Answer:
Average velocity will be 58.181 m/sec
Explanation:
Distance between station A and station B = 1188 meters
As the train starts from station A its initial velocity u = 0 m/sec
Final velocity is when it reaches at station B is 20 m/sec
Acceleration 
From first equation of motion 
20 = 0+2.41×t
t = 8.298 sec
Now from station train began to deaccelerate and finaly stop so final velocity v = 0 m /sec
Initial velocity u = 20 m/sec
We know that v = u+at
Deacceleration 
So 0 =20 -1.65×t
t = 12.12 sec
So total time = 8.298 + 12.12 = 20.419 sec
So average velocity 
So average velocity will be 58.181 m/sec
I am absolutely sure that if a car accelerates on a horizontal road due to the force exerted by road. I consider the last option from the list represented above to be the correct one because the <span>road oses as a nice example of a frictional force. Hope you will agree with me and find it helpful! Regards!</span>
We know that:
p=mv
p=89kg*9m/s
p=801Ns
Explanation:
It is given that,
Slit width, 
A diffraction pattern is formed on a flat screen located 4.0 m away, L = 4 m
The distance between the middle of the central bright fringe and the first dark fringe is 4.2 mm, y = 4.2 mm = 0.0042 m
Let 
is the wavelength of the light.
Using condition of diffraction as,
..............(1)
Also, 
So, the wavelength of the light is 596 nm. Hence, this is the required solution.
