You receive orders to sail due east for 24.6 mi to put your salvage ship directly over a sunken pirate ship. However, when your divers probe the ocean floor at that location and find no evidence of a ship, you radio back to your source of information, only to discover that the sailing distance was supposed to be 24.6 nautical miles, not regular miles. Use the Length table below to calculate how far horizontally you are from the pirate ship. Length cm МЕTER km in. ft mi 1 centimeter = 1 102 105 0.3937 3.281 x 102 6.214 × 106 1 METER = 100 1 103 39.37 3.281 6.214 x 104 1 kilometer = 10° 1000 1 3.937 x 104 3281 0.6214 1 inch =2.540 2.540 x 102 2..540 × 105 1 8.333 x 102 1.578 × 105 1 foot = 30.48 0.3048 3.048 × 104 12 1 1.894 x 10* 1 mile = 1.609 x 105 1609 1.609 6.336 x 104 5280 1 1 angström = 10-10 1 fermi = 1015 m 1 fathom = 6 ft 1 rod = 16.5 ft m 1 nautical mile = 1852 m 1 light-vear = 9.461 × 1012 km 1 Bohr radius = 5.292 x 101" m 1 mil = 10* in. = 1.151 miles = 6076 ft 1 parsec = 3.084 × 1013 km 1 yard = 3 ft 1 nm = 10* m Number i 3.9E+1 Units km

You receive orders to sail due east for 24.6 mi to put your salvage ship directly over a sunken pirate ship. However, when your divers probe the ocean floor at that location and find no evidence of a ship, you radio back to your source of information, only to discover that the sailing distance was supposed to be 24.6 nautical miles, not regular miles. Use the Length table below to calculate how far horizontally you are from the pirate ship. Length cm МЕTER km in. ft mi 1 centimeter = 1 102 105 0.3937 3.281 x 102 6.214 × 106 1 METER = 100 1 103 39.37 3.281 6.214 x 104 1 kilometer = 10° 1000 1 3.937 x 104 3281 0.6214 1 inch =2.540 2.540 x 102 2..540 × 105 1 8.333 x 102 1.578 × 105 1 foot = 30.48 0.3048 3.048 × 104 12 1 1.894 x 10* 1 mile = 1.609 x 105 1609 1.609 6.336 x 104 5280 1 1 angström = 10-10 1 fermi = 1015 m 1 fathom = 6 ft 1 rod = 16.5 ft m 1 nautical mile = 1852 m 1 light-vear = 9.461 × 1012 km 1 Bohr radius = 5.292 x 101" m 1 mil = 10* in. = 1.151 miles = 6076 ft 1 parsec = 3.084 × 1013 km 1 yard = 3 ft 1 nm = 10* m Number i 3.9E+1 Units km

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter1: Units And Measurement

Section: Chapter Questions

Problem 83AP: A marathon runner completes a 42.188-km course in 2 h, 30 min, and 12s. There is an uncertainty of...

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(a) A car speedometer has a 5% uncertainty. What is the range of possible speeds when it reads 90 km/h? (b) Convert this range to miles per how. Note 1 km = 0.6214 mi.
The first atomic bomb was detonated on July 16, 1945, at the Trinity test site about 200 mi south of Los Alamos. In 1947, the U.S. government declassified a film reel of the explosion. From this film reel, British physicist G.I. Taylor was able to determine the rate at which the radius of the fireball from the blast grew. Using dimensional analysis, he was then able to deduce the amount of energy released in the explosion, which was a closely guarded secret at the time. Because of this, Taylor did not publish his results until 1950. This problem challenges you to recreate this famous calculation. (a) Using keen physical insight developed from years of experience, Taylor decided the radius rof the fireball should depend only on time since the explosion, t, the density of the air, , and the energy of the initial explosion, E. Thus, he made the educated guess that r=kEabtcfor some dimensionless constant kand some unknown exponents a,b, and c. Given that [E]=ML2T-2 , determine the values of the exponents necessary to make this equation dimensionally consistent. (Hint: Notice the equation implies that k=rEabtcand that [k]=1 ). (b) By analyzing data from high-energy conventional explosives, Taylor found the formula he derived seemed to be valid as long as the constant khad the value 1.03. From the film reel, he was able to determine many values of rand the corresponding values of t. For example, he found that after 25.0 ms, the fireball had a radius of 130.0 m. Use these values, along with an average air density of 1.25kg/m3 , to calculate the initial energy release of the Trinity detonation in joules (J). (Hint: To get energy in joules, you need to make sure all the numbers you substitute in are expressed in terms of SI base units.) (c) The energy released in large explosions is often cited in units of “tons of TNT” (abbreviated “t TNT”), where 1 t TNT is about 4.2 GJ. Convert yow answer to (b) into kilotons of TNT (that is, kt TNT). Compare your answer with the quick-and-duty estimate of 10 kt TNT made by physicist Enrico Fermi shortly after witnessing the explosion from what was thought to be a safe distance. (Reportedly, Fermi made his estimate by dropping some shredded bits of paper right before the remnants of the shock wave hit him and looked to see how far they were carried by it.)
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