Chapter 16-17 - (5.976 X 10^19 particles /year) X (1 yr /...

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Astronomy –odd- numbered –Homework Chapter 16 1./ (angular diameter) / (205,265") = (linear diameter) / (distance) So, (angular separation) = (205,265") X (linear diameter) / (distance) The linear diameter in this case is the distance between Jupiter and the sun: (linear diameter) = 778.3 X 10^6 km = 7.78 X 10^11 m. The distance is that from the sun to the nearest star: (distance) = (1.3 pc) X (3.08 X 10^16 m / 1 pc ) = 4 X 10^16 m. (angular diameter) = (205,265") X { (7.78 X 10^11 m) / (4 X 10^16 m) } angular diameter = 4 seconds of arc 3./ The mercury diameter is smallest one. Its diameter is : 2439 km 7./ The mass of the earth is 5.976 X 10^24 kg. In order to build up this mass with 100 g ( = 0.1 kg) particles, requires (5.976 X 10^24 kg) X (1 particle / 0.1 kg ) = 5.976 X 10^25 particles. If these particles hit the earth over a period of 10^6 years We have: (5.976 X 10^25 particles) / (10^6 years) = 5.976 X 10^19 particles /year hit.
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Unformatted text preview: (5.976 X 10^19 particles /year) X (1 yr / 365 d ) X (1 d / 24 h) X (1 h / 3600 s) =1.89 X 10^12 particles / second So, the answer is : 1.89 X 10^12 particles / second 9./ The standard equation for a problem involving velocity and distance is v = d / t (v = velocity; d = distance; t = time ) => t = d / v; where v = 400 km /s and d = 5900 X 10^6 km. t = (5900 X 10^6 km) / (400 km /s) = 1.475 X 10^7 sec (1.475 X 10^7 sec) X (1 hr / 3600 s ) X (1 d / 24 h ) X (1 y / 365 d )=0.46 years = 5.6 months Chapter 17: 1./ 6400 km= 6,400,000 meters 6,400,000 meters=6,400,000,000 mm 6,400,000,000/ 3= 213,333,333 years ago. 5./ The speed of light is 299,792 km/s. Distance from Earth to Venus is 41,400,000 km. Divide the speed into the distance to get the time for a one way trip in seconds: 41,400,000/299,792=138.1 seconds. Round trip is: 138.1 X 2=276.2 seconds.= 4 min, 36 s The answer is 4 minutes, 36 seconds....
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Chapter 16-17 - (5.976 X 10^19 particles /year) X (1 yr /...

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