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The space shuttle travels at a speed of about \(7.6 \times 10^{3} \mathrm{m} / \mathrm{s} .\) The blink of an astronaut's eye lasts about \(110 \mathrm{ms}\). How many football fields (length \(=91.4 \mathrm{m})\) does the shuttle cover in the blink of an eye?

Short Answer

Expert verified
The shuttle covers about 9.15 football fields in the blink of an eye.

Step by step solution

01

Convert Blink Duration to Seconds

The blink duration is given in milliseconds (ms). Since 1 second is 1000 milliseconds, convert the blink time to seconds by dividing by 1000: \( 110 \, \text{ms} = \frac{110}{1000} \, \text{s} = 0.11 \, \text{s} \).
02

Calculate Distance During Blink

To find out how far the shuttle travels in the blink of an eye, use the formula for distance: \( \text{Distance} = \text{speed} \times \text{time} \). With a speed of \( 7.6 \times 10^3 \, \text{m/s} \) and a time of 0.11 seconds, calculate the distance: \[ \text{Distance} = 7.6 \times 10^3 \times 0.11 = 836 \, \text{meters} \].
03

Calculate Number of Football Fields

A standard football field is 91.4 meters long. To find out how many full football fields the distance represents, divide the total distance traveled by the length of one field: \[ \text{Number of football fields} = \frac{836}{91.4} \approx 9.15 \].

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Key Concepts

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

Velocity
Velocity is a fundamental concept in physics that defines the speed of an object in a specified direction. It is crucial to motion calculations. Unlike speed, which is scalar and only measures how fast an object is moving, velocity is a vector quantity. This means it takes into account both magnitude and direction. For example, saying an object travels at 20 meters per second north*.* *You can express velocity in various units. Commonly, it is given in meters per second (m/s). When dealing with astronomical or high-speed entities like the space shuttle, velocity may be expressed in terms like \( 7.6 \times 10^3 \text{ m/s} \).* The formula to calculate velocity is fairly straightforward:
  • \( v = \frac{d}{t} \)
Where:
  • \( v \) is velocity
  • \( d \) is distance traveled
  • \( t \) is the time taken
Understanding velocity helps in predicting motion trajectories, essential for space travel and understanding sports science.
Distance-Time Relationship
The relationship between distance and time is key to understanding how objects move. It tells us how far an object has traveled over a specific period. This relationship is woven into the equations of motion, where distance plays a crucial role.

In cases where motion is uniform, the distance an object covers can be calculated simply by multiplying its velocity by the time period considered. The formula is:

  • \( ext{Distance} = ext{Velocity} \times ext{Time} \)
The concept can be used to glean insights into various applications, such as:
  • Predicting the travel path of vehicles
  • Estimation in sports for scoring or positioning
  • Understanding astronomical distances over time.
For instance, in the exercise given, the space shuttle's distance covered during a blink was determined by applying this relationship. By knowing the velocity is \( 7.6 \times 10^3 \text{ m/s} \), and the blink duration is \( 0.11 \text{ s} \), it's straightforward to find the shuttle travels 836 meters in one blink.
Unit Conversion in Physics
Physics often involves calculations requiring unit conversions for accuracy and coherence. Converting between units is crucial for comparing measurements or making precise calculations across different scales. For example, time can be expressed in different units like milliseconds, seconds, or hours. When exactness is important, you must convert to the unit used in your calculations. As part of our example exercise:
  • The blink duration of 110 milliseconds is converted to seconds for easier calculation, using: \( 1 \text{ s} = 1000 \text{ ms} \), hence, \( 110 \text{ ms} = 0.11 \text{ s} \).
In distance measurements, you might convert between meters, kilometers, miles for surface travel, or astronomical units for space travel. In contexts like the space shuttle's travel, converting distance into familiar concepts, such as football fields, aids understanding. Effective use of unit conversions allows clearer communication of scientific concepts and results, ensures consistency, and avoids errors in computation.

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Most popular questions from this chapter

A car is traveling along a straight road at a velocity of \(+36.0 \mathrm{m} / \mathrm{s}\) when its engine cuts out. For the next twelve seconds the car slows down, and its average acceleration is \(\bar{a}_{1} .\) For the next six seconds the car slows down further, and its average acceleration is \(\bar{a}_{2} .\) The velocity of the car at the end of the eighteen-second period is \(+28.0 \mathrm{m} / \mathrm{s}\). The ratio of the average acceleration values is \(\bar{a}_{1} / \bar{a}_{2}=1.50 .\) Find the velocity of the car at the end of the initial twelve-second interval.

One afternoon, a couple walks three-fourths of the way around a circular lake, the radius of which is \(1.50 \mathrm{km}\). They start at the west side of the lake and head due south to begin with. (a) What is the distance they travel? (b) What are the magnitude and direction (relative to due east) of the couple's displacement?

Two cars cover the same distance in a straight line. Car A covers the distance at a constant velocity. Car B starts from rest and maintains a constant acceleration. Both cars cover a distance of \(460 \mathrm{m}\) in \(210 \mathrm{s}\). Assume that they are moving in the \(+x\) direction. Determine (a) the constant velocity of car A, (b) the final velocity of car \(\mathrm{B},\) and (c) the acceleration of car B.

The leader of a bicycle race is traveling with a constant velocity of \(+11.10 \mathrm{m} / \mathrm{s}\) and is \(10.0 \mathrm{m}\) ahead of the second-place cyclist. The secondplace cyclist has a velocity of \(+9.50 \mathrm{m} / \mathrm{s}\) and an acceleration of \(+1.20 \mathrm{m} / \mathrm{s}^{2}\). How much time elapses before he catches the leader?

A car is traveling at a constant speed of \(33 \mathrm{m} / \mathrm{s}\) on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is \(2.5 \mathrm{km}\) away?

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