hw 18 - homework 18 HETTIGER, CHRISTOF Due: Mar 1 2008,...

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homework 18 – HETTIGER, CHRISTOF – Due: Mar 1 2008, 4:00 am 1 Question 1, chap 8, sect 5. part 1 of 1 10 points A hot rod of mass 1600 kg, starting from rest reaches a speed of 110 m / s in only 17 . 2 s. What is the average output power? 1. 0 . 000297458 MW 2. 0 . 0327204 MW 3. 1 . 12558 MW 4. 0 . 562791 MW correct 5. 0 . 0102326 MW Explanation: The average output power is equal to the kinetic energy acquired divided by the total time, so P = 1 2 mv 2 t = 1 2 (1600 kg) (110 m / s) 2 17 . 2 s · mW 10 6 W = 0 . 562791 MW Question 2, chap 8, sect 5. part 1 of 1 10 points A car with a mass of 1 . 72 × 10 3 kg starts from rest and accelerates to a speed of 14.1 m/s in 12.1 s. Assume that the force of resistance remains constant at 433.8 N during this time. What is the average power developed by the car’s engine? Correct answer: 17188 . 6 W (tolerance ± 1 %). Explanation: Basic Concepts: v f = a Δ t and Δ x = 1 2 a t ) 2 since v i = 0 m/s. P = W Δ t = F Δ x Δ t F = ma + F r Given: m = 1 . 72 × 10 3 kg v f = 14 . 1 m / s Δ t = 12 . 1 s F r = 433 . 8 N Solution: a = v f Δ t so P = F Δ x Δ t = F · 1 2 a t ) 2 Δ t = Fa Δ t 2 = ( ma + F r ) v f 2 = p m v f Δ t + F r P · v f 2 = b (1720 kg)(14 . 1 m / s) 12 . 1 s + 433 . 8 N B · 14 . 1 m / s 2 = 17188 . 6 W Question 3, chap 8, sect 5. part 1 of 1
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hw 18 - homework 18 HETTIGER, CHRISTOF Due: Mar 1 2008,...

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