/*! 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 4 About how far apart must you hol... [FREE SOLUTION] | 91Ó°ÊÓ

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About how far apart must you hold your hands for them to be separated by 1.0 nano-light-second (the distance light travels in 1.0 ns)?

Short Answer

Expert verified
Your hands should be about 29.98 cm apart.

Step by step solution

01

Understanding the Problem

Light travels at a speed of approximately 299,792,458 meters per second. We need to find out how far light travels in 1.0 nanosecond (ns). A nanosecond is 1 billionth of a second.
02

Converting Time to Seconds

Since a nanosecond is \((1/1,000,000,000) \) of a second, we start by considering this time in seconds.
03

Applying the Speed of Light

We use the formula \( ext{distance} = ext{speed} imes ext{time} \). Given that the speed of light is approximately 299,792,458 m/s, we substitute the time as 1 x \( 10^{-9} \) seconds.
04

Calculating the Distance

Substituting the values into the formula, we have \( ext{distance} = 299,792,458 imes 10^{-9} \). Performing the multiplication gives us approximately \( 0.299792458 \) meters.
05

Concluding the Distance in a More Usable Unit

To convert meters to centimeters (since it's a more practical unit for holding hands), recall there are 100 centimeters in a meter. Thus, \( 0.299792458 \) meters is equal to approximately \( 29.9792458 \) centimeters.

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Key Concepts

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

Distance Calculation
Calculating the distance that light travels in a certain amount of time can be an intriguing exercise. It involves understanding the relationship between speed and time. In this context, the speed of light is fundamental. Light travels at an astonishing speed of approximately 299,792,458 meters per second. To calculate the distance light travels during a specified time period, you will use the formula \( \text{distance} = \text{speed} \times \text{time} \). In our exercise, we are interested in how far light travels in 1.0 nanoseconds. By substituting the values: speed as 299,792,458 m/s and time as 1 nanosecond (or \( 1 \times 10^{-9} \) seconds), we can determine the distance traveled by light, which turns out to be roughly 0.299792458 meters. Understanding and applying this formula is a core part of solving problems involving distance and speed.
Unit Conversion
Unit conversion is a crucial skill in physics and everyday life. It allows us to switch from one measurement unit to another, making data easier to understand and work with. In the exercise, after calculating the distance in meters, we convert it to centimeters for practicality. To convert meters to centimeters, recall that 1 meter equals 100 centimeters. Converting the light-traveled distance of \( 0.299792458 \) meters results in approximately \( 29.9792458 \) centimeters. This conversion helps us visualize the distance better and is essential for comprehensively understanding measurements. Always ensure you understand which conversion factors to use and practice doing so in various contexts, as this skill will enhance your precision and accuracy in problem-solving.
Nanoseconds
Nanoseconds, though a tiny fraction of time, play a significant role in high-speed calculations and physics. Specifically, a nanosecond is one-billionth of a second (\( \frac{1}{1,000,000,000} \) seconds). This minuscule duration is often used to measure the time it takes for light and other fast-moving objects to travel short distances. When you're dealing with such rapid measurements, understanding the size of a nanosecond is critical. The exercise involved understanding how far light travels within this tiny time frame, illustrating the significance of nanoseconds in calculating swift phenomena. Being comfortable with such small time measurements is important in fields like computer science, telecommunications, and physics, where precision timing is paramount.

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Most popular questions from this chapter

A small spaceship with a mass of only \(1.5 \times 10^{3} \mathrm{~kg}\) (including an astronaut) is drifting in outer space with negligible gravitational forces acting on it. If the astronaut turns on a \(10 \mathrm{~kW}\) laser beam, what speed will the ship attain in 1.0 day because of the momentum carried away by the beam?

An electromagnetic wave is traveling in the negative direction of a \(y\) axis. At a particular position and time, the electric field is directed along the positive direction of the \(z\) axis and has a magnitude of \(100 \mathrm{~V} / \mathrm{m}\). What are the (a) magnitude and (b) direction of the corresponding magnetic field?

A plane electromagnetic wave traveling in the positive direction of an \(x\) axis in vacuum has components \(E_{x}=E_{y}=0\) and \(E_{z}=(2.0 \mathrm{~V} / \mathrm{m}) \cos \left[\left(\pi \times 10^{15} \mathrm{~s}^{-1}\right)(t-x / c)\right] .\) (a) What is the amplitude of the magnetic field component? (b) Parallel to which axis does the magnetic field oscillate? (c) When the electric field component is in the positive direction of the \(z\) axis at a certain point \(P\), what is the direction of the magnetic field component there?

Light in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of \(32.0^{\circ}\) with the normal to the surface, while in the glass it makes an angle of \(21.0^{\circ}\) with the normal. What is the index of refraction of the glass?

Someone plans to float a small, totally absorbing sphere \(0.500 \mathrm{~m}\) above an isotropic point source of light, so that the upward radiation force from the light matches the downward gravitational force on the sphere. The sphere's density is \(19.0 \mathrm{~g} / \mathrm{cm}^{3},\) and its radius is \(2.00 \mathrm{~mm}\). (a) What power would be required of the light source? (b) Even if such a source were made, why would the support of the sphere be unstable?

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