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Genrish500 [490]
3 years ago
8

While you are in a bus that moves at 100 km/h you walk from the back to the front at 10 km/h. What is your speed relative to the

road outside?
Physics
1 answer:
aleksklad [387]3 years ago
5 0

Answer: 110km/h

100+10=110km/h

Explanation:

Motion is defined as a change of position. The frame of reference is usually assumed to be at rest. If one is sitting on a bus, the road appears to be moving backwards relative to the observer. If he/she now walks to the front of the bus, he has a speed relative to the earth which is now greater than that of the bus hence the answer.

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Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

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$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

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Acceleration due to gravity at 2000 km depths is :

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$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

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Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

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Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

  $= 153.84 \times 2371 \times 10^{-8}$

  $=0.364 \times 10^{-2} \ km/s$

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3 years ago
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