An Introduction to Physical Science
An Introduction to Physical Science
14th Edition
ISBN: 9781305079137
Author: James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Publisher: Cengage Learning
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Chapter 19, Problem 2E

(a)

To determine

The height of the pikes peak in km and mi .

(a)

Expert Solution
Check Mark

Answer to Problem 2E

The height of the pikes peak is 4.27 km or 2.65mi .

Explanation of Solution

Given info: The height of the pikes peak is 14000 ft .

Explanation:

The height of the pikes peak is 14000 ft .

Convert the height from ft into km .

1 ft = 0.000305 km

Use unitary method to find the height of the pikes peak in km .

Therefore, the height of the pikes peak in km is as follows,

14000 ft = (0.000305 km)(14000)=4.27 km

The height of the pikes peak is 14000 ft .

Convert the height from km into mi .

1.61 km = 1 mi

Use unitary method to find the height of the pikes peak in mi .

1 km = 11.61 mi

Therefore, the height of the pikes peak in mi is as follows,

4.27 km = (11.61 mi)(4.27)=2.65mi

Conclusion:

Therefore, the height of the pikes peak is 4.27 km or 2.65mi .

(b)

To determine

The height of the top of the Mt. Everest in km and mi .

(b)

Expert Solution
Check Mark

Answer to Problem 2E

The height of the top of the Mt. Everest is 8.845 km or 5.49mi .

Explanation of Solution

Given info: The height of the top of the Mt. Everest is 29000 ft .

Explanation:

The height of the top of the Mt. Everest is 29000 ft .

Convert the height from ft into km .

1 ft = 0.000305 km

Use unitary method to find the height of the top of the Mt. Everest in km .

Therefore, the height of the top of the Mt. Everest in km is as follows,

29000 ft = (0.000305 km)(29000)=8.845 km

The height of the top of the Mt. Everest is 14000 ft .

Convert the height from km into mi .

1.61 km = 1 mi

Use unitary method to find the height of the top of the Mt. Everest in mi .

1 km = 11.61 mi

Therefore, the height of the top of the Mt. Everest in mi is as follows,

8.845 km = (11.61 mi)(8.845)=5.49mi

Conclusion:

Therefore, the height of the top of the Mt. Everest is 8.845 km or 5.49mi .

(c)

To determine

The height of the commercial airline flight in km and mi .

(c)

Expert Solution
Check Mark

Answer to Problem 2E

The height of the commercial airline flight is 10.675 km or 6.63mi .

Explanation of Solution

Given info: The height of the commercial airline flight is 35000 ft .

Explanation:

The height of the commercial airline flight is 35000 ft .

Convert the height from ft into km .

1 ft = 0.000305 km

Use unitary method to find the height of the commercial airline flight in km .

Therefore, the height of the commercial airline flight in km is as follows,

35000 ft = (0.000305 km)(35000)=10.675 km

The height of the commercial airline flight is 35000 ft .

Convert the height from km into mi .

1.61 km = 1 mi

Use unitary method to find the height of the top of the Mt. Everest in mi .

1 km = 11.61 mi

Therefore, the height of the commercial airline flight in mi is as follows,

10.675 km = (11.61 mi)(10.675)=6.63mi

Conclusion:

Therefore, the height of the commercial airline flight is 10.675 km or 6.63mi .

(d)

To determine

The height of the supersonic transport flight in km and mi .

(d)

Expert Solution
Check Mark

Answer to Problem 2E

The height of the supersonic transport flight is 19.825 km or 12.31mi .

Explanation of Solution

Given info: The height of the supersonic transport flight is 65000 ft .

Explanation:

The height of the supersonic transport flight is 65000 ft .

Convert the height from ft into km .

1 ft = 0.000305 km

Use unitary method to find the height of the supersonic transport flight in km .

Therefore, the height of the supersonic transport flight in km is as follows,

65000 ft= (0.000305 km)(65000)=19.825 km

The height of the supersonic transport flight is 35000 ft .

Convert the height from km into mi .

1.61 km = 1 mi

Use unitary method to find the height of the supersonic transport flight in mi .

1 km = 11.61 mi

Therefore, the height of the supersonic transport flight in mi is as follows,

19.825 km = (11.61 mi)(19.825)=12.31mi

Conclusion:

Therefore, the height of the supersonic transport flight is 19.825 km or 12.31mi .

