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Question Number 85534 by liki last updated on 22/Mar/20

Commented by liki last updated on 22/Mar/20

$$....pleasehelpideaquestionno14\left(a\right)$$

Commented by liki last updated on 22/Mar/20

$$....ineedhelppleaseemergenceqnno14a$$

Answered by MJS last updated on 22/Mar/20

$$\mathrm{easy}:\mathrm{earth}\mathrm{is}\mathrm{sphere}\mathrm{with}\mathrm{radius}R$$$$\mathrm{cut}\mathrm{through}\mathrm{earth}\mathrm{through}\mathrm{both}\mathrm{poles}$$$$\Rightarrow \mathrm{circle}\mathrm{with}\mathrm{radius}R$$$$\mathrm{coordinate}\mathrm{of}\mathrm{a}\mathrm{point}\mathrm{at}\mathrm{angle}23.5°\mathrm{N}\mathrm{is}$$$$P=\left(\begin{array}{c}R\cos 23.5°\\ R\sin 23.5°\end{array}\right)$$$$\mathrm{in}3\mathrm{dimensions}P\mathrm{fulfills}\mathrm{a}\mathrm{circle}\mathrm{of}\mathrm{radius}$$$$r=R\cos 23.5°\mathrm{within}24\mathrm{hours}$$$$\mathrm{the}\mathrm{perimeter}\mathrm{of}\mathrm{this}\mathrm{circle}\mathrm{is}$$$$2\pi r=2\pi R\cos 23.5°$$$$\mathrm{the}\mathrm{speed}\mathrm{then}\mathrm{is}\frac{2\pi R\cos 23.5°}{24}\mathrm{km}/\mathrm{hr}$$$$\mathrm{if}\mathrm{we}\mathrm{take}R\approx 6370\mathrm{km}\mathrm{we}\mathrm{get}\approx 1530\mathrm{km}/\mathrm{hr}$$

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