/*! 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 41 How long does a laser beam take ... [FREE SOLUTION] | 91Ó°ÊÓ

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How long does a laser beam take to travel from the Earth to a reflector on the Moon and back? Take the distance from the Earth to the Moon to be \(2.4 \times 10^{5} \mathrm{mi} .\) (This experiment was done when the Apollo flights of the earlv 1970 s left laser reflectors on the lunar surface. \()\)

Short Answer

Expert verified
It takes approximately 2.577 seconds for the laser beam to travel to the Moon and back.

Step by step solution

01

Understand the Problem

We need to determine the time it takes for a laser beam to travel from Earth to the Moon and back. Given the one-way distance is \(2.4 \times 10^5\) miles, we need to incorporate both the forward and return journey to compute the total distance traveled by the laser beam.
02

Determine the Total Distance

The laser beam travels to the Moon and then returns to Earth. Therefore, the total distance is twice the one-way distance. \[\text{Total Distance} = 2 \times 2.4 \times 10^5 \text{ miles} = 4.8 \times 10^5 \text{ miles}\]
03

Know the Speed of Light

The speed of light in a vacuum is approximately \(186,282\) miles per second (mi/s). This constant will be used to determine how long the laser beam takes to travel the calculated total distance.
04

Calculate Travel Time

Use the formula for time, which is distance divided by speed, to find the travel time.\[\text{Time} = \frac{\text{Total Distance}}{\text{Speed of Light}} = \frac{4.8 \times 10^5 \text{ miles}}{186,282 \text{ mi/s}}\]Calculate this to find the time.
05

Perform the Calculation

Calculate the division:\[\frac{4.8 \times 10^5}{186,282} \approx 2.577 \, \text{seconds}\]Thus, it takes approximately 2.577 seconds for the laser beam to travel to the Moon and back to Earth.

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Key Concepts

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

Speed of Light
The speed of light is a fascinating constant in physics, and it's incredibly fast. Light travels at approximately 186,282 miles per second (mi/s) in a vacuum. This speed means that light can circumnavigate the Earth about seven and a half times in just one second! Knowing this constant is crucial when dealing with problems related to light travel over vast distances, like in our exercise with the laser beam heading to the Moon.

Because the speed of light is so immense, it allows us to calculate how fast information and energy can travel across the universe. It also plays a role in technologies like fiber optics and satellite communication, where precise calculations are necessary to ensure signals arrive when they're supposed to.

So next time you flick a light switch, remember that you are engaging with one of the most fundamental constants in nature!
Distance Calculation
Distance calculation plays a significant role in figuring out how long it takes for light, like a laser beam, to make a round trip. In our exercise, the distance from the Earth to the Moon is given as \(2.4 \times 10^5\) miles.

To find the total distance the laser beam needs to travel, we must consider both the journey to the Moon and its return journey back to Earth. Therefore, we double the initial distance:
  • Total Distance = \(2 \times 2.4 \times 10^5 = 4.8 \times 10^5\) miles
Calculating the total distance requires an understanding of simple arithmetic and unit measurement. This principle is fundamental in physics as distances often need to be converted or adjusted depending on the problem's requirements.
Time Calculation
Time calculation is an essential concept when dealing with motion and distance. In our scenario, the amount of time it takes for the laser beam to travel to the Moon and back is solved using the formula for time:\[\text{Time} = \frac{\text{Distance}}{\text{Speed}} \]In this case, we insert the total round trip distance and the speed of light:
  • \(\text{Total Distance} = 4.8 \times 10^5\) miles
  • \(\text{Speed of Light} = 186,282\) mi/s
  • \(\text{Time} = \frac{4.8 \times 10^5}{186,282} \approx 2.577\) seconds
Time calculations are invaluable in accurately measuring how long it takes for light or other objects to travel specific distances in space. This allows scientists to predict and understand the dynamics of celestial movements, communication signals, and many phenomena.
Apollo Missions
The Apollo missions were a landmark series of spaceflights conducted by NASA during the 1960s and early 1970s. One of the noteworthy accomplishments of the Apollo missions was placing laser reflectors on the lunar surface.

