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Chapter 1: Problem 25

A person on a diet might lose \(2.3 \mathrm{~kg}\) per week. Express the mass loss rate in milligrams per second, as if the dieter could sense the second- by-second loss.

Short Answer

Expert verified
3.80 mg/s

Step by step solution

01

Determine the Mass Loss Rate in Kilograms per Second

First, convert the weekly mass loss rate to a per-second rate. Given the person loses \(2.3\,\text{kg}\) per week, we need the number of seconds in a week. There are 7 days in a week, 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute. So, the total number of seconds in a week is \(7 \times 24 \times 60 \times 60 = 604800\) seconds. Now, divide the mass loss in kilograms by the number of seconds: \( \frac{2.3 \text{ kg}}{604800 \text{ s}} \approx 3.80 \times 10^{-6} \text{ kg/s} \).
02

Convert Kilograms per Second to Milligrams per Second

One kilogram equals \(10^6\) milligrams. Therefore, to convert \(3.80 \times 10^{-6} \text{ kg/s}\) to milligrams per second, multiply by \(10^6\): \(\begin{align*}3.80 \times 10^{-6} \text{ kg/s} \times 10^6 = 3.80 \text{ mg/s}\rightarrow 2.3 \text{ kg/week} \rightarrow 3.80 \text{ mg/s}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Conversion
Unit conversion is a key skill in physics and engineering. It involves changing a quantity from one unit to another. For example, to understand mass loss on a diet, we may need to convert kilograms to milligrams. Different situations use different units, so being able to interchange them ensures better comprehension of physical quantities.
Let's take a look at how unit conversion works with our exercise. The person loses 2.3 kilograms per week. We start by converting seconds in a week which is 604800 seconds. Now, divide 2.3 kilograms by 604800 seconds. Then convert the result from kilograms/second to milligrams/second. This completes the unit conversion.
Rate Calculation
Understanding how to calculate rates is essential. A 'rate' tells us how one quantity changes relative to another. In our case, it's about how much mass is lost per unit of time.
To find the rate of mass loss in seconds, we knew the weekly loss (2.3 kg) and had to convert this weekly loss to a per-second basis. By dividing the total mass loss by the number of seconds in a week, we got a rate in kilograms per second. This shows the importance of handling rates when tackling such physics problems.
Physics Education
Physics education involves helping students grasp complex concepts like mass, time, and their interrelationships. By breaking down problems step-by-step, students can better understand each part.
Looking at exercises like the one above, it becomes clear how to solve real-world problems through systematic approaches. Each step relies on solid physics knowledge, ensuring students can transfer these skills to other problems.
Dimensional Analysis
Dimensional analysis is a method to check the consistency of units in equations and conversions. It ensures that calculations make sense from a physical standpoint.
In our exercise, we used dimensional analysis to verify our conversions: starting with weekly mass loss in kilograms, converting weeks to seconds, and then changing kilograms to milligrams. Following this method helps avoid errors and confirms that our resulting unit (milligrams per second) is correct.
Such techniques are crucial in solving and understanding various physical phenomena.

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Most popular questions from this chapter

Grains of fine California beach sand are approximately spheres with an average radius of \(50 \mu \mathrm{m}\) and are made of silicon dioxide. A solid cube of silicon dioxide with a volume of \(1.00 \mathrm{~m}^{3}\) has a mass of \(2600 \mathrm{~kg}\). What mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a cube \(1 \mathrm{~m}\) on an edge?

Historically the English had a doubling system when measuring volumes; 2 mouthfuls equal 1 jigger, 2 jiggers equal 1 jack (also called a jackpot); 2 jacks equal 1 jill; 2 jills \(=\) 1 cup; 2 cups \(=1\) pint; 2 pints \(=1\) quart; 2 quarts \(=1\) pottle; 2 pottles \(=1\) gallon; 2 gallons \(=1\) pail. (The nursery rhyme "Jack and Jill" refers to these units and was a protest against King Charles I of England for his taxes on the jacks of liquor sold in the tavern. (See A. Kline, The World of Measurement, New York: Simon and Schuster, 1975, pp. \(32-39 .\) American and British cooks today use teaspoons, tablespoons, and cups; 3 teaspoons \(=1\) tablespoon; 4 tablespoons \(=1 / 4\) cup. Assume that you find an old English recipe requiring 3 jiggers of milk. How many cups does this represent? How many tablespoons? You can assume that the cups in the two systems represent the same volume.

Until 1883 , every city and town in the United States kept its own local time. Today, travelers reset their watches only when the time change equals \(1.0 \mathrm{~h}\). How far, on the average, must you travel in degrees of longitude until your watch must be reset by \(1.0 \mathrm{~h}\) ? (Hint: Earth rotates \(360^{\circ}\) in about 24 h.)

Dose We know from our dimensional analysis that if an object maintains its shape but changes its size, its area changes as the square of its length and its volume changes as the cube of its length. Suppose you are a parent and your child is sick and has to take some medicine. You have taken this medicine previously and you know its dose for you. You are \(5^{\prime} 10^{\prime \prime}\) tall and weigh \(180 \mathrm{lb}\), and your child is \(2^{\prime} 11^{\prime \prime}\) tall and wcighs \(30 \mathrm{lb}\). Estimate an appropriate dosage for your child's medicine in the following cases. Be sure to discuss your reasoning. (a) The medicine is one that will enter the child's bloodstream and reach cvery cell in the body. Your dose is \(250 \mathrm{mg}\). (b) The medicine is one that is meant to coat the child's throat. Your dose is \(15 \mathrm{ml}\).

A fortnight is a charming English measure of time equal to \(2.0\) weeks (the word is a contraction of "fourteen nights"). That is a nice amount of time in pleasant company but perhaps a painful string of microseconds in unpleasant company. How many microseconds are in a fortnight?
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