A car traveling at 27 m/s slams on its brakes to come to a stop. It decelerates at a rate of 8 m/s2 . What is the stopping dista
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speed of the car is given on the road
reaction time is given as
now we can find the distance that it will move during this reaction time
now the deceleration of the car is 10 m/s^2
so the distance that it will move before stop is given by
so the total distance that it require to stop is given as
while the deer is standing at distance 38 m
so the car will stop 2.8 m before the position of deer
Answer: 0.077 M
Explanation:
Expression for rate law for first order kinetics is given by:
where,
k = rate constant =
t = time taken for decay process = 10 minutes
a = initial amount of the reactant= 0.859 M
a - x = amount left after decay process =?
Putting values in above equation, we get:
Thus the concentration of a after 10.0 minutes is 0.077 M.