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trapecia [35]
3 years ago
15

I was driving along at 20 m/s , trying to change a CD and not watching where I was going. When I looked up, I found myself 46 m

from a railroad crossing. And wouldn't you know it, a train moving at 29 m/s was only 56 m from the crossing. In a split second, I realized that the train was going to beat me to the crossing and that I didn't have enough distance to stop. My only hope was to accelerate enough to cross the tracks before the train arrived. If my reaction time before starting to accelerate was 0.42 s , what minimum acceleration did my car need for me to be here today writing these words?
Physics
1 answer:
vampirchik [111]3 years ago
5 0

Answer:

6.46393559312 m/s²

Explanation:

Time taken to cover 56 m

t=\dfrac{56}{29}\ s

Distance covered in 0.42 seconds

0.42\times 20=8.4\ m

From equation of linear motion

s=ut+\frac{1}{2}at^2

8.4+20(\dfrac{56}{29}-0.42)+\dfrac{1}{2}a(\dfrac{56}{29}-0.42)^2=46\\\Rightarrow a=(46-8.4-20(\dfrac{56}{29}-0.42))\times\dfrac{2}{(\dfrac{56}{29}-0.42)^2}\\\Rightarrow a=6.46393559312\ m/s^2

The minimum acceleration is 6.46393559312 m/s²

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How much energy in Joules (J) would an electric heater that draws 9.5 A when connected to a 120 V supply use if the heater were
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Explanation:

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what is the value of the constant for a second order reaction if the reactant concentration drops from .657 M to ,0981 M in 17 s
yaroslaw [1]

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Explanation :

The expression used for second order kinetics is:

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