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3 years ago

14

A ball is dropped and falls with an acceleration of 9.8m/s2 downward. it hit the ground with a velocity of 49 m/s downward. how

long did it take the ball to fall to the ground?

1 answer:

Lelechka [254]3 years ago

4 0

V=u+at
49=0+9.8t
t=49/9.8
t = 5 (seconds)

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$k_2 = Ce ^{-\frac{Q}{RT_2} }$

Now the ratio between the two given rate constant is

$\frac{k_1 }{k_2} = e^{(\frac{Q}{R} [\frac{1}{\frac{T_2 - 1}{T_1} } ] )}$

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