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The following times are given in seconds. Use metric prefixes to rewrite them so the numerical value is greater than one but less than \(1000 .\) For example, \(7.9 \times 10^{-2} \mathrm{s}\) could be written as either 7.9 cs or 79 ms. (a) \(9.57 \times 10^{5}\) s; (b) \(0.045 \mathrm{s} ;\) (c) \(5.5 \times 10^{-7} \mathrm{s} ;\) (d) \(3.16 \times 10^{7} \mathrm{s}\)

Short Answer

Expert verified
The converted time units are: (a) 957 ks, (b) 45 ms, (c) 0.55 μs, (d) 31.6 Ms.

Step by step solution

01

Convert (a) \(9.57 \times 10^{5}\) s

For unit conversion, we need to find a metric prefix that will result in a numerical value between 1 and 1000. The 'kilo-' prefix stands for \(10^3\), and is commonly used for time to represent 'kiloseconds'. We divide \(9.57 \times 10^{5}\) by \(10^3\) to obtain the numerical value in kiloseconds. This results in \(957\) ks.
02

Convert (b) 0.045 s

To work with seconds, milli is a commonly used metric prefix representing \(10^{-3}\) seconds. We multiply 0.045 by \(10^3\) to obtain the numerical value in milliseconds (ms). This results in \(45\) ms.
03

Convert (c) \(5.5 \times 10^{-7}\) s

The micro- prefix stands for \(10^{-6}\) and is also commonly used with time units. Multiply \(5.5 \times 10^{-7}\) by \(10^6\) to obtain the numerical value in microseconds (μs). This results in \(0.55\) μs.
04

Convert (d) \(3.16 \times 10^{7}\) s

The s stands for seconds, but in this case, it's a large number, therefore we can use the prefix 'Mega-' which denotes \(10^6\). Dividing \(3.16 \times 10^{7}\) by \(10^6\) gives the value in Megaseconds (Ms). This results in \(31.6\) Ms.

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Key Concepts

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

Unit Conversion
Understanding unit conversion is essential for scientists, engineers, and students alike, as it allows for clear communication and standardization of measurements. Unit conversion involves changing the representation of a quantity without altering the actual value. Imagine you're baking and you need to measure ingredients. Just as you might convert tablespoons to cups for convenience, in physics, we convert units for clarity and simplicity.

When converting units, like time from seconds to hours, you're not changing the duration, only how it's expressed. In the exercise given, converting seconds to an appropriate unit with a metric prefix ensures the value stays between 1 and 1000, making it easier to read and understand.

For example, converting large numbers of seconds into kiloseconds (ks) or smaller fractions into milliseconds (ms) follows the same principle as turning a vast number of pennies into more manageable dollars. This simplification is vital in presenting data in a digestible format and avoids large, unwieldy numbers that can complicate calculations and understanding.
Scientific Notation
Scientific notation is a method of representing numbers that are too large or too small to be conveniently written in decimal form. It expresses numbers as a product of two terms: a coefficient and a power of 10. This format is particularly useful in sciences like physics, where measurements can span many orders of magnitude.

In the context of our exercise, scientific notation helps to neatly express time measurements, whether they are very large or very small. It's akin to packing a suitcase: just as you would neatly fold and organize clothes to make them fit, scientific notation allows us to neatly 'fold' numbers into a more compact form. The coefficient (the 'clothes' in this analogy) provides the significant figures, while the exponent (the 'suitcase') determines the scale or size. This process not only makes numbers easier to work with but also facilitates easier unit conversion. Converting from one prefix to another often just involves adjusting the exponent, similar to shifting items into a different compartment within a suitcase.
Time Measurement
Time measurement is a fundamental concept that requires an understanding of both units and the increments in which time is divided. From the smallest microseconds to the vast spans of megaseconds, being able to quantify and work with different scales of time is crucial in many aspects of science and everyday life.

For instance, in the exercise provided, recognizing the appropriate magnitude of time aids in applying the correct metric prefix. Consider the range from the tick of a clock to the rotation of the Earth around the Sun. Each event matches a certain time scale and, thereby, a particular unit – microseconds for rapid processes and kiloseconds or megaseconds for prolonged events. It's like having different clocks for different purposes; a stopwatch for athletes, a calendar for historians. Each 'clock' is suited for its purpose, providing the most useful information for the scale of time being measured.

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Most popular questions from this chapter

(a) How many significant figures are in the numbers 99 and \(100 . ?\) (b) If the uncertainty in each number is 1 , what is the percent uncertainty in each? (c) Which is a more meaningful way to express the accuracy of these two numbers: significant figures or percent uncertainties?

Suppose your bathroom scale reads your mass as 65 kg with a \(3 \%\) uncertainty. What is the uncertainty in your mass (in kilograms)?

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The first atomic bomb was detonated on July 16,1945 at the Trinity test site about \(200 \mathrm{mi}\) south of Los Alamos. In 1947 , the U.S. government declassified a film reel of the explosion. From this film reel, British physicist G. I. Taylor was able to determine the rate at which the radius of the fireball from the blast grew. Using dimensional analysis, he was then able to deduce the amount of energy released in the explosion, which was a closely guarded secret at the time. Because of this, Taylor did not publish his results until 1950. This problem challenges you to recreate this famous calculation. (a) Using keen physical insight developed from years of experience, Taylor decided the radius \(r\) of the fireball should depend only on time since the explosion, \(t,\) the density of the air, \(\rho,\) and the energy of the initial explosion, \(E .\) Thus, he made the educated guess that \(r=k E^{a} \rho^{b} t^{c}\) for some dimensionless constant \(k\) and some unknown exponents \(a, b,\) and \(c .\) Given that \([\mathrm{E}]=\mathrm{ML}^{2} \mathrm{T}^{-2}\) determine the values of the exponents necessary to make this equation dimensionally consistent. (Hint: Notice the equation implies that \(k=r E^{-a} \rho^{-b} t^{-c}\) and that \([k]=1\) ) (b) By analyzing data from high-energy conventional explosives, Taylor found the formula he derived seemed to be valid as long as the constant \(k\) had the value \(1.03 .\) From the film reel, he was able to determine many values of \(r\) and the corresponding values of \(t .\) For example, he found that after \(25.0 \mathrm{ms}\), the fireball had a radius of \(130.0 \mathrm{m}\). Use these values, along with an average air density of 1.25 \(\mathrm{kg} / \mathrm{m}^{3},\) to calculate the initial energy release of the Trinity detonation in joules (J). (Hint: To get energy in joules, you need to make sure all the numbers you substitute in are expressed in terms of SI base units.) (c) The energy released in large explosions is often cited in units of "tons of TNT" (abbreviated "t TNT"), where 1 t TNT is about 4.2 GJ. Convert your answer to (b) into kilotons of TNT (that is, kt TNT). Compare your answer with the quickand-dirty estimate of \(10 \mathrm{kt}\) TNT made by physicist Enrico Fermi shortly after witnessing the explosion from what was thought to be a safe distance. (Reportedly, Fermi made his estimate by dropping some shredded bits of paper right before the remnants of the shock wave hit him and looked to see how far they were carried by it.)

Soccer fields vary in size. A large soccer field is 115 \(\mathrm{m}\) long and \(85.0 \mathrm{m}\) wide. What is its area in square feet? (Assume that \(1 \mathrm{m}=3.281 \mathrm{ft}\) )

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