Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 3 - Kinematics in Two Dimensions - Problems: 11

Answer

a) 2.99 x $10^{4}$ m/s b) a) 2.69 x $10^{4}$ m/s

Work Step by Step

a) Distance = $\frac{2\pi\times r}{4}$ = $\frac{2\pi\times 1.50 \times 10^{11}}{4}$ $\approx$ 2.3562 x $10^{11}$ m Average speed = $\frac{2.3562 \times 10^{11}}{7.89 \times 10^{6}}$ $\approx$ 2.99 x $10^{4}$ m/s b) displacement = $\sqrt 2 \times (1.50 \times 10^{11})^{2}$ $\approx$ 2.1213 m Average velocty = $\frac{2.1213 \times 10^{11}}{7.89 \times 10^{6}}$ $\approx$ 2.69 x $10^{4}$ m/s
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.