HW18 - homework 18 PAPAGEORGE MATT Due Mar 1 2008 4:00 am...

Info icon This preview shows pages 1–2. Sign up to view the full content.

homework 18 – PAPAGEORGE, MATT – 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 1300 kg, starting from rest reaches a speed of 180 m / s in only 28 . 4 s. What is the average output power? 1. 0 . 000145061 MW 2. 1 . 4831 MW 3. 0 . 0261109 MW 4. 0 . 00823944 MW 5. 0 . 741549 MW correct 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 (1300 kg) (180 m / s) 2 28 . 4 s · mW 10 6 W = 0 . 741549 MW Question 2, chap 8, sect 5. part 1 of 1 10 points A car with a mass of 1 . 30 × 10 3 kg starts from rest and accelerates to a speed of 12.3 m/s in 14.1 s. Assume that the force of resistance remains constant at 492.2 N during this time. What is the average power developed by the car’s engine? Correct answer: 10001 . 4 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 . 30 × 10 3 kg v f = 12 . 3 m / s Δ t = 14 . 1 s F r = 492 . 2 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 = parenleftBig m v f Δ t + F r parenrightBig · v f 2 = bracketleftbigg (1300 kg)(12 . 3 m / s) 14 . 1 s + 492 . 2 N bracketrightbigg · 12 . 3 m / s 2 = 10001 . 4 W Question 3, chap 8, sect 5. part 1 of 1 10 points A car weighing 7200 N moves along a level highway with a speed of 74 km / h. The power of the engine at this speed is 65 kW. The car
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern