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3 years ago

11

An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

?

1 answer:

Arisa [49]3 years ago

4 0

Explanation:

Distance = (intial speed)X(Time) + 1/2(acceleration)X(Time) [Third equation of motion]

As initial speed is zero, therefore;

Distance = 1/2(acceleration)X(Time)

= 1/2 (6 X 15)

= 1/2 (90)

= 45 meters

Hence, the object traveled 45 meters.

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According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

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For satellite Y, the orbital period $T_{Y}$ is:

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Where $r_{Y}=1.9(10)^{5}m$

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This means $T_{X}>T_{Y}$

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$V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}$

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$V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}$

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