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Chapter 31: Problem 27

Determine the distance in feet that light can travel in vacuum during \(1.00 \mathrm{~ns}\).

Short Answer

Expert verified
Answer: Approximately 0.98 feet.

Step by step solution

01

Convert the speed of light to feet per second

Since the speed of light is given in meters per second, we need to convert it to feet per second. To do this, we will use the conversion factor: \(1 \mathrm{~m} = 3.2808 \mathrm{~ft}\). So, we have: \(c_{ft} = c_{m} \times 3.2808\) \(c_{ft} = (3.00\times10^8) \mathrm{~(m/s) } \times 3.2808 \mathrm{~(ft/m)}\) \(c_{ft} = 9.8425 \times 10^8 \mathrm{~ft/s}\)
02

Convert the time from nanoseconds to seconds

Now we need to convert the given time, 1.00 ns, into seconds by using the conversion factor: \(1 \mathrm{~s} = 10^9 \mathrm{~ns}\). \(t_s = \dfrac{1.00 \mathrm{~ns}}{10^9}\) \(t_s = 1.00\times10^{-9} \mathrm{~s}\)
03

Calculate the distance using the speed of light and time

Now that we have the speed of light in feet per second and the time in seconds, we can use the equation: \(distance= speed \times time\) In our case: \(distance = c_{ft} \times t_s\) \(distance = (9.8425\times10^8 \mathrm{~ft/s}) (1.0\times10^{-9} \mathrm{~s})\) \(distance = 9.8425 \times 10^{-1} \mathrm{~ft}\)
04

Write down the final answer

The distance that light can travel in a vacuum during 1.00 ns is: \(distance = 9.8425 \times 10^{-1} \mathrm{~ft}\) It can be approximated to: \(distance \approx 0.98 \mathrm{~ft}\)

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Most popular questions from this chapter

.An industrial carbon dioxide laser produces a beam of radiation with average power of \(6.00 \mathrm{~kW}\) at a wavelength of \(10.6 \mu \mathrm{m}\). Such a laser can be used to cut steel up to \(25 \mathrm{~mm}\) thick. The laser light is polarized in the \(x\) -direction, travels in the positive \(z\) -direction, and is collimated (neither diverging or converging) at a constant diameter of \(100.0 \mu \mathrm{m} .\) Write the equations for the laser light's electric and magnetic fields as a function of time and of position \(z\) along the beam. Recall that \(\vec{E}\) and \(\vec{B}\) are vectors. Leave the overall phase unspecified, but be sure to check the relative phase between \(\vec{E}\) and \(\vec{B}\) .

Unpolarized light of intensity \(I_{0}\) is incident on a series of five polarizers, each rotated \(10.0^{\circ}\) from the preceding one. What fraction of the incident light will pass through the series?

An electric field of magnitude \(200.0 \mathrm{~V} / \mathrm{m}\) is directed perpendicular to a circular planar surface with radius \(6.00 \mathrm{~cm}\). If the electric field increases at a rate of \(10.0 \mathrm{~V} /(\mathrm{m} \mathrm{s}),\) determine the magnitude and the direction of the magnetic field at a radial distance \(10.0 \mathrm{~cm}\) away from the center of the circular area.

A \(14.9-\mu F\) capacitor, a \(24.3-\mathrm{k} \Omega\) resistor, a switch, and a 25.-V battery are connected in series. What is the rate of change of the electric field between the plates of the capacitor at \(t=0.3621 \mathrm{~s}\) after the switch is closed? The area of the plates is \(1.00 \mathrm{~cm}^{2}\) .

What is the wavelength of the electromagnetic waves used for cell phone communications in the 850 -MHz band?
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