How do massive stars change the atmosphere?

1 answer:

6 0

Not sure what you're referring to... our atmosphere is mostly nitrogen/oxygen - both gases were created by very massive stars, many times greater than our sun - our sun isn't massive enough to create the gases needed for our atmosphere...

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4. Locate the data and observations collected in your lab guide. What are the key results? How would you best summarize the data

melisa1 [442]

Answer:

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Explanation:

5 0

3 years ago

The seismic activity density of a region is the ratio of the number of earthquakes during a given time span to the land area aff

Natalija [7]

Answer:

0.0059

Explanation:

According to the question the seismic activity density is given by

$\text{Seismic activity density}=\frac{\text{Number of Earthquakes over a given time span}}{\text{The land area affected}}$

Here,

Number of Earthquakes over a given time span = 424

The land area affected = 71300 mi²

So,

$\text{Seismic activity density}=\frac{424}{71300}\\\Rightarrow \text{Seismic activity density}=0.0059$

The seismic activity density is 0.0059

8 0

3 years ago

Read 2 more answers

A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?

anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\ \end{gathered}$

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

$\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}$

now, replace

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}$

hence, the answer is 0.495 m/s