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2 years ago

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A centrifuge used in DNA extraction spins at a maximum rate of 7000rpm producing a "g-force" on the sample that is 6000 times th

e force of gravity. If the centrifuge takes 10 seconds to come to rest from the maximum spin rate: a) What is the angular acceleration of the centrifuge? b)what is the angular displacement of the centrifuge during this time

1 answer:

defon2 years ago

8 0

Answer:

A) a = 73.304 rad/s²

B) Δθ = 3665.2 rad

Explanation:

A) From Newton's first equation of motion, we can say that;

a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.

Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s

Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s

We are given; t = 10 s

Thus;

a = 733.04/10

a = 73.304 rad/s²

B) From Newton's third equation of motion, we can say that;

ω² = ω_o² + 2aΔθ

Where Δθ is angular displacement

Making Δθ the subject;

Δθ = (ω² - ω_o²)/2a

At this point, ω = 0 rad/s while ω_o = 733.04 rad/s

Thus;

Δθ = (0² - 733.04²)/(2 × 73.304)

Δθ = -537347.6416/146.608

Δθ = - 3665.2 rad

We will take the absolute value.

Thus, Δθ = 3665.2 rad

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Answer:

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2). Toward us

3). $v = - 493421m/s$

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$Redshift: \lambda_{measured} > \lambda_{0}$

$Blueshift: \lambda_{measured} < \lambda_{0}$

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Star 1: $\lambda_{measured} = 120.4nm$

Star 2: $\lambda_{measured} = 121.4nm$

Star 3: $\lambda_{measured} = 122.2nm$

Star 4: $\lambda_{measured} = 122.9nm$

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Star 2:

$Blueshift: 121.4nm < 121.6nm$

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Star 3:

$Redshift: 122.2nm > 121.6nm$

Away from us

Star 4:

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Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

$v = c\frac{\Delta \lambda}{\lambda_{0}}$ (1)

Where $\Delta \lambda$ is the wavelength shift, $\lambda_{0}$ is the wavelength at rest, v is the velocity of the source and c is the speed of light.

$v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}})$ (2)

<em>Case for star 1 $\lambda_{measured} = 120.4 nm$ :</em>

<em></em>

$v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})$

$v = - 2960526m/s$

Notice that the negative velocity means that is approaching to the observer.

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$v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})$

$v = - 493421m/s$

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$v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})$

$v = 1480263m/s$

<em>Case for star 4 $\lambda_{measured} = 122.9 nm$ :</em>

$v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})$

$v = 3207236m/s$

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