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Alex777 [14]
4 years ago
6

You record the spectrum of a star and find that a calcium absorption line has an observed wavelength of 394.0 nm. This calcium a

bsorption line has a rest wavelength of 393.3 nm. What is the approximate radial velocity of this star?
Physics
1 answer:
saveliy_v [14]4 years ago
8 0

Answer:

radial velocity = 0.533 × 10^{6} m/s

Explanation:

given data

observed wavelength = 394.0 nm

rest wavelength = 393.3 nm

solution

we get here radial velocity  by the Doppler effect of light that is

\frac{\triangle \lambda }{\lambda} = \frac{v}{c}     ......................1

v = \frac{c \times \triangle \lambda }{\lambda}    

put here value

v = \frac{c \times (\lambda ' - \lambda  }{\lambda}

v = \frac{3 \times 10^8  \times (394 - 393.3}{393.3}

v = 533943.55

v = 0.533 × 10^{6} m/s

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A man standing 1.54 m in front of a shaving mirror produces a real, inverted image 15.2 cm from it. What is the focal length of
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Answer:

The focal length is 16.86 cm and the distance of the man  if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

Explanation:

Given that,

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Using formula of mirror

\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}

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\dfrac{1}{f}=\dfrac{1}{15.2}+\dfrac{1}{-154}

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Using formula of magnification

m= \dfrac{-v}{u}

Put the value into the formula

2=\dfrac{v}{u}

v = -2u

Using formula of for focal length

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

\dfrac{1}{16.86}=\dfrac{1}{u}-\dfrac{1}{2u}

\dfrac{1}{16.86}=\dfrac{1}{2u}

2u=16.86

u=\dfrac{16.86}{2}

u=8.43\ cm

Hence, The focal length is 16.86 cm and the distance of the man  if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

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This problem could be solved using this equation:

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