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.
![\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7B%5B1%20%2B%20%28%5Cfrac%7Bdy%7D%7Bdx%7D%29%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B%5Cfrac%7Bd%5E%7B2%7Dy%7D%7Bdx%5E%7B2%7D%7D%7D)
= ![\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B1%20%2B%200.2e%5E%7B%5Cfrac%7B400%7D%7B1000%7D%7D%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B0.2%2810%5E%7B-3%7D%29e%5E%7B%5Cfrac%7B400%7D%7B1000%7D%7D%7D)
= 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
.
A. Solid
Solid structures are actually made of repeating patterns.
Answer:
The time it will take for the car to reach a velocity of 28 m/s is 7 seconds
Explanation:
The parameters of the car are;
The acceleration of the car, a = 4 m/s²
The final velocity of the car, v = 28 m/s
The initial velocity of the car, u = 0 m/s (The car starts from rest)
The kinematic equation that can be used for finding (the time) how long it will take for the car to reach a velocity of 28 m/s is given as follows;
v = u + a·t
Where;
v = The final velocity of the car, v = 28 m/s
u = The initial velocity of the car = 0 m/s
a = The acceleration of the car = 4 m/s²
t = =The time it will take for the car to reach a velocity of 28 m/s
Therefore, we get;
t = (v - u)/a
t = (28 m/s - 0 m/s)/(4 m/s²) = 7 s
The time it will take for the car to reach a velocity of 28 m/s, t = 7 seconds.
The frequency of a wave becomes higher due to the object moving at a fast pace coming towards you with shorter wavelengths (depending on the speed) aka the Doppler Effect.
Hope this helps
I would say our solar sustem will die when the sun dies or a black hole sucks it up.