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Brut [27]
3 years ago
14

If the force of gravity between a book of mass 0.50 kg and a calculator of 0.100 kg is 1.5 × 10-10 N, how far apart are they?

Physics
1 answer:
Nady [450]3 years ago
5 0
m_1=0,5kg \\ m_2=0,1kg \\ F=1,5*10^{-10}N \\ G=6,67*10^{-11}m^3/kg*s^2 \\  \\ r=? \\  \\  \\ F= \frac{Gm_1m_2}{r^2 }  \\  \\ Fr^2=Gm_1m_2 \\  \\r^2=\frac{Gm_1m_2}{F}  \\  \\ r= \sqrt{\frac{Gm_1m_2}{F} } = 


=\sqrt{ \frac{6,67*10^{-11} \frac{m^3}{kg*s^2}*0,5kg*0,1kg }{1,5*10^{-10}N} }  = \sqrt{ \frac{6,67*10^{-11} \frac{m^3}{kg*s^2}*0,05kg^2 }{1,5*10^{-10} \frac{kg*m}{s^2} } }  =  \\  \\ =\sqrt{ \frac{6,67 *10^{-11}*0,05 }{1,5*10^{-10}}m^2 }  \approx\boxed{0,15m}  \\  \\  \\  \\  \\  \\ "Non\ nobis,\ Domine,\ non\ nobis,\ sed\ Nomini\ tuo\ da\ gloriam." \\  \\  \\  \\ \underline{\underline{Regards\ M.Y.}}

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Explanation:
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b) Now that we have v final = 0, v initial = 100, a = -10, t = 10s (10s because maximum displacement means the displacement from the ground to the max height) we can easily find the displacement by applying the second formula of UARM:

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= 500 m

Hope this helps, brainliest would be appreciated :)
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