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evablogger [386]
4 years ago
12

When light travels from a medium it moves faster in (such as air) to a medium it moves slower in (such as glass) its velocity de

creases. What else happens? a) The frequency increases but the wavelength remains the same. b) The wavelength increases but the frequency remains the same. c) The wavelength decreases but the frequency remains the same. d) The frequency decreases but the wavelength remains the same.
Physics
1 answer:
shutvik [7]4 years ago
5 0

Answer:

c) The wavelength decreases but the frequency remains the same.

Explanation:

Light travels at different speed in different mediums.

Refractive index is equal to velocity of the light 'c' in empty space divided by the velocity 'v' in the substance.

Or ,

n = c/v.

<u>The frequency of the light does not change but the wavelength of the light changes with change in the speed.</u>

c = frequency × Wavelength

Frequency is constant,

The formula can be written as:

n = λ / λn.

Where,  

λn is the wavelength in the medium

λ is the wavelength in vacuum

<u>When the light travels to glass, it speed slows down and also the wavelength decreases as both are directly proportional. There will be no effect on frequency.</u>

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Answer:

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Explanation:

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3 years ago
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Which light will produce white​
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Answer:

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2 years ago
A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to
Nina [5.8K]

Answer:

1) \Delta s=1000\ ft

2)  \Delta s'=998.11\ ft.s^{-1}

3) t\approx125\ s

t'\approx463.733\ s

Explanation:

Given:

width of river, w=500\ ft

speed of stream with respect to the ground, v_s=8\ ft.s^{-1}

speed of the swimmer with respect to water, v=4\ ft.s^{-1}

<u>Now the resultant of the two velocities perpendicular to each other:</u>

v_r=\sqrt{v^2+v_s^2}

v_r=\sqrt{4^2+8^2}

v_r=8.9442\ ft.s^{-1}

<u>Now the angle of the resultant velocity form the vertical:</u>

\tan\beta=\frac{v_s}{v}

\tan\beta=\frac{8}{4}

\beta=63.43^{\circ}

  • Now the distance swam by the swimmer in this direction be d.

so,

d.\cos\beta=w

d\times \cos\ 63.43=500

d=1118.034\ ft

Now the distance swept downward:

\Delta s=\sqrt{d^2-w^2}

\Delta s=\sqrt{1118.034^2-500^2}

\Delta s=1000\ ft

2)

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

v'=v.\cos37

v'=4\times \cos37

v'=3.1945\ ft.s^{-1}

<u>Now the net effective speed of stream sweeping the swimmer:</u>

v_n=v_s-v'

v_n=8-3.1945

v_n=4.8055\ ft.s^{-1}

<u>The  component of swimmer's velocity heading directly towards the opposite bank:</u>

v'_r=v.\sin37

v'_r=4\sin37

v'_r=2.4073\ ft.s^{-1}

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

\tan\phi=\frac{v_n}{v'_r}

\tan\phi=\frac{4.8055}{2.4073}

\phi=63.39^{\circ}

  • Now let the distance swam in this direction be d'.

d'\times \cos\phi=w

d'=\frac{500}{\cos63.39}

d'=1116.344\ ft

<u>Now the distance swept downstream:</u>

\Delta s'=\sqrt{d'^2-w^2}

\Delta s'=\sqrt{1116.344^2-500^2}

\Delta s'=998.11\ ft.s^{-1}

3)

Time taken in crossing the rive in case 1:

t=\frac{d}{v_r}

t=\frac{1118.034}{8.9442}

t\approx125\ s

Time taken in crossing the rive in case 2:

t'=\frac{d'}{v'_r}

t'=\frac{1116.344}{2.4073}

t'\approx463.733\ s

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Yuliya22 [10]

Answer:

case 1 of physics is the answer

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castortr0y [4]
The first factor is wind speed, the second factor is wind duration, and the last factor is the fetch, the distance over which the wind blows without a change in direction.

all these factors determines the strength of a wave.

hope this helps :)
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3 years ago
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