Answer:four times
Explanation:
Given
mass of both cars A and B are same suppose m
but velocity of car B is same as of car A
Suppose velocity of car A is u
Velocity of car B is 2 u
A constant force is applied on both the cars such that they come to rest by travelling certain distance
using to find the distance traveled
where, v=final velocity
u=initial velocity
a=acceleration(offered by force)
s=displacement
final velocity is zero
For car A
For car B
divide 1 and 2 we get
thus 
distance traveled by car B is four time of car A
Answer:
1.4m/s
Explanation:
Average velocity is the total distance covered divided by the total time taken.
Average velocity = 
Total time taken = 5s + 6s = 11s
The first distance covered = velocity x time = 1.4 x 5 = 7m
second distance covered = velocity x time = 1.4 x 6 = 8.4m
So;
Average velocity = 
= 1.4m/s
Answer:
1.08
Explanation:
This is the case of interference in thin films in which interference bands are formed due to constructive interference of two reflected light waves , one from upper layer and the other from lower layer . If t be the thickness and μ be the refractive index then
path difference created will be 2μ t.
For light coming from rarer to denser medium , a phase change of π occurs additionally after reflection from denser medium, here, two times, once from upper layer and then from the lower layer , so for constructive interference
path diff = nλ , for minimum t , n =1
path diff = λ
2μ t. = λ
μ = λ / 2t
= 626 / 2 x 290
= 1.08
