/*! 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} Q8E While driving in an exotic forei... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

While driving in an exotic foreign land, you see a speed limit sign that reads 180,000 furlongs per fortnight. How many miles per hour is this? (One furlong is 1/8 mile, and a fortnight is 14 days. A furlong originally referred to the length of a plowed furrow.)

Short Answer

Expert verified

Answer

The speed is 67 miles/hour

Step by step solution

01

Conversion of furlong and fortnight

Here,

Furlong and fortnight are the units used in olded days to measure the distance and the time travelled by a person respectively.

According to the data one furlong was considered approximately 1/8thmiles.

Similarly 1 fortnight is considered almost equal to two weeks.

Given data:

Speed limit in furlongs/fortnight = 180,000 furlongs/fortnight

Other necessaryconversions given

1furlong=18=miles=0.125mile

1fornight=14days

02

Conversion into mile/ hour

We know that 1 day contains 24 hour.

Hence 14 days contain =24 x 14 hours =336 hour.

We need to convert furlong into mile so we need to multiply the value with 0.125 to convert it into hours. Similarly to convert fortnight into hours we need to divide it with 14 days (two week contains) and also by 24 because each day contain 24 hours.

=(180000furlongsfornight)×(0.125miles1forlong)×(1fornight14days)×(1day24h)=67mile/hour

It is therefor concluded that the speed limit is 67 miles/hour.

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 91Ó°ÊÓ!

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

A bird is flying due east. Its distance from a tall building is given by

xt=28.0m+12.4m/st-0.0450 m/s3t3.What is the instantaneous velocity of the bird when t = 8.00 s?

Water flows steadily from an open tank as in Fig. P12.81. The elevation of point 1 is 10.0 m, and the elevation of points 2 and 3 is 2.00 m. The cross-sectional area at point 2 is 0.0480 m2; at point 3 it is 0.0160 m2. The area of the tank is very large compared with the cross-sectional area of the pipe. Assuming that Bernoulli’s equation applies, compute (a) the discharge rate in cubic meters per second and (b) the gauge pressure at point 2.

A physics professor leaves her house and walks along the sidewalk toward campus. After 5 min, it starts to rain, and she returns home. Her distance from her house as a function of time is shown in Fig. E2.10 At which of the labeled points is her velocity (a) zero? (b) constant and positive? (c) constant and negative? (d) increasing in magnitude? (e) decreasing in magnitude?

A cube of oak wood with very smooth faces normally floats in water. Suppose you submerge it completely and press one face flat against the bottom of a tank so that no water is under that face. Will the block float to the surface? Is there a buoyant force on it? Explain.

A hammer with mass m is dropped from rest from a height h above the earth’s surface. This height is not necessarily small compared with the radiusof the earth. Ignoring air resistance, derive an expression for the speed v of the hammer when it reaches the earth’s surface. Your expression should involve h,, and(the earth’s mass).

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.