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59. We apply Eq. 435 to solve for speed
v
and Eq. 434 to find centripetal acceleration
a
.
(a)
v
= 2
π
r
/
T
= 2
(20 km)/1.0 s = 126 km/s = 1.3
×
10
5
m/s.
(b) The magnitude of the acceleration is
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Acceleration

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