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Gemiola [76]
3 years ago
12

Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 1.80 cm from the axis to equal

6.00×105 g (where g is the acceleration due to gravity).
Physics
1 answer:
Arturiano [62]3 years ago
7 0

Answer:

The required angular speed ω of an ultra-centrifuge is:

ω = 18074 rad/sec

Explanation:

Given that:

Radius = r = 1.8 cm

Acceleration due to g = a = 6.0 x 10⁵ g

Sol:

We know that

Angular Acceleration = Angular Radius x Speed²

a = r x ω ²

Putting the values

6 x 10⁵ g = 1.8 cm x ω ²

Converting 1.8 cm to 0.018 m, also g = 9.8 ms⁻²

6 x 10⁵ x 9.8 = 0.018 x ω ²

ω ² = (6 x 10⁵ x 9.8) / 0.018

ω ² =  5880000 / 0.018

ω ² =  326666667

ω = 18074 rad/sec

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The correct answer is

Air resistance

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We have that the spring constant is mathematically given as

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Generally, the equation for angular velocity is mathematically given by

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And

\omega =\frac{2\pi}{T}

Therefore

\frac{2\pi}{T}=\sqrt{k}{n}

Hence giving spring constant k

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Generally

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Period for on complete resolution of Earth around the Sun

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T=365*24*3600

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In conclusion

The effective spring constant of this simple harmonic motion is

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For more information on this visit

brainly.com/question/14159361

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If do 50 J of Work in 20 seconds, what is my Power ?
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