hw6sol - homework 06 – ARMSTRONG DOMINIC – Due 3:00 am...

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Unformatted text preview: homework 06 – ARMSTRONG, DOMINIC – Due: Oct 16 2007, 3:00 am 1 Question 1, chap 8, sect 1. part 1 of 1 10 points A 69 g block is released from rest and slides down a frictionless track that begins 1 . 8 m above the horizontal, as shown in the figure. At the bottom of the track, where the surface is horizontal, the block strikes and sticks to a light spring with a spring constant of 10 N / m. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 8 m 69 g 10 N / m Find the maximum distance the spring is compressed. Correct answer: 0 . 493388 m (tolerance ± 1 %). Explanation: Let : m = 69 g = 0 . 069 kg , h i = 1 . 8 m , and k = 10 N / m . From conservation of mechanical energy, we have mg h i = 1 2 k x 2 x = radicalbigg 2 mg h i k = radicalBigg 2 (0 . 069 kg) (9 . 8 m / s 2 ) (1 . 8 m) 10 N / m = . 493388 m . Question 2, chap 9, sect 1. part 1 of 1 10 points Given: G = 6 . 67259 × 10- 11 N m 2 / kg 2 A distant star has a single planet circling it in a circular orbit of radius 7 . 36 × 10 11 m. The period of the planet’s motion about the star is 692 day. What is the mass of the star? Correct answer: 6 . 59871 × 10 31 kg (tolerance ± 1 %). Explanation: According to Newton’s explanation of Ke- pler’s third law R 3 B T 2 B = GM s 4 π 2 = const. where G is universal constant, R B is the dis- tance between the star and its planet, T B is the period of the motion. The mass of the star can be given as M s = 4 π 2 G R 3 B T 2 B = 4 (3 . 1415926) 2 (6 . 67259 × 10- 11 N m 2 / kg 2 ) × (7 . 36 × 10 11 m) 3 (5 . 97888 × 10 7 s) 2 = 6 . 59871 × 10 31 kg Question 3, chap -1, sect -1. part 1 of 1 10 points Two telecommunication satellites A and B have a circular orbit around the Earth in the same plane. Their masses and radii have the relationships 6 m B = 5 m A and 16 r B = 9 r A ....
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hw6sol - homework 06 – ARMSTRONG DOMINIC – Due 3:00 am...

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