Problem 27
Question
Determine the distance in feet that light can travel in vacuum during \(1.00 \mathrm{~ns}\).
Step-by-Step Solution
Verified Answer
Answer: Approximately 0.98 feet.
1Step 1: 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:
\(c_{ft} = c_{m} \times 3.2808\)
\(c_{ft} = (3.00\times10^8) \mathrm{~(m/s) } \times 3.2808 \mathrm{~(ft/m)}\)
\(c_{ft} = 9.8425 \times 10^8 \mathrm{~ft/s}\)
2Step 2: 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: \(1 \mathrm{~s} = 10^9 \mathrm{~ns}\).
\(t_s = \dfrac{1.00 \mathrm{~ns}}{10^9}\)
\(t_s = 1.00\times10^{-9} \mathrm{~s}\)
3Step 3: 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:
\(distance = c_{ft} \times t_s\)
\(distance = (9.8425\times10^8 \mathrm{~ft/s}) (1.0\times10^{-9} \mathrm{~s})\)
\(distance = 9.8425 \times 10^{-1} \mathrm{~ft}\)
4Step 4: 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:
\(distance \approx 0.98 \mathrm{~ft}\)
Key Concepts
Unit ConversionNanosecondsDistance Calculation
Unit Conversion
Unit conversion is a fundamental skill in solving physics problems. It involves changing a measurement from one unit to another without altering the value. This comes in handy when measurements are given in units that are not immediately usable in equations.
For example, the speed of light is often given in meters per second (m/s), but in certain situations, such as the exercise above, it might be more practical to use feet per second (ft/s). The conversion factor between meters and feet is:
\[ c_{ft} = c_{m} \times 3.2808 \]
This step ensures that we have a consistent measurement system fit for our calculations. The converted speed was found to be approximately 9.8425 x 10^8 ft/s, which greatly facilitates solving the problem at hand.
For example, the speed of light is often given in meters per second (m/s), but in certain situations, such as the exercise above, it might be more practical to use feet per second (ft/s). The conversion factor between meters and feet is:
- 1 meter = 3.2808 feet
\[ c_{ft} = c_{m} \times 3.2808 \]
This step ensures that we have a consistent measurement system fit for our calculations. The converted speed was found to be approximately 9.8425 x 10^8 ft/s, which greatly facilitates solving the problem at hand.
Nanoseconds
Nanoseconds are an incredibly small unit of time, and understanding them is key for tackling high-speed calculations like those involving the speed of light. A nanosecond is one billionth of a second, symbolized by 'ns'. This means:
\[ t_s = \dfrac{1.00 \, \text{ns}}{10^9} = 1.00 \times 10^{-9} \, \text{s} \]
By converting to seconds, we align our units with the speed given and can proceed with calculations accurately.
- 1 second = 1,000,000,000 nanoseconds
- 1 nanosecond = \(10^{-9}\) seconds
\[ t_s = \dfrac{1.00 \, \text{ns}}{10^9} = 1.00 \times 10^{-9} \, \text{s} \]
By converting to seconds, we align our units with the speed given and can proceed with calculations accurately.
Distance Calculation
Calculating distance when provided with speed and time is a straightforward process once unit conversions are completed. The fundamental equation used is:
\[ \text{distance} = \text{speed} \times \text{time} \]
In this context, after converting the speed of light to feet per second and time to seconds, we can determine how far light travels in a given time interval. Using the values calculated:
- Speed of light in feet per second: 9.8425 x 10^8 ft/s
- Time: 1.00 x 10^{-9} seconds
The calculation becomes:
\[ \text{distance} = (9.8425 \times 10^8 \, \text{ft/s}) \times (1.0 \times 10^{-9} \, \text{s}) \]
This results in a distance of 9.8425 x 10^{-1} feet, or approximately 0.98 feet. This exercise demonstrates the often surprising efficiency of light travel over extremely short periods, illustrating the breathtaking speed of light in the context of very brief time intervals like a nanosecond.
\[ \text{distance} = \text{speed} \times \text{time} \]
In this context, after converting the speed of light to feet per second and time to seconds, we can determine how far light travels in a given time interval. Using the values calculated:
- Speed of light in feet per second: 9.8425 x 10^8 ft/s
- Time: 1.00 x 10^{-9} seconds
The calculation becomes:
\[ \text{distance} = (9.8425 \times 10^8 \, \text{ft/s}) \times (1.0 \times 10^{-9} \, \text{s}) \]
This results in a distance of 9.8425 x 10^{-1} feet, or approximately 0.98 feet. This exercise demonstrates the often surprising efficiency of light travel over extremely short periods, illustrating the breathtaking speed of light in the context of very brief time intervals like a nanosecond.
Other exercises in this chapter
Problem 22
The current flowing in a solenoid that is \(20.0 \mathrm{~cm}\) long and has a radius of \(2.00 \mathrm{~cm}\) and 500 turns decreases from \(3.00 \mathrm{~A}\)
View solution Problem 25
The voltage across a cylindrical conductor of radius \(r\), length \(L\), and resistance \(R\) varies with time. The timevarying voltage causes a time-varying c
View solution Problem 28
How long does it take light to travel from the Moon to the Earth? From the Sun to the Earth? From Jupiter to the Earth?
View solution Problem 29
Alice made a telephone call from her home telephone in New York to her fiancé stationed in Baghdad, about \(10,000 \mathrm{~km}\) away, and the signal was carri
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