Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

How many nanoseconds does it take light to travel 1.00 ft in vacuum? (This result is a useful quantity to remember.)

Short Answer

Expert verified
Approximately 1.017 nanoseconds.

Step by step solution

01

Understand the Speed of Light

First, note that the speed of light in a vacuum is a fundamental constant, approximately \( c = 299,792,458 \) meters per second (m/s).
02

Convert 'feet' to 'meters'

We need to convert 1.00 ft to meters. Using the conversion, \(1 \text{ ft} = 0.3048 \text{ m}\), we find: \[1.00 \text{ ft} = 0.3048 \text{ meters}.\]
03

Calculate Time in Seconds

Using the formula for speed, \( v = \frac{d}{t} \), where \( d \) is distance and \( t \) is time, we rearrange to find time:\[ t = \frac{d}{v}.\]Substituting the values, \(d = 0.3048 \text{ m}\) and \(v = 299,792,458 \text{ m/s} \), we get:\[ t = \frac{0.3048}{299,792,458} \approx 1.0167038 \times 10^{-9} \text{ seconds}.\]
04

Convert Time to Nanoseconds

Since we want the time in nanoseconds, remember that \(1 \text{ ns} = 10^{-9} \text{ s}\). Therefore, \[ t = 1.0167038 \text{ ns}.\]

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 Vaia!

Key Concepts

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

Speed of Light
The speed of light is one of the fundamental constants in physics, denoted by the symbol \( c \). It's known to be approximately \( 299,792,458 \) meters per second (m/s) when traveling through a vacuum. Light travels incredibly fast, and this speed is crucial in areas like astronomy and electronics.
It represents the maximum speed at which information and matter can travel. Understanding this speed helps us calculate a variety of scenarios, from everyday phenomena to complex scientific processes.
Fun fact: because light is so fast, it can travel around the Earth's equator 7.5 times in just one second!
Unit Conversion
Unit conversion is a vital skill in physics, enabling us to switch between different measuring systems. For this exercise, we convert feet to meters because the speed of light is given in meters per second.
  • 1 foot is equivalent to 0.3048 meters
  • By multiplying 1.00 ft by 0.3048, we convert it to meters
Understanding unit conversion allows you to work seamlessly across various problems and ensures the accuracy of your scientific calculations.
Always double-check conversion factors, as mistakes can easily lead to errors in final answers.
Time Calculation
To find out how long it takes for light to travel in a vacuum, we use the formula \( t = \frac{d}{v} \), where \( d \) is the distance and \( v \) is the velocity or speed.
In our exercise, we had to determine the time for light to travel 0.3048 meters (converted from feet) at its incredible speed of \( 299,792,458 \) m/s. Plugging these values into the formula gives us the time:
\[ t = \frac{0.3048}{299,792,458} \approx 1.0167038 \times 10^{-9} \text{ seconds} \]
This quick calculation tells us the journey for light over a very short distance takes only a billionth of a second.
Scientific Notation
Scientific notation is a way to express very large or very small numbers in a more manageable form. It's especially useful in physics when dealing with extremes like the speed of light or atomic scales.
In this case, the time for light to travel a foot is approximately \( 1.0167038 \times 10^{-9} \) seconds. This notation helps simplify numbers by indicating how many times to multiply by ten.
  • The exponent \( -9 \) tells us to move the decimal point nine places to the left.
  • It makes complex calculations easier to read and understand.
Grasping scientific notation is crucial for tackling many physics problems efficiently.

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 density of air under standard laboratory conditions is 1.29 kg/m\(^3\), and about 20% of that air consists of oxygen. Typically, people breathe about \\(\frac{1}{2}\\) L of air per breath. (a) How many grams of oxygen does a person breathe in a day? (b) If this air is stored uncompressed in a cubical tank, how long is each side of the tank?

A fence post is 52.0 m from where you are standing, in a direction 37.0\(^{\circ}\) north of east. A second fence post is due south from you. How far are you from the second post if the distance between the two posts is 68.0 m?

The most powerful engine available for the classic 1963 Chevrolet Corvette Sting Ray developed 360 horsepower and had a displacement of 327 cubic inches. Express this displacement in liters (L) by using only the conversions 1 L = 1000 cm\(^3\) and 1 in. = 2.54 cm.

A spelunker is surveying a cave. She follows a passage 180 m straight west, then 210 m in a direction \(45^{\circ}\) east of south. and then \(280 \mathrm{~m}\) at \(30^{\circ}\) east of north. After a fourth displacement,she finds herself back where she started. Use a scale drawing to determine the magnitude and direction of the fourth displacement. (See also Problem 1.61 for a different approach.)

An explorer in Antarctica leaves his shelter during a whiteout. He takes 40 steps northeast, next 80 steps at 60\(^{\circ}\) north of west, and then 50 steps due south. Assume all of his steps are equal in length. (a) Sketch, roughly to scale, the three vectors and their resultant. (b) Save the explorer from becoming hopelessly lost by giving him the displacement, calculated by using the method of components, that will return him to his shelter.

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.

Sign-up for free