Explanation Given information: The spirogram trace is V ( t ) = 3 + 0.05 sin ( 160 π t − π 2 ) where t is measured in minutes and V ( t ) is the lung volume in liters The graph on the spirogram trace device is shown in the figure below: Explanation: The spirogram trace is V ( t ) = 3 + 0.05 sin ( 160 π t − π 2 ) There will be 80 inspirations in one minute. Each inspiration represents V ( 1 160 ) − V ( 0 ) = ( 3 + 0.05 sin ( 160 π ( 1 160 ) − π 2 ) ) − ( 3 + 0.05 sin ( 160 π ( 0 ) − π 2 ) ) = ( 3 + 0.05 sin ( π − π 2 ) ) − ( 3 + 0.05 sin ( 0 − π 2 ) ) = ( 3 + 0.05 sin ( π 2 ) ) − ( 3 + 0.05 sin ( − π 2 ) ) = ( 3 + ( 0

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Chapter 8, Problem 58RE

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Whether the assertions that when a person undergoes spontaneous hyperventilation, the peak inspiratoryflow equals π times the minute volume and the mean inspiratory flow equals twice the minute volume is true or not by using the data from exercise 57.

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Chapter 8, Problem 57RE

Chapter 8, Problem 59RE

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Whether the assertions that when a person undergoes spontaneous hyperventilation, the peak inspiratoryflow equals π times the minute volume and the mean inspiratory flow equals twice the minute volume is true or not by using the data from exercise 57 .

Solution Summary: The author analyzes whether the assertions that when a person undergoes spontaneous hyperventilation, the peak inspiratory flow equals pi times the minute volume are true or not.

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Whether the assertions that when a person undergoes spontaneous hyperventilation, the peak inspiratoryflow equals π times the minute volume and the mean inspiratory flow equals twice the minute volume is true or not by using the data from exercise 57.

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Chapter 8 Solutions