Homework Help is Here – Start Your Trial Now!
Engineering
Chemical Engineering
|d²y Given the L 10 +9y = 5t with IVP y(0) = -1 and y'(0) = 2 di dt 4. What is the initial Y(s) after substituting the initial values? 5-s +12s? 5-s' +3s? a. Y(s)(s² +10s +9) = Y(s)(s² – 10s +9) =- 5+s +12s? 5-s +12s? b. Y(s)(s -10s+9) = d. Y(s)(s – 10s +9) =

|d²y Given the L 10 +9y = 5t with IVP y(0) = -1 and y'(0) = 2 di dt 4. What is the initial Y(s) after substituting the initial values? 5-s +12s? 5-s' +3s? a. Y(s)(s² +10s +9) = Y(s)(s² – 10s +9) =- 5+s +12s? 5-s +12s? b. Y(s)(s -10s+9) = d. Y(s)(s – 10s +9) =

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
See similar textbooks
icon
Related questions
Question

Answer 13-17 and 4-9

Question 4-9 Sd²y Given the L di +9y = 5t > with IVP y(0) = -1 and y'(0) = 2 dt 4. What is the initial Y(s) after substituting the initial values? 5- s° +12s? 5-s' +3s? а. Y(s)(s +10s +9) = c. Y(s)(s – 10s +9) = s² 5+ s³ +12s? 5-s' +12s? b. Y(s)(s -10s +9) D d. Y(s)(s –10s +9): = 5. What is the Y(s) before identifying the arbitrary constants? s (12- s)+5 s(s – 9)(s – 1) 6. How is the evaluation of the partial fraction expressed? A Bs s(12+s)+5 s (s – 9)(s – 1) (12-s)+5 (12-s) +5 Y(s) = b. Y(s) =- c. Y(s) = d. Y(s) = а. s (s - 9)(s – 1) (s - 9)(s – 1) 12s? – s +5 12s2 - s +5 с. s(s -9)(s – 1) CD A B C D a. %3D %3D s' (s - 9)(s – 1) S-9 S-1 s -9 s-1 12s + s +5 s'(s - 9)(s – 1) 12s? – s' +5 d. A B D В A + C D b. s (s -9)(s –1) S -9 7. What is the value of A in the partial faction? S-1 S-9 s-1 50 3 b. A =- 81 5 A 81 d. A= 81 а. с. 81 8. What is the value of D? D=9 b. D=2 D=-9 d. D=-2 а. с. 9. What is the transformation of the given f(t)? 50, 5 +21+e" - 2e' 81 31 31 50 c. Y(t) = 81 5 -e' 81 Y(1) = -2t + et а. 9. 81 81 50 31 + t+=e" – 9e' 81 81 3 5 5 d. Y(t) = +t+2e" 81 81 b. Y(s) = - -- 81
Transcribed Image Text:Question 4-9 Sd²y Given the L di +9y = 5t > with IVP y(0) = -1 and y'(0) = 2 dt 4. What is the initial Y(s) after substituting the initial values? 5- s° +12s? 5-s' +3s? а. Y(s)(s +10s +9) = c. Y(s)(s – 10s +9) = s² 5+ s³ +12s? 5-s' +12s? b. Y(s)(s -10s +9) D d. Y(s)(s –10s +9): = 5. What is the Y(s) before identifying the arbitrary constants? s (12- s)+5 s(s – 9)(s – 1) 6. How is the evaluation of the partial fraction expressed? A Bs s(12+s)+5 s (s – 9)(s – 1) (12-s)+5 (12-s) +5 Y(s) = b. Y(s) =- c. Y(s) = d. Y(s) = а. s (s - 9)(s – 1) (s - 9)(s – 1) 12s? – s +5 12s2 - s +5 с. s(s -9)(s – 1) CD A B C D a. %3D %3D s' (s - 9)(s – 1) S-9 S-1 s -9 s-1 12s + s +5 s'(s - 9)(s – 1) 12s? – s' +5 d. A B D В A + C D b. s (s -9)(s –1) S -9 7. What is the value of A in the partial faction? S-1 S-9 s-1 50 3 b. A =- 81 5 A 81 d. A= 81 а. с. 81 8. What is the value of D? D=9 b. D=2 D=-9 d. D=-2 а. с. 9. What is the transformation of the given f(t)? 50, 5 +21+e" - 2e' 81 31 31 50 c. Y(t) = 81 5 -e' 81 Y(1) = -2t + et а. 9. 81 81 50 31 + t+=e" – 9e' 81 81 3 5 5 d. Y(t) = +t+2e" 81 81 b. Y(s) = - -- 81
Question 13-17. A breeder reactor converts relatively stable Uranium-238 into the isotope Plutonium-239. After 5 years it was determined that 0.043% of the initial amount A, of plutonium has disintegrated. 13. What will be the differential equation for this problem? dA = kA dt dP b. = kdt c. Both a and b d. Neither a nor b а. 14. If the atoms of A. have disintegrated, how much is remained? c. 0.043% d. None of them a. 99.957% 15. What is the value of the decay constant? a. -0.20977/y 16. What is the corresponding value of A(t) at half-life? b. 99.043% b. -0.000028672 c. -0.20977 d. -0.000028672/y а. 3/2 Ао b. ½ Ao с. 2 Ао d. None of them 17. What is the half-life of the isotope? a. 24180 years b. 24174.35 уears 18.454 years d. 27,617.995 years с.
Transcribed Image Text:Question 13-17. A breeder reactor converts relatively stable Uranium-238 into the isotope Plutonium-239. After 5 years it was determined that 0.043% of the initial amount A, of plutonium has disintegrated. 13. What will be the differential equation for this problem? dA = kA dt dP b. = kdt c. Both a and b d. Neither a nor b а. 14. If the atoms of A. have disintegrated, how much is remained? c. 0.043% d. None of them a. 99.957% 15. What is the value of the decay constant? a. -0.20977/y 16. What is the corresponding value of A(t) at half-life? b. 99.043% b. -0.000028672 c. -0.20977 d. -0.000028672/y а. 3/2 Ао b. ½ Ao с. 2 Ао d. None of them 17. What is the half-life of the isotope? a. 24180 years b. 24174.35 уears 18.454 years d. 27,617.995 years с.
Expert Solution
steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions
Recommended textbooks for you
Introduction to Chemical Engineering Thermodynami…
Introduction to Chemical Engineering Thermodynami…
Chemical Engineering
ISBN:
9781259696527
Author:
J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:
McGraw-Hill Education
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemical Engineering
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
Elements of Chemical Reaction Engineering (5th Ed…
Elements of Chemical Reaction Engineering (5th Ed…
Chemical Engineering
ISBN:
9780133887518
Author:
H. Scott Fogler
Publisher:
Prentice Hall
Process Dynamics and Control, 4e
Process Dynamics and Control, 4e
Chemical Engineering
ISBN:
9781119285915
Author:
Seborg
Publisher:
WILEY
Industrial Plastics: Theory and Applications
Industrial Plastics: Theory and Applications
Chemical Engineering
ISBN:
9781285061238
Author:
Lokensgard, Erik
Publisher:
Delmar Cengage Learning
Unit Operations of Chemical Engineering
Unit Operations of Chemical Engineering
Chemical Engineering
ISBN:
9780072848236
Author:
Warren McCabe, Julian C. Smith, Peter Harriott
Publisher:
McGraw-Hill Companies, The
SEE MORE TEXTBOOKS