/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 A person on a diet might lose \(... [FREE SOLUTION] | 91Ó°ÊÓ

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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
The mass loss rate is approximately 3.8 mg/s.

Step by step solution

01

Convert kilograms to milligrams

First, remember that 1 kilogram is equal to 1,000,000 milligrams. Therefore, if a person loses 2.3 kg per week, we first convert this mass to milligrams by multiplying by 1,000,000: \[ 2.3 \times 1,000,000 = 2,300,000 \text{ mg} \]
02

Convert weeks to seconds

One week consists of 7 days, and each day has 24 hours, each hour has 60 minutes, and each minute has 60 seconds. Therefore, the total number of seconds in a week is:\[ 7 \times 24 \times 60 \times 60 = 604,800 \text{ seconds} \]
03

Calculate the rate of mass loss

To find the rate of mass loss per second, divide the total mass loss in milligrams by the total number of seconds in a week.\[ \frac{2,300,000 \text{ mg}}{604,800 \text{ seconds}} \approx 3.8036 \text{ mg/s} \]
04

Round and express the final result

Typically, it is suitable to express this rate with a reasonable number of decimal places. Therefore, the mass loss rate can be approximated as:\[ 3.8 \text{ mg/s} \]

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Key Concepts

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

Mastering Mass Conversion
Mass conversion is the process of changing a unit of mass from one form to another. It is often necessary when dealing with different measurement systems or when you need to express a mass in a specific unit. In our exercise, we converted kilograms to milligrams.
Understanding the relationships between metric units is important:
  • 1 kilogram (kg) = 1,000 grams (g)
  • 1 gram = 1,000 milligrams (mg)
Therefore, to convert kilograms to milligrams, you multiply by 1,000,000, because 1 kg equals 1,000,000 mg. For instance, to convert 2.3 kg to milligrams, you would calculate:\[2.3 \times 1,000,000 = 2,300,000 \text{ mg}\]Make sure to always double-check your conversion factors and calculations to avoid mistakes. It's a crucial step in ensuring that your results are accurate.
Time Conversion Made Easy
Time conversion involves changing a time measurement from one unit to another, such as converting weeks to seconds. This skill is handy when calculating rates over different periods.
To convert weeks into seconds:
  • 1 week = 7 days
  • 1 day = 24 hours
  • 1 hour = 60 minutes
  • 1 minute = 60 seconds
By multiplying these values, you can find the total seconds in a week:\[7 \times 24 \times 60 \times 60 = 604,800 \text{ seconds}\]Breaking down these conversions into smaller steps can make the process seem less complex and help prevent errors. Remembering the basics of how to switch between units is key to mastering time conversion.
Understanding Rate Calculation
Rate calculation is the process of finding out how one quantity changes in relation to another over a particular period. It is often expressed as a unit per time, such as milligrams per second in our example. Calculating rates involves dividing the total quantity by the total time.
In the exercise, we calculated the rate of mass loss per second by dividing the amount of mass lost in milligrams by the total seconds in a week:\[\frac{2,300,000 \text{ mg}}{604,800 \text{ seconds}} \approx 3.8036 \text{ mg/s}\]Rounding the result to a sensible number of decimal places, we get:\[3.8 \text{ mg/s}\]This way of representing change over time helps in understanding the efficiency and speed of a process. By mastering rate calculations, you can better interpret various real-world situations.

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Most popular questions from this chapter

One molecule of water \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) contains two atoms of hydrogen and one atom of oxygen. A hydrogen atom has a mass of \(1.0 \mathrm{u}\) and an atom of oxygen has a mass of \(16 \mathrm{u}\), approximately, (a) What is the mass in kilograms of one molecule of water? (b) How many molecules of water are in the world's oceans, which have an estimated total mass of \(1.4 \times 10^{21} \mathrm{~kg}\) ?

One cubic centimeter of a typical cumulus cloud contains 50 to 500 water drops, which have a typical radius of \(10 \mu \mathrm{m}\). For that range, give the lower value and the higher value, respectively, for the following. (a) How many cubic meters of water are in a cylindrical cumulus cloud of height \(3.0 \mathrm{~km}\) and radius \(1.0 \mathrm{~km} ?\) (b) How many 1-liter pop bottles would that water fill? (c) Water has a density of \(1000 \mathrm{~kg} / \mathrm{m}^{3} .\) How much mass does the water in the cloud have?

Go The record for the largest glass bottle was set in 1992 by a team in Millville, New Jersey - they blew a bottle with a volume of 193 U.S. fluid gallons. (a) How much short of \(1.0\) million cubic centimeters is that? (b) If the bottle were filled with water at the leisurely rate of \(1.8 \mathrm{~g} / \mathrm{min}\), how long would the filling take? Water has a density of \(1000 \mathrm{~kg} / \mathrm{m}^{3}\).

Suppose that, while lying on a beach near the equator watching the Sun set over a calm ocean, you start a stopwatch just as the top of the Sun disappears. You then stand, elevating your eyes by a height \(H=1.70 \mathrm{~m}\), and stop the watch when the top of the Sun again disappears. If the elapsed time is \(t=11.1 \mathrm{~s}\), what is the radius \(r\) of Earth?

Time standards are now based on atomic clocks. A promising second standard is based on pulsars, which are rotating neutron stars (highly compact stars consisting only of neutrons). Some rotate at a rate that is highly stable, sending out a radio beacon that sweeps briefly across Earth once with each rotation, like a lighthouse beacon. Pulsar PSR \(1937+21\) is an example; it rotates once every \(1.55780644887275 \pm 3 \mathrm{~ms}\), where the trailing \(\pm 3\) indicates the uncertainty in the last decimal place (it does not mean \(\pm 3 \mathrm{~ms}\) ). (a) How many rotations does PSR \(1937+21\) make in \(7.00\) days? (b) How much time does the pulsar take to rotate exactly one million times and (c) what is the associated uncertainty?

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