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3 years ago

12

What happens to light waves from a star as the star moves toward Earth?

2 answers:

Ksivusya [100]3 years ago

4 0

Answer:

The other person is right, the answer is they appear to shift towards blue.

Explanation:

Musya8 [376]3 years ago

3 0

This shifts the star’s spectral lines toward the blue end of the spectrum. If the star is moving away from us, it’s waves are effectively stretched out when they reach earth, increasing their wavelength. This shifts the star’s spectral lines toward the red end of the spectrum.

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After a displacement of 17 m, a train on a straight track is at the position xf = –2.5 m

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-19.5m

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2 years ago

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DEFINE UNIFORM AND NON UNIFORM VELOCITY

Inessa05 [86]

Explanation:

Uniform velocity is when an object goes an equal amount of space in an equal amount of time whereas non uniform velocity is when the object covers an unequal amount of distance in an equal amount of time.

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2 years ago

What do you mean by kinetic energy and potential energy ?

Verizon [17]

Kinetic Energy of an object is the measure of the work an object can do by virtue of its motions..

Formula - KE = 1/2 mv^2

Where KE is the kinetic energy, m is the body’s mass, and v is the body’s velocity.

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1 year ago

If the frequency of a periodic wave is doubled, the period of the wave will be

Lunna [17]

The period of the wave would be halved

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2 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

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2 years ago