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Unformatted text preview: Husain, Zeena – Homework 4 – Due: Sep 23 2003, 4:00 am – Inst: H L Berk 1 This printout should have 24 questions, check that it is complete. Multiplechoice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. 001 (part 1 of 2) 10 points Two masses m 1 and m 2 are connected in the manner shown. m 2 m 1 T 1 T 2 a = g 2 The system is accelerating downward with acceleration of magnitude g 2 . Determine T 2 . 1. T 2 = µ m 2 1 2 m 1 ¶ g 2. T 2 = m 2 g 3. T 2 = 1 2 m 2 g correct 4. T 2 = µ 1 2 m 2 m 1 ¶ g 5. T 2 = 3 2 m 2 g 6. T 2 = 1 2 ( m 2 m 1 ) g Explanation: m 2 T 2 a = g 2 m 2 g From Newton’s Second Law, m 2 g T 2 = m 2 g 2 . So T 2 = m 2 g m 2 g 2 = 1 2 m 2 g . 002 (part 2 of 2) 10 points Determine T 1 . 1. T 1 = 5 2 ( m 1 + m 2 ) g 2. None of these. 3. T 1 = 2( m 1 + m 2 ) g 4. T 1 = 3 2 ( m 1 + m 2 ) g 5. T 1 = 1 2 ( m 1 + m 2 ) g correct 6. T 1 = ( m 1 + m 2 ) g Explanation: m 1 + m 2 T 1 a = g 2 ( m 1 + m 2 ) g Consider m 1 and m 2 as a whole object with mass ( m 1 + m 2 ), then ( m 1 + m 2 ) g T 1 = ( m 1 + m 2 ) g 2 . So T 1 = 1 2 ( m 1 + m 2 ) g . 003 (part 1 of 1) 10 points Given: g = 9 . 8 m / s 2 . A man weighs 987 N on Earth. What would he weigh on Jupiter, where the freefall acceleration is 25 . 9 m / s 2 ? Correct answer: 2608 . 5 N. Explanation: The man’s weight on earth is given by W e = mg e Solving for his mass gives m = W e g e = 987 N 9 . 8 m / s 2 = 100 . 714 kg His weight on Jupiter will be W J = mg J = (100 . 714 kg)(25 . 9 m / s 2 ) = 2608 . 5 N Husain, Zeena – Homework 4 – Due: Sep 23 2003, 4:00 am – Inst: H L Berk 2 004 (part 1 of 2) 10 points The average speed of a Nitrogen molecule in air is about 469 m / s, and its mass is about 4 . 68 × 10 26 kg. If it takes 2 × 10 13 s for a nitrogen molecule to hit a wall and rebound with the same speed but in the opposite direction, what is the magnitude of the average acceleration of the molecule during this time interval? Correct answer: 4 . 69 × 10 15 m / s 2 . Explanation: If you take the initial velocity of the molecule to be positive, its initial velocity is 469 m / s and its final velocity is 469 m / s. Therefore, its change in velocity is k Δ ~v k = k ~v f ~v i k = 2 v i This gives the magnitude of the average ac celeration as k ~a avg k = k Δ ~v k Δ t = 2 v i Δ t = 2(469 m / s) Δ t = 4 . 69 × 10 15 m / s 2 . 005 (part 2 of 2) 10 points What is the magnitude of the average force that the molecule exerts on the wall? Correct answer: 2 . 19492 × 10 10 N. Explanation: From Newton’s 3 rd law, the average force on the wall due to the molecule is equal to the average force on the molecule due to the wall....
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 Fall '08
 Turner
 Friction, Work, kg, Husain

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