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3 years ago

Interstellar space is filled with radiation of frequency 160.23 GHz.

Physics

1 answer:

nasty-shy [4]3 years ago

8 0

Answer:

1.548K

Explanation:

Given that f= 160.23GHz

c= 3E8m/s

b= 2.898*10^-3mk

So using

(Lambda)m x T= b

So T = b/ lambda

But wavelength ( lambda) = c/f

So T = bf/c

= 2.898E-3x 160.23E9/3E8

=1.548K

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MrRa [10]

Answer:

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This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

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$v^2 - u^2 = 2\, a\, x$ ,

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$v$ is the final velocity of the moving object,
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$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
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For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
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The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .

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