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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

The voltage across a cylindrical conductor of radius \(r\), length \(L\), and resistance \(R\) varies with time. The timevarying voltage causes a time-varying current, \(i\), to flow in the cylinder. Show that the displacement current equals \(\epsilon_{0} \rho d i / d t,\) where \(\rho\) is the resistivity of the conductor.

A house with a south-facing roof has photovoltaic panels on the roof. The photovoltaic panels have an efficiency of \(10.0 \%\) and occupy an area with dimensions \(3.00 \mathrm{~m}\) by \(8.00 \mathrm{~m} .\) The average solar radiation incident on the panels is \(300 . \mathrm{W} / \mathrm{m}^{2}\), averaged over all conditions for a year. How many kilowatt hours of electricity will the solar panels generate in a 30 -day month?

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 continuous-wave (cw) argon-ion laser beam has an average power of \(10.0 \mathrm{~W}\) and a beam diameter of \(1.00 \mathrm{~mm}\). Assume that the intensity of the beam is the same throughout the cross section of the beam (which is not true, as the actual distribution of intensity is a Gaussian function). a) Calculate the intensity of the laser beam. Compare this with the average intensity of sunlight at Earth's surface \(\left(1400 . \mathrm{W} / \mathrm{m}^{2}\right)\) b) Find the root-mean-square electric field in the laser beam. c) Find the average value of the Poynting vector over time. d) If the wavelength of the laser beam is \(514.5 \mathrm{nm}\) in vacuum, write an expression for the instantaneous Poynting vector, where the instantaneous Poynting vector is zero at \(t=0\) and \(x=0\) e) Calculate the root-mean-square value of the magnetic field in the laser beam.

31.8 According to Gauss's Law for Magnetic Fields, all magnetic field lines form a complete loop. Therefore, the direction of the magnetic field \(\vec{B}\) points from _________ pole to ________ pole outside of an ordinary bar magnet and from ____ pole to pole _______ inside the magnet. a) north, south, north, south b) north, south, south, north c) south, north, south, north d) south, north, north, south
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