(e)

To determine

The height of the communications satellite in km .

(e)

Expert Solution
Check Mark

Answer to Problem 2E

The height of the communications satellite is 644 km .

Explanation of Solution

Given info: The height of the communications satellite is 400mi .

Explanation:

The height of the communications satellite is 400mi .

Convert the height from mi into km .

1 mi =1.61 km

Use unitary method to find the height of the communications satellite in km .

Therefore, the height of the communications satellite in km is as follows,

400mi =(1.61 km)(400)=644 km

Conclusion:

Therefore, the height of the communications satellite is 644 km .

(f)

To determine

The height of the E and F ion layers in mi .

(f)

Expert Solution
Check Mark

Answer to Problem 2E

The height of the E layer is 62.1 mi . The height of the F layer is 217.4 mi .

Explanation of Solution

Given info: The height of the E layer is 100 km . The height of the F layer is 350 km .

Explanation:

The height of the E layer is 100 km .

Convert the height from km into mi .

1.61 km = 1 mi

Use unitary method to find the height of the E layer in mi .

1 km = 11.61 mi

Therefore, the height of the E layer is as follows,

100 km = (11.61 mi)(100)=62.1 mi

The height of the F layer is 350 km .

Convert the height from km into mi .

1.61 km = 1 mi

Use unitary method to find the height of the F layer in mi .

1 km = 11.61 mi

Therefore, the height of the F layer is as follows,

350 km = (11.61 mi)(350)=217.4 mi

Conclusion:

Therefore, the height of the E layer is 62.1 mi . The height of the F layer is 217.4 mi .

(g)

To determine

The height of the aurora displays in mi .

(g)

Expert Solution
Check Mark

Answer to Problem 2E

The height of the E layer is 62.1 mi . The height of the F layer is 217.4 mi .

Explanation of Solution

Given info: The height of the aurora displays is 500 km .

Explanation:

The height of the aurora displays is 500 km .

Convert the height from km into mi .

1.61 km = 1 mi

Use unitary method to find the height of the aurora displays in mi .

1 km = 11.61 mi

Therefore, the height of the aurora displays is as follows,

500 km = (11.61 mi)(500)=310.6 mi

Conclusion:

Therefore, the height of the aurora displays is 310.6 mi .

(h)

To determine

The height of the syncom satellite in km .

(h)

Expert Solution
Check Mark

Answer to Problem 2E

The height of the syncom satellite is 37030 km .

Explanation of Solution

Given info: The height of the syncom satellite is 23000mi .

Explanation:

The height of the syncom satellite is 23000mi .

Convert the height from mi into km .

1 mi =1.61 km

Use unitary method to find the height of the syncom satellite in km .

Therefore, the height of the syncom satellite in km is as follows,

23000mi =(1.61 km)(23000)=37030 km

Conclusion:

Therefore, the height of the syncom satellite is 37030 km .

To determine

The comparison of different heights provided in the problem.

Expert Solution
Check Mark

Answer to Problem 2E

In ascending order the heights can be arranged as follows,

pikes peak, the top of the Mt. Everest, the commercial airline flight, the supersonic transport flight, E layer, F layer, Aurora, communications satellite and syncom satellite.

Explanation of Solution

Given info: The heights of the pikes peak, Mt. Everest, commercial airline flight, supersonic transport flight, communications satellite, E layer, F layer, aurora displays and

syncom satellite are provided.

Explanation:

The height of the pikes peak is 4.27 km or 2.65mi . The height of the top of the Mt. Everest is 8.845 km or 5.49mi . The height of the commercial airline flight is 10.675 km or 6.63mi . The height of the supersonic transport flight is 19.825 km or 12.31mi .

The height of the communications satellite is 644 km . The height of the E layer is 62.1 mi . The height of the F layer is 217.4 mi . The height of the aurora displays is 310.6 mi . The height of the syncom satellite is 37030 km .

Conclusion:

Therefore, in ascending order the heights can be arranged as follows,

pikes peak, the top of the Mt. Everest, the commercial airline flight, the supersonic transport flight, E layer, F layer, Aurora, communications satellite and syncom satellite.

