Explanation Given: Function A ( t ) = − 2 cos ( 2 π 5 t ) + 2 Formula Used: Formulas of derivatives are as follows: d d x ( cos x ) = − sin x d d x ( x n ) = n x n − 1 d d x { c . f ( x ) } = c . d d x f ( x ) Calculation: It is given that adult takes about 12 breaths per minute. Thus, we have 12 breaths in 1 min 1 breath in 1 12 min Also, we know 1 min = 60 sec Thus, 1 breath&# To determine (b) To find: The derivative at t = 1 , where A ( t ) = − 2 cos ( 2 π 5 t ) + 2.

6th Edition

ISBN: 9781119399353

Author: DA

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Question

Chapter 3.5, Problem 26P

To determine

(a)

To find:

One breathing cycle of an adult for the below function:

A(t)=−2cos(2π5t)+2

To determine

(b)

To find:

The derivative at t=1, where

A(t)=−2cos(2π5t)+2.

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Chapter 3.5, Problem 25P

Chapter 3.5, Problem 27P

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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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(a) To find: One breathing cycle of an adult for the below function: A ( t ) = − 2 cos ( 2 π 5 t ) + 2

Solution Summary: The author explains how one breathing cycle of an adult is of 5mathrmsec.

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(a)

To find:

One breathing cycle of an adult for the below function:

A(t)=−2cos(2π5t)+2

(b)

To find:

The derivative at t=1, where

A(t)=−2cos(2π5t)+2.

Want to see the full answer?

Chapter 3 Solutions

(a) To find: One breathing cycle of an adult for the below function: A ( t ) = − 2 cos ( 2 π 5 t ) + 2

Solution Summary: The author explains how one breathing cycle of an adult is of 5mathrmsec.

Concept explainers

Videos

(a) To find: One breathing cycle of an adult for the below function: A(t)=−2cos(2π5t)+2

(b) To find: The derivative at t=1, where A(t)=−2cos(2π5t)+2.

Want to see the full answer?

Chapter 3 Solutions

(a)

To find:

One breathing cycle of an adult for the below function:

A(t)=−2cos(2π5t)+2

(b)

To find:

The derivative at t=1, where

A(t)=−2cos(2π5t)+2.