DIFFERENTIAL EQUATIONS-NEXTGEN WILEYPLUS

3rd Edition

ISBN: 9781119764564

Author: BRANNAN

Publisher: WILEY

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Textbook Question

Chapter 1.2, Problem 5P

Phase Line Diagrams. Problems 1 through 7 involve equations of the form d y / d t = f ( y ) . In each problem, sketch the graph of f ( y ) versus y , determine the critical (equilibrium) points, and classify each one as asymptotically stable or unstable. Draw the phase line, and sketch several graphs of solutions in the t y − plane.

d y / d t = y 2 ( y + 1 ) ( y − 3 ) , − ∞ < y 0 < ∞ .

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Chapter 1.2, Problem 4P

Chapter 1.2, Problem 6P

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Question 1: If a barometer were built using oil (p = 0.92 g/cm³) instead of mercury (p = 13.6 g/cm³), would the column of oil be higher than, lower than, or the same as the column of mercury at 1.00 atm? If the level is different, by what factor? Explain. (5 pts) Solution: A barometer works based on the principle that the pressure exerted by the liquid column balances atmospheric pressure. The pressure is given by: P = pgh Since the atmospheric pressure remains constant (P = 1.00 atm), the height of the liquid column is inversely proportional to its density: Step 1: Given Data PHg hol=hgx Poil • Density of mercury: PHg = 13.6 g/cm³ Density of oil: Poil = 0.92 g/cm³ • Standard height of mercury at 1.00 atm: hμg Step 2: Compute Height of Oil = 760 mm = 0.760 m 13.6 hoil = 0.760 x 0.92 hoil = 0.760 × 14.78 hoil = 11.23 m Step 3: Compare Heights Since oil is less dense than mercury, the column of oil must be much taller than that of mercury. The factor by which it is taller is: Final…

Question 3: A sealed flask at room temperature contains a mixture of neon (Ne) and nitrogen (N2) gases. Ne has a mass of 3.25 g and exerts a pressure of 48.2 torr. . N2 contributes a pressure of 142 torr. • What is the mass of the N2 in the flask? • Atomic mass of Ne = 20.1797 g/mol • Atomic mass of N = 14.0067 g/mol Solution: We will use the Ideal Gas Law to determine the number of moles of each gas and calculate the mass of N2. PV = nRT where: • P = total pressure • V volume of the flask (same for both gases) n = number of moles of gas • R 0.0821 L atm/mol K • T = Room temperature (assume 298 K) Since both gases are in the same flask, their partial pressures correspond to their mole fractions. Step 1: Convert Pressures to Atmospheres 48.2 PNe = 0.0634 atm 760 142 PN2 = = 0.1868 atm 760 Step 2: Determine Moles of Ne nNe = mass molar mass 3.25 nNe 20.1797 nne 0.1611 mol Step 3: Use Partial Pressure Ratio to Find n

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Chapter 1 Solutions

DIFFERENTIAL EQUATIONS-NEXTGEN WILEYPLUS

Ch. 1.1 - Newton’s Law of Cooling. A cup of hot coffee has...Ch. 1.1 - Blood plasma is stored at . Before it can be...Ch. 1.1 - At 11:09p.m. a forensics expert arrives at a crime...Ch. 1.1 - The rate constant if the population doubles in ...Ch. 1.1 - The field mouse population in Example 3 satisfies...Ch. 1.1 - Radioactive Decay. Experiments show that a...Ch. 1.1 - A radioactive material, such as the isotope...Ch. 1.1 - Classical Mechanics. The differential equation for...Ch. 1.1 - For small, slowly falling objects, the assumption...Ch. 1.1 - Mixing Problems. Many physical systems can be cast...

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