Answer
$F_{tan} = 3.0 \times 10^3~N$
$F_{R} = 430~N$
Work Step by Step
$F_{tan} = ma_{tan} = (950 ~kg)(3.2 ~m/s^2) = 3.0 \times 10^3~N$
$F_{R} = ma_{R} = (950 ~kg)(0.45 ~m/s^2) = 430~N$
Physics: Principles with Applications (7th Edition)
by Giancoli, Douglas C.
Published by
Pearson
ISBN 10:
0-32162-592-7
ISBN 13:
978-0-32162-592-2
Chapter 5 - Circular Motion; Gravitation - Problems - Page 133: 24
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