Answer:
Part a)
the time after which both cars will have same speed is t = 4.05 s
Part b)

Part c)
t = 5.12 s
Part d)

Explanation:
Part a)
As we know that Car A is moving with initial speed 3.1 cm/s with deceleration 2.10 cm/s/s while car B is moving with constant 5.40 cm/s speed
Now if the speed of Car A and Car B will be same after some time
So we will have



so the time after which both cars will have same speed is t = 4.05 s
Part b)
After this time when both cars have same speed
this speed will be same as that of speed of Car B as it is moving at constant speed
so we have

Part c)
if final position of both cars will be same
then we will have


also we have


now we have


t = 5.12 s
Part d)
Final positions of both the particles is given as



<h2><em>so </em><em>2</em><em> </em><em>is </em><em>the </em><em>answer </em><em>of</em><em> </em><em>this </em><em>question</em></h2>
Chemical energy. Elimination is useful to answer this question.
Answer:
146 neutrons.
Explanation:
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Need to see the statements mentioned in the q posted.