Explanation:
At point B, the velocity speed of the train is as follows.
=
= 26.34 m/s
Now, we will calculate the first derivative of the equation of train.
y =
Now, second derivative of the train is calculated as follows.
Radius of curvature of the train is as follows.
=
= 3808.96 m
Now, we will calculate the normal component of the train as follows.
=
= 0.1822
The magnitude of acceleration of train is calculated as follows.
a =
=
=
Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is .
Answer:
Electromagnetic waves consist of both electric and magnetic field waves. These waves oscillate in perpendicular planes with respect to each other, and are in phase. The creation of all electromagnetic waves begins with an oscillating charged particle, which creates oscillating electric and magnetic fields.
Explanation:
Frost will disturb the smooth flow of air over the wing, unpleasantly
distressing its lifting competence. In other words, this spoils the even flow
of air over the wings, by this means decreasing lifting capability. Also, frost
may avoid the airplane from becoming flying at normal departure speed.