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Murrr4er [49]
2 years ago
5

Good Luck! And Happy Weekend!

Physics
1 answer:
Anastaziya [24]2 years ago
7 0
<h3><u>Question</u> :</h3>

Circle the scenario that shows balanced forces and will result in no movement of the gold bar.

<h3><u>Answer</u> :</h3><h3 />
  • 1st one is the scenario that shows balanced forces and will result in no movement of the gold bar. (circled in attachment)

<h3><u>Explanation</u> :</h3>

In the first scenario there are equal forces of 5N acting on both sides of the gold bar, but this is not the case for the second and third scenarios. Equal forces are acting on both sides of the bar in the first scenario, hence showing balanced forces and resulting in no movement of the gold bar.

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Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr
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This question can be solved from the Kepler's law of planetary motion.

As per this law the square of time period of a planet  is proportional to the cube of semi major axis.

Mathematically it can be written as   T^{2} \alpha R^{3}

                                                          ⇒T^{2} = KR^{3}

Here K is the proportionality constant.

If T_{1} andT_{2} are the orbital periods of the planets and

R_{1} and R_{2} are the distance of the planets from the sun, then Kepler's law can be written as-

          \frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{2} }

      ⇒ R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }

  Here we are asked to calculate the the distance of Saturn from sun.It can solved by comparing it with earth.

Let the distance from sun and orbital period of Saturn is denoted as R_{1} and T_{1} respectively.

Let the distance  from sun and orbital period of earth is denoted as R_{2} and T_{2} respectively.

we are given thatT_{1} =29.46 years

we know that R_{2} = 1 AU and T_{2} = 1 year.

1 AU is the mean distance of earth from the sun which is equal to 150 million kilometre.

Hence distance of Saturn from sun  is calculated as -

From Kepler's law as mentioned above-

                                    R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }

                                             =[1 ]^{3} *\frac{[29.46]^{2} }{[1]^{2} } AU

                                    =867.8916 AU^{3}

                                        ⇒R_{1} =\sqrt[3]{867.8916}

                                           =9.5386 AU [ans]

5 0
3 years ago
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