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Orlov [11]
3 years ago
15

How does the density and distribution of your "stars" change as the circle gets larger (expands)?

Biology
1 answer:
Vilka [71]3 years ago
7 0

Density and distribution of stars are directly affected when the change occur in the size of the circle.

<h3>Effect on density </h3>

The density and distribution of "stars" changed as the circle gets larger because when the circle gets larger, its volume increase which results in the decrease in the density. The distribution of stars also changed due to more space or volume present in the medium.

<h3>Distribution</h3>

If the universe is more condensed, then the stars are more closer together but when the universe is more expand then the stars are present farther away from each other so we can conclude that density and distribution of stars are directly affected when the change occur in the size of the circle.

Learn more about density here: brainly.com/question/6838128

Learn more: brainly.com/question/26243555

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6 0
3 years ago
What type of chromosomal mutation has taken place?
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5 0
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Why do we want to test one variable at a time? a. Because it is hard to setup a data table with more b. It is what has always be
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C

Explanation:

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7 0
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Which of the following must be true of the structures of axons and muscle cells an order for a nerve and passes to stimulate mus
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7 0
3 years ago
Gold has a specific heat of 0.129 J/(g×°C). How many joules of heat energy are required to raise the temperature of 15 grams of
Elodia [21]

Answer:

\boxed {\boxed {\sf 121.905 \ J  }}

Explanation:

We are asked to find the energy given mass, specific heat, and change in temperature. Therefore, we must use this formula;

q= mc \Delta T

The mass is 15 grams and the specific heat is 0.129 J/(g×°C). Let's calculate the change in temperature.

  • ΔT= final temperature - initial temperature
  • ΔT= 85 °C- 22°C = 63°C

Now we know all the values:

m= 15 \ g \ \\c= 0.129 \ J / (g* \textdegree C) \\\Delta T= 63 \ \textdegree C

Substitue the values into the formula.

q= (15 \ g)( 0.129 \ J / (g* \textdegree C)) ( 63 \ \textdegree C)

Multiply the first numbers together. The grams will cancel.

q= (1.935 \ J/ \textdegree C) ( 63 \ \textdegree C)

Multiply again, this time the degrees Celsius cancels.

q= 121.905 \ J

<u>121.905 Joules</u> are required.

5 0
2 years ago
Read 2 more answers
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