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If earth contracts to half of its present radius what would be length of the day at equator? |
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Answer» \(I_1= \frac{2}{5}MR^2,I_2=\frac{2}{5}M(\frac{R}{2})^2\) or \(I_2=\frac{I_1}{4}\) \(I_1w_1=I_2w_2\) \(I(\frac{2π}{T_1})=\frac{1}{4}I(\frac{2π}{T_2})\) \(T_2= \frac{T_1}{4}=\frac{24}{4}=6\, hours\) |
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