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3 years ago

Explain why Earth is not spherical in shape, but bulges at theEquator?

Physics

2 answers:

Strike441 [17]3 years ago

8 0

Explanation:

Every rotating body experiences centrifugal force. Due to this force the body tends to bulge out around it mid point and gets flattened at the poles. Same is applicable to Earth as well. Since the Earth is rotating at a very high speed, its equator gets bulged out due to centrifugal force. Because of this bulged equator, Earth's pole to pole diameter and equatorial diameter has difference of around 42.76 km. It is flatter on the poles. This also proves that Earth is not a perfect sphere.

vlada-n [284]3 years ago

8 0

Answer and Explanation:

The reason for the not being perfectly spherical ad bulging out at the equator is that The centripetal force acting toward's the earth gravitational center tries to keep the Earth in perfect spherical shape.

Also the angular momentum of the orbiting planet influences the bulge,

The greater angular momentum results in more bulge while the lower value of it results in lesser bulge and more perfect spherical shape.

Also, a greater amount of force directed towards the center and acting on the object at the equator results in the bulges at the equator whereas at poles this force is not required and hence radius is lower in that region.

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Luba_88 [7]

Answer:

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Explanation:

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3 years ago

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3 years ago

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harina [27]

Answer:

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2 years ago

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Two parallel slits are illuminated by light composed of two wavelengths. One wavelength is λA = 622nm. The other wavelength is λ

Sergio039 [100]

Answer:

$\lambda_{B}=414.67 nm$

Explanation:

In this question we have given

$\lambda_{A}=622nm$

we have to find

$\lambda_{B}=?$

We know that

optical path difference for bright fringe is given as $=n\lambda$

Here,

n is order of fringe

and optical path difference for dark fringe is given as $=(n+.5)\lambda$

since the light with wavelength $\lambda_{A}$ produces its third-order bright fringe at the same place where the light with wavelength $\lambda_{B}$ produces its fourth dark fringe