3 years ago

Please help!! I'm stuck on this, could someone also explain if they can!! Will give brainiest if I can!!

Physics

1 answer:

Ann [662]3 years ago

8 0

The answer is going to be c. hope that helped

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Someone please help!! Will mark brainliest

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(a) Suppose that your measured weight at the equator is one-half your measured weight at the pole on a planet whose mass and dia

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7160.2812 s or 1.988 hours

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m = Mass of person

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$\omega$ = Angular speed

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$F_p=mg$

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$F_e=\dfrac{1}{2}F_p\\\Rightarrow m(g-\omega^2R)=\dfrac{1}{2}F_p\\\Rightarrow \omega=\sqrt{\dfrac{g}{2R}}$

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$T=\dfrac{2\pi}{\omega}\\\Rightarrow T=2\pi\sqrt{\dfrac{2R}{g}}\\\Rightarrow T=2\pi\sqrt{\dfrac{2\times 6.37\times 10^6}{9.81}}\\\Rightarrow T=7160.2812\ s=1.988\ hours$

The rotational period of the planet is 7160.2812 s or 1.988 hours

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