Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
The required equation f(10) = 13.52 mg remains.
We have given that ,
m is the initial mass and h is the half-life in years. cobalt-60 has a half-life of about 5.3 years. which equation gives the mass of a 50 mg cobalt-60
What is the fromula for he amount of a radioactive substance remaining after t years?
The amount of a radioactive substance remaining after t years is given by the function
[tex]f(t)=m(0.5)^{t/h}[/tex]............ (1),
where m = initial mass and h= half-life in years.
Now, for Cobalt-60, h = 5.3 years, m = 50 mg and t = 10 years,
then from equation (1) we get,
[tex]f(10)=50(0.5)^{10/5.3}[/tex]
Therefore the required equation f(10) = 13.52 mg .
To learn more about the mass of radioactive substance visit:
https://brainly.com/question/25793075
Answer:
The required equation f(10) = 13.52 mg remains.
We have given that ,
m is the initial mass and h is the half-life in years. cobalt-60 has a half-life of about 5.3 years. which equation gives the mass of a 50 mg cobalt-60
What is the fromula for he amount of a radioactive substance remaining after t years?
The amount of a radioactive substance remaining after t years is given by the function
............ (1),
where m = initial mass and h= half-life in years.
Now, for Cobalt-60, h = 5.3 years, m = 50 mg and t = 10 years,
then from equation (1) we get,
Therefore the required equation f(10) = 13.52 mg .
To learn more about the mass of radioactive substance visit:
brainly.com/question/25793075
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
