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Chapter 22: Problem 16

How long does it take sunlight to travel from the Sun to Earth?

Short Answer

Expert verified
Answer: Approximately 8.31 minutes.

Step by step solution

01

Convert the distance into the same unit as the speed

We need to convert the distance from the Sun to Earth to the same unit as the speed of light. Let's convert the distance into miles. Distance = 93,000,000 miles
02

Rearrange the speed formula to find time

The formula of speed is Speed = Distance/Time. We need to rearrange the formula to find time: Time = Distance / Speed
03

Substitute the distance and speed values into the formula

Now, we will substitute the distance and speed values into the formula to find the time: Time = (93,000,000 miles) / (186,282 miles/second)
04

Calculate the time

After substituting the values, we can now calculate the time it takes for sunlight to travel from the Sun to Earth: Time = (93,000,000 miles) / (186,282 miles/second) = 498.67 seconds
05

Convert the time into a more convenient unit

Since 498.67 seconds is not easy to understand, let's convert it into minutes: Time = 498.67 seconds / 60 seconds/minute = 8.31 minutes So, it takes approximately 8.31 minutes for sunlight to travel from the Sun to Earth.

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Most popular questions from this chapter

Prove that, in an EM wave traveling in vacuum, the electric and magnetic energy densities are equal; that is, prove that,$$\frac{1}{2} \epsilon_{0} E^{2}=\frac{1}{2 \mu_{0}} B^{2}$$ at any point and at any instant of time.,
The solar panels on the roof of a house measure \(4.0 \mathrm{m}\) by $6.0 \mathrm{m} .\( Assume they convert \)12 \%$ of the incident EM wave's energy to electric energy. (a) What average power do the panels supply when the incident intensity is \(1.0 \mathrm{kW} / \mathrm{m}^{2}\) and the panels are perpendicular to the incident light? (b) What average power do the panels supply when the incident intensity is \(0.80 \mathrm{kW} / \mathrm{m}^{2}\) and the light is incident at an angle of \(60.0^{\circ}\) from the normal? (W) tutorial: solar collector) (c) Take the average daytime power requirement of a house to be about 2 kW. How do your answers to (a) and (b) compare? What are the implications for the use of solar panels?
An electric dipole antenna used to transmit radio waves is oriented vertically.At a point due south of the transmitter, what is the direction of the wave's magnetic field?
The electric field in a radio wave traveling through vacuum has amplitude $2.5 \times 10^{-4} \mathrm{V} / \mathrm{m}\( and frequency \)1.47 \mathrm{MHz}$. (a) Find the amplitude and frequency of the magnetic field. (b) The wave is traveling in the \(+x\) -direction. At \(x=0\) and \(t=0,\) the electric field is \(1.5 \times 10^{-4} \mathrm{V} / \mathrm{m}\) in the \- y-direction. What are the magnitude and direction of the magnetic field at \(x=0\) and \(t=0 ?\).
By expressing \(\epsilon_{0}\) and \(\mu_{0}\) in base SI units $(\mathrm{kg}, \mathrm{m}, \mathrm{s}, \mathrm{A})$ prove that the only combination of the two with dimensions of speed is $\left(\boldsymbol{\epsilon}_{0} \mu_{0}\right)^{-1 / 2}$.
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