It is given that an<span> airplane is flying through a thundercloud at a height of 2000 m.
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Since the parity of charges is opposite and the airplane lies between the two charges and both the electric fields are in the same direction at the plane. Therefore, the magnitudes of the electric field at the aircrafts will add up.
Now, check the image to see the calculations:
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Answer:
The first graph is showing the constant acceleration (1 m/s)
Explanation:
The second graph showing the flexible velocity therefore a in the graph is different at t1, t2, t3, t4
The last graph is showing constant velocity therefore there is no acceleration (a = 0)
1=6, 2=8
I hope this helped
Answer:
32km per hour
Explanation:
Explanation:
In first case v = a t
==> a t = 40 km p h
Now distance covered S1 + S2 + S3
S1 = 1/2 a t^2 and S3 = 1/2 a t^2
But S2 = 3t * 40 = 120 t km
Hence total distance = at^2 + 120 t
Time taken (total) = t + 3t + t = 5 t
Hence average speed = at^2 + 120 t / 5 t
Cancelling t we have at + 120 / 5 = 40 + 120 / 5 = 160/5 = 32 km per hour