = 0.64

x = \sqrt{0.64}

\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark

Answer:  

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

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