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2 years ago

6

Anne has a sample of a substance. Its volume is 20 cm3, and its mass is 100 grams. What is the sampleâ€™s density? The sampleâ€™

s density is g/cm3.

1 answer:

Tanzania [10]2 years ago

6 0

Answer:

<h2>5 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

$d = \frac{m}{v} \\$

m is the mass

v is the volume

From the question

m = 100 g

v = 20 cm³

We have

$d = \frac{100}{20} = 5 \\$

We have the final answer as

<h3>5 g/cm³</h3>

Hope this helps you

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Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

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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Where;

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Where $r_{X}=1.2(10)^{6}m$

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

$T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}$ (5)

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

$T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}$ (6)

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

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<u>For Satellite X:</u>

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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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$V_{Y}=\sqrt{\frac{GM}{r_{Y}}}$ (10)

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

$V_{Y}= 45801.13 m/s$ (11)

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

Therefore:

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