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Physics 15 Online
OpenStudy (anonymous):

What is the magnitude of the speed of a person at the equator due to rotational speed of the Earth?

OpenStudy (ujjwal):

Firstly calculate the angular speed of rotation of earth: \[\omega=\frac{2\pi}{T}=\frac{2\pi}{24\times 60\times 60}=\frac{\pi}{43200}\]Now, velocity at equator can be obtained by simply using the relation, V=R\(\omega\) where R is radius of earth So, \[V=6400000\times\frac{\pi}{43200}=465.42ms^{-1}\]

OpenStudy (anonymous):

can you write out this math work using the draw function.. my computer isn't putting it into the equation.. its all jumbled.

OpenStudy (ujjwal):

Keep pressing F5

OpenStudy (anonymous):

? it did nothing?

OpenStudy (ujjwal):

see this! Its a screenshot!

OpenStudy (anonymous):

thanks! :)

OpenStudy (ujjwal):

Welcome!

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