Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 2 - Describing Motion: Kinematics in One Dimension - General Problems: 76

Answer

The belt moves at a speed of 0.43 meters per minute. The rate of burger production is 1.7 burgers/minute.

Work Step by Step

We first find the required speed of the conveyor belt: $v = \frac{x}{t}$ $v = \frac{1.2 ~m}{2.8 ~min}$ $v = 0.43 ~m/min$ We can then find the number of burgers $N$ on the 1.2-meter belt: $N = \frac{1.2~m}{0.25~m} = 4.8~burgers$ On average, 4.8 burgers will be cooked every 2.8 minutes. We can find the rate of burger production in units of burgers/minute: $rate = \frac{4.8~burgers}{2.8~min} = 1.7 ~burgers/min$ The belt moves at a speed of 0.43 meters per minute. The rate of burger production is 1.7 burgers/minute.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.