Answered

Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

Sketch the graph to show the relationship between tand mfor t≥ 0 days.Show that the value of A is 120.Show that the value of k is 0.02877, given to 4 significant figures.Using the value of k as 0.02877, find the mass of the substance after 16 days,giving the answer to 3 significant figures,Using the value of k as 0.02877, find the day during which the mass of thesubstance first reaches 50 milligrams.

Sketch The Graph To Show The Relationship Between Tand Mfor T 0 DaysShow That The Value Of A Is 120Show That The Value Of K Is 002877 Given To 4 Significant Fig class=

Sagot :

Given the mass of a radioactive substance after time t days is given by

[tex]m(t)=Ae^{-kt}[/tex]

The initial mass of the substance is 120 milligrams, which means m(0)=120. So,

[tex]\begin{gathered} m(0)=120 \\ \Rightarrow Ae^0=120 \\ \Rightarrow A=120 \end{gathered}[/tex]

The mass of the substance is 90 milligrams after 10 days, which means m(10)=90. So,

[tex]\begin{gathered} m(10)=90 \\ \Rightarrow120e^{-10k}=90 \\ \Rightarrow e^{-10k}=\frac{3}{4} \end{gathered}[/tex]

Now, take natural logarithm on both the side and use the property

[tex]\ln (e^n)=n[/tex]

So, the above equation will become

[tex]\begin{gathered} \ln (e^{-10k})=\ln (\frac{3}{4}) \\ \Rightarrow-10k=-0.2877 \\ \Rightarrow k=\frac{-0.2877}{-10} \\ \Rightarrow k=0.02877 \end{gathered}[/tex]

Therefore, the relation between the mass of the substance at time t days is given by t

[tex]m(t)=120e^{-0.02877t}[/tex]

The mass of the substance after 16 days will be

[tex]m(16)=120e^{-0.02877(16)}=120(0.631)=75.729\text{ milligrams}[/tex]

The day during which the mass of the substance reaches 50 milligrams can be obtained as follows:

[tex]\begin{gathered} m(t)=50 \\ 120e^{\mleft\{-0.02877t\mright\}}=50 \\ \Rightarrow e^{-0.02877t}=\frac{50}{120} \\ \Rightarrow-0.02877t=\ln (\frac{50}{120}) \\ \Rightarrow t=\frac{-0.8754}{-0.02877} \\ \Rightarrow t=30.43\text{ days} \end{gathered}[/tex]

The graph that shows the relationship between m and t is given below:

View image KhalidN543835
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.