Understanding Our Universe
3rd Edition
ISBN: 9780393614428
Author: PALEN, Stacy, Kay, Laura, Blumenthal, George (george Ray)
Publisher: W.w. Norton & Company,
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Chapter 1, Problem 1QAP
To determine
Whether the statement is true or false.
Expert Solution & Answer
Answer to Problem 1QAP
The given statement is false.
Explanation of Solution
Light travels at a speed of
Such large distances are measured in units of light years denoted by
Here, the distance given is
Conclusion:
Thus, one mega-light-year is not equal to 1,000 light years. Hence, the given statement is false.
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Chapter 1 Solutions
Understanding Our Universe
Ch. 1.1 - Prob. 1.1CYUCh. 1.2 - Prob. 1.2CYUCh. 1.3 - Prob. 1.3CYUCh. 1 - Prob. 1QAPCh. 1 - Prob. 2QAPCh. 1 - Prob. 3QAPCh. 1 - Prob. 4QAPCh. 1 - Prob. 5QAPCh. 1 - Prob. 6QAPCh. 1 - Prob. 7QAP
Ch. 1 - Prob. 8QAPCh. 1 - Prob. 9QAPCh. 1 - Prob. 10QAPCh. 1 - Prob. 11QAPCh. 1 - Prob. 12QAPCh. 1 - Prob. 13QAPCh. 1 - Prob. 14QAPCh. 1 - Prob. 15QAPCh. 1 - Prob. 16QAPCh. 1 - Prob. 17QAPCh. 1 - Prob. 18QAPCh. 1 - Prob. 19QAPCh. 1 - Prob. 20QAPCh. 1 - Prob. 21QAPCh. 1 - Prob. 22QAPCh. 1 - Prob. 23QAPCh. 1 - Prob. 24QAPCh. 1 - Prob. 25QAPCh. 1 - Prob. 26QAPCh. 1 - Prob. 27QAPCh. 1 - Prob. 28QAPCh. 1 - Prob. 29QAPCh. 1 - Prob. 30QAPCh. 1 - Prob. 31QAPCh. 1 - Prob. 32QAPCh. 1 - Prob. 34QAPCh. 1 - Prob. 35QAPCh. 1 - Prob. 36QAPCh. 1 - Prob. 37QAPCh. 1 - Prob. 38QAPCh. 1 - Prob. 39QAPCh. 1 - Prob. 40QAPCh. 1 - Prob. 41QAPCh. 1 - Prob. 42QAPCh. 1 - Prob. 43QAPCh. 1 - Prob. 44QAPCh. 1 - Prob. 45QAP
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- If you start at the equator and travel to 10 degrees north approx. how many kilometers (or miles) north of the equator will you be? Take the circumference of earth to be 40,000 kilometers (24,900) miles).arrow_forwardIt is important to have an idea about the distances between and relative sizes of celestial objects in the solar system. In Part 1 we will pretend to shrink the solar system until its center piece, the Sun, is 67.3 cm in diameter. This will represent the Sun which is 1,390,000 km in diameter. The scale of our model is thus: 67.3 cm = 4.84 x 10-5 cm km Scale 1, 390, 000 km To find the size or distance between objects in centimeters for the model, simply multiply the actual size or distance in kilometers by the scale factor above. 1. Fill in following table: Quantity Actual Distance (km) Model Distance (cm) Diameter of Sun 1,390,000 Diameter of Earth 12,760 Diameter of Moon 3,480 Distance Between Earth and Sun 1.5 x 108 Distance Between Earth and Moon 384,000 Distance to Proxima Centauri 3.97 x 1013arrow_forwardUse the equation E = mc^2 where E is energy in Joules (J), m is mass in kilograms (kg) and c is the speed of light 3 x 10^8 m/s to answer the following: a) One ton of TNT releases 4.18 gigajoules of energy. The metric prefix giga means billion. a) How much mass would be required to release an equivalent amount of energy? b) How much energy (J) is equivalent to 1 kilogram of mass?arrow_forward
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