WebAssign Printed Access Card for Larson's Calculus: An Applied Approach, 10th Edition, Single-Term
WebAssign Printed Access Card for Larson's Calculus: An Applied Approach, 10th Edition, Single-Term
10th Edition
ISBN: 9781337652308
Author: Ron Larson
Publisher: Brooks Cole
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Chapter A5, Problem 42E

(a)

To determine

The expression for cost as a single fraction if the cost C (In dollars) of ordering and storing x units of product is given as,

C=6x+900,000x

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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…
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