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aev [14]
3 years ago
8

Two creatures sit on a horizontal frictional rotating platform. The platform rotates at a constant speed. The creatures do not s

lip off as it rotates.
ASSUME:

Red has a mass of 5 kg

Red is 1.5 m from the center

Red has a speed of 9 m/s

Blue has a mass of 25 kg

Blue has a speed of 1.8 m/s

The force of friction on Red is EQUAL to the force of friction on Blue





DETERMINE:

How far from the center is Blue
Physics
1 answer:
ddd [48]3 years ago
6 0

Answer:

M v^2 / R = centripetal force

For Red: M v^2 / R = 5 * 9^2 / 1.5 = 270

For Blue M v^2 / R = 270 = 25 * 1.8^2 / Rb

So Rb = 25 * 1.8^2 / 270 = .3 m

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3 years ago
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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Answer:

Answer:

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τ = 0.95 x 29.5

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Thus, the torque is 28.025 Nm.

Explanation:

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Explain how work is related to energy
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