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3 years ago

I need help guyssss plssss

Physics

1 answer:

Kipish [7]3 years ago

8 0

Answer:

Cd(NO3)2 + Na2S --> CdS + 2 NaNO3

Explanation:

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c) kinetic energy

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How much force would be needed to cause a 4.6kg object to accelerate at 9.2m/s/s? *

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3 years ago

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

vivado [14]

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 $T_{1}$ and $T_{2}$ 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 that $T_{1} =29.46 years$

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]

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3 years ago