Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
Given:
Time = 8.30 s ahead of the hemoglobin
For the equation of both glucose and hemoglobin:
[tex]\begin{gathered} \sqrt{2D_gt_g}=\sqrt{2D_h(t_g+\Delta t)} \\ \\ D_gt_g=D_h(t_g+\Delta t) \\ \\ D_gt_g=D_ht_g+D_h\Delta_t \end{gathered}[/tex]Rewrite the equation for tg:
[tex]\begin{gathered} D_gt_g-D_ht_g=D_h\Delta_t \\ \\ t_g(D_g-D_h)=D_h\Delta_t \\ \\ t_g=\frac{D_h\Delta_t}{D_g-D_h} \\ \\ \end{gathered}[/tex]Where:
Dh is the diffusion coefficient of hemoglobin = 6.9 x 10⁻¹¹ m²/s
Dg is the diffusion coefficient of glucose = 6.7 x 10⁻¹⁰ m²/s
Δt = 8.30 seconds
Substitute these values for the variables in the equation and solve:
[tex]\begin{gathered} t_g=\frac{6.9\times10^{-11}*8.30}{6.7\times10^{-10}-6.9\times10^{-11}} \\ \\ t_g=\frac{5.727\times10^{-10}}{6.01\times10^{-10}} \\ \\ t_g=0.953\text{ seconds} \end{gathered}[/tex]Therefore, the time that passes before the glucose is 8.30 s ahead of the hemoglobin is 0.953 seconds.
ANSWER:
0.953 seconds
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
