Answer:
[tex]f=5*10^1^4kg/year[/tex]
Explanation:
[tex]T=m/f\\f=m/T\\f=\frac{3*10^1^5kg}{6 years}[/tex]
The mass flow rate of CO2 into and out of the atmosphere is [tex]f = 5 * 10^{14} kg/ year[/tex]
What is the method ?
[tex]T = m/f[/tex]
Therefore, [tex]f = m/ t[/tex]
[tex]f=3 * 10^{15} / 6 years[/tex]
So, the correct answer is : [tex]f = 5 * 10^{14} kg/ year[/tex]
What is mass flow rate?
Learn more about mass flow rate below,
https://brainly.com/question/15877818
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