Answer:
L= 2 mH
Explanation:
Given that
Frequency , f= 10 kHz
Maximum current ,I = 0.1 A
Maximum energy stored ,E= 1 x 10⁻⁵ J
The maximum energy stored in the inductor is given as follows
Where ,L= Inductance
I=Current
E=Energy
Now by putting the values in the above equation
L=0.002 H
L= 2 mH
We know that frequency f is given as
C=Capacitance , f=frequency ,L=Inductance
Now by putting the values
Therefore the inductance and capacitance will be 2 mH and 1.26 x 10⁻⁷ F respectively.
1. 40-0=40
3. 40/5=8
8 ml/s
you find the range of acceleration(step one)
divide by the time(step two)
Its a, metal is a good conductor of heat so yea
Hope this helps :)
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 .
