/*! 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} Q20Q How much longer (percentage) is ... [FREE SOLUTION] | 91影视

91影视

How much longer (percentage) is a one-mile race than a 1500-mile race (鈥渢he metric mile鈥)?

Short Answer

Expert verified

A one-mile race is 7.27% longer than a 1500-mile race.

Step by step solution

01

Conversion factor

The unit of a quantity is changed either by multiplying or dividing the quantity with the conversion factor.

For example, if the length of a table is 7.00 inches and you want to measure it in centimeters, you can do this by multiplying the length of the table by its conversion factor. The conversion factor to convert inches into centimeters is \(1\,\,{\rm{in}}{\rm{.}} = 2.54\,\,{\rm{cm}}\). Thus, the length of the table in centimeters will be \(\left( {7.00\,\,{\rm{in}}{\rm{.}}} \right) \times \left( {2.54\,\,\frac{{{\rm{cm}}}}{{{\rm{in}}{\rm{.}}}}} \right) = 17.8\,\,{\rm{cm}}\)

02

Conversion of a mile into a meter

Here, you have to compare two quantities; i.e., a one-mile race and a 1500-mile race. The two quantities can only be compared when they have the same unit. Thus, you need to convert one mile into meters.

You know that there are 5280 feet in a mile, 12 inches in a foot, 2.54 cm in an inch, and 100 cm in a meter. Using these conversion factors, one mile can be written as follows:

\(\begin{aligned}{c}1\,\,{\rm{mile}} = \left( {5280\,\,{\rm{ft}}} \right) \times \left( {12\,\,\frac{{{\rm{in}}{\rm{.}}}}{{{\rm{ft}}}}} \right) \times \left( {2.54\,\,\frac{{{\rm{cm}}}}{{{\rm{in}}{\rm{.}}}}} \right) \times \left( {\frac{{1\,\,{\rm{m}}}}{{100\,\,{\rm{cm}}}}} \right)\\ = 1609\,\,{\rm{m}}\end{aligned}\)

Thus, there are 1609 meters in one mile.

03

Comparison of a one-mile race and a 1500-m race

A one-mile race is equivalent to a 1609-meter race. Therefore, a one-mile race is longer than a 1500-meter race. Thus, the difference in lengths\( = 1609 - 1500 = 109\,\,{\rm{m}}{\rm{.}}\)

The percentage difference in the lengths of a one-mile race and a 1500-m race is:

\(\begin{aligned}{c}{\rm{Diff}}\% = \frac{{109}}{{1500}} \times 100\\ = 7.27\% \end{aligned}\)

Thus, a one-mile race is 7.27 % longer than a 1500-mile race.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The two pulses shown in Fig. 11鈥54 are moving toward each other. (a) Sketch the shape of the string at the moment they directly overlap. (b) Sketch the shape of the string a few moments later. (c) In Fig.11鈥37a, at the moment the pulses pass each other, the string is straight. What has happened to the energy at this moment?

A clock pendulum oscillates at a frequency of 2.5 Hz. At t= 0, it is released from rest starting at an angle of\(12^\circ \)to the vertical. Ignoring friction, what will be the position (angle in radians) of the pendulum at (a)t= 0.25 s, (b) t = 1.60 s, and (c) t = 500 s?

(III) A 1.60-kg object oscillates at the end of a vertically hanging light spring once every 0.45 s. (a) Write down the equation giving its position y (+upward) as a function of time t. Assume the object started by being compressed 16 cm from the equilibrium position (where y = 0), and released. (b) How long will it take to get to the equilibrium position for the first time? (c) What will be its maximum speed? (d) What will be the object鈥檚 maximum acceleration, and where will it first be attained?

A sailor strikes the side of his ship just below the surface of the sea. He hears the echo of the wave reflected from the ocean floor directly below 2.4 s later. How deep is the ocean at this point?

An earthquake-produced surface wave can be approximated by a sinusoidal transverse wave. Assuming a frequency of 0.60 Hz (typical of earthquakes, which actually include a mixture of frequencies), what amplitude is needed so that objects begin to leave contact with the ground? (Hint: Set the acceleration a > g.)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.