Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 1 - Chemical Tools: Experimentation and Measurement - Section Problems - Page 31: 90


1400 kg car moving at 115 km/h.

Work Step by Step

We find: Kinetic energy $K.E=\frac{1}{2}mv^{2}$ For the car, $K.E=\frac{1}{2}\times1400\times(115)^{2}=9.3\times10^{6}$ For the truck, $K.E=\frac{1}{2}\times12000\times(38)^{2}=8.7\times10^{6}$ Kinetic energy is more for the car.
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.