These reflectors have played a vital role in scientific experiments, like the one described in our exercise. By measuring the time it takes for a laser beam to travel to these reflectors and back to Earth, scientists can accurately calculate the Earth-Moon distance. This data helps in various aspects, such as verifying Einstein's theory of general relativity and improving our understanding of lunar motions.

The Apollo missions not only advanced human space exploration but also provided indispensable tools and data for ongoing scientific inquiry. They continue to influence space technology and exploration strategies today.

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Most popular questions from this chapter

A boy is traveling due north at a constant speed while carrying a metal rod. The rod's length is oriented in the east-west direction and is parallel to the ground. (a) There will be no induced emf when the rod is (1) at the equator, (2) near the Earth's magnetic poles, (3) somewhere between the equator and the poles. Why? (b) Assume that the Earth's magnetic field is \(1.0 \times 10^{-4} \mathrm{~T}\) near the North Pole and \(1.0 \times 10^{-5} \mathrm{~T}\) near the equator. If the boy runs with a speed of \(1.3 \mathrm{~m} / \mathrm{s}\) northward near each location, and the rod is \(1.5 \mathrm{~m}\) long, calculate the induced emf in the rod in each location.

Microwave ovens can have cold spots and hot spots due to standing electromagnetic waves, analogous to standing wave nodes and antinodes in strings (v Fig. 20.32). (a) The longer the distance between the cold spots, (1) the higher the frequency of the waves, (2) the lower the frequency of the waves, (3) the frequency of the waves is independent of this distance. Why? (b) In your microwave the cold spots (nodes) occur approximately every \(5.0 \mathrm{~cm},\) but your neighbor's microwave produces them at every \(6.0 \mathrm{~cm}\). Which microwave operates at a higher frequency and by how much?

(a) To increase the output of an ac generator, a student has the choice of doubling either the generator's magnetic field or its frequency. To maximize the increase in emf output, (1) he should double the magnetic field, (2) he should double the frequency, (3) it does not matter which one he doubles. Explain. (b) Two students display their ac generators at a science fair. The generator made by student A has a loop area of \(100 \mathrm{~cm}^{2}\) rotating in a magnetic field of \(20 \mathrm{mT}\) at \(60 \mathrm{~Hz}\). The one made by student \(\mathrm{B}\) has a loop area of \(75 \mathrm{~cm}^{2}\) rotating in a magnetic field of \(200 \mathrm{mT}\) at \(120 \mathrm{~Hz}\). Which one generates the largest maximum emf? Justify your answer mathematically.

Suppose you wanted to build an electric generator using the Earth's magnetic field. Assume it has a strength of \(0.040 \mathrm{mT}\) at your location. Your generator design calls for a coil of 1000 windings rotated at exactly \(60 \mathrm{~Hz}\). The coil is oriented so that the normal to the area lines up with the Earth's field at the end of each cycle. (a) What must the coil diameter be to generate a maximum voltage of \(170 \mathrm{~V}\) (required in order to average \(120 \mathrm{~V}\) )? Does this seem like a practical design? (b) Some generators operate at \(50 \mathrm{~Hz}\). How would this change the coil diameter? (c) What number of windings would make this coil arrangement a "manageable" size?

The flux through a loop of wire changes uniformly from \(+40 \mathrm{~Wb}\) to \(-20 \mathrm{~Wb}\) in \(1.5 \mathrm{~ms}\). (a) What is the significance of the negative number attached to the final flux value? (b) What is the average induced emf in the loop? (c) If you wanted to double the average induced emf by changing only the time, what would the new time interval be? (d) If you wanted to double the average induced emf by changing only the final flux value, what would it be?

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