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Chapter 19 Solutions

An Introduction to Physical Science

Ch. 19.5 - Prob. 2PQCh. 19 - KEY TERMS 1. atmospheric science (Intro) 2....Ch. 19 - Prob. BMCh. 19 - Prob. CMCh. 19 - Prob. DMCh. 19 - Prob. EMCh. 19 - KEY TERMS 1. atmospheric science (Intro) 2....Ch. 19 - Prob. GMCh. 19 - Prob. HMCh. 19 - Prob. IMCh. 19 - Prob. JMCh. 19 - Prob. KMCh. 19 - Prob. LMCh. 19 - Prob. MMCh. 19 - Prob. NMCh. 19 - Prob. OMCh. 19 - Prob. PMCh. 19 - Prob. QMCh. 19 - KEY TERMS 1. atmospheric science (Intro) 2....Ch. 19 - Prob. SMCh. 19 - Prob. TMCh. 19 - Prob. UMCh. 19 - Prob. VMCh. 19 - Prob. WMCh. 19 - Prob. XMCh. 19 - Prob. YMCh. 19 - Prob. ZMCh. 19 - Prob. AAMCh. 19 - Prob. BBMCh. 19 - Prob. CCMCh. 19 - Prob. DDMCh. 19 - Prob. 1MCCh. 19 - Prob. 2MCCh. 19 - Photosynthesis is responsible for the atmospheric...Ch. 19 - What regulates the Earths average temperature?...Ch. 19 - Prob. 5MCCh. 19 - Prob. 6MCCh. 19 - Prob. 7MCCh. 19 - Prob. 8MCCh. 19 - Prob. 9MCCh. 19 - Prob. 10MCCh. 19 - What is the cloud root name that means heap?...Ch. 19 - Prob. 12MCCh. 19 - Prob. 1FIBCh. 19 - Prob. 2FIBCh. 19 - Prob. 3FIBCh. 19 - Prob. 4FIBCh. 19 - One standard atmosphere of pressure is ___ lb2....Ch. 19 - Prob. 6FIBCh. 19 - Prob. 7FIBCh. 19 - Prob. 8FIBCh. 19 - Prob. 9FIBCh. 19 - Prob. 10FIBCh. 19 - Prob. 11FIBCh. 19 - Prob. 12FIBCh. 19 - Prob. 1SACh. 19 - Prob. 2SACh. 19 - Prob. 3SACh. 19 - Prob. 4SACh. 19 - Prob. 5SACh. 19 - Prob. 6SACh. 19 - Prob. 7SACh. 19 - Prob. 8SACh. 19 - Prob. 9SACh. 19 - In terms of Rayleigh scattering, why is it...Ch. 19 - Prob. 11SACh. 19 - Prob. 12SACh. 19 - What are the four fundamental atmospheric...Ch. 19 - Prob. 14SACh. 19 - Prob. 15SACh. 19 - When is the relative humidity 100%? It may be...Ch. 19 - Which way, relative to the wind direction, does a...Ch. 19 - Prob. 18SACh. 19 - Prob. 19SACh. 19 - What is a convection cycle, and what are the...Ch. 19 - Prob. 21SACh. 19 - Why does weather generally move from west to east...Ch. 19 - Prob. 23SACh. 19 - Prob. 24SACh. 19 - Name the cloud family for each of the following:...Ch. 19 - Prob. 26SACh. 19 - Prob. 27SACh. 19 - Why are clouds generally white, while some are...Ch. 19 - Visualize the connections and give answers in the...Ch. 19 - (a) Why does the land lose heat more quickly at...Ch. 19 - Prob. 2AYKCh. 19 - Prob. 3AYKCh. 19 - Prob. 4AYKCh. 19 - Water leaks from a cup with a hole, as shown in ...Ch. 19 - Express the approximate thicknesses of the (a)...Ch. 19 - Prob. 2ECh. 19 - If the air temperature is 70F at sea level, then...Ch. 19 - If the air temperature is 20C at sea level, then...Ch. 19 - On a day when the air temperature is 85F, the...Ch. 19 - Prob. 6ECh. 19 - On a very hot day with an air temperature of 105F,...Ch. 19 - On a winter day, a psychrometer has a dry-bulb...Ch. 19 - The dry-bulb and wet-bulb thermometers of a...Ch. 19 - Prob. 10ECh. 19 - On a day when the air temperature is 70F, a fellow...Ch. 19 - On another day with the same air temperature...
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