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AleksandrR [38]
3 years ago
15

A roller coaster starts down a hill at 10 m/s. Three seconds later, its speed is 32 m/s. What is the roller coaster’s accelerati

on?
-7.3 m/s2

7.3 m/s

-7.3 m/s

7.3 m/s 2
Chemistry
1 answer:
erica [24]3 years ago
8 0
Final velocity(v) = 32 m/s
Initial velocity(u) = 10 m/s

Using kinematic equation v = u + at, 

32 = 10 + a(3)

32-10
---------  = a
    3

a = 22
      ----
       3

a = 7.3 m/s^2.

Hence acceleration of the roller coaster is 7.3 m/s^2.

Hope this helps!!
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Rearrange the formula is as follows.

k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)

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\epsilon =y_{A_{o}}\delta

\delta = change in total number of moles per mole of A reacte.

=1(2-1)=1

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k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}

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The rate constant in case of the CSTR can be calculated by using the formula.

\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)

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=8.1\,kcal\,mol^{-1